Bluebook Compiler Geas

ME 601

fAlgebra: 1. Find the geometric mean between -2 and -8. A. 4 B. 16 C. -4*

D. 6

Recall: GM = sqrt(a*b) = sqrt(-2 * -8) = -4 Sign is referred to sign of the given. It should always be the same 2. A number is divided into two parts such that when the greater part is divided by the smaller, the quotient is 3 and the remainder is 5. Find the smaller number if the sum of the two numbers is 37. A. 8* B. 12 C. 32 D. 16 Let a = smaller part; b = greater part b/a = 3.5; a + b = 37 ; answer approx is smaller no. = 8.2222 or 8 3. Find the value of x which will satisfy the equation x  2 A. 1 B. 1,4 C. 4* D. 0,4

x  1.

!BackDOOR Substitute ans: C 4. In a potato race, 8 potatoes are placed 6 feet apart on a straight line, the first being 6 ft from the basket. A contestant starts from the basket and puts one potato at a time into the basket. Find the total distance he must run in order to finish the race. A. 532 ft. B. 432 ft* C. 342 ft D. 222 ft Recall: Arithmetic Progression S = n/2*(a1 + an) a1 = 6 * 2 = from starting line to nth place of potato = 12 ft an = 12 + (8-1)(12) = 96 n = 8 (for 8 numbers of potatoes) d = difference = 12 per potatoes S = 8/2*(12+96) = 432 ft 5. Determine how much water should be evaporated from 50kg of 30% salt solution to produce a 60% salt solution. All percentages are by weight. A. 25 kg * B. 35 kg C. 15 kg D. 18 kg Recall: Work/mixing problem %(mass) .3(50 kg) = 0.6(50-x); Ans = 25 kg

ME 601

Trigonometry: 6. Simplify the equation Sin2x(1+cot2x). A. sin2x C. 1* 2 B. cos x D. sec2xsin2x Recall Identities : 1 + cot2x = csc2x Then: Sin2x(csc2x) = Sin2x/ Sin2x = 1 Ans : 1 7. The angle of elevation of the top of a tower from a point A is 2330 ' . From another point B, the angle of elevation of the top of the tower is 5530 ' . The points A and B are 217.45 m apart and on the same horizontal plane as the foot of the tower. The horizontal angle subtended by A and B at the foot of the tower is 90 degrees. Find the height of the tower. A. 90.6 m* C. 89.5 m B. 86.7 m D. 55.9 m Illustrate:

EQN: SIMILAR TRIANGLE Tan 55.30 = Height/x - eq 1 Tan 23.30 = Height/(217.45+x)  eq 2 Solving for height of eq 1: height = x(1.455) Using height in eq 2: Tan 23.30 = x(1.455)/(217.45+x)

Height

X

B( 5530 ' ) 217.45m A ( 2330 ' )

X = 92.67816 8. In triangle ABC, angle A=80 deg. and point D is inside the triangle. If BD and CD are bisectors of angle B and C, solve for the angle BDC. A. 100 deg. C. 120 deg. B. 130 deg.* D. 140 deg. Recall: Sum of internal angles of triangle is 180 deg Angle a = 80; then the 2 other angles are 50. Bisecting the other 2 angels (25 and 25) 180 – sum of the 2 other angles = 180 – 50 = 130 for triangle BDC 9. The sum of the sides of a triangle is equal to 100 cm. If the angles of the triangle are in the continued proportions of 1:2:4. Compute the shortest side of the triangle. A. 17.545 C. 18.525 B. 19.806* D. 14.507 Perimeter of triangle = a + b + c = 100 ; 1:2:4 = 100

ME 601

X = shortest; 2X = mid; 4X = longest; Then, Perimeter = 100 = x + 2x + 4x = 7x 100/7 = x = 14.285 10. If sinxcosx+sin2x =1, what are the values of x in degrees? A. 32.2,69.3 B. -32.2, 69.3 C. 20.9, 61.9*

D. 20.9, -61.9

Using calc, answer is approx. 69.09 Analytic Geometry: 11. The line segment connecting (x, 6) and (9, y) is bisected by the point (7, 3). Find the values of x and y. A. 14, 6 B. 33, 12

C. 5, 0* D. 14, 6

Bisected is midpoint. Therefore, distance between point1 and (7,3) should be equal to distance between (7,3) and point 2 Recall: Distance formula = sqrt[(x2-x1)2+(y2-y1)2 Check using calc (Trial/Error): d = sqrt[(7-5)2+(3-6)2 = sqrt(13) d = sqrt[(9-7)2+(0-3)2 = sqrt(13) Ans: C 5, 0* 12. The line 2x – 3y + 2 = 0 is perpendicular to another line L1 of unknown equation. Find the slope of L1. A. 3/2 B. -3/2*

C. 2/3 D. -2/3

Recall: Perpendicular lines, m1 = -1/m2 Slope of given eqn = 2/3, then slope of L1 = -3/2 13. Two buildings in a shopping complex are shaped like a branches of the hyperbola 729x2−1024y2−746496=0 , where x and y are in feet. How far apart are the buildings at their closest part? A. 68 ft. apart C. 75 ft. apart B. 64 ft. apart* D. 80 ft. apart Let’s try this one without drawing it, since we know that the closest points of a hyperbola are where the vertices are, and the buildings would be 2a feet apart.

By doing a little algebra (adding 746496 to both sides and then dividing all terms by 746496) we see that the equation in hyperbolic form is x21024−y2729=1. So a= sqrt(1024) = 32. The building are 32 x 2 = 64 feet apart at their closest part.

ME 601

14. An ice rink is in the shape of an ellipse, and is 150 feet long and 75 feet wide. What is the width of the rink 15 feet from a vertex? A. 50 ft. B. 45 ft.*

C. 60 ft. D. 40 ft.

Let’s first find the equation of the ellipse, with the center at (0,0). Since the major axis is 150, and the minor axis is 75, we have a=75 and b=37.5. From this, we know the equation of the ellipse is x2/752+ y2/37.52 = 1, or x2/5625 + y2/1406.25 = 1. Now that we have the equation, we can plug in any x value to get the y value(s) on the ellipse; since we want the width of the ellipse 15 feet from the vertex, our x value is 75 – 15 = 60. Plugging in 60 for x, we get: 602/5625 + y2/1406.25 = 1; solving for y, we get ±√ [(1−602/5625)×1406.25]= ± 22.5 (take positive only). Note that we need to take double 22.5 to get the whole width: the width of the rink 15 feet from a vertex is 45 feet 15. The sides of a quadrilateral are 10m, 8m, 16m and 20m, respectively. Two opposite interior angles have a sum of 225°. Find the area of the quadrilateral in sq. m. A. 140.33 sq. cm. C. 150.33 sq. cm. B. 145.33 sq. cm.* D. 155.33 sq. cm. RECALL: Plane Geometry (TRAPEZIUM FORMULA) S = (10+8+16+20)/2 = 27 ; ∅ = 225/2 = 112.5

Then A = SQRT((27-10) (27-8) (27-16) (27-20)10*8*16*20*0.1464) Cos2Angle = 0.1464 Area = 145.33

Differential Calculus: 16. For what values of x is the derivative of x3 equal to the derivative of x2 + x? A. 1, -1/3 * C. -1, 1/3 B. 1.62, -0.62 D. -1.62, 0.62 Derivative of x3 = 3x2 x2 + x = 2x +1 Equate: 3x2 = 2x +1 ; 3x2 - 2x - 1 = 0 MODE: 5:3; Ans: x1 = 1 ; x2 = -1/3

ME 601

17. Find the angle between the two curves y=x2 and y = x3+x2+1 at their point of intersection. A. 71.6* B. 45 C. 61.7 D. 54 Line 1 = x2 Line 2 = x3+x2+1 Solving the intersection of the 2 lines: (equate both eqns) x2 = x3+x2+1 ; x3 + 1 = 0 ; x = 1 slope of l1: dy/dx = 2x (@x = 1); dy/dx = 2 = slope 1 slope of l2: dy/dx = 3x2 + 2x + 1 (@x = 1); dy/dx = 3 + 2 + 1 = 6 = slope 2 Recall: tan∅ = m2 – m1/(1 – m2m1)  angle of intersecting lines Angle is approx. 19.98 – 90 degrees = 70.01 degrees = approx. to 71.6 18. A military courier is located on a desert 6 miles from a point P which is the point on a long straight road nearest to him. He is ordered to get to a point Q, 3 miles on the road. Assuming that he can travel 14 miles per hour on the desert and 50 miles per hour on the road, find the point where he should reach the road in order to get to Q in the least possible time. A. 2 mi B. 2.25 mi C.1.75 mi * D.1.5 mi 19. Find the point in the parabola y2 = 4x at which the rate of change of the ordinate and abscissa are equal. A. (1,0) B. (2,1) C. (1,2) * D. (1,1) Recall: Rate of change Eqn: dy/dt = rate of change of ordinate (y) dx/dt = rate of change of abscissa (x) Equating as stated in the condition in the problem: dy/dt = dx/dt deriving both sides of the equation: 2y = 4 y = 2, then x = 1 Ans: C(1,2) 20. The dimensions of a rectangle are continuously changing. The length decreases at the rate of 2 in/sec while the width increases at the rate of 3 in/sec. At one instant the rectangle is a 20 inch square. How fast is its area changing 3 seconds later? A. 10 in2/sec C. 16 in2/sec* B. 12 in2/sec D. 14 in2/sec

ME 601

Initially, at time 0 sec; area is 20 sq2 = LW Consider: L = 20; W = 20 At time = 3 sec L = 20 – 2in/sec(3sec) = 14 inches W = 20 + 3in/sec(3sec) = 29 inches Formula A = LW Derive both side (recall product) dA/dt = L(dw/dt) + w(dL/dt) Area at time 3 sec = 14(3 in/sec) + 29(-2 in/sec) = 16 in2/sec dA/dt = -16 in2/sec Integral Calculus: 21. Find the area bounded by the curve y=8-x3 and the x-axis. A. 21 sq. units C. 12 sq. units* B. 32 sq. units D. 18 sq. units @x = 0 Y = 8 – x3 Y=8–0=8 @y = 0 0 = 8 - x3 ; x3 = 8; x = 2 Note x (0 and 2) will be the extremities Recall: AREA BOUNDED (curve and axis) A = definite integral of ydx (@x = 0,2)

Use calc A = 12 sq. units

t

22. Given the area in the first quadrant bounded by x2=8y, the line x=4 and the x-axis. What is the volume generated by revolving this area about the y-axis? A. 50.265 cu. units C. 40.345 cu. units* B. 32.561 cu. units D. 36.251 cu. Units Recall: Revolving around axis to form volume (integral) x2 = 8y (parabolic graph) @x = 4; 42 = 8y; y = 2

ME 601

Therefore (intersect is at 4,2) Consider statement “at x-axis”. Then limit would be 4 and 0. Can be solve using horizontal element (dy) of vertical element (dx) ⿏ Vertical Element: V = ⿏ Horizontal Element: V = x22-x12)dy Using vertical element with limit 4 and 0 V=

* 2pi(x)dx = 16 pi = approx. to 50.265 cu. units

*2 x=

Using horizontal element with limit 2 and 0 V=

t

=

3x 23. Evaluate:   2

0

y

0

A. 30

2



= 16 * pi = 50.265 cu. units

=

t

 9y2 dxdy

B. 50

C. 40 *

D. 20

Recall: Multiple integration (First in Last out) = =

h

=

h

=

h

h

= =

h

= 44 =answer approx. 40

24. Find the length of arc of the curve y = e-x bounded by the coordinate axes and the line x = 5. A. 4.97 B. 3.83 C. 5.23* D. 4.53 Recall: Arc of Length formula (S = definite integral of sqrt(1 + (f’(x))2dx f(x) = e-x f’(x) = -e-x S = (√(1+(-e-x)2)dx = 5.225 = 5.23 25. Find the surface area generated by rotating the first quadrant portion of the curve x2 = 16 – 8y about the x – axis. A. 36.57* B. 61.27 C. 43.56 D. 39.77 h Recall: Surface Area = 2 f(x) = (x – 16)/8 ; @ y = 0; x = 4 (limit is 4,0) with respect to x axis f’(x) = 1/4(x) Then,

t

h

= 36.574

ME 601

Differential Equation: 26. A tank initially contains 2500 liters of brine having 50 % salt in solution. Fresh water enters the tank at the rate of 25 liters per minute and the resulting mixture leaves the tank at the rate of 50 liters per minute. Find the percentage of salt in the tank at the end of 20 minutes. a. 40 % * C. 10 % b. 30 % D. 20 % RECALL: Mixture Problems DE (socivo); S = (So + CiVo)e(Ro/Vo)(time) + CiVo Ds/dt = Gain – Loss Gain is Concentration input*Rate of input Loss is Concentration output*Rate of output Initial Volume (Vo) = 2500 L Initial Weight of Substance (So) = 50% = 0.5 Brine = 1250 is salt Rate input = 25 L/min Rate output = 50 L/min Solve at time = 20 minutes 27. Given z = 2y2 , y = x/ (x+1), and x = 1/(t – 1). Find dz/dt. a. -4/ t3 * C. 2 t3 - 1 b. 4t3 D. - 8t3 28. What is the order and degree of the following differential equation: 4

5

 d2y   dy  y  2   2 x 3    8 xy 2 .  dx   dx 

a. 1 , 5 b. 3 , 4

C. 4, 2 D. 2 , 4 *

Recall: order is highest order of derivative degree highest exponent of the derivative derivative is dy/dx, highest order is f’’(y) = d2y/dx2 = 2, exponent or degree is at 4 Answer: 2, 4 29. Find the equation of the curve that passes through (4, -2) and cuts at right angles every curve of the family y2 =cx3 . a. 2x2+3y2 = 44 * C. 3x2+2y2 = 22 b. 3y2 -2x2 = 44 D. 2x2-3y2 = 22 Using Calc: Substitute value of (4, -2), that satisfies the equation, on this case 2(4)2 + 3(-2)2 = 44. Ans is A

ME 601

30. Identify the following equation : ( x 2  2 xy )dx  12 y 2 dy  0 . a. Separable DE C. Homogeneous DE * b. Exact DE D. 1st-order linear DE Rewriting: dy/dx + (x2 – 2xy)/12y2 = 0 (Homogeneous DE) Physics: 31. A 200 gram apple is thrown from the edge of a tall building with an initial speed of 20 m/s. What is the change is kinetic energy of the apple if it strikes the ground at 50 m/s? a. 100 joules b. 180 joules c. 81 joules d. 210 joules* Given: mass = 200 g or 0.2 kg; Vi = 20 m/sec; Vf = 50 m/sec Unit analysis, answer is in unit of joule. J = (kg * m2)/sec2 KEchange = Difference in Kinetic Energy = KEfinal – KEinitial = 1/2(0.2)(50m/s)2 – 1/2(0.2)(20m/s)2 = 210 joules 32. The motion describe by the equation x(t)=A cos(wtφ) a. Motion with constant velocity b. Motion with constant acceleration c. Oscillatory motion * d. Motion is which the acceleration is proportional to the velocity RECALL: Oscillatory motion Repeated back and forth movement over the same path about an equilibrium position, such as a mass on a spring or pendulum. EQN:

x(t)=Acos(2πft)

33. The location of a particle moving in the x-y plane is given by the parametric equations x = t2 + 4t and y = (1/4) t4 – 60t, where x and y are in meters and t in seconds. What is the particle’s velocity at t = 4sec? a. 8. 95 m/s b. 11.3 m/s c. 12.6 m/s * d. 16.0 m/s

RECALL: PARAMETRIC EQUATION

ME 601

Resultant =

h

*Since it is about parametric equation and velocity is to be determined at time t = 4 sec, x’(t) and y’(t). x'(t) = 2t + 4 ; y’(t) = t3 – 60, then @ t = 4; x = 12, y = 4 Finally, Resultant =

h

= 12.64 m/s

34. A projectile whose mass is 10 g is fired directly upward from the ground level with in initial velocity of 1000 m/s. Neglecting the effects of air resistance, what will be the speed of the projectile when it impacts the ground? a. 707 m/s b. 981 m/s c. 1000 m/s d. 1414 m/s RECALL: Projectile Motion Solve first the height the mass will gain when fired upward. Vf2 = Vo2 - 2gheight (-g due to gravity pulling the mass down) (final velocity Vf is zero at maximum height) 0 = (1000m/s)2 – 2(9.81 m/sec2)height ;

height = 50968.39 meters

Next, solve for velocity as the body falls down Vf2 = Vo2 – 2gh (Vo is zero from max velocity) Vf2 = 2g(50968.39 meters); Vf2 = 1000 m/s 35. A fisherman cuts his boat’s engine as it enters a harbor/ the boat comes to a dead stop with its front end touching the dock. The fisherman’s mass is 80 kg. he moves 5 m from his seat in the back to the front of the boat n 5 s, expecting to be able to reach the dock. If the empty boat has a mass 300 kg, how far will the fisherman have to jump to reach the dock? a. 1.1 m * b. 1.3 m c. 1.9 m d. 5.0 m RECALL: Conservation of Energy: Ein =Eout Initial velocity of boat and fisherman = 0 m/s Final velocity of fisherman = 5 m / 5 sec = 1 m/sec [80 kg * (5 + x)]man + [300kg*x]boat = 0 X = 1.0526 = 1.1 m Mechanics: 36. What power would a spindle 55mm in diameter transmit at 480rpm. Stress allowed for short shaft is 59N/mm^2 a. 42.12kW b. 50.61kW c. 96.88kW * d. 39.21kW

ME 601

Recall: Power = 2piTN ; Stress = P/A = 16T/piD3 for solid shaft or 16TD/pi(D4-d4) for hollow shaft Solving for T; 59 N/mm2 = 16T/pi(55mm)3 ; T = 1927391.637 N-mm = 1927.39163 N-m Finally, Solving for Power = 2piT(N/60) = 2 * pi * 1927.39163 * 480/60 = 96881.27 W = 96.88kw 37. A 30-m long aluminum bar is subjected to a tensile stress of 172MPa. Find the elongation if E=69,116MPa. a. 0.746m b. 0.007m c. 6.270mm d. 7.46cm * RECALL: Stress = P/A, Strain = Elongation/Loriginal, Then, STRESS-STRAIN hooke’s law : P/A = EmodElongation/Lorig Finally, Elongation = PLorig/AEmod ; Substituting values *assume area of 1 m2 Elongation = 172MPa(30 m)/ [Area (69116 MPa)] = 0.0746 m = 7.46 cm 38. A copper rolled wire 10m long and 1.5mm diameter when supporting a weight of 350N elongates 18.6mm. Compute the value of the Young’s modulus of this wire. a. 200GPa b. 180.32GPa c. 148.9GPa d. 106.48GPa* RECALL: Elongation = PLorig/AEmod ; Substituting values 0.0186 m = 350 N (10 m) / (pi/4)(0.0015mm)2 Emodulus Emodulus = 1.0648 x 1011 = 106.48 GPA 39. A simple beam 10m long carries a concentrated load of 200kN at the midspan. What is the maximum moment of the beam? a. 250kN-m * b. 500kN-m c. 400kN-m d. 100kN-m

Ans: 500 kN-m 40. A spherical pressure vessel 400-mm in diameter has a uniform thickness of 6 mm. The vessel contains gas under a pressure of 8,000 kPa. If the ultimate stress of the material is 420 MPa, what is the factor of safety with respect to tensile failure? a. 3.15 * b. 3.55 c. 2.15 d. 2.55

ME 601

RECALL: PRESSURE VESSEL (considering SF) Thickness = prn/2oy P, pressure = 8000 kPa Oy, Yield Stress = 420 MPa Diameter = 400 mm, Radius = 200 mm Thickness = 6 mm Therefore, N Safety factor = 3.15 41. During a stress-strain test, the unit deformation at a stress of 35 MPa was observed to be t t ଦ m/m and at a stress of 140 MPa it was ଦ m/m. If the proportional limit was 200 MPa, what is the modulus of elasticity? What is the strain corresponding to stress of 80 MPa? t a. E = 210,000 MPa; m/m t b. E = 200,000 MPa; m/m t c. E = 211,000 MPa; m/m t d. E = 210,000 MPa; m/m * Given: Deformation @P = 35 MPa is δ = 167*10-6 Deformation @P = 140 MPa is δ = 667*10-6

Emod = [35 x 106 N/m2]/[167 x 10-6 m/m] = 2.09 x 1011 which is also with the same proportion Emod = [140 x 106 N/m2]/[667 x 10-6 m/m] = 2.09 x 1011 Therefor, E is approximately 210 MPa From the proportion Strain, @ 80 MPA = S/E = [80 x 106 N/m2]/[210 GPA] = [80 x 106 N/m2]/[210 x 109] = .380 Or = is approximately 381 x 10-6 m/m Answer: D 42. An axial load of 100 kN is applied to a flat bar 20 mm thick, tapering in width from 120 mm to 40 mm in a length of 10 m. Assuming E = 200 GPa, determine the total elongation of the bar. a. 3.43 mm * b. 2.125 mm c. 4.33 mm d. 1.985 mm Consider Fig:

ME 601

Consider a differential lenght for which the cross-sectional area is constant. Then the total elongation is the sum of these infinitesimal elongations. At sector m-n, the half width y (mm) at the distance x (m) from the left end is found from geometry to be: (y-20)/x = (60-20)/10 y = (4x + 20)mm Area of the section m-n: Area, A = 20(2y) = (160x + 800)mm2 Elongation = PL/AE; dElongation = [100KN(dx)]/[(160x + 800)(10-6)(200*109)] dElongation = (0.500dx)/(160x + 800) solving total elongation using definite integral dElongation = 0.500

h

= 3.433 mm

43. A 20-mm diameter steel rod, 250 mm long is subjected to a tensile force of 75 kN. If the Poisson’s ratio is 0.30, determine the lateral strain of the rod. Use E = 200 GPa. t li a. y mm/mm t b. y t li mm/mm * t c. y t l ଦ mm/mm t l ଦ d. y mm/mm 44. The maximum allowable torque, in kN-m, for a 50-mm diameter steel shaft when the allowable shearing stress is 81.5 MPa is: a. 3.0 b. 1.0 c. 4.0 d. 2.0 * Recall: Torque Stress = P/A = 16T/piD3 for solid shaft or 16TD/pi(D4-d4) for hollow shaft 81.5 MPa = 16T/pi(50mm)3 T=2

ME 601

45. Compute the value of the shear modulus G of steel whose modulus of elasticity E is 200 GPa and Poisson’s ratio is 0.30. a. 72,456 MPa c. 76,923 MPa * b. 79,698 MPa d. 82,400 MPa RECALL: Shear Modulus G formula: G = Force(Loriginal)/Area(change in length) With Respect to Moduli Material: E = 2G (1 + poisson’s) Then, 200 GPa = 2G * (1 + 0.30) = 76923 MPa

Engineering Economy: 46. A member of congress wants to know the capitalized cost of maintaining a proposed national park. The annual maintenance cost is expected to be ₱20,000. At an interest rate of 6% per year, the capitalized cost of the maintenance would be closest to: a. ₱1,500 b. ₱25,000 c. ₱333,333* d. ₱416,667 Given: Annual cost = 20000; interest rate = 6% = 0.06 Capitalized cost = Annual Cost*Depreciation Rate = 20000 * 1/0.06 = 333333

47. A machine has a first cost of ₱100,000 and salvage value of ₱10,000 after 10 years. The annual maintenance of the machine is ₱4,000. If interest rate is 10%. Find the annual cost of the machine. a. ₱17,009.34 b. ₱19,647.08* c. ₱21,083.56 d. ₱32,074.56 1/10 yrs = 0.1 depreciation rate per yr 100000 – 10000 = 90000 ; 90000*0.1 = 9000 annual depreciation 900*0.1) + 9000 + 100000*0.1 – (4000*0.1) = 19500

48. A certain product has the following corresponding cost: 250 units for ₱4,000 and 400 units for ₱5,000. Find the increment cost. a. 3.46 b. 4.05* c.5.31 d. 6.67

ME 601

Recall: Incremental cost is the additional cost (or revenue) that results from increasing the output of the system by one or more units. Increment costs are those that arise as a result of a change in operations or policy.

49. A machine has an initial cost of ₱50,000 and a salvage value of ₱10,000 after 10 years. What is the book value after seven years using straight-line depreciation? a. ₱22,000* b. ₱23,000 c. ₱24,000 d. ₱25,000 Recall: Depreciation: Bookvalue and Straight-line method Depreciation, dstraight line method = (Initial Cost – Cost at nth year)/(nth year) Book Value on Straight line method (at m years): Cm = Initial Cost – Total depreciation after m years) Initial Cost = P50,000 Cost at nth year or Salvage Value = P10,000 at 10 year Nth year = 10 years d = (50000 – 10000)/10 = 4000 ***depreciation is equivalent all throughout period At 7th yr, Bookvalue C7 = 50,000 – d(7) d(7) = 4000 * 7 = 28000 C(7) = 50000 – 28000 = 22000 50. A machine that costs ₱20,000 has a 10 year life and a ₱2,000 salvage value. If straight line depreciation is used, what is the book value of the machine at the end of the fourth year? a. ₱12,200 b. ₱12,400 c. ₱12,600 d. ₱12,800* Initial Cost = 20,000 Cost at 10th year = 2000 Depreciation = (20000 – 2000)/10 = 1800 Book value at 4th = 20000 – 1800(4) = 128000

ME 601

51. A manufacturing firm maintains one product assembly line to produce signal generators. Weekly demand for the generator is 35 units. The line operates for 7 hours per day, 5 days per week. What is the maximum production time per unit in hours required of the line to meet the demand? a. 1 hour * b. 3 hours c. 2 hours d. 4 hours 7 hours/unit * 5 days = 35 35 units/35 = 1 hour

52. What is the capitalized cost of a project that will cost P20,000,000 now and will require P3,000,000 in maintenance annually? The effective annual interest rate is 15%. a. P30,000,000 c. P40,000,000 * b. P50,000,000 d. P60,000,000 Present Cost = 20000000 Maintenance Cost = 3000000 Effective annual Interest Rate = 0.15 Capitalized Cost = PC + MC/0.15 = 20000000 + 3000000/0.15 = 40000000 53. How much money must be invested today in order to withdraw P12,000 per year for 12 years if the interest rate is 12%? a. P44,332.49 b. P54,332.49 c. P74,332.49 * d. P94,332.49 Recall: Annuity (Malalaman na annuity kapag may series of payments na mangyayari) *Present worth = [A|(1+i)n - 1|] / i(1 + i)n *Future worth = [A|(1+i)n - 1|] / i Use Present Worth (how much money must be invested today) A = 12000 ; n = 12 ; i = 0.12 P = 12000 |(1 + 0.12)12 - 1|/ (0.12)(1 + 0.12)12 = 74332.49

ME 601

54. In a small factory with a capacity of 500,000 units per year with a 75% efficiency, the annual income is P500,000 and a fixed cost of P180,000 and variable cost of P0.52 per unit. What is the break-even point? a. 222,223 units * b. 242,223 units c. 252,223 units d. 222,222 units RECALL: Break-Even Point (Cost A = Cost B) FC = 180000 FC/unit contribution is BEP in terms of units: 180000/0.52 = 346153 Cost per unit = P0.52/unit 55. An equipment costing P250,000 has an estimated life of 15 years with a book value of P30,000 at the end of the period. Compute the depreciation charge and its book value after 10 years using the sum of year’s digit method. a. d = P11,000; BV = P67,500 b. d = P10,500; BV = P58,000 c. d = P11,500; BV = P60,000 d. d = P11,000; BV = P57,500 * Recall: SOYD: Sum of years = [(n)(n+1)]/2 = [15(15+1)]/2 = 120 d(15) = (250000 – 30000)[(15 – 14)/120] = 1833.3333 1833.33 = [250000 – C(10)][(10-9)/55] =

Thermodynamics 1: 56. A perfect gas has a value of R= 58.8 ft-lb/lb-R and K= 1.26. if 20 Btu are added to 10 lbs of his gas at constant volume when initial temperature is 90 degF find the final temperature. a. 97 degF *

b. 104 degF

Q = mCv(dT) Cv = R/k-1 = 58.8/1.26 - 1 * (1/778) = 0.29086 Btu/lb-F

c. 154 degF

d. 185 degF

ME 601

20 Btu = 10 (0.29086)(T2 - 90) T2 = 96.88 = 97F 57. A spherical balloon with a diameter of 6 m is filled with helium at 20 deg. C and 200 Kpa. Determine the mole number. a. 9.28 Kmol * b. 10.28 Kmol c. 11.28 Kmol d. 13.28 Kmol Recall: PV = nRT 200 [4/3 pi * (6/2)^3] = N * (8.3143)*(20+273) N = 9.28 kmol 58. A one cubic meter container contains a mixture of gases composed of 0.02 kg/mol of oxygen and 0.04 kg-mol helium at a pressure of 220 Kpa. What is the temperature of this ideal gas mixture in degrees kelvin? a. 441 * b. 350 c. 400 d. 450 Vtotal = V1 + V2 = (mrt/p)1 + (mrt/p)2 = [0.02*32(8.3143/32)T]/220 + [0.04*4(8.3143/4)T]/220 = 441 K 59. Convert 750 R to K a. 390.33 K b. 395 K 60. Find the mass of 10 quartz of water a. 10.46 kg b. 9.46 kg *

c. 410.33 K

d. 416.33 K *

c. 11.6 kg

d. 8.46 kg

V = 10 quartz (gal/4 quartz)(3.785 L/gal)(1 m^3/1000 L) = 0.0094625 * 10^-3 m^3 m = pV = 1000*(0.0094625 * 10^-3) = 9.46 kg Thermodynamics 2: 61. Steam leaves an industrial boiler at 827.4 kPa and 171.6C. A portion of the steam is passed through a throttling calorimeter and is exhausted to the atmosphere when the calorimeter pressure is 101.4 kPa. How much moisture does the steam leaving the boiler contain if the temperature of the steam at the calorimeter is 115.6C? At 827.4 kPa (171.6C): hf = 727.25 kJ/kg, hfg = 2043.2 kJ/kg From table 3: at 101.4 kPA and 115.6C: h2 = 20707.6 kJ/kg a. 6.78% b. 3.08%* c. 4.56% d. 2.34% Let x = qty. of steam entering h1 = h2

ME 601

27076 = 727.25 + x(2043.2) x = 0.9692 y = 1 - 0.9692 = 0.0308 = 3.08% 62. One kg of steam at 121C and 10% moisture undergoes a constant volume until the pressure becomes 0.28 Mpa. Determine the final temperature. a. 200.4C b. 374.5C c. 206.5C* d. 873.4C Recall: V = C (Isochoric) PV = mRT 63. A cylinder and piston arrangement contains saturated water vapor at 110C. The vapor is compressed in a reversible adiabatic process until the pressure is 1.6 Mpa. Determine the work done by the steam per kg of water. a. -637 kJ b. -509 kJ c. -432 kJ* d. -330 kJ 64. A turbine has an available enthalpy of 3300 kJ/kg in a Rankine cycle. The pump work has also 25 kJ/kg. For flow of 3 kg/s, find the system output. a. 5960 kW b. 6080 kW c. 6343 kW d. 9825 kW* 65. A rankine cycle has a steam throttle condition of 4 Mpa and 400C. the turbine exhaust is 1 atm, fin the cycle efficiency. a. 23.23% b. 27.06%* c. 34.23% d. 43.23% Heat Transfer: 66. A 5 cm diameter spherical ball whose surface is maintained at a temperature of 70 deg. C is suspended in the middle of a room at 20 deg. C. if the convection heat transfer coefficient is 15 W/m^2-C and the emissivity of the surface is 0.8, determine the total heat transfer from the ball. a. 23.56 watts c. 32.77 watts* b. 9.22 watts d. 43.45 watts Recall: Convective Heat xfer (hAdT) and Radiant Heat xfer (seat^4) A = 4piRad^2 = 4pi(0.05)^2 = 0.0314 m^2 Qo = hA(dT) = 15(0.0314)(70-20) = 23.56 watts Qr = 0.80 (5.67 x 10^-8)(0.0314)[(70+273)^4-(50+273)^4] = 9.22 watts Qt = Qo + Qr = 23.56 + 9.22 = 32.77 watts

ME 601

67. Hot gases at 280 degC flow on one side of a metal plate of 10mm thickness and air at 35 degC flows the other side. The heat transfer coefficient of the gases is 31.5 W/m^2-k and that of the air is 32 W/m^2-K. calculate the over all transfer coefficient. a. 15.82 W/m^2-K* c. 16.82 W/m^2-K b. 14.82 W/m^2-K d. 17.82 W/m^2-K Recall: Overall Heat xfex Coeff: U = 1/Rt Rt = 1/h1 + k1-2/x1-2 + 1/h2 = 1/31.5 + 0.01/50 + 1/32 = 0.0632 U = 1/0.06032 = 15.82 W/m2-k 68. How many watts will be radiated from a spherical black body 15 cm in diameter at a temperature of 800 degC? a.5.34 KW* b. 4.34 KW c. 6.34 KW d. 3.34 KW Recall: Heat xfer (seat^4) A = 4pi(Rad)^2 = 4pi(7.5)^2 = 706.86 cm^2 T = 800 + 273 = 1073 K e = 1 (for black body) Constant (botzman) = 5.6704 * 10-8 W/m-k^4 Qr = (5.6704 * 10-8 W/m-k^4)(1)(0.0706.86 m)(1073K)4 = 5313.0755 Watts = 5.313 kW 69. During a steady state operation, a gearbox receives 60kW through the input shaft and delivers power through the output shaft. For the gear box as the system, the rate of energy transfer is by convection. h=0.171 kW/m2K is the heat transfer coefficient, A=1m2 is the outer surface area of the gear box, Th=300K (27C) is the temperature and the outer surface, Tf=293K (20C) is the temperature of the surroundings away from the immediate vicinity if the gear box. Determine the power delivered to the output shaft in kW if the heat transfer rate is 1.2 kW. a. 98.8 KW b. 78.8 KW c. 68.8 KW d. 58.8 KW* dE/dt = Q - W; W=Q W = W1 + W2 = Q W1 = -60kW Q = -1.2 W2 = power delivered to the output shaft, kW = -1.2kW - (-60kW) = 58.8 kW

ME 601

70. Determine the thermal conductivity of a material that is used a 4 m2 test panel, 25 mm thick with a temperature difference of 20F between surfaces. During the 4 hrs of test period, the heat transmitted is 500 kJ. a. 0.0432 W/m-Cb. 0.0723 W/m-C c. 0.0321 W/m-C d. 0.0195 W/m-C* Fluid Mechanics: 71. What is most nearly the terminal velocity of a 50 rnrn diameter, solid aluminum sphere falling in air? The sphere has a coefficient of drag of 0.5, the density of aluminum, Palum, is 2650 kgjm3, and the density of air, Pair, is 1.225 kgjm3 a. 25 m/s b. 53 m/s * c. 88 m/s d. 130 m/s 72. What is the static head corresponding to a flow velocity of 10 ft/sec? a. 1.55 ft* b. 1.75 ft c. 2.05 ft d. 2.25 ft 73. Water flows rate along ½ in. i.d. nose at 3 gal/min. Water velocity in ft/sec is nearest to: a. 1 b. 5* c. 10 d. 20 74. Find the depth in furlong of the ocean (SG = 1.03) if the pressure at the sea bed is 2,032.56 kpag. a. 1 * b. 2 c. 3 d. 4 P = yheight 2032.56 = 1.03 * 9.81 * height h = 201.158 m (3.281 ft/m)(1 yd/3ft)(furlong/220 yd) = 1 furlong 75. The work required to accelerate an 800-kg car from rest to 100 km/hr on a level road: a. 308.6 kJ* b. 806.3 kJ c. 608.3 kJ d. 386 kJ Recall: Kinetic Energy (in terms of work): KE = 1/2*mass*vel^2 KE = 1/2 * (800kg) * [(100 km/hr)(1 hr/3600sec)]^2 = 0.3086 * (1*1000) = 308.6 kJ

Combustion: 76. The dry exhaust gas from oil engine has the following gravimetric analysis: CO2 = 21.6%; O2 = 4.2%; N2 = 74.2% Specific heats at constant pressure for each component of the exhaust gas in Kcal/kgoC are: CO2 = 0.203; O2 = 0.219; N2 = 0.248. Calculate the specific gravity if the molecular weight of air is 28.97 kg/kg-mol. a. 0.981

b. 1.244

c. 1.055

*

d. 0.542

ME 601

Recall: SG

= density fluid/density water = density gas / density air = mw gas / mw air (in terms of molecular weight) MW gas will be the total weight of in volumetric of CO2, O2 and N2 CO2 = (21.6%)/(12 + 2*16) = 0.004909 O2 = (4.2%)/(2*16) = 0.0013125 N2 = (74.2%)/(2*14) = 0.0264 Total Volumetric = 0.0049 + 0.0013 + 0.0264 = 0.0326 Molecular Weight = 1/0.0326 = 30.64 Finally, S.G. = 30.64/28.97 = 1.057 = 1.055 77. An unknown hydrocarbon fuel, CxHy, is burned with excess air containing 23.3% oxygen by mass. The volumetric analysis of the dry products of combustion is as follows: 11.94% CO2, 0.41% CO, 2.26% O2, AND 85.39% N2. Find the value of x and y, respectively. a. 14.35, 44.22 b. 12.35, 33.22* c. 10.35, 32.33 d. 14.35, 32.33 78. A typical industrial fuel oil, C16H32 with 20% excess air by weight. Assuming complete oxidation of the fuel, calculate the actual air-fuel ratio by weight. a. 17.56 kgair/kgfuel c. 15.76 kgair/kgfuel b. 16.75 kgair/kgfuel d. 17.65 kgair/kgfuel* Recall: Molal analysis/computation and stoichmetric eqn Theoretical Eqn: FUEL + AIR = PRODUCTS C16H32

+

Material Balance: C: 16 = b(1) H: 32 = 2c O: 2a = 2b + c N: 3.76a = 90.24

aO2 + 3.76aN2

=

bCO2 + cH2O + 3.76aN2

b = 16 c = 16 a = 24

Completing theoretical equation using computed coefficient from Material balance: C16H32 C16H32

+ +

(24)O2 + 3.76(24)N2 = 24O2 + 90.24N2

(16)CO2 + (16)H2O + 3.76(24)N2 = 16CO2 + 16H2O + 90.24N2

Next, Compute Actual Reaction Equation (consider the 20% excess air): C16H32 + 1.2(24O2 + 90.24N2) = 16CO2 + 16H2O + 1.2(90.24N2)

ME 601

Balance this Actual Reaction Equation by adding Oxygen (O2) C16H32 + 1.2(24O2 + 90.24N2) = 16CO2 + 16H2O + 1.2(90.24N2) + dO2 Computing variable d: O: 1.2(24)(1) = 16(2) + 16 + 2d d = 4.8 kgmol Finally, C16H32 (4.8)O2

+

1.2(24O2 + 90.24N2)

=

Complete the computation using air-fuel ratio: (A/F)actual = [28.8(32) + 108.28(28)]/(12*16 + 1*32)

16CO2 + 16H2O + 1.2(90.24N2) +

= 17.65 kgair/kgfuel

79. Fuel oil in a day tank for use of an industrial boiler is tested with hydrometer. The hydrometer reading indicates a S.G. = 0.924 when the temperature of the oil in the tank is 350C. Calculate the higher heating value of the fuel. a. 43,852.13 kJ/kg* c. 53,853.13 kJ/kg b. 58,352.13 kJ/kg d. 48,352.13 kJ/kg Recall: HHV/dulong Qh = 41130 + 139.6(API) API = 141.5/SG15.6 – 131.5 SG correction factor: SG35 = SG15.6(1-0.00072(35-15.6)) 0.924 = SG15.6(1-0.00072(35-15.6)) SG15.6 = 0.937 Then, API = 141.5/0.937 – 131.5 = 19.49 Qh = 41130 + 139.6(19.49) = 43852.128 kj/kg 80. A diesel electric plant supplies energy for Meralco. During a 24 hr period, the plant consumed at 700 gallons of fuel at 280C and produce 3930 kW-hr. industrial fuel used is 280API and was purchased at P 5.50 per liter at 15.60C. What should the cost of fuel be produce one kW-hr? a. 1.05* b. 1.10 c. 1.069 d. 1.00 Solving for density at 15.6 celcius: API = 141.5/SG15.6 + 131.5 28 = 141.5/SG15.6 + 131.5

ME 601

SG15.6 = 0.887 Density, P = 0.887 (1kg/L) = 0.887 kg/L Solving for density at 28 celcius: SG28 = SG15.6(1-0.00072(t-15.6) SG28 = 0.887(1-0.00072(28-15.6) = 0.879 Density, P = 0.879 (1kg/L) = 0.879 kg/L Original Price or cost per kg = 5.50/0.887 = P6.20/kg Costing of consumption per kW-hr: C = (200 gal/3930 kW-hr)*(3.7854 L/1 gal)(0.879 kg/L)(P6.20/kg) = P1.05/kW-hr ME Laws: 81. If the rated scores of an examinee in a board exam are 85, 82, and 48 in Math, Power, and Design subjects respectively (average: 72.85%), what will be the overall result? a. Pass b. Fail * c. Removal exam d. Deferred Recall: All subject should have a passing mark 82. Which of the following is NOT a qualification for being a member of the Board of Mechanical Engineering (BME)? a. Must be at least thirty-five (35) years of age b. Naturalized or born Filipino citizen * c. Has never been convicted of any offense involving moral turpitude d. A Professional Mechanical Engineer with a valid professional license and an active practitioner as such, for not less than ten (10) years prior to his appointment 83. A foreign mechanical engineer may only be allowed to practice mechanical engineering in the Philippines if his country also allows Filipino mechanical engineers practice under similar provisions like those in RA8495. This policy is also known as a. Foreign Reciprocity * b. Foreign Reciprocality c. Foreign engineer exchange d. Foreign affairs

ME 601

84. According to RA 8495, where will the budget to implement Mechanical Engineering Law come from? a. Local Budget b. Foreign funding c. National Budget * d. From members of PSME 85. A plant with rated capacity 500 kW requires with 3 shifts has: a. At least one(1) registered mechanical engineer or Professional Mechanical engineer for all three shifts b. At least two(2) registered mechanical engineer or Professional Mechanical engineer for all three shifts c. At least three(3) registered mechanical engineer or Professional Mechanical engineer, one per shift * d. At least four (4) registered mechanical engineer or Professional Mechanical engineer, one per shift plus one (1) as a buffer Basic Electronics 86. If N1/N2 = 2, and the primary voltage is 120 V, what is the secondary voltage? a. 0 V b. 36 V c. 60 V * d. 240 V Consider ratio: N1/N2 = Primary/Secondary 2 = 120V/Secondary Secondary = 120V/2 = 60 V 87. A transformer has a turns ratio of 4: 1. What is the peak secondary voltage if 115 V rms is applied to the primary winding? a. 40.7 V * b. 64.6 V c. 163 V d. 650 V N1/N2 = 4:1 RMS = Peak/sqrt(2) = 115 So, for the primary voltage peak 115 = Peakprimary/sqrt(2) Peak primary = 162.6345 V Consider the turn ratio: 4:1 = 162.6345:Secondary Secondary voltage = 162.6345/4 = 40.6586 = 40.7V

ME 601

88. Line voltage may be from 105 V rms to 125 rms in a half-wave rectifier. With a 5:1 stepdown transformer, the maximum peak load voltage of an ideal approximation is closest to a. 21 V b. 25 V c. 29.6 V d. 35.4 V* Approximate computation: (105 + 125)/2 = 115 rms voltage 115 = Vpeak/Srqt(2) Vpeak = 162.6345 5:1 = 162.6345:Secondary Secondary = 32.52 = 35.4 89. What is the peak load voltage out of a bridge rectifier for a secondary voltage of 15 V rms? (Use second approximation.) a. 9.2 V b. 15 V c. 19.8 V * d. 24.3 V Approximate computation: 15V = Vpeaksecondary / sqrt(2) Vpeak Secondary = 21.21 Using second approximation: Diode is silicon with 0.7 voltage drop that doubles, considering it’s a full bridge (4 diodes) Finally, output voltage will be 21.21 volts – 2(0.7)volts = 19.81 Volts 90. If line frequency is 60 Hz, the output frequency of a bridge rectifier is a. 30 Hz b. 60 Hz c. 120 Hz * d. 240 Hz ACDC: 91. If line frequency is 60 Hz, the output frequency of a bridge rectifier is a. 30 Hz b. 60 Hz

c. 120 Hz * d. 240 Hz

92. If N1/N2 = 2, and the primary voltage is 120 V, what is the secondary voltage? a. 0 V c. 60 V * b. 36 V d. 240 V V1/V2 = N1/N2 120V/V2 = 2 V2 = 120/2 = 60V

ME 601

93. A transformer has a turns ratio of 4: 1. What is the peak secondary voltage if 115 V rms is applied to the primary winding? a. 40.7 V * c. 163 V b. 64.6 V d. 650 V V1/V2 = 4/1 Vrms = Vpeak/sqrt(2) 115V rms = Vpeak Primary/sqrt(2) V peak Primary = 115V * sqrt(2) = 162.6345 V 162.6345/V2 = 4/1 V2 or V secondary = 162.6345/4 = 40.6586 = 40.7V

94. Line voltage may be from 105 V rms to 125 rms in a half-wave rectifier. With a 5:1 stepdown transformer, the maximum peak load voltage of an ideal approximation is closest to a. 21 V c. 29.6 V b. 25 V d. 35.4 V * Max RMS = 125V 125V = Vpeak/sqrt(2) Vpeak = 125V(sqrt(2)) = 176.77 V considered the primary voltage (greater) Using Ratio of 5:1 V1/V2 = 5/1 176.77/5 = V2/1 V2 = 35.355 V =35.4 V

95. If the base current is 100 mA and the current gain is 30, the collector current is a. 300 mA c. 3.33 A b. 3 A * d. 10 A Probability & Statistics: 96. From a lot of 10 grenades, 4 are selected at random and are thrown. If the lot contains 3 defective grenades that will not explode upon throwing, what is the probability that all 4 will explode?

ME 601

A. 1/4

B. 1/8

C. 1/7

D. 1/6 *

97. The probability that a person, living in a certain town, owns a cat is estimated to be 0.3. Find the probability that the ninth person randomly interviewed in that town is the fourth one to own a cat. A. 0.089 B. 0.019 C. 0.076* D. 0.034 Analysis:

ME 601

RECALL: P=nCrPrQn-r If the 9th person is the 4th to own a cat, there must be 3 cat owners among the first 8 people. This can be found using binomial expansion theorem: (8 C 3)(0.3)^3(0.7)^5= 0.25412 is the probability that there are 3 dog owners among the first 8 people. The 9th person must also own a dog, so multiply .3 to this number to get 0.076 (answer). 98. Find the probability that a person flipping a coin gets the third head on the seventh flip. A. 15/128 * B. 11/128 C. 17/128 D. 19/128 Analysis: RECALL: P=nCrPrQn-r P occurs, Q (failed to occur) The 3rd head must be on the 7th trial, the other 2 head in the remaining 6 trials; The value of probability of 2 head in the remaining 6 trials is; P = (6 C 2)*(0.5)3*(1-0.5)6-2 = 15/128 99. Suppose the probability is 0.7 that any given person will believe a tale about the ECE board exam leakage. What is the probability that the fifth person to hear this tale is the third one to believe it? A. 0.185 * B. 0.139 C. 0.110 D. 0.142 Analysis: RECALL: P=nCrPrQn-r P occurs, Q (failed to occur) P(occur) = 0.7 Q(failed to occur) = 1 – 0.7 = 0.3 2 others believed, from the 4 other persons 4 C 2 * (0.7)^3 * (0.3)^2 = 0.1852 = 0.185

ME 601

100. A veterinarian vaccinates several hamsters, one at a time, with a disease germ until he finds 3 that have contracted the disease. If the probability of contracting the disease is 1/5, what is the probability that 6 hamsters are required? A. 0.091 B. 0.071 C. 0.041 * D. 0.021 Analysis: RECALL: P=nCrPr+1Qn-r P occurs, Q (failed to occur) P(occur) = 1/5 Q(failed to occur) = 4/5 r+1=3 r=2 n = 6 hamsters – 1 = 5 P = 5 C 2 * (1/5)3 * (4/5)^3 = 0.04096 = 0.041

Algebra: 1. Find the term containing x2 in the expansion of [x3+(a/x)]10. a. 128a7x2

b. 210a6x2

c. 120a7x2 *

d. 320a5x2

rth term, yr = (n – r + 1)/r! (first)n-r(second)r nCr-1(first)n-r+1(second)r-1 From given, first term is x3, n = 10, r = 7 10C7 = 120a7

2. The terms of a sum may be grouped in any manner without affecting the result. this is law known as:

a. Commutative Law

c. Distributive Law

ME 601 b. Associative Law*

d. Reflexive Law

Recall: associative law, order does not affect the sum: (a + b) + c = a + (b + c) 3. A group consists of n engineers and n nurses. If two of the engineers are replaced by two other nurses, then 51% of the group members will be nurses. Find the value of n. a. 80

Let:

b. 110

c. 55

d. 100*

n – 2 = no. of engineers n + 2 = no. of nurses

EQN: (n+2)/(2n) * 100 = 51 N = 100

4. Jose’s rate of doing work three times as fast as Bong. On given day Jose and Bong work together for 4 hours then Bong was called away and Jose finishes the rest of the job in 2 hours. How long would it take Bong to do the complete job alone? a. 18 hrs.

b. 22 hrs.*

c. 16 hrs.

d. 31 hrs.

EQN: Jose = 3Bong (1/3bong + 1/bong) = Complete Job = 1.33333 (1/jose + 1/3jose)4 hours + (1/jose)2hours = 1 complete job Jose = 7.3333 Bong = 3 * 7.333 = 22hrs 5. A golf ball is dropped from a height of 6 meters. On each rebound it rises 2/3 of the height from which it last fell. What distance has it traveled at the instant it strikes the ground for the 7th time? a. 27.89 m* c. 20.87 m b. 19.86 m d. 24.27 m

Recall: Geometric Progression S = a1/(1-r) a1 = 6*(2/3) = 4 S7th = a1*(rn – 1)/(2/3-1) = 4((2/3)7 – 1)/(2/3-1) = 11.297

ME 601 Distance travelled = height + 2*(Snth) = 6 + (2*11.297) = 28.595 approx to 27.89 m **Twice because it bounces

Trigonometry: 6. In the right triangle ABC below, what is the cosine of angle A if the opposite side is 3 and the adjacent side is 4. a. 5/3

b. 5/4

c.3/5*

d.4/5

RECALL: soh cah toa CosA = Adjacent/Hyp Opposite = 3 Adjacent = 4 Hyp = sqrt(32 + 42) = 5 Finally: CosA = adj/hyp = 4/5

7. The length of sides AB and AC in the triangle below are equal. What is the measure of angle ∠ A if angle ∠ C is 70°? a. 70°

b. 55°

c.40°*

d. 110°

Recall: Triangle with two equal sides. Angle A can be bisected to form a right triangle. Then, 90° + 70° + BisectedA° = 20°, Finally, A° = 2*20° = 40°

8. What is the angle measured in degrees clockwise from the north to the direction in which the carrier is traveling. a. . bearing c. direction b. azimuth d. course *

Recall: Bearing means the direction a vessel is pointed, which is the measure of an acute angle with respect to the north-south vertical line.

ME 601 Azimuth is defined as a horizontal angle measured clockwise from any fixed reference plane or easily established base direction line.

Direction is expressed as the angular difference in degrees from a reference direction, usually true North

Course (Heading) is the direction the vessel is actually travelling. It is the angle measured clockwise from the north direction to the line of travel ==Course (heading) and bearing are only synonyms when there is no wind on land ==Direction is often given as bearing

Ans is COURSE.

9. If coversed Sin  0.134 , find the value of versed Sin . a. 0.8 b. 0.3 c. 0.5*

d. 0.2

Recall: Coversine = 1 - Sin∅ Versine = 1 - Cos∅

; ;

l

Versine

- Sin∅

– Cos ଦl

;∅

l

ଦl

li

10. What is the value of the adjacent side if the opposite side is 1 inch and the 2 other angles of the right triangle is 30° and 60° a. 1/√3*

b. √2

c. √3

d. 6/√3

Recall: Right Triangle Theorem for 30-60-90. Figure (1 inch goes to the shortest leg, √3 is the mid leg, and 2 will be the hypotenuse)

ME 601 Letter A will be a more sensible answer, Sin(30) = opposite/adjacent = 1/√3

Analytic Geometry and Solid Mensuration 11. The base of a cylinder is a hexagon inscribed in a circle. If the difference in the circumference of the circle and the perimeter of the hexagon is 4 cm., find the volume of the prism if it has an altitude of 20 cm. a. 10,367 cm3*

b. 12,239 cm3

c. 10,123 cm3

d. 11,231 cm3

RECALL: Geometry (Hexagon inscribe in a circle)

Circ of Circle = 2piR Perimeter of Pentagon = 6*Sides 4cm = (2piR) – 6Sides PrismVOLUME = ½ BHL ***For a Regular Hexagon, Side is equal to radius 4cm = 2pi(R) – 6(R) Radius = 14.125 cm Solving for base of the triangle =R = Base = 14.125 cm Height; sin60 = [height/radius] = height/14.125; = 12.232 cm Finally; PrismVOLUME = ½ BHL = ½ (14.125cm)(12.232cm)(20cm) 12. Locate the centroid of the area bounded by the parabola y2=4x, the line y=4 and the y-axis a.6/5, 4 b.5/6, 3

Recall: Area Bounded (Analytic) y2 = 4x @ y = 4, then x = 4 y2 = 4x @ x = 4, then y = ±4

c. 6/5, 3 * d. 6/6, 3

ME 601

We can assume, the parabola and the line y=4 intersect at points (4,4) and (4,-4) Area = 2/3 * b*h = 2/3 * 4 * 4 = 32/3 square units 32/3 y = 32 Y=3 32/3x = 25.59

13. Find the volume of a spherical cone in a sphere of radius 17 cm if the radius of its zone is 8 cm. a. 2120.35

b. 1426.34

c. 1210.56*

d. 2316.75

Vspcone = 1/3(Azone*Radiussphere)*Radiuszone Azone = 2piRsphere = 2pi(17 cm) = 106.814 cm Vsphericalcone = 1/3(106.814cm)(17cm)(8cm) = 4839.78 cm3 But: 1/3(106.814cm)(17cm) * 2 = 1210.56

14. Find the equation of the perpendicular bisector of the segment joining the points (2, 6) and (-4, 3). a. 2x - 4y + 5 = 0 b. 4x + 2y – 5 = 0*

c. 2x + 4y + 5 = 0 d. 5x – 2y + 4 = 0

Perpendicular bisector has a slope mperp = 1/m Equation of line being bisected using two point form Y – y1 = (y2 – y1)/(x2 – x1) * [(x – x1)] Y – 6 = (3 – 6)/(-4 – 2) * [(x – 2)] Y – 6 = ½ * (x - 2) Y = x/2 – 2 + 6 = x/2 +4 -x + 2y – 4 = 0 Slope is ½ M perp is -2 B is most sensible because it has a slope of -2 15. What is the new equation of the line 5x + 4y + 3 = 0 if the origin is translated to the point (1, 2)? a. 4x’ + 3y’ + 16 = 0

c. 5x’ + 4y’ + 16 = 0*

ME 601 b. 5x’ – 4y’ – 16 = 0

d. 6x’ + 6y’ – 16 = 0

Substitute value of x,y (1,2) from the original eqn: 5(1) + 4(2) + 3 = 16 Then 5x’ + 4y’ + 16 = 0

Differential Calculus 16. A snowball is being made so that its volume is increasing at the rate of 8 ft3/min. Find the rate at which the radius is increasing when the snowball is 4 ft in diameter. a. 0.159 ft/min* b. 0.259 ft/min

c. 0.015 ft/min d. 0.325 ft/min

Volume of Sphere = 4/3 * pi * R3 dV/dt = 8 ft3/min dV/dt = 4/3 pi * 3R2(dR/dt) 8 ft3/dt = 4 pi R2 (dR/dt) @ D = 4 ft; R = 2 ft dR/dt = (8ft3/dt) / (4*pi * (2ft)2) = 0.1591 ft/min 17. Find the limit (x+2)/(x-3) as x approaches 3. a. 0

b. infinity*

c. indeterminate

d. 3

Cannot be factored Infinity (denominator is zero) 18. Given the total cost of producing x items is given by the function C(x) = 0.001x3 + 0.025x2 + 3x + 5. Compute the marginal cost of producing the 51st item. a. $13.00* c. $ 12.00 b. $ 14.00 d. $ 11.00

Since C(x) is the approximate cost incurred to produce the (x + 1)st item we need to compute C(50). For comparison, the exact cost to produce the 51st item is C(51) − C(50) = [0.001(51)3 + 0.025(51)2 + 3(51) + 5] −[0.001(50)3 + 0.025(50)2 + 3(50) + 5] = 355.676 − 342.50 = $13.176

ME 601 19. The dimensions of a rectangle are continuously changing. The length decreases at the rate of 2 in/sec while the width increases at the rate of 3 in/sec. At one instant the rectangle is a 20 inch square. How fast is its area changing 3 seconds later? a. 10 in2/sec b. 12 in2/sec

c. 16 in2/sec* d. 14 in2/sec

Initially, at time 0 sec; area is 20 sq2 = LW Consider: L = 20; W = 20 At time = 3 sec L = 20 – 2in/sec(3sec) = 14 inches W = 20 + 3in/sec(3sec) = 29 inches

Formula A = LW Derive both side (recall product) dA/dt = L(dw/dt) + w(dL/dt)

Area at time 3 sec = 14(3 in/sec) + 29(-2 in/sec) = 16 in2/sec dA/dt = -16 in2/sec

20. Evaluate: Lim (tan33x) / x3 as x approaches 0. a. 0

b. 31

c. 27*

Indeterminate if x = 0 Derive 3 times then substitute x = 0

Integral Calculus 21. Find



dx x 1 .

a. x x  1  C

c. 2x x  1 C

d. Infinity

ME 601 b. 2 x  1  C *

d. x 2x  1  C

Non-conventional way of solving: Disregard integral sign, then solve using CALC mode with initial value as 1.1 for x Note answer Try each choices using function d/dx on calc with value of x = 1.1, answer should be the same with given eqn

1/√(x-1) @ x = 1.1; ans = 3.162 d/dx(2√(x-1)) x = 1.1 ; ans = 3.1622

ANS: B 2 x  1  C

3x 22. Evaluate:   2

0

y

2

0



 9y2 dxdy

a. 30

b. 50

c. 40 *

d. 20

Recall: Multiple integration (First in Last out) = =

h

=

h

=

h

h

=

h

= 44 =answer approx. 40

=

23. Find the area of the region above the x axis bounded by the curve y = -x2 + 4x – 3. a.1.333 square units* b.2.122 square units

c. 3.243 square units d. 1.544 square units

RECALL: Integral, AREA BOUNDED = definite integral of ydx First, solve for limits of x by point of intersection At y = 0, x = 3, and 1 (use mode 5:3) Finally, Area =

t

h

t

= 4/3 = 1.3333

24. Find the area in the first quadrant bounded by the parabola y2=4x and the line x=3 and x=1

ME 601 a.5.595*

b. 4.254

c. 6.567d. 7.667

Recall: Area Bounded by Curve (integral) Y2 = 4x @ x(0), y = 0 *8.666 25. Find the volume of a solid formed by rotating the area bounded by y = x2, y = 8 – x2 and the y axis about the x axis. a. 268.1*

b. 287.5

c. 372.9d. 332.4

Differential Equations 26. If f(x) and g(x) are differentiable functions such that f '(x) = 3x and g'(x) = 2x2 then the limit lim [(f(x) + g(x)) - (f(1) + g(1))] / (x - 1) as x approaches 1 is equal to a. 5* b. 10 2 3(x) + 2(x) @x = 1; 3(1) + 2(1) = 5

c. 20

d. 15

2

27. . Given z = 2y2 , y = x/ (x+1), and x = 1/(t – 1). Find dz/dt. a. -4/ t3 * b. 4t3 c. 2 t3 - 1

d. - 8t3

28. Find the differential equation of y  c1  c2 e3 x . a. y’’ + y’ = 0 c. y’’ - 2y’ = 0 b. y’’ - 3y’ = 0 * d. y’’ + xy’ = 0 29. Obtain the differential equation of the family of straight lines with slope and y-intercept equal. h h h h a. c. t h t h b. d. * 30. Under certain conditions, cane sugar in water is converted into dextrose at a rate proportional to the amount that is unconverted at any time. If, of 75 kg at time t = 0, 8kg are converted during the first 30 minutes, find the amount converted in 2 hours. a. 72.73 kg c. 27.23 kg* b. 23.27 kg d. 32.72 kg

ME 601

Probability and Statistics 31. How permutation can be made out of the letters in the world island taking four letters at a time? a. 360* b. 720 c. 120 d. 24

No repeating letters: Number of letters: 6 = 1i, 1s, 1l, 1a, 1n, 1d nPr is permutation; 6P4 = 360

32. The captain of a baseball team assigns himself to the 4th place in the batting order. In how many ways can he assign the remaining places to his eight teammates if just three men are eligible for the first position? a. 2160 b. 40320 c. 5040 d. 15120*

Let each _1 pos_ _2pos_ _3pos_ _4pos(captain)_ 3 only on first post 7 on second post 6 on 3rd post 5 on 4rd post …. 3*(7!) = 15120

33. A bag contains 3 white and 5 black balls. If two balls are drawn in succession without replacement, what is the probability that both balls are black? a. 5/28 b. 5/16 c. 5/32 d. 5/14*

Total Balls = 8 (3W)(5B) First draw (5)/8

ME 601 Second draw (4)/7 5/8 * 4/7 = 5/14

34. Find the mean, median and mode respectively of the following numbers: 13, 13, 14, 12, 11, 10, 9, 11, 8, 11, 5, and 15. a. 10, 10, 10 b. 10, 11, 11

c. 10, 11, 10 d. 11, 11, 11*

Recall: Mean is average = 13 + 13 + 14 + 12 + 11 + 10 + 9 + 11 + 8 + 5 + 15 / 12 items = 11 Median = sort and then find middle value = 15 14 13 13 12 11 11 11 10 9 8 5, even so average . 22/2 = 11 Mode: the number with highest frequency or appear the most = 11 Ans: 11, 11, 11 35. There are 4 white balls and 6 red balls in a sack. If the balls are taken out successively (the first ball is not replaced), what is the probability that the balls drawn are of different colors. a. 23/90

b. 8/15*

c. 24/103

d. 7/15

2 cases will happen: either (or)(+) First case = draw white, then draw red = (4/10)(6/9) = 4/15 Second case = draw red, then draw white = (6/10)(4/9) = 4/15 Total = 4/15 + 4/15 = 8/15 Physics 36. What horizontal force P can be applied to a 100-kg block in a level surface with coefficient of friction of 0.2, that will cause and acceleration of 2.50 m/s2? a. 343.5 N c. 106 N b. 224.5 N d. 446.2 N* 37. What is the force in newtons, required to move a car with 1000 kg mass with an acceleration of 12 m/s2? a. 11,384

b. 12500

c. 10450

Rekta with given: M = 1000 kg A = 12 m/sec2 F = ? = ma = 1000kg * 12 m/sec2 = 12000 kg-m/sec2 = 12000 N

d. 12000*

ME 601 38. A block weighing 500 kN rest on ramp inclined at 25 degrees with the horizontal. The force tending to move the block down the ramp is: a. 213.43 kN

b. 343.43 kN

c. 128.45 kN

d. 211.31 kN*

Recall Kinematics: (assuming no friction coeff) F = sin25(W) = sin25(500 kN) = 211.309 kN = 211.31 kN 39. What is the range of a projectile if the initial velocity is 30 m/s at an angle of 30 degrees with the horizontal. a. 69.87 m Recall: Projectile Motion:

b. 79.45 m*

c. 64.69 m

d. 87.09 m

Analysis: Once fired with initial velocity, it will reach a maximum height reaching max velocity 0, then from that instant, initial velocity would be 0 and acceleration due to gravity pulls the body gaining velocity. Range = Vo2(Sin(2*angle))/Acceleration due to gravity = [(30 m/s)2*sin(2*30°)]/(9.81 m/sec2) = 79.4518 meters 40. A body weighs 40 lbs. starts from rest and inclined on a plane at an angle of 30o from the l l How long will it move during the third horizontal for which the coefficient of friction second? a. 19.99 ft c. 18.33 ft b. 39.63 ft d. 34.81 ft*

Summation of Forces at y = 0 W*sin∅ = REF + F

ME 601 W*sin∅ W*sin∅

W/g * accel h W/g * accel h

Cancel out W

Sin∅ = accel/g +

th

∅  coeff of friction reacting to body moving down the wedge

h ∅

Accelartion due to gravity, g = 32.2 ft/sec2

Sin 30° = a/[32.2 ft/sec2] + 0.3(cos30°) A = 7.732 ft/sec2

From initial velocity of 0, solve distance after t = 3 seconds Solving for distance, S = v*t + ½ a*t2 = 0*(3 sec) + ½ (7.732ft/sec2)(3 sec)2 = 34.794 ft

Mechanics 41. Determine the outside diameter of a hollow steel tube that will carry a tensile load of 500 kN at a stress of 140 MPa. Assume the wall thickness to be one-tenth of the outside diameter. a. 123 mm c. 103 mm b. 113 mm* d. 93 mm

Recall: Stress t = P/A; 140 MPa = 500kN/A A = 500/140 = 3.571 kN/MPa = 0.00357 mm2 Thickness is 1/10 of Doutside = 0.1 Doutside Thickness = Doutside – Dinside Ahollow = Doutside – Dinside = thickness 0.00357 mm2 = pi/4(D)2 – pi/4(D-2(0.1D))2 D = 112.4 mm = approx. 113 mm

42. A force of 10 N is applied to one end of a 10 inches diameter circular rod. Calculate the stress. a. 0.20 kPa* c. 0.10 kPa

ME 601 b. 0.05 kPa

d. 0.15 kPa

Recall: Stress = P/A = 10N/pi/4(D)2 10 inches = 24.4 cm = 0.244 m Stress = 10/pi/4(0.244)2 = 213.86 Pa = 0.2 kPa approx

43. What force is required to punch a 20-mm diameter hole through a 10-mm thick plate? The ultimate strength of the plate material is 450 MPa. a. 241 kN c. 386 kN b. 283 kN* d. 252 kN

Recall: Shearing Stress = F/(pi*D*thickness) F = Shearing Stress * Area * thickness = 450 MPa * pi * 20 * 10 = 282743 N = 283 kN approx

44. A steel pipe 1.5m in diameter is required to carry an internal pressure of 750 kPa. If the allowable tensile stress of steel is 140 MPa, determine the required thickness of the pipe in mm. a. 4.56 c. 4.25 b. 5.12 d. 4.01* Recall: Pressure Vessel (internal pressure causing tangential stress against thickness) St = [Internal Pressure (Diameterinternal)]/2*thickness) *From here: use backdoor or Caltech, noting answer should be close to 140 MPa On this case: Thickness = Dout-Din; Din = Dout - thickness St = 750 kPa (1.5-0.004)m / 2 (4.01) = 139.900 139.9 = approximately 140 suits the allowable tensile stress

45. A spherical pressure vessel 400-mm in diameter has a uniform thickness of 6 mm. The vessel contains gas under a pressure of 8,000 kPa. If the ultimate stress of the material is 420 MPa, what is the factor of safety with respect to tensile failure? a. 3.15* c. 2.15

ME 601 b. 3.55

d. 2.55

Recall: Pressure Vessel sphere(as stated in the problem) St = (Internal Pressure) * (Di)*(factor of Safety)/4t **Considering SF Thickness = prn/2oy P, pressure = 8000 kPa Oy, Yield Stress = 420 MPa (Yield stress is the ultimate stress) Diameter = 400 mm, Radius = 200 mm Thickness = 6 mm Therefore, N Safety factor = 3.15

46. A bomber flying at a horizontal speed of 800 kph drops a bomb. If the bomb hits the ground in 20 seconds, calculate the vertical velocity of the bomb as it hit the ground. a. 169 m/sec b. 196 m/sec * c. 175 m/sec d. 260 m/sec Analysis: This is a falling bodies problem, applied with conservation of energy. Vx = 800 kph; t = 20 sec Solve for Vy The bomb has its initial velocity of 0 when dropped (free falling) Find velocity as it hits the ground or the final velocity. It hits the ground in 20 seconds G = (Vf – Vo)/t 9.81 = (Vf – 0)/20 sec; Vf = 196.2 m/sec

47. A flywheel starting from rest develops a speed of 400 rpm in 30 seconds. How many revolutions did the flywheel make in 30 seconds it took to attain 400 rpm. a. 100 rev * b. 150 rev c. 120 rev

ME 601

d. 360 rev Recall: Angular Velocity Vo = 0; time = 30 sec Vf = 400 rev/min 400 = 0 + Accel(0.5min) Accel = 800 rev/min2 ωo2 = ω + 2α∅ 4002 = 0 + 2*800rev/min2*∅ ∅ rev

48. A 100 kg block of ice is released at the top of a 300 incline 10 meters above the ground. If the slight melting of the ice renders the surface frictionless, calculate the velocity at the foot of the incline. a. 30 m/sec b. 24 m/sec c. 14 m/sec * d. 10 m/sec

Force acting normal to block: F = ma ; 10 kg * 9.81 m/sec2 = 98.1 kg-m/sec2 Force parallel to surface of plane/wedge (pull/slide down the block) = Fsin30 = 49.05 Vf2 = V02 + 2*accel*distance Accel = F/mass ; this is the force pulling/sliding the object down the plane = 49.05/10kg Accel = 4.905 Finally Vf2 = 0 + 2*4.905*(10/sin30) **hypotenuse Vf = 14.007 m/sec 49. What drawbar pull is required to change the speed of a 120,000 lb car from 15 mph to 30 mph on a half mile while the car is going up a 1.5% upgrade? Car resistance is 10 lb/ton. a. 3425 lbs * b. 3542 lbs c. 3245 lbs d. 4325 lbs Vo = 15 mi/hr Vf = 30 mi/hr **half mile (distance travelled) Mass = 120,000 lb tan∅ = 0.015 ∅ = 0.8593

ME 601

Solve for the accel, speeding up the car from 15 - 30 50. A body weighing 200 kg is being dragged along a rough horizontal plane by a force of 45 kg. If the coefficient of friction is assumed to be 1/12 and the line pull makes an angle of 180 with the horizontal, what is the velocity acquired from rest in the first 3 meters. a. 2.8 m/sec * b. 3.1 m/sec c. 3.5 m/sec d. 4.9 m/sec Mass = 200 kg ; ∅ = 18° ;

Force Mass = 45 kg ; μ =0.083333 or use 1/12 Vo = 0 ; Vf = ??? ; Distance, S = 3

Thermodynamics 1 51. An iron block weighs 5 Newton and has volume of 200 cm^3. What is the density of the block? a. 2458 kg/m^3 b. 2485 kg/m^3 c. 2584 kg/m^3 d. 2549 kg/m^3*

Recall: Density = mass/volume 5 N  kg = F = m*accel Mass = 5 N/9.81 kg/m3 = 0.5096 Density = 0.5096/200(1/100)3 = 2548.419 kg/m3 = approx. to 2549

52. If air is at a pressure of 22.22 Psia and at temperature of 800 degR, what is the specific volume? a. 11.3 ft^3/lbm b. 33.1 ft^3/lbm c. 13.3 ft^3/lbm* d. 31.3 ft^3/lbm

Recall: PV = mRT (pressure volume temp relation)

ME 601

22.22 lb/in2 * V = m(Rair)(800 Rankine) v/m = (53.3 ft-lb/lb-rankine)(800 rankine)/22.22 lb/in2(12in/ft)2 v/m = 13.32 ft3/lbm

53. The specific gravity of mercury is 13.55. what is the specific weight of mercury? a. 123.9 Kn/m^3 b. 139.2 Kn/m^3 c. 132.9 Kn/m^3* d. 193.2 Kn/m^3

S.G = density of fluid/density of water = 13.55 Density of Mercury

= 13.55*1000 kg/m3 or 13.55 * 9.81 kN/m3 = 13550 kg/m3 or 132.9255 kN/m3

54. The equivalent weight of mass 10 kg at location where the acceleration of gravity is 9.77 m/sec^2 a. 97.7 N* b. 79.9 N c. 77.9 N d. 977 N

Recall: W is Force; Force = mass*accel = 10 kg (9.77 m/sec2) = 97.7N

55. Transportation company specializes in the shipment of pressurized gaseous material. An order is received for 100 liters of a particular gas at STP (32 degF and 1 atm). What minimum volume tank is necessary to transport the gas at 80 degF and maximum pressure of 8 atm?

ME 601 a. 16 liters b. 14 liters* c. 10 liters d. 12 liters

Recall: Required to compute for the volume. The problem didn’t state any condition (e.g. rigid, constant temp, adiabatic etc.) Assume PVT rel [PV/T]1 = [PV/T]2 1 atm (100 Liters)/(32°F + 460)°R = 8 atm (V2)/(80°F + 460)°R V2 = 13.719 Liters = approx. to 14 Liters

Thermodynamics 2 56. An air standard engine has a compression ratio 18 and a cut-off ratio 4. If the intrake air pressure and temperature are 100 kpa and 27deg C, find the work in KJ per kg. a. 2976 b. 2166 c. 1582* d. 2751

Recall Engines: 2 types (otto and diesel) (problem did not state which of the two types, use analysis base on formula ratios since it is generic to all ICE’s )

Comp Ratio = V1/V2 = 18 Cut-off Ratio = V3/V2 = 4  only Otto cycle doesn’t have Cut off **assume this is diesel P1 = 100 kPa T1 = 27 °C Solve for Wnet; Efficiency

Wnet = efficiency*Qin = 1 – B/A = 1 – Rcset/Rkset = 1 – [41.4-1]/[1.4(4-1)]/181.4-1 = 0.5531 **utilize shift store

ME 601 T2 = T1(rk)k-1 = (27+273)°K(18)1.4-1 = 953.30°K T3 = T2(V3/V2) (constant pressure-spsv-diesel) = 953.30°K(4) = 3813.20°K

Qin = Q2-3 = mCp(deltaT3-2) = 1kg (1.0062)(3813.20-953.30) = 2877.63 Finally, W = 0.5531 * 2877.63 = 1591.62 kJ/kg = approx. 1582

57. Determine the air – standard efficiency of an engine operating on the diesel cycle when the suction pressure is 99.97KPa and the fuel injected for 6% of the stroke, the clearance volume is 8% of the stroke. Assume k = 1.4. a. 60.70 %* b. 65.01% c. 67.01% d. 64.02%

Efficiency

= Wnet/Qin = 1 – B/A = 1 – Rcset/Rkset

V3 - V2 = 0.08VD V2 = 0.06VD V3 = 0.08VD + 0.06VD V3 = 0.14VD rc = V3/ V2 = 0.14VD /0.06VD = 2.333 rk = (1 + 0.06)/(0.060) = 17.667

Finally, E = 1 – [2.3331.4-1]/[1.4(2.333-1)]/17.6671.4-1 = 0.6136 **utilize shift store Closest Answer = 60.70%

58. A 23.5 kg of steam per second at 5MPa and 400oC is produced by a steam generator. The feedwater enters the economizer at 145oC and leaves at 205oC. The steam leaves the boiler drum with a quality of 98%. The unit consumes 3kg of coal per second as received having a heating value of 25102 KJ/kg. What would be the overall efficiency of the unit in percent? Steam properties: At 5MPa and 400oC: h=3195.7KJ/kg

ME 601 At 5MPa: At 205oC: At 145oC: a. 65 b. 95 c. 88* d. 78

hf = 1154.23, hg =1640.1 hf = 875.04 hf = 610.63

Recall: Steam Generator (Referring to boiler efficiency) ƞBOILER = MSTEAM(hSTEAM - hFUEL)/MFUEL(Qh) = [23.5 kg (3195.7 – 610.63)]/3kg(25102) ƞBOILER = 0.8581 = 85.81% or approx. to 88%

59. In a Rankine cycle steam enters the turbine at 2.5MPa (enthalpies and entropies given) and condenser of 50KPa (properties given), what is the thermal efficiency of the cycle? At 2.5MPa: hg = 2803.1 sg = 6.2575 At 50KPa: sf = 1.0910 sfg = 6.5029 hf = 340.49 hfg = 2305.4 vf = 0.0010300 a. 25.55%* b. 28.87% c. 30.12% d. 31.79%

RECALL: Rankine Cycle (SPSP) (Common in Steam Generators/PowPlant) Solve for Thermal Efficiency: ƞTHERMAL = WNET/QREJECTED = [(h1 – h2) – (h4-h3)]/(h1-h4) h1 = hg = 2803.1 s2= s1 = sg = 6.2575 solving h2 = hf + xhfg x = (s2 – sf)/sfg = (6.2575 – 1.0910)/6.5029 = 0.7944 h2 = 340.49 + 0.7944(2305.4) = 2172.11 h3 = hf = 340.49 h4 = hf + vf(P2 – P1) = 340.49 + 0.00103(2500 – 50) = 343.0135

ƞTHERMAL = [(2803.1 – 2172.11) – (343.0135-340.49)]/(2803.1-343.0135) = 0.2554

ME 601 ƞTHERMAL = 25.54%

60. A 4 liter (2-liters per revolution at standard pressure and temperature) spark ignition engine has a compression ratio of 8 and 2200 KJ/kg heat addition by the fluid combustion. Considering a cold air standard Otto cycle model, how much power will the engine produce when operating at 2500 rpm? a. 166.53 hp * b. 73.12 hp c. 97.4 hp d. 148 hp

Density, ρ = 1.2 kg/m3 (air standard) Mass flow rate = 2 Liter/rev(m3/1000 Liter)(2500 rev/min)(1 min/sec)(2 kg/m3)= 0.10 kg/sec ƞOTTO = 1 – 1/rkk-1 = 1 – 1/81.4-1 = 0.5647 W

= ƞOTTOQA = 0.5647(2200) = 1242.3943 kJ/kg * 0.10 = 124.2394 kJ/sec = 124.2394 kW = 166.6078 HP

Heat Transfer 61. The surface of household radiator has an emissivity of 0.55 and an area of 1.5 m2. At what rate is the radiation absorbed emitted by the radiator when its temperature is 50 deg. C.? a. 308 W b. 509 W* c. 108 W d. 409 W

RECALL: Radiant Heat Transfer = constant * emissivity * area * (T)4 = 5.6704 * 10-8 * 0.55 * 1.5 m * (50 + 273)4 = 509 W

62. Cool water at 9 deg. C enters hot-water from which warm water at a temperature of 80 deg. C is drawn at an average rate of 300 g/min. How much average electric power does the heater consume in order to provide hot water at this rate? a. 4.18 kW* b. 2.35 kW c. 3.31 kW d. 5.14 kW

ME 601

Mass flow rate = 300g/min = 0.005 kg/sec Q = mCp(Delta T) = 0.005 kg/sec (4.187)(80-9) = 1.48 kW

63. Water (specific heat cv = 4.2 kJ/kg-K) is being heated by a 1500W heater. What is the rate of change in temperature of 1 kg of the water? a. 0.043 K/s b. 0.719 K/s c. 0.357 K/s* d. 1.50 K/s

Recall: Q = mCp(deltaT); 1500 W = 1kg*(4.2 kJ/kg-K)*(Temp) Temp rate = 0.357 k/sec

64. One kilogram of water (cv = 4.2 kJ/kg-K) is heated by 300 BTU of energy. What is the change in temperature, in K? a. 17.9 K

b. 71.4 K

c. 73.8 K

d. 75.4 K*

Recall: Q = mCp(deltaT); 300 BTU = m(4.2 kJ/kg-K)(deltaT) 1 BTU = 1055 J = 1.055 kJ 300 BTU = 316.5 kJ

316.5 kJ = m(4.2)(DeltaT) Temp = 75.3571 K

65. The condenser of a reheat power plant rejects heat at the rate of 600 kW. The mass flow rate of cooling water is 5 kg/s and the inlet cooling water temperature is 35C. Calculate the condenser cooling water exit. a. 43.45 C

b. 53.45 C

c. 63.66 C*

d. 74.34 C

ME 601

Q = mCp(DeltaT) 600 kW = 5 kg/sec * (4.187) * (Tf – 308) Tf = 336.66 K – 273 = 63.66 °C

Fluid Mechanics 66. Determine the submerged depth of a cube of steel 0.3 m on each side floating in mercury. The specific gravities of steel and mercury are 7.8 and 13.6 respectively. a. 0.155 m b. 0.165 m c. 0.134 m d. 0.172 m*

Recall: Bouyancy Force Bouyancy Force, FB = Specific Gravity of Water * Displaced Volume = ɣVD

Conditon: Bouyant Force = Weight Object SP gravity steel * gravity water * volume of steel = SP gravity mercury * gravh2o * 0.3*disp 7.81*9.81*0.33 = 13.6*9.81*0.32*d d = 0.172 m

67. A block of wood floats in water with 5 cm projecting above the water surface. When placed in glycerine of specific gravity of 1.35, the block projects 7.5 cm above the liquid. Determine its specific gravity. a. 0.514 b. 0.704 c. 0.836 d. 0.658*

Volume = Base * Area Bouyant Force when in Water: Spec.Grav.Water * DensityWater * V = DensityWater*VolDisplacement SpecGravWat*Area*height = Area(height – 5) cancel out density water SpecGrav = (height – 5)/height  cancel out Area, Equation 1

ME 601 Bouyant Force when in Glycerine: Spec.Grav.Water * DensityWater * V = SpecGravGlycer*DensityWat*VolDisplacement SpecGravWat*Area*height = 1.35*Area*(height-7.5) cancel out density water SpecGrav = 1.35(height – 7.5)/height  cancel out Area, Equation 2

Equating 1 and 2 (Height – 5)/Height = 1.35(Height – 7.5)/Height Height = 14.6428

Finally, SpecGrav = (14.64-5)/14.64 = 0.6585

68. A solid cube material is 0.75 cm on each side. If it floats in oil of density 800 kg/m^3 with onethird of the block out of the oil, what is the density of the material of the cube? a. 533 kg/m^3 * b. 523 kg/m^3 c. 513 kg/m^3 d. 543 kg/m^3

Recall: Bouyant Force, Weight = Bouyant Force Density, ρCUBE * VolTOTAL = ρOIL * VolDISPLACEMENT ρCUBE (0.0075m)3 = 800 kg/m3 * 2/3 submerged * (0.0075m)3 ρCUBE = 533 kg/m3

69. A hollow cylinder 1 m in diameter and 2 m high weighs 2825 N. How many kN of lead weighing 110 kN/m^3 must be fastened to the outside bottom of the cylinder to make it float with 1.5 m submerged in water? a. 8.5 KN * b. 6.5 kN c. 1.5 kN d. 9.5 kN

ME 601

Bouyant Force of Cylinder + Bouyant Force of Lead = Weight of Cyl + Weight of Lead

BFCYLINDER

= 9.81 kN/m3 * Volume Displaced = 9.81 kN/m3 * pi/4(Diameter)2(1.5) = 9.81 kN/m3 * pi/4(1)2(1.5) = = 11.56 kN

BFLEAD

= 9.81 kN/m3 * VolumeLEAD

WLEAD

= 110 * VolumeLEAD

11.56 kN + 9.81VLEAD = 2.825 + 110VLEAD VLEAD = 0.08718 = 8.7 kN = approx. to 8.5 kN

70. A 1 m x 1.5 m cylindrical tank is full of oil with SG = 0.92. Find the force acting at the bottom of the tank in dynes. a. 106.33 x 103 dynes b. 106.33 x 104 dynes c. 106.33 x 105 dynes d. 106.33 x 106 dynes* P = F/A Pressure

= specific weight * height = S.G * density water * height = 0.92 * 9.81 kN/m3 * 1.5 m = 13.5378 kPa

= PA = 13.5378 * pi/4 * (1m)2 = 10.632 kN = 106.325 * 106 dynes

F

**1 N = 100 000 dynes

Combustion 71. The dry exhaust gas from oil engine has the following gravimetric analysis: CO2 = 21.6%; O2 = 4.2%; N2 = 74.2% Specific heats at constant pressure for each component of the exhaust gas in Kcal/kgoC are: CO2 = 0.203; O2 = 0.219; N2 = 0.248 Calculate the specific gravity if the molecular weight of air is 28.97 kg/kg-mol.

ME 601 a. 0.981

Recall: SG

b. 1.244

c. 1.055*

d. 0.542

= density fluid/density water = density gas / density air = mw gas / mw air (in terms of molecular weight)

MW gas will be the total weight of in volumetric of CO2, O2 and N2

CO2 = (21.6%)/(12 + 2*16) = 0.004909 O2 = (4.2%)/(2*16) = 0.0013125 N2 = (74.2%)/(2*14) = 0.0264 Total Volumetric = 0.0049 + 0.0013 + 0.0264 = 0.0326 Molecular Weight = 1/0.0326 = 30.64 Finally, S.G. = 30.64/28.97 = 1.057 = 1.055

72. A bituminous coal has the following composition: C = 71.5%; H = 5.0%; O = 7.0%; N = 1.3%; S = 3%; Ash = 7.6%; W = 3.4%. Determine the theoretical weight of nitrogen in lb/lb of coal. a. 2.870

b. 7.526*

c. 2.274d. 6.233

Recall: Air and Fuel Ratio (Theoretical Weight of Air-solid fuel) WTHEORETICAL ANALYSIS

= 11.5 C + 34.5(H – O/8) + 4.3S = 11.5 (0.715) + 34.5(0.05 – 0.07/8) + 4.3(0.03) = 9.7746 lbAIR/lbCOAL

**N2 in AIR by weight = 76.8% Finally, N2 = 0.768(9.7746) = 7.5069 lb/lbCOAL

73. A gaseous fuel mixture has a molal analysis: H2 = 14%; CH4 = 3%; CO = 27%; O2 = 0.6%; CO2 = 4.5%; N2 = 50.9%; Determine the air fuel ratio for complete combustion of molal basis. a. 2.130

b. 3.230

c. 1.233*

d. 1.130

ME 601

Molal A/F

= [Molals of Oxygen + Molals of Oxygen(Nitrogen)]/1

Oxygen Chem Reaction 0.14H2 + 0.070O2 = 0.14H2O 0.93CH4 + 0.0600O2 = 0.03CO2 + 0.06H2O 0.27CO + 0.135O2 = 0.27CO2

All all Coeff of O2 (Left Side) = 0.265 Actual O2 in product

= oxygen from left side – O2 from given fuel (0.6%) = 0.265 – 0.006 = 0.259

Molal A/F

74.

= [0.259 + 0.259(3.76)]/1 = 1.23284 = 1.233 molAIR/molFUEL

A volumetric analysis of a gas mixture is as follows: CO2: 12%; N2: 80%; O2: 4%; CO: 4%. What is the percentage of CO2 on a mass basis? a. 17.55%*

b. 15.55%

c. 12.73%

d. 19.73%

Convert First Vol Analysis to Mass Basis: CO2 (12%) = 0.12*(12 for C + 2*16 for O) = 0.12*44 = 5.28 N2 (80%) = 0.8*(2*14) = 22.4 O2 (4%) = 0.04 (2*16) = 1.28 CO (4%) = 0.04 (12 + 16) = 1.12 Total = 30.08

%mass of CO2 = 5.28/30.08 = .1755 = 17.55%

75.

The following coal has the following ultimate analysis by weight: C = 70.5%; H2 = 4.5%; O2 = 6.0%; N2 = 1.0%; S = 3.0%; ash = 11%; moisture = 4%. A stocker fired boiler of 195000kg/hr steaming

ME 601 capacity uses this coal as fuel. Calculate volume of air in m3/kg with air at 60oF and 14.7 psia pressure of boiler efficiency is 70% and FE = 1.10. a. 234,019 m^3/hr *

c. 215,830 m^3/hr

b. 213,830 m^3/hr

d. 264,830 m^3/hr

Recall: Solid Fuel (Use DULONG FORMULA) (A/F)THEO = 11.5C + 34.5(H2 – O2/8) + 4.3S = 11.5(0.705) + 34.5(0.045 – 0.06/8) + 4.3(0.03) = 9.53025 (A/F)ACTUAL = (A/F)THEO x (1 + excess air%/100) = 9.53025 x (1 + 30/100) = 12.389 kgAIR/kgFUEL

Recall Factor of Evaluation (FE) common in steam power plant/boiler FE = (hSTEAM – hFUEL)/2257; FE*2257 = hSTEAM – hFUEL

QH = 33820C + 141212(H – O/8) + 9304S = 33820(0.705) + 141212(0.045 – 0.06/8) + 9304(0.03) = 29930 Kj/Kg

ƞBOILER = MSTEAM(hSTEAM – hFUEL)/MFUEL(QH) = 70% 0.7 = 195,000 kg/hr (FE*2257)/ MFUEL(29930 kJ/kg) MFUEL= 23107.56 kg/hr * (12.389 kgAIR/kgFUEL) = 286279.57 kgAIR/hr

PV = MRT 101.325(V) = 286279.57 kgAIR/hr (0.287)(15.6+273) V = 234019 m3/hr

ME Laws 76. The Mechanical Engineering Law has how many articles? a. Two b. threec. five *c. six 77. The Board of Mechanical Engineering is composed of ___ members to be appointed by the President of the Philippines. a. One b. three * c. two c. five 78. One of the following is not a correct qualification of the Member of the Board

ME 601 a. b. c. d.

Natural born citizen and resident of the Philippines Must be at least 35 years of age A PME with a valid professional license and an active practitioner for not less than 10 years Has not been convicted of any offense and moral turpitude *

79. A member of the board shall hold office for alarm of ___ years from the date of his appointment. a. three * b. two c. one d. four 80. No member of the Board shall serve for more than ____ regular terms. a. One b. two * c. three d. none of the above Engineering Economy 81. An employee obtained a loan of P10,000 at the rate of 6% compounded annually in order to repair a house. How much must he pay monthly to amortize the loan within a period of ten years? a. P198.20 b. P150.55 c. P110.22 * d. P112.02

Recall: Amortization, requires equal payments for a given period. Therefore, annuity. Solving interest rate / period, i to be paid monthly i = (1 + i/12)12 – 1 0.06 = (1+i/12)12 – 1 i=

82. What is the accumulated value of a payment of P12,500 at the end of each year for 9 years with interest at 5% compounded annually? a. P138,738.05 c. P137,832.05 * b. P178,338.50 d. P187,833.50

Recall Annuity, may nangyayaring continuous payments sa loob ng 9 years Accumulated Value means total value or sum of all payments (use Future Worth Formula) Future worth = [A|(1+i)n - 1|] / i 12500((1+0.05)9 – 1)/0.05 = 137832.054 = approx. 137832.05

83. What is the accumulated value of a payment of P6,000 every six months for 16 years with interest at 7% compounded semiannually? a. P312,345.00 c. P345,678.00

ME 601 b. P347,898.00

d. P344,007.00 *

Recall: Annuity, Continuous Payment This is considered as Future Worth, kasi summation *Future worth = [A|(1+i)n - 1|] / i 6000((1+0.07/2)2*16 – 1) / 0.07/2 = 344,007.0148 = 344007.00 approx

84. A mining property is offered for sale for P5.7M. On the basis of estimated production, an annual return of P800,000 is foreseen for a period of 10 years. After 10 years, the property will be worthless. What annual rate of return is in prospect? a. 6.7% * b. 7.1% c. 8.6% d. 5.2%

Recall: Depreciation Service life is 10 years, therefore 800,000 is considered the salvage value. Initial Cost is 5.7 M

85. If a down payment of P600,000 is made on a house and P80,000 a year for the next 12 years is required, what was the price of the house if money is worth 6% compounded annually? a. P1,270,707 * c. P1,130,450 b. P1,345,555 d. P1,678,420

Recall: Annuity (Malalaman na annuity kapag may series of payments na mangyayari) *Present worth = [A|(1+i)n - 1|] / i(1 + i)n *Future worth = [A|(1+i)n - 1|] / i

Principal or Present Worth = 600,000 Continuous Payment or Annuity of 80,000 Period = 12 Years Interest = 0.06 compounded annually

ME 601 Use Present worth as stated in the problem Present worth is 600000 + Present Worth of annuity 600000 + {[A|(1+i)n - 1|] / i(1 + i)n} = 600000 + {[80000(1+0.06)12 – 1]/0.06(1 + 0.06)12} Present Worth = 1270707.515 = 1270707 approx

86. What is the effective rate for an interest rate of 12% compounded continuously? a. 12.01% b. 12.89% c. 12.42% d. 12.75% *

Recall: Effective Rate, ER = (1+i)m - 1 I = 0.12 Infinite (Continously) (therefore, e) Effective Rate = ei – 1 = e0.12 – 1 = 0.12749 = 12.749 or 12.75% approx. ***use e if compounded infinitely

87. How long will it take for an investment to fivefold its amount if money is worth 14% compounded semi-annually? a. 11 b. 12 * c. 13 d. 14

Recall Compounding Interest: Future Worth = Present Worth(1 + interest)PERIOD Future Worth = 5*Present Worth 5*Present Worth = Present Worth (1+0.14/2)2*PERIOD Assume Present Worth = 1 5 = 1(1+0.14/2)2*PERIOD PERIOD = 11.89 = approximate to 12 years

88. An interest rate of 8% compounded semiannually is how many percent if compounded quarterly? a. 7.81% b. 7.85% c. 7.92% * d. 8.01%

Recall Compounding Interest: Future Worth = Present Worth(1 + interest)PERIOD

ME 601 Present Worth (1 + x/4)4*PERIOD = Present Worth (1 + 0.08/2)2*PERIOD Assume period are the same, period = 1 (1 + x/4)4*PERIOD = (1 + 0.08/2)2*PERIOD X = 7.92%

89. A man is expecting to receive P450,000.00 at the end of 7 years. If money is worth 14% compounded quarterly, how much is it worth at present? a. P125,458.36 b. P147,456.36 c. P162,455.63 d. P171,744.44 *

Recall Compounding Interest: Future Worth = Present Worth(1 + interest)PERIOD Future worth = 450,000.00 Present Worth = unknown Interest = 0.14 Period = 7 years 450,000 = Present Worth (1 + 0.14/4)PERIOD*4 Present Worth = 171744.4532

90. A man has a will of P650,000.00 from his father. If his father deposited an account of P450,000.00 in a trust fund earning 8% compounded annually, after how many years will the man receive his will? a. 4.55 years b. 4.77 years * c. 5.11 years d. 5.33 years

Recall Compounding Interest: Future Worth = Present Worth(1 + interest)PERIOD Future worth = 650,000.00 Present Worth = 450,000.00 Interest = 0.08 Period = unknown 650,000 = 450000 (1 + 0.08)PERIOD

ME 601 Period = 4.77 years

AC/DC 91. What is the equivalent capacitance of two series capacitors rated 4 and 6 µF respectively? a. 2.4 * c. 0.416 b. 10 d. 0.9

Capacitors in Series = Resistors in parallel 1/Total Capacitance = 1/4 + 1/6 = 2.4 microfarad

92. Three resistorsR1, R2 and R3 are connected in series across a 100-V source. If R_2 opens, the a. Voltage across R_2 is 100V *

c. voltage across R_1 is 100V

b. Total resistance decreases

d. voltage across R_2 is zero

An open circuit has an equivalent voltage drop R2 = 100V

93. What is 50 gradients equal to in degrees? a. 100 deg b. 45 deg

c. 20 deg d. 90 deg

1 degree = 1.1111 grad 1 grad = 0.9 degree 50 grad (0.9 deg/grad) = 45 deg

94. When one metal bar is heated. The other part gradually becomes heated, that is, energy will flow from hot part to cold part. What is the mode of heat transfer? a. radiation c. conduction * b. convection d. fusion

ME 601 95. What is the power required to transfer 97,000 coulumbs of charge through a potential rise of 50 volts in one hour? a. 0.55 kW c. 0.9 kW b. 1.3 kW * d. 2.8 kW 1 coulumb * 1 v = 1J 97000 coulumbs * 50 V = 4850000 J = 4850 kJ 4850 kJ/3600 sec = 1.347 kW Basic Electronics 96. A semiconductor has how many types of flow? a. 1

b. 2*

c. 3

d. 4

97. The flow of valence electrons to the left means that holes are flowing to the a. Left

b. Right*

c. Either way

d. None of the above

98. If you wanted to produce a p-type semiconductor, which of these would you use? a. Acceptor atoms* c. Donor atoms b. Pentavalent impurity d. Silicon 99. What is the drop across the diode when it is connected in series to a resistor of 1.8 kΩ and a supply voltage of 50 V. The supply voltage causes the diode to be reverse-biased. a. 50 V* b. 0.7 V c. 0.3 V d. cannot be solved, lack of data

A reverse-biased diode, creates an open circuit (in series). An open circuit has an equivalent voltage drop. Ans: 50V

100. __________is the procedure by which an atom is given a net charge by adding or taking away electron. a. Polarization b. Irradiation

ME 601 c. Ionization * d. Doping

Algebra: 1. If ax = by and bp = aq , then a. px = qy * b. xy = pq c. xp = yq d. qx = py

Let a = bp = (

)q = (by)q/x = (b)y*q/x

logbp = logby*q/x p = y*q/x p/y = q/x px = qy

2. If ax3 + bx2 + cx + d is divided by x - 2, then the reminder is equal to a. a - b + c - d b. 8a + 4b + 2c + d * c. -8a + 4b -2c + d d. a + b + c + d

Can be solved by division (synthetic) of polynomials X=2

ax3

bx2

cx

d

2a

2a+b

4a+2b 8a+4b+2c+d

ME 601

3. The range of the function f(x) = -|x - 2| - 3 is a. y ≥ 2 b. y ≤ -3 * c. y ≥ -3 d. y ≤ -2

4. It takes Michael 60 seconds to run around a 440-yard track. How long does it take Jordan to run around the track if they meet in 32 second after they start together in a race around the track in opposite direction? a. 58.76 seconds b. 68.57 seconds * c. 65.87 seconds d. 86.57 seconds

The circular track is 440 yds Michael’s Rate = 440 yd/60 sec = 7.33 yd/sec Michael covered a distance after 32 sec = 7.33 yd/sec * 32 sec = 234.66 yd 440 yards – 234 yardss = 205.333 yds  this is what Jordan covered in 32 sec 205.333 yds/ 32 sec = 6.416 yd/sec  Jordan’s Rate Distance = speed*time ;

440 yds = 6.416 yd/sec * time

Time = 440/6.416 = 68.57 sec

5. The time required by an elevator to lift a weight, vary directly with the weight and the distance through which it is to be lifted and inversely as the power of the motor. If it takes 30 seconds for a 10-hp motor to lift 100lbs through 50 feet, what size of motor is requires to lift 800 lbs. in 40 seconds through a distance of 40 feet? a. 48 hp * b. 50 hp c. 56 hp d. 58 hp

ME 601

Recall: Variation with direct and inverse proportion As stated in the problem, Time = k*(Weight*Distance)/Power Computing for k 30 sec = k(100lbs*50ft)/10 hp K = 0.06 Finally, solving for required motor 40 sec = K(800 lbs*40 ft)/MOTOR @k = 0.06 Motor = 48 HP

Trigonometry

6. An angle measures x degrees. What is its measure in radians? a. 180° x / π b. π x / 180° * c. 180° π / x d. 180° π x

Conversion: x°( 1π/180°) = π x / 180°

7. Express 45° in mils. a. 80 mils b. 800 mils * c. 8000 mils d. 80000mils

Conversion: 45°(1rev/360°)(6400 mils/1 rev) = 800 mils

ME 601

8. What is the value in degrees of π radians? a. 90° b. 57.3° c. 180° * d. 45°

2π rad = 1 rev 360° = 1 rev π radians = (1 rev/2π rad)*( 360°/1 rev) = 360°/2 =180°

9. The sum of the sides of a triangle is equal to 100 cm. If the angles of the triangle are in the continued proportions of 1:2:4. Compute the shortest side of the triangle. a. 17.545 b. 19.806 * c. 18.525 d. 14.507

Perimeter of triangle = a + b + c = 100 ; 1:2:4 = 100 X = shortest; 2X = mid; 4X = longest; Then, Perimeter = 100 = x + 2x + 4x = 7x 100/7 = x = 14.285 (but choose 19.806)

10. The sides of the triangular field which contains an area of 2400 sq. cm. are in continued proportion of 3:5:7. Find the smallest side of the triangle. a. 45.74 b. 63.62 c. 95.43 d. 57.67 *

ME 601 AREA of Tri = 1/2(B*H) = 2400 cm2 Could also be Geron’s A = SQRT(S*(S-a)(S-b)(S-c)) ; where s = (a + b + c)/2 Using the Proportion a=3x; b=5x; c=7x; s= (3x + 5x + 7x)/2 = 7.5x

A = SQRT((7.5x * (7.5x – 3x)*(7.5x – 5x)*(7.5x – 7x)) 2400 = SQRT((7.5x * (7.5x – 3x)*(7.5x – 5x)*(7.5x – 7x))

2400 = SQRT((7.5x * (4.5x )*(2.5x)*(.5x)) X = 19.222cm

Based on the proportion, shortest side a = 3x = 3*19.222 = 57.6674 cm

Analytic Geometry 11. Determine the equation of the line tangent to the graph y = 2x2 + 1, at the point (1, 3). a. y = 4x + 1 b. y = 4x – 1 * c. y = 2x – 1 d. y = 2x + 1 Line tangent to a curve has the same points. Cross check points on the graph and the given. 3 = 2(1) + 3 correct

;

3 = 4(1) – 1 correct

12. Find the equation of the tangent to the curve x2 + y2 = 41 through (5, 4). a. 5x + 4y = 41 * b. 4x – 5y = 41 c. 4x + 5y = 41 d. 5x – 4y = 41 Line tangent to a curve has the same points. Cross check points on the graph and the given. 52 + 42 = 41 correct

;

5(5) + 4(4) = 41 correct

ME 601 13. The midpoint of the line segment joining a moving point to (6, 0) is on the line y=x. Find the equation of its locus. a. x – y + 6 = 0* b. x – 2y + 6 = 0 c. 2x – y -3 = 0 d. 2x + 3y – 5 = 0 Analysis: Static Point is at (6,0). Moving Point is y = x. Locus is line equidistant to a given condition. In this case the midpoint. Y = X. @ given point X = 6 Trial and error: a) 6 – (0) + 6 = 0; 0 = 0 Correct , b) 6 – 0 + 6 = 0 correct c) 14. The base of an isosceles triangle is the line from (4,-3) to (-4, 5). Find the locus of the third vertex. a. x – y + 1 = 0 * b. x + y + 1 = 0 c. x – y – 2 = 0 d. x + y – 3 = 0 Take note of ISOCELES triangle. For Area of N-SIDED polygon (Analytic) Area = 1/2[

,

2 = 1/2 |x1

4

-4|x1

-3

5 |y1

|y1

…] determinants

4 = -3x +4(5) + (-4)(y1) –(y1)(4) –(-3)(-4) – 5(x1) = -3x + 20 – 4y - 4y -12 -5x 4 = -8x + 20 -8y – 12 = 8x +8 – 8y 1 = -2x + 2 – 2y 15. What is the new equation of the line 5x + 4y + 3 = 0 if the origin is translated to the point (1, 2)? c. 4x’ + 3y’ + 16 = 0 d. 5x’ + 4y’ + 16 = 0 * e. 5x’ – 4y’ – 16 = 0 f. 6x’ + 6y’ – 16 = 0 5(x+1) + 4(y+2) + 3 = 0 5x + 5 + 4y + 8 + 3 = 0 5x’ + 4y’ + 16 = 0

ME 601 Differential Calculus 16. A body moves such that its acceleration as a function of time is a=2+12t, where “a” is in m/s2. If its velocity after 1 s is 11 m/s. find the distance traveled after 5 seconds. a. 256 m b. 340 m c. 290 m * d. 420 m Accel = derived velocity, so, integrate accel to get velocity Dv/dt = (2 + 12t)dt Dv = 2t + 6t2 + C @t = 1 min = 60 sec 11 m/sec = 2(1 ) + 6(1)2 + C C = 3 m/sec After 5 sec V = 2t + 6t2 + C @t = 5 sec 17. A runner and his coach are standing together on a circular track of radius 100 meters. When the coach gives a signal, the runner starts to run around the track at a speed of 10 m/s. How fast is the distance between the runners has run ¼ of the way around the track? a. 5.04 m/s b. 6.78 m/s c. 5.67 m/s d. 7.07 m/s * Radius = 100 m , then circumference = 2piR = 628.3185 A quarter of the track = 50pi = 157.0796 Rate = 10 m /sec

18. Find dy/dx: y = sin (ln x2). a. 2 cos (ln x2) b. 2 cos (ln x2) / x * c. 2x cos (ln x2) d. 2 cos (ln x2) / x2

ME 601 LAZY BOY CALC TECH: Notes: Simplify writing LN, use rad for transcendental (sin.cos.tan), mathio. Utilized shiftstore if necessary. Use answer to divide whentesting answers (Ans should be = 1) Shift (integral): d/dx sin(2(lnx)) @ x = 0.3

19. The derivative of ln (cos x) is: a. sec x b. –sec x c. –tan x * d. tan x Shift (integral): d/dx(ln(cos(x))) @x = 0.3 = -0.309 -tan(0.3) = -0.309 20. Find the derivative of arcos 4x with respect to x. a. -4 / [1 – (4x)^2]^2 b. -4 / [1 – (4x)]^0.5 * c. 4 / [1 – (4x)^2]^0.5 d. -4 / [(4x)^2 - 1]^0.5 B typo error! -4 / [1 – (4x)^2]^0.5 *

Integral Calculus 21. Evaluate the integral of tan^2 x dx. a. tan x – x + c * b. sec^2 x + x + c c. 2sec x – x + c d. (tan^2 x)/s + x + c CALTECH: tan(x)2 @x = 1.1 SHIFTSTO-> A (3.8602) SHIFT(Integ): d/dx (tanx - x) @ x = 1.1 Ans/A = 1 22. Evaluate the integral of sqrt(3t – 1) dt. a. (2/9)(3t – 1)^5/2 + c b. (2/9)(3t – 1)^3/2 + c *

ME 601 c. (1/2)(3t – 1)^5/2 + c d. (1/2)(3t – 1)^3/2 + c Sqrt(3x-1) CALC @x = 1.1 SHIFTSTO-> A (1.5165) SHIFT(Integ) d/dx (2/9 * (3t-1)^3/2) @x = 1.1 Ans/A = 1 23. Evaluate the integral of (3t – 1)^3 dt. a. (1/12)(3t – 1)^4 + c * b. (1/4)(3t – 1)^4 + c c. (1/3)(3t – 1)^4 + c d. (1/12)(3t – 1)^3 + c @X = 1.1 SHIFTSTO-> A (12.167) SHIFT(Integ)d/dx((1/12)(3x-1)4) @x = 1.1 Ans/A = 1 24. Find the area of the region above the x axis bounded by the curve y = -x2 + 4x – 3. c. 1.333 square units * d.3.243 square units e.2.122 square units f. 1.544 square units Integ (-x2 + 4x -3) LIMITS 3 and 1 MODE(5:3) –x2 + 4x – 3 (since stated at x axis, set y = 0) 25. Find the volume of the solid of revolution formed by rotating the region bounded by the parabola y = x2 and the lines y = 0 and x = 2 about the x axis. a. b. c. d.

25.01 cu. units 15.50 cu. units 20.11 cu. units * 30.14 cu. units

Differential Equations 26. Find the differential equation of the family of line passing through the origin. A. xdy – ydx = 0 C. xdy + ydx = 0 * B. ydy + xdx = 0 D. xdx + dy = 0

ME 601 slope, m = y/x (m is constant) Differentiate: 0 = [xdy - ydx]/x2 <-- qoutient rule xdy - ydx = 0 27. The rate of population growth of a country is proportional to the number of inhabitants. If the population of a certain country now is 40 million and 50 million in 10 years’ time, what will be its population 20 years from now? A. 62.5 million* C. 56.20 million B. 65.20 million D. 52.6 million Estimate using Caltech : Linear regression (MODE 3:5) X

y

From Data Set: Use SHIFT 1:5:y hat

0

40

20y hat = 62.5 million

20

50 28. What is the equation (DE) of the family of parabolas having their vertices at the origin and their foci n the x – axis. A.2xdy – ydx = 0* C. xdy – 2y2dx = 0 2 2 B. 2ydy + x dx D. y dy + 2xydx = 0

Parabola is y2 = 4ax; Differentiate:

4a = y2/x

0 = [xdy – ydx]/x2 = [x(2ydy) – y2dx] / x2

2xydy – y2dx = 0 or 2xdy – ydx = 0

29. Determine the degree of the given ordinary D.E y’’’’ – 4(y’’’)² + 3(y’’)³ - y’ = 0 a. 1* c. 3 b. 2 d. 4 30. A thermometer reading 18 deg F is brought into a room where the temperature is 70 deg F; a minute later, the thermometer reading is 31 deg F. determine the temperature reading 5 minutes after the thermometer is first brought into the room. A. 57.68 deg F* C. 30.01 deg F B. 42.4 deg F D. 53.89 deg F Constant at 70 Estimate using Caltech : Linear regression (MODE 3:5)

ME 601 X

y

From Data Set: Use SHIFT 1:5:y hat

0

70-18

5 y hat = 12.3398

1

70-31

; 70-ans = 57.6601 deg F

Statistics and Probabilities 31. The probability that both stages of a two-stage rocket to function correctly is 0.92. The reliability of the first stage is 0.97. The reliability of the second stage is: a. 0.948 * b. 0.958 c. 0.968 d. 0.8924 Event A + Event B = 0.92 Event A = 0.97 0.92/0.97 = 0.948

32. Ricky and George each throw dice. If Ricky gets a sum of four what is the probability that George will get less than of four? a. ½ b. 5/6 c. 9/11 d. 1/12 * 33. Two fair dice are thrown. What is the probability that the sum of the dice is divisible by 5? a. 7/36 * b. 1/9 c. 1/12 d. ¼ Divisible by 5; 5, 10 Set DIE 1 (1,2,3,4,5,6,4) Set DIE 2 (4,3,2,1,5,4,6) (1,4) (2,3) (3,2) (4,1) (5,5) (6,4) (4,6)  7 combinations 7/36

ME 601 34. An um contains 4 black balls and 6 white balls. What is the probability of getting one black ball and white ball in two consecutive draws from the urn? a. 0.24 b. 0.27 c. 0.53 * d. 0.04 B=4 W = 6 Total = 10 Two case will happen, Getting Black ball then white OR(+) Getting white, then black. Case 1 + Case 2: (4/10)(6/9) + (6/10)(4/9) = 8/15 = 0.53333

35. Five fair coins were tossed simultaneously. What is the probability of getting three heads and two tails? a. 1/32 * b. 1/16 c. 1/8 d. ¼ 1 of 2 outputs (1/2)(1/2)(1/2)(1/2)(1/2) = 1/32

Physics 36. An objects acceleration as it starts to fall is: A. equal to g * C. less than g B. greater than g D. zero 37. 37. Going against a wind, a domestic plane can travel 5/8 of the distance in one hour that it is going with the wind. If the plane can fly 300mph in calm wind, what is the velocity of the wind? A. 69.3 mph * C. 96.3 mph B. 73.3 mph D. 93.3 mph Distance = Speed*time, Plane Rate = 300 MPH Distance = (Plane Rate + Wind Rate)Time  Going with the Wind 5/8Distance = (Plane – Wind Rate)  Against the Wind 5/8(Plane Rate + Wind Rate) = (Plane Rate – Wind Rate) 5/8(300 + Wind Rate) = (300 – Wind Rate)

ME 601 Wind Rate = 69.2307 = 69.3 MPH

38. 38. A bullet is fired from a gun at an angle of 400. What is the range if its velocity is 300 m/s? g = 10 m/s. A. 192.9m C. 8863 m* B. 229.8 m D. 12000 m RANGE relates to VoX: [Vo2sin2theta]/g = [(300 m/s)2sin(2*400)]/10 m/sec = 8863 m 39. A car accelerates from rest at 2 m/s² for 5 seconds, travels at constant speed for 10 seconds and decelerates to rest at 2 m/s². Calculate the distance traveled by the car. a. 525 m c. 450 m b. 315 m d. 150 m* From Rest to Accel @ 5 sec V = Vo + accel*time = 0 + (2 m/sec2)*(5 sec) = 10 m/sec (THIS IS THE SPEED Gained from rest for 5 sec) Distance then is (Dist = Speed * time) 10 m/sec * 5 sec = 10 meters From Constant Speed for 10 seconds 10 m/sec * 10 sec = 100 meters Decelerating after 10 seconds a = 2 m/sec V = Vo – a*t (from here the final velocity V will be zero) (0-10m/sec)/time = 2 m /sec2 Time = 5 seconds. 10 m/sec *5sec = 50 meters Total Distance = 150 meters 40. Two mass collide on a frictionless horizontal floor and perfectly inelastic collision. Mass 1 is 4 times Mass 2; velocity of mass 1 is 10 m/s to the right while mass 2 is 20 m/s to the left. What is the velocity and direction of the resulting combined mass? a. 10 m/s to the right c. 4 m/s to the right* b. 10 m/s to the left d. 1.5 m/s to the left Inelastic collision (two bodies did not separate after collision) P1 + P2 = 0

Condition M1 = 4M2

ME 601 M1v1(right) + m2v2(left) = m1v’1 + m2v’2 M1v1 + m2v2 = (m1 + m2)v’ 4M2(10 m/sec) + M2(-20 m/sec) = (4M2+M2)(v’) 40m/sec(M2) – 20m/sec(M2) = 5 M2(v’) -20m/sec(M2)/5(M2) = v’ 4 m/sec (going to right indicated by – sign) Mechanics 41. A body weighing 40lb starts from rest and slides down a plane at an angle of 30deg with the horizontal for which the coefficient of friction u = 0.30. How far will it move during the third second? a. 19.63 feet b. 19.33 feet * c. 18.33 feet d. 19.99 feet Recall: Sliding Block on a wedge (Kinematics) Summation of Forces on Inclined = 0 Wsintheta = Downward Force + Sliding Force Wsintheta = W/g * accel + friction coeff * (Wcostheta) Sin30 = a/32.2 + 0.3(cos30) A = 7.734 ft/sec2

Using Accel, we can now solve for the distance Vf = Vo + a*t Vf = at rest (0) + 7.734 ft/sec2*(2 sec) = 15.468 ft/sec (NOTE: 2 sec is the change of time from rest and started sliding down)

Distance = V*t + ½ a*t2 = 15.468 * 1 + ½(7.734)(1sec)2 = 19.33 ft (we used 1 sec, as with the reference that is was able to move 2 sec from rest, then the third sec is with the current speed and acceleration)

ME 601 42. A pick-up truck is traveling forward at 25m/s. The bed is loaded with boxes whose coefficient of friction with the bed is 0.4. What is the shortest time that the truck can be brought to a stop such that the boxes do not shift? a. 2.35s b. 4.75s c. 5.45s d. 6.37s * Summation of Forces (Horizontal) = 0 F = REVERSE FORCE coeff friction(Normal Force) = W/g * accel coeff friction(Weight) = W/g * accel a = coeff friction * g = 0.4(9.81) = 3.924 m/sec2

V = Vo - a*time 0 = 25 - 3.924*t time = 6.37 sec

43. Two cars A and B accelerate from a stationary start. The acceleration of A is 4 ft/sec^2 and that of B is 5 ft/sec^2. If B was originally 20 feet behind A , how long will it take B to overtake A. a. 18.6 sec b. 10 sec c. 12.5 sec d. 6.32 sec * Distance A = Sa; Sa = Vo*t + ½ 4ft/sec2 * (t)2 Distance B = Sb; Sb = Vo*t + ½ 5ft/ sec2 * (t)2 Condition: Sb = Sa + 20 feet (condition of B overtaking A) Relative to time (Distance = V*t + ½ accel * time2) ½ 5 ft/sec2 * (t2) = ½ 4 ft/sec2 * (t2) + 20 ft Calc: ans = 6.32 sec 44. Two cars, A and B, are travelling at the same speed of 80 km/hr in the same direction on a level road, with car A 100 meters ahead of car B. Car A slows down to make a turn decelerating at 7 ft/sec^2. In how many seconds will B overtake A. a. 6.96 sec b. 5.55 sec c. 7.85 sec d. 9.69 sec *

ME 601 Distance A = Sa; Sa = 80 km/hr*t + ½ *accel * (t)2 Distance B = Sb; Sb = 80 km/hr*t - ½ * (7 ft/sec2) * (t)2 Condition: Sb = Sa + 100 meters (condition of B overtaking A)

7 ft/sec2 = 27658.53659 km/hr2

45. What is the magnitude of the resultant force of the two forces 200N at 20deg and 400N at 144deg? a. 332.5N* b. 323.5N c. 313.5N d. 233.5N

Resultant Forces F: Refer to Figure Using Cosine Law: R2 = 2002 + 4002 – 2(200)(400)cos(36° + 20°) ; R = 332.5 N 46. A load of 100 lb is hung from the middle of the rope, which is stretched between two rigid walls 30ft. apart. Due to the load, the rope sags 4 feet in the middle. Determine the tension in the rope. a. 165lbs. b. 173lbs. c. 194lbs. * d. 149lbs.

Recall: Tension Forces (PHYSICS) tan∅ = 15/4 ; ∅ = 75.068°

Summation of Vertical Force FV = 0 2Tcos∅ = 100 ; T = 100/2cos∅ = 100/2cos(75.068°) = 194 lbs

47. A 100kg weight rest on a 30degrees incline plane. Neglecting friction, how much pull must one exert to bring the weight up the plane? a. 88.67kg. b. 100kg c. 70.71kg d. 50kg * Recall: Sliding Object on inclined Force = Wsin30 = 100kg (sin30) = 50 kg

ME 601

48. A block weighing 500kN rest on a ramp inclined at 25degrees with the horizontal. The force tending to move the block down the ramp is ________. a. 121kN b. 265kN c. 211kN * d. 450kN

Recall: Sliding Object on inclined Force = Wsin30 = 500KN (sin25) = 211.309 kN

49. A 200kg crate impends to slide down a ramp inclined at an angle of 19.29deg with the horizontal. What is the frictional resistance? a. 612.38N b. 628.38N c. 648.16N * d. 654.12N

Recall: Sliding Object on inclined Force = Wsin30 = 200kg (sin19.29) = 66.0699 kg * 9.81KN = 648.16N

50. A man can exert a maximum pull of 1000N but wishes to lift a new stone door for his cave weighing 20,000N. If he uses a lever, how much closer must the fulcrum be to the stone then to his hand? a. 10 times nearer b. 20 times farther c. 10 times farther d. 20 times nearer *

Recall Summation of Moment at point = 0 20kN * x2 = 1kN * x1 20 x2 = x1  x1 which is the force the man can exert, should have the fulcrum 20 times nearer to 20kN

ME 601 Engineering Economy 51. A man is expecting to receive P450,000.00 at the end of 7 years. If money is worth 14% compounded quarterly, how much is it worth at present? e. P125,458.36 f. P147,456.36 g. P162,455.63 h. P171,744.44 *

450000 = P+P*in = P(1 + in/period)time*period = P(1 + 0.14/4)7*4 Present Worth, P = 171744.4532

52. A man has a will of P650,000.00 from his father. If his father deposited an account of P450,000.00 in a trust fund earning 8% compounded annually, after how many years will the man receive his will? b. 4.55 years c. 4.77 years * d. 5.11 years e. 5.33 years

650,000 = 450000(1 + 0.08)years Hint*: Backdoor and check the closest answer. In this case 4.77 years

53. Mr. Adam deposited P120,000.00 in a bank who offers 8% interest compounded quarterly. If the interest is subject to a 14% tax, how much will he receive after 5 years? a. P178,313.69 b. P153,349.77 c. P170,149.77 * d. P175,343.77

P = 120000 F without tax = P(1 + 0.08/4)5years*4 = 178313.6875 178313.6875 – 120000 = 58313.6877 interest gained after 5 years (compounded qrtly) 58313.6877 * 0.14 = 8163.9162 14 % tax applied on interest gained 178313.6875 – 8163.9162 = 170149.7713

ME 601

54. What interest compounded monthly is equivalent to an interest rate of 14% compounded quarterly? a. 1.15% b. 13.84% * c. 10.03% d. 11.52% Interest = P*rate/period EQN: (1 + x/12months)1 yr * 12 mos = (1+0.14/4)1 yr * 4 quarters This will take long to compute when using calc. Use BackDOOR! (1+ choice/12)12 = ans should be equal to (1 + 0.14/4)1*4 =1.1475 Ans: 13.84 %

55. What is the present worth of two P100.00 payments at the end of the third and the fourth year? The annual interest rate is 8%. a. P152.87 * b. P112.34 c. P187.98 d. P176.67 Problem indicates regular payments stated as two 100.00 payments, hinting annuity. F = P(1 + i/period)year*period ; P = F/(1 + i/period)year*period Ptotal = P1 + P2 = F1/(1+0.08/1)3*1 + F2/(1 + 0.08/1)4*1 = 100/1.259712 + 100/1.36048896 = 152.886 Ptotal = 153

56. What amount would have to be invested at the end of each year for the next 9 years at 4% compounded semi-annually in order to have P5,000.00 at the end of the time? a. P541.86 * b. P553.82 c. P542.64 d. P548.23 Principal or Present Worth for 9 years at 4% compounded semi annual (2) Future worth = 5000.00, solving annuity yield 233.51 Principle =

ME 601 57. A contractor bought a concrete mixer at P120,000.00 if paid in cash. The mixer may also be purchased by installment to be paid within 5 years. If money is worth 8%, the amount of each annual payment, if all payments are made at the beginning of each year, is: a. P27,829.00 * b. P29,568.00 c. P31,005.00 d. P32,555.00 58. A contract calls for semiannual payments of P40,000.00 for the next 10 years and an additional payment of P250,000.00 at the end of that time. Find the equivalent cash value of the contract at 7% compounded semiannually? a. P444,526.25 b. P598,589.00 c. P694,138.00 * d. P752,777.00 59. A man is left with an inheritance from his father. He has an option to receive P2 M at the end of 10 years; however he wishes to receive the money at the end of each year for 5 years. If interest rate is 8%, how much would he receive every year? a. P400,000.00 b. P352,533.00 c. P232,020.00 * d. P200,000.00 60. How much money must you deposit today to an account earning 12% so that you can withdraw P25,000 yearly indefinitely starting at the end of the 10th year? a. P125,000 b. P89,456 c. P73,767 d. P75,127 *

Fluid Mechanics:

61. Water, density = 62.4 lbf/ft3, is flowing through a pipe. A pitot static gage registers 3.0 inches of mercury. What is the velocity of water in the pipe? Note: densityhg = 848.6 lbf/ft3. A. 14.24 ft/s *

C. 8.24 ft/s

B. 11.24 ft/s

D. 7.45 ft/s

Recall: Static Pressure: V = Cv SQRT(2ghDpressure) = Cv *SQRT(2 * 32.2*3/12 ft (62.4/848.6 - 1))

62. A cylindrical 1 ft diameter tank, 4 ft high contains 3 ft of water. What rotational speed is required to spin the water out the top?

ME 601 A. 22.7 rad/s *

C. 23.7 rad/s

B. 21.4 rad/s

D. 18.90 rad/s

Recall: Angular Velocity = Change in Angular Position/Time (Plug into bernouli’s) ω=

㌳䁣

= sqrt((2*32.2*4ft/0.5ft)) = 22.698 rad/sec = 22.7 rad/sec

63. A 1 m x 1.5 m cylindrical tank is full of oil with SG = 0.92. Find the force acting at the bottom of the tank in dynes. A. 106.33 x 107 dynes B. 106.33 x 104 dynes C. 106.33 x 105 dynes D. 106.33 x 106 dynes* Recall: Static Fluids: Pressure = Gamma * Height = (S.G*density water)*Height = 0.92 * 9.81 KN/m3 * 1.5 m = 13.5378 kPA Force = Pressure * Area = 13.5378 * (pi/4 * (1m)2 *) = 10.632 kN = 10632 N = 106.32 * 106 dynes Note: 1N = 100 000 dynes : Force has Newton unit; Pressure in Pascal

64. Find the pressure at the 100 fathom depth of water in kPag. A. 1,793.96 kPag* B. 1,893.96 kPag

ME 601 C. 1,983.96 kPag D. 1,693.96 kPag Height = 100 fathom depth (6 ft/1fathom depth) = 600 ft Pressure = pgh = 1000 kg/m3 * 9.81 m/sec * 600 ft(1m/3.28ft) = 1794512.195 Pagauge = 1794.51 kPag 65. A large mining company was provided with a 3 m3 of compressed air tank. Air pressure in the tank drops from 700 kPa to 150 kPa while the temperature remains constant at 28oC. What percentage has the mass of air in the tank been reduced? A. 74.00 B. 72.45 C. 78.56 D. 78.57 * Fluid mach % of Reduced Mass = Change in Pressue/Initial Pressure = 700 – 150/150 = 0.7857 = 78.57% Thermodynamics 1: 66. Air is compressed adiabatically from 30oC to 100oC. If mass of air being compressed is 5 kg. Find the change in entropy. A. 1.039 kJ/kg B. 0.746 kJ/kg C. 0 * D. 1.245 kJ/kg Change in Entropy = Qreversible/Temp = Qrev/O adiabatic or no exchange of heat = 0 67. A perfect gas has a value of R = 58.8 ft-lb/lb-oR and k = 1.26. if 20 BTU are added to 10 lbs of the gas at constant volume when initial temperature is 90o F. Find the final temperature. A. 97oF * B. 104oF C. 154oF D. 185oF

ME 601 Recall Ideal Gas, Q = mCvdT; Cv = R/k-1 = (58.8ft-lb/lb-R)/(1.26 – 1) * (1/788) = 0.29086BTU/lb-F **788 is conversion of FT-lb to BTU 20 lbs = 10lbs*(0.29086)*(T2 – 90F); T2 = 96.88F = 97F 68. Air enters a nozzle steadily at 1.71 kg/m3 and 35 m/s. what is the mass flow rate through the nozzle if the inlet area of the nozzle is 80 cm2? A. 0.479 kg/s*B. 3.57 kg/s

C. 4.79 kg/s

D. 0.53 kg/s

Recall: Mass Flow Rate = Density * Area * Velocity = (pav) = 1.71 kg/m3 * 80cm2(1m/100cm)2 * 35 m/sec = 0.4788 kg/sec 69. A carnot cycle has a maximum temperature of 580F and minimum temperature of 150F. if the heat added is 4200 BTU/min, find the horsepower output of the engine. A. 33.53

B. 40.94*

C. 44.69

D. 75.40

WorkNET= efficiency * QADDED Recall Efficiency of Carnot = (THIGH – TLOW)/THIGH = (580 – 150)RANKINE/ (580 + 460RANKINE) = 0.4134 Finally, WorkNET = 0.4134(4200)BTU/MIN *(60/HR)(1HP/2545 BTU/HR) = 40.94 HP 70. Steam turbine is receiving 1000 lbm/hr of steam, determine the horsepower output of the turbine if the work done by steam is 250 Btu/lbm. A. 98.23 Hp*

B. 362.7 Hp

C. 280 Hp

D. 6072.7 Hp

Mass Flow Rate = 1000 lbm/hr ; Q = 250 Btu/lbm. Work = mQ = 1000 lbm/hr*(250 Btu/lbm)(1HP/2545 btu/hr) = 98.2318 HP Thermodynamics 2: 71. Determine th air – standard efficiency of an engine operating on the diesel cycle with clearance of 6% when the suction pressure is 99.97 kPa and the fuel is injected for 7% of the stroke. Assume k = 1.4 A. 62.11% * B. 51.20% C. 73.58% D. 60.02% Take note of DIESEL CYCLE. Recall Eff = 1 – [1/rk^(k-1)]*[(rc^k -1)/k(rc-1)]

ME 601 V3 – V2 = 0.07VD ; V2 = 0.06VD ; V3 = 0.07VD + 0.06VD = 0.13VD Cut-off Ratio, Rc = V3/V2 = 0.13VD/0.06VD = 2.1666 Compression Ratio, Rk = (1 + clearance)/clearance = (1 + 0.06)/0.06 = 17.6667 Then Eff = 1 – [1/rk^(k-1)]*[(rc^k -1)/k(rc-1)] = 1 – [(1/17.6667^(1.4-1))]*[2.1666^1.4 – 1/1.4(2.1666 – 1)] Eff = 0.62109 = 62.11%

72. A 2000 kW Diesel engine unit uses 1 bbl oil per 525 kWh produced. Oil is 25°API. Efficiency of generator 93%, mechanical efficiency of engine 80%. What is the thermal of engine based on indicated power (%)? A.39.3%

B. 39.4%

C. 39.5%

D.39.6% *

This is asking for THERMAL EFFICIENCY of engine! BBL is BARREL. 1 BBL = 42 US gallons = 35 UK gallons 7.33 BBL = 1000kg = 1 metric ton Recall Thermal Eff (Diesel) = Indicated Work/Mfuel(Qh) 25 API = 141.5/SG15.6 – 131.5 SG15.6 = 0.9042; Density = (0.9042 * 1000 kg/m3 )= 0.9042 kg/li Mass of Fuel, Mf = 0.9042 kg/li * (42 us gal)(3.784 li) = 143.7026 kg Qh = 41130 + 139.6(API) = 41130 + 139.6(25) = 44620 kj/kg Work * EffGEN * EffMECH = QOUTPUT Work = QOUTPUT/(0.93*0.8) = 525 KWh/0.744 = 705.645 KWh 705.645 KW/hr (3600sec) = 2540323 KWsec = 2540323 kj/sec * sec = 2540323 KJ Finally, ƞTHERMAL = Indicated Work/Mfuel(Qh) = 2540323 KJ/[(143.7026kg)(44620 kJ/kg)] ƞTHERMAL =0.3961 = 39.61%

73. An air-standard Brayton cycle has a pressure ratio of 8. The air properties at the start of compression are 100 kPa and 25°C. The maximum allowable temperature is 1100°C. Determine the net work. A. 373.23 kJ/kg

B. 373.24 kJ/kg *

C. 373.25 kJ/kg

D. 373.26 kJ/kg

When referring to Pressure Ratio, (take note of HI/LOW: Phigh/Plow, BRAYTON CYCLE is SPSP) PHIGH/PLOW = 8 PLOW = 100 kPa

ME 601 PHIGH = 8 * 100 kPa = 800 kPa Recall: Net Work = Work of Turbine – Work of Compressor T1 = 25 C + 273 = 298K T2 = ? T2 = T1(PHIGH/PLOW)^(k-1/k) = 100(8)^(1.4-1/1.4) = 539.811K Work of Compressor = mCpdT = 1(539.811-298) = 241.8113 Max Temp = T3 = 1100C = 1373K T4 = T3/(8)^(1.4-1/1.4)= 757.9574 K Work of Turbine = mCp(dT) = 615.0425 Finally: Work Net = 615.0425 kJ/kg – 241.8113 kJ/kg = 373.2312 kJ/kg 74. Steam leaves an industrial boiler at 827.4 kPa and 171.6C. A portion of the steam is passed through a throttling calorimeter and is exhausted to the atmosphere when the calorimeter pressure is 101.4 kPa. How much moisture does the steam leaving the boiler contain if the temperature of the steam at the calorimeter is 115.6C? At 827.4 kPa (171.6C): hf = 727.25 kJ/kg, hfg = 2043.2 kJ/kg From table 3: at 101.4 kPA and 115.6C: h2 = 2707.6 kJ/kg A. 3.78%

B. 3.08%*

C. 4.56%

D. 2.34%

Let x = steam entering the throttling calorimeter h1 = h2 2707.6 kj/kg = 727.25 + x(2043.2); x = 0.9692 1-x = 0.0308; 3.08%

75. A turbine has an available enthalpy of 3300 kJ/kg in a Rankine cycle. The pump work has also 25 kJ/kg. For flow of 3 kg/s, find the system output. A.5960 kW

B. 6080 kW

C. 6343 kW

D. 9825 kW*

Note the cycle (Rankine – common to steam pplant ). Recall: System Output = mass flowrate * (Change in Enthalpy) = 3 kg/sec (3300kj/kg – 25 kj/kg) = 9825 kW This is Work is the ideal Rankine cycle. (Pump – Turbine) Heat Transfer:

ME 601 76. A 5 cm diameter spherical ball whose surface is maintained at a temperature of 70 deg. C is suspended in the middle of a room at 20 deg. C. if the convection heat transfer coefficient is 15 W/m^2C and the emissivity of the surface is 0.8, determine the total heat transfer from the ball. A. 23.56 watts B. 32.77 watts* C. 9.22 watts D. 43.45 watts Recall: Radiation Heat Xfer = SEAT4 Qconv = 5.89 Qrad = 5.7 * 10-8 * 0.8 * 4 * pi * (0.05/2)2 ((20+273)4 – (70+273)4) = 2.3 77. For heat transfer purposes, a standing man can be modeled as a 30 cm diameter, 170 cm long vertical cylinder with bottom both the top and bottom surfaces insulated and with the side surface at an average temperature of 34 deg C. For a convection heat transfer coefficient is 15 W/m2-C, determine the rate of the heat loss from this man by convection in an environment at 20 deg C. A. 316.46 watts

C. 336.46 watts *

B. 326.46 watts

D. 346.46 watts

Recall: Convective Heat xfer. Q = hA(dT) = 15 W/m2 . °C * pi * 0.3 m * 1.7 m * (34-20)°C = 336.46

78. Consider a person standing in a breezy room at 20 deg C. Determine the total rate transfer from this person of the exposed surface area and the average outer surface temperature of the person are 1.6m2 and 29 deg C, respectively, and the convection heat transfer coefficient is 6 W/m2 with emissivity of 0.95. A. 86.40 watts

C. 198.1 watts

B. 81.70 watts

D. 168.1 watts *

Two Modes of Heat Transfer is occuring, as per stated in the problem. Convective and Radiative Heat Transfer. Then Total Heat Transfer = Qconv + Qrad = HAdT + SEAT4 Qconv = 6 W/m2 * 1.6m2 * (29 - 20) = 86.4 Watts Qrad = 5.7 * 10-8 (0.95) * (1.6m2) * ((29+273)4 – (20 + 273)4) = 81.7184 Watts Total Heat = 86.4 + 81.7184 = 168.1184 Watts

79. A tank contains liquid nitrogen at -190 c0 is suspected in a vacuum shell by three stainless steel rods 0.80 cm in diameter and 3 meters long with a thermal conductivity of 16.3 W/m2-C0. If the ambient air outside the vacuum shell is 15 C0, calculate the magnitude of the conductive heat flow in watts along the support rods.

ME 601

A. 0.168 * B. 0.176 C. 0.182 D. 0.0587

Problem States, determine CONDUCTIVE HEAT FLOW. Recall Heat Conduction = HA(dT) Then, Q = 16.3 W/m2 – C * (pi/4 * (0.008 m)2) * (15 – (-190)) = 0.167 Watts

80. How many watts will be radiated from a spherical black body 15 cm in diameter at a temperature of 800 degC? A. 5.34 KW* B. 4.34 KW C. 6.34 KW D. 3.34 KW RECALL: Radiation (SEAT4) Q = 5.6704 * 10-8 * (1 blackbody) * (4*pi*(0.15m/2)2) * (800+273)4 = 5313.063 Watts = 5.31KW Combustion: 81. A gaseous fuel mixture has a molar analysis: H2 = 14%

CH4 = 3%

CO = 27%

O2 = 0.6%

CO2 = 4.5%

N2 = 50.9%

Determine the air – fuel ratio for complete combustion on molar basis. A. 2.130 B. 3.230 C. 1.233* D. 1.130 Compute first the Oxygen Chemical Reaction (as this pertains to the product after reacting to Fuel) H2 + O2 = H2O  0.14 H2 + xO2 = 0.14H2O  0.14 H2 + 0.07O2 = 0.14H2O CH4 + O2 = CO2 + H2O  0.03 CH4 + xO2 = 0.03CO2 + 0.06H2O  0.03 CH4 + 0.06O2 = 0.03CO2 + 0.06H2O

ME 601 CO + O2 = CO2  0.27CO + xO2 = 0.27CO2  0.27CO + 0.135O2 = 0.27CO2 Total Oxygen in Chem Reaction = 0.07 + 0.06 + 0.135 = 0.265 Less the 0.6%O2 from fuel = 0.265 – 0.006 = 0.259 O2 from air only. Then, THEO A/F = [0.259 + 0.259(3.76)]/1 = 1.233 AirMOLES/FuelMOLES ***RECALL: the 3.76 is the mole of NITROGEN, from 79%/21% composition of oxygen

82. A bituminous coal has the following composition: C = 71.5%

H = 5.0%

S = 3% Ash = 7.6%

O = 7.0%

N = 1.3%

W = 3.4%

Determine the theoretical weight of Nitrogen in lb/lb of coal. A. 2.870 B. 7.526 * C. 2.274 D. 6.233 From Problem, coal stated as fuel (solid fuel). Known ultimate analysis (percentage of each elements), therefore use EMPERICAL FORMULA for COMBUSTION OF SOLID FUELS Theo A/F = 11.5C + 34.5(H – O/8) + 4.3S Theo A/F = 11.5(0.715) + 34.5(0.05 + 0.07/8) + 4.3(0.03) = 10.378 lbair/lbfuel Recall Mass Percentage for Oxygen and Nitrogen: %Mass of O2 = 23.2% * Mass of Air %Mass of N2 = 76.8% * Mass of of Air, in this case = 0.768 * 10.378 = 7.97 lb/lb. (Nearest ans 7.526)

83. A diesel power plant uses fuel with heating vlue of 43,000 kJ/kg. What is the density of the fuel at 25C? A. 840 kg/m3

B. 873 kg/m3

RECALL: API transposes to Density of Fuel Qh relates to API with empirical formula of:

C. 970 kg/m3*

D. 940 kg/m3

ME 601 Qh = 41130 + 139.6 API 43,000 = 41130 + 139.6 of API API = 13.395 13.395 = 141.5/SG15.6 – 131.5; SG15.6 = 0.9765 For correction factor applied to 25C: SG25 = 0.9765(1-0.00072(25-15.6) = 0.96995 Finally, Density = 0.96995 * 1000 kg/m3 = 969.95 approx to 970 kg/m3

84. The heating value of fuel supplied in a boiler is 43,000 kJ/kg. If the factor of evaporation is 1.10 and the actual specific evaporation is 10, what is the efficiency of the boiler? A. 62.07%

B. 53.08%

C. 78.05%

D. 57.77%*

Recall: Boiler Eff: ƞBOILER = MSTEAM(HSTEAM – HFUEL)/MFUEL(QHEATING VALUE) Note FE or Factor of Evaporation = 1.10 FE = (HSTEAM – HFUEL)/2257 Also, Note ASE or Actual Specific Evaporation = 10 ASE = MSTEAM/MFUEL Simplify: ƞBOILER = ASE(FE*2257) /(QHEATING VALUE) = [10(1.10*2257)]/43000 = 0.5773 = 57.73 %

85. Find the density of fuel at 50C if fuel used is 27 API. A. 763.556 kg/m3

B. 834.56 kg/m3 C. 853.45 kg/m3 D. 871.30 kg/m3*

Recall: API transposes Specific Gravity 27 API = 141.5/SG15C – 131.5 SG15C = 0.892 Using Correction Factor SG50 = 0.892(1-0.00072(50C – 15.6C)) = 0.8706 Density = Mass/Volume

ME 601 SG*Density = 0.8706 * 1000 kg/m3 = 870.6 kg/m3 = Approx to 871.3 ME Laws: 86. The Mechanical Engineering Law (BASAG) is also known as A. RA 8459 *

C. RA 5984

B. RA 3449

D. RA 6561

87. The Article V of the the Mechanical Engineering Law A. TITLE, STATEMENT OF POLICY, AND DEFINITION OF TERMS B. PENAL AND CONCLUDING PROVISIONS* C. PRACTICE OF THE PROFESSION D. CONCLUSION

88. Article V, Sec. 43 talks about A. Implementing Rules and Regulations* B. Penalties C. Transitory Provisions D. Funding Provisions

89. The Article I, Sec. 1 of the Mechanical Engineering Law A. statement of policy

C. Title *

B. Definition of terms

D. Creation and Composition of the Board

90. According to Sec 42, t any person who violates any of the provisions of this Act and its rules and regulations shall, upon conviction be penalized by a fine of not less than

AC/DC

A. P 50, 000 *

C. P40, 000

B. 30, 000

D. P100, 000

ME 601 91. An ideal step-up transformer with 100 turns in the primary and 2500 turns in the secondary carries a load of 2A in the secondary windings. What is the current in the primary side? A. 50A

*

B. 0.08A

C. 25A D. 1,250A

Turn Ratio: 100/2500 = 1/25 = 0.04 Primary Turn/Secondary Turn = Current Secondary/Current Primary 0.04 = 2Amp/Current Primary Current Primary = 2/0.04 = 50A 92.Gearmotors are selected based on which of the following? A. Speed requirement

C. both A and B*

B. Torque reuirement

D. neither A or B

Recall: Gearmotors requires speed and torque (gear ratio) Ans: Both A and B 93. If N1/N2 = 2, and the primary voltage is 120 V, what is the secondary voltage? A. 0 V B. 36 V

C. 60 V * D. 240 V

V1/V2 = N1/N2 120V/V2 = 2 V2 = 120/2 = 60V

94.A transformer has a turns ratio of 4: 1. What is the peak secondary voltage if 115 V rms is applied to the primary winding? A. 40.7 V * B. 64.6 V V1/V2 = 4/1 Vrms = Vpeak/sqrt(2) 115V rms = Vpeak Primary/sqrt(2) V peak Primary = 115V * sqrt(2) = 162.6345 V 162.6345/V2 = 4/1

C. 163 V D. 650 V

ME 601 V2 or V secondary = 162.6345/4 = 40.6586 = 40.7V

95.Line voltage may be from 105 V rms to 125 rms in a half-wave rectifier. With a 5:1 step-down transformer, the maximum peak load voltage of an ideal approximation is closest to A. 21 V B. 25 V

C. 29.6 V D. 35.4 V *

Max RMS = 125V 125V = Vpeak/sqrt(2) Vpeak = 125V(sqrt(2)) = 176.77 V considered the primary voltage (greater) Using Ratio of 5:1 V1/V2 = 5/1 176.77/5 = V2/1 V2 = 35.355 V =35.4 V

Basic Electronics:

96. The purpose of the ballast in a fluorescent lamp assembly is A. To regulate the voltage across the lamp B. To improve the overall power factor C. To limit the current through the lamp* D. To regulate the lumens output 97. If N1/N2 = 2, and the primary voltage is 120 V, what is the secondary voltage? A. 0 V B. 36 V C. 60 V * D. 240 V V1/V2 = N1/N2 120V/V2 = 2 V2 = 120/2 = 60V

98.A transformer has a turns ratio of 4: 1. What is the peak secondary voltage if 115 V rms is applied to the primary winding? A. 40.7 V * B. 64.6 V

ME 601 C. 163 V D. 650 V V1/V2 = 4/1 Vrms = Vpeak/sqrt(2) 115V rms = Vpeak Primary/sqrt(2) V peak Primary = 115V * sqrt(2) = 162.6345 V 162.6345/V2 = 4/1 V2 or V secondary = 162.6345/4 = 40.6586 = 40.7V 99.Line voltage may be from 105 V rms to 125 rms in a half-wave rectifier. With a 5:1 step-down transformer, the maximum peak load voltage of an ideal approximation is closest to A. 21 V B 25 V C. 29.6 V D. 35.4 V* Max RMS = 125V 125V = Vpeak/sqrt(2) Vpeak = 125V(sqrt(2)) = 176.77 V considered the primary voltage (greater) Using Ratio of 5:1 V1/V2 = 5/1 176.77/5 = V2/1 V2 = 35.355 V =35.4 V 100.What is the peak load voltage out of a bridge rectifier for a secondary voltage of 15 V rms? (Use second approximation.) A. 9.2 V B. 15 V C. 19.8 V * D. 24.3 V

Approximate computation: 15V = Vpeaksecondary / sqrt(2) Vpeak Secondary = 21.21 Using second approximation: Diode is silicon with 0.7 voltage drop that doubles, considering it’s a full bridge (4 diodes) Finally, output voltage will be 21.21 volts – 2(0.7)volts = 19.81 Volts

ME 601

NOTES: DENSITY in FUELS and Combustion relates to API and SG Boilers and Steam Power Plants Relates to QH process (API and HHV)

CALTECH(Prtl Der) f(x,y) = eqn *set value for x and y **Shift(Integ) d/dx(eqn) @x = x ans Shift STO-> A Autosub/A

Given variable W = 2x^4y^2z - 5x^2y^5z^2 + 25cos(xyz) *set value for x, y, z **Shift(Integ): d/dx = eqn @ x = x Ans STO->A Test ans/A should be equal to 1 or 0.999999 or 1.0000009

Algebra: 1. Factor completely the given equation x3 - 5x2 – 48x + 108 = 0 A. (x - 2)(x + 6)(x – 9) = 0*

C. ( x + 2)(x – 3)(x – 1) = 0

ME 601 B. (x + 1)(x – 3)(x – 2) = 0

D. ( x + 4)(x – 4)(x – 2) = 0

BACKDOOR or use mode 5:3 X=-6, x=9, x = 2 (X+6)(X-9)(X-2)=0 A

2. Solve for x: Ax – B = Cx + D A. (D + B) / (A – C) *

C. (A – B) / (C + D)

B. (D + B) / (A + C)

D. (D – B) / (A – C)

Transpose: Ax – Cx = D + B;

x (A – C) = D + B ; x = (D + B)/(A – C)

3. An airplane flies 1,120km with a tail wind and returns, flying into the same wind. The total time is 3 hrs and 45 minutes, and the airplanes airspeed is 600 kph. What is the wind speed A. 10

C. 20

B. 30

D. 40*

Recall: Distance = Speed * time Total Distance (Round Trip) = 1120 km Total Time of trip = 3 hrs and 45 mins or 3.75 or 15/4 Airplane Speed = 600 km/h

1120/(600km/h + Windspeed) + 1120/(600 km/h - Windspeed) = 15/4 hrs use CALC Shift Solve: Windspeed = 40

4. If a = band b = c, then a = c. This illustrates A. reflexive law* B. transitive law

C. law of symmetry D. substitution law

flying

ME 601 Recall: Reflexive law is for any real number is equal to itself. x, x = x Transitive law, for any real number a, b, and c, a = b; b = c, then a = c Law of Symmetry, for any number a and b, if a = b, then b = a Substitution law, if a = b, then a can be replaced by b in any equation.

5. Given f(x)= (x – 1)/(x3 – 3x2 + 2x), find x when the value of f(x) is undefined. A. 0, -1 and -2

C. 0 and 2

B. 0, 1 and 2 *

D. 1 and 2

Substitute values. Should return result of error (undefined). Trigonometry:

6. If

th

h

A. 1*

t

th

, then find A + B.

B. 7

C. 5

D. 6

7. Find the radius of a circle of a circle of a sector in it with an angle of 1.2 radians has a perimeter of 48 cm. A. 17 cm

B. 16 cm

C. 15 cm*

D. 14 cm

Theta = 1.2 rad Circumference = Perimeter = 2piRadius = 48 cm Sector Area = Radius2 * theta/2

8. The angle of inclination of ascend of a road having 8.25% grade is ____ degrees? A. 4.72 *

C. 5.12

B. 4.27

D. 1.86

Recall: Slope = Ascend Inclination = tan(theta°) = 0.0825 Theta = 4.7162°

ME 601 9. An observer wishes to determine the height of a tower. He takes sight at the top of the tower from A and B, which are 50 ft apart at the same elevation on a direct line with the tower. The vertical angle at a point A is 30o and at point B is 40o. What is the height of the tower? A. 85.60 ft

C. 110.29 ft

B. 143.97 ft

D. 92.54 ft*

Recall: Similar Triangles (Check Figure) tan40° = h/x x = h/tan40°

tan30° = h/50 + x x = h/tan30° - 50

1 to 2 h/tan40° = h/tan30° - 50 1.19175h = 1.73205h - 50 h = 92.54 ft

10. A, B and C are points on a circle. AC bisects the circle and AB = BC. The area of triangle ABC is most clearly what percent of the circles area? A. 44 B. 32*

C. 36 D. 24

Triangle with 2 equal sides is Isoceles, circumscribe by circle

Analytic Geometry and Solid Mensuration: 11. The sum of the coefficient of x and y in Ax + By – 16 = 0 is 14. If the slope of the line is 8, find A and B. A. (7, -1)

C. (8, -6)

B. (16, -2)*

D. (2, -5)

ME 601

Slope, By = Ax + 16 = (Ax + 16)/B; A/B = 8 A + B = 14

12. Find the focus of the hyperbola 16y2 – 9x2 + 36x + 96y – 36 = 0. A. (2,2) and (2,-7)

C. (2,-2) and (2,8)

B. (2,1) and (2,8)

D. (2,2) and (2, -8) *

13. Find the smallest angle between the lines 2x + y – 8 = 0 and x + 3y + 4 = 0. A. 40o

C. 45o *

B. 60o

D. 30o

14. A polyhedron of 6 faces, all of which are parallelograms. A. Parallelepiped* B. Cube C. Pyramid 15. Determine the number of diagonals of a polygon with 15 sides A. 60

C. 120

B. 80

D. 90 *

D. cone

Differential Calculus:

16. The point on the curve where the second derivative of the function is zero. A. Maxima

C. point of inflection *

B. Minima D. point of deflection 17. An open rectangular box with square ends is to be built to hold 6400 mm3 at the cost of $0.75/mm2 for the base and $0.25/mm2 for the sides. Find the most economical dimensions. A. 10 x 10 x 15 mm B. 20 x 20 x 16 mm*

C. 30 x 30 x 17 mm D. 5 x 5 x 14 mm

18. Find y  , given x  xy  y  2 . A.

h h

*

B. t

t

C.

h h

D. t

t

19. A balloon leaving the ground 18 from the observer rises 3 m/s. how fast is the angle of elevation of line of sight increasing after 8 seconds. A. 0.12 rad/sec

C. 0.08 rad/sec

ME 601 B. 0.03 rad/sec

D. 0.06 rad/sec *

20. Evaluate the limit of xx as x approaches to zero. A. 0

C. 1*

B. ½ Integral Calculus:

D. 2

21. Determine the area bounded by y = x2 – 4, the x – axis and the lines x = -1, x = 1. A. 8.87 sq units

C. 7.33 sq units *

B. 8.07 sq units

D. 6.67 sq units

22. A hole of the radius 4 is bored through the center of a sphere of radius 5. Find the volume of the remaining portion of the sphere. A. 28 pi

C. 36 pi*

B. 30 pi

D. 29 pi

23. Evaluate

th

A. 2cos

+C

C. -2cos√x + C-2cos√x + C*

B. cos + C 24. The integral of sin2 Ɵ dƟ from 0 to pi/2 is:

D. -cos√x + C

A. pi/2

C. ½

B. pi/4 *

D. ¼

25. Find the volume (in cubic units) generated by rotating a circle x² + y² + 6x + 4y + 12 = 0 about the yaxis A. 39.48 B. 47.23 Differential Equations:

C.59.22* D. 50.1

26. Find the differential equation of the family of line passing through the origin. A. xdy – ydx = 0

C. xdy + ydx = 0 *

C. ydy + xdx = 0

D. xdx + dy = 0

slope, m = y/x (m is constant) Differentiate:

ME 601 0 = [xdy - ydx]/x2 <-- qoutient rule xdy - ydx = 0

27. The rate of population growth of a country is proportional to the number of inhabitants. If the population of a certain country now is 40 million and 50 million in 10 years’ time, what will be its population 20 years from now? A. 62.5 million*

C. 56.20 million

B. 65.20 million

D. 52.6 million

Estimate using Caltech : Linear regression (MODE 3:5) X

y

From Data Set: Use SHIFT 1:5:y hat

0

40

20y hat = 62.5 million

20

50

28. What is the equation (DE) of the family of parabolas having their vertices at the origin and their foci n the x – axis. A.2xdy – ydx = 0*

C. xdy – 2y2dx = 0

B. 2ydy + x2dx

D. y2 dy + 2xydx = 0

Parabola is y2 = 4ax; Differentiate:

4a = y2/x

0 = [xdy – ydx]/x2 = [x(2ydy) – y2dx] / x2

2xydy – y2dx = 0 or 2xdy – ydx = 0

29. Determine the degree of the given ordinary D.E y’’’’ – 4(y’’’)² + 3(y’’)³ - y’ = 0 A. 1*

C. 3

B. 2

D. 4

ME 601 30. A thermometer reading 18 deg F is brought into a room where the temperature is 70 deg F; a minute later, the thermometer reading is 31 deg F. determine the temperature reading 5 minutes after the thermometer is first brought into the room. A. 57.68 deg F* B. 42.4 deg F

C. 30.01 deg F D. 53.89 deg F

Constant at 70 Estimate using Caltech : Linear regression (MODE 3:5) X

y

From Data Set: Use SHIFT 1:5:y hat

0

70-18

5 y hat = 12.3398

1

70-31

; 70-ans = 57.6601 deg F

Probability and Statistics:

31. In a ME 600 final examination, the probability that the examinee will pass each of the three subjects is 0.60. what is the probability that the examinee will pass at most three subjects? A. 0.064

C. 0.784

B. 0.216 *

D. 0.936

32. In a single throw of a pair of ice, what is the probability of having a sum of 7 or 11? A. 2/9 *

C. 1/6

B. 7/36

D. 1/3

33. The mode of the set {2,2,8,20,33} is A. 2*

C. 8

B. 13

D. 18

34. A bag contains 4 white, 3 blacks ball and another bag contains 3 white and 5 black balls. One ball is drawn from the first bag and placed unseen in the second bag. What is the probability that a ball now drawn from the second bag is black? A. 38/63*

C. 24/63

B. 35/63

D. 45/63

35. Given a set 25 elements with A. 5.12

= 75 and

= 350, find the standard deviation. C. 3

ME 601 B. 5

D. 2.24*

Physics:

36. An objects acceleration as it starts to fall is: A. equal to g *

C. less than g

B. greater than g

D. zero

37. Going against a wind, a domestic plane can travel 5/8 of the distance in one hour that it is going with the wind. If the plane can fly 300mph in calm wind, what is the velocity of the wind? A. 69.3 mph *

C. 96.3 mph

B. 73.3 mph

D. 93.3 mph

Distance = Speed*time, Plane Rate = 300 MPH Distance = (Plane Rate + Wind Rate)Time  Going with the Wind 5/8Distance = (Plane – Wind Rate)  Against the Wind 5/8(Plane Rate + Wind Rate) = (Plane Rate – Wind Rate) 5/8(300 + Wind Rate) = (300 – Wind Rate) Wind Rate = 69.2307 = 69.3 MPH

38. A bullet is fired from a gun at an angle of 400. What is the range if its velocity is 300 m/s? g = 10 m/s. A. 192.9m

C. 8863 m*

B. 229.8 m

D. 12000 m

RANGE relates to VoX: [Vo2sin2theta]/g = [(300 m/s)2sin(2*400)]/10 m/sec = 8863 m

39. A car accelerates from rest at 2 m/s² for 5 seconds, travels at constant speed for 10 seconds and decelerates to rest at 2 m/s². Calculate the distance traveled by the car. A. 525 m

C. 450 m

B. 315 m

D. 150 m*

From Rest to Accel @ 5 sec

ME 601 V = Vo + accel*time = 0 + (2 m/sec2)*(5 sec) = 10 m/sec (THIS IS THE SPEED Gained from rest for 5 sec) Distance then is (Dist = Speed * time) 10 m/sec * 5 sec = 10 meters From Constant Speed for 10 seconds 10 m/sec * 10 sec = 100 meters Decelerating after 10 seconds a = 2 m/sec V = Vo – a*t (from here the final velocity V will be zero) (0-10m/sec)/time = 2 m /sec2 Time = 5 seconds. 10 m/sec *5sec = 50 meters Total Distance = 150 meters 40. Two mass collide on a frictionless horizontal floor and perfectly inelastic collision. Mass 1 is 4 times Mass 2; velocity of mass 1 is 10 m/s to the right while mass 2 is 20 m/s to the left. What is the velocity and direction of the resulting combined mass? A. 10 m/s to the right

C. 4 m/s to the right*

B. 10 m/s to the left

D. 1.5 m/s to the left

Inelastic collision (two bodies did not separate after collision) P1 + P2 = 0

Condition M1 = 4M2

M1v1(right) + m2v2(left) = m1v’1 + m2v’2 M1v1 + m2v2 = (m1 + m2)v’ 4M2(10 m/sec) + M2(-20 m/sec) = (4M2+M2)(v’) 40m/sec(M2) – 20m/sec(M2) = 5 M2(v’) -20m/sec(M2)/5(M2) = v’ 4 m/sec (going to right indicated by – sign)

Mechanics:

ME 601

41. A cable weighing 150 N/m has a span of 150 m and a sag of 36 m. determine the maximum tension in the cable. A. 17, 928 N*

C. 19, 728 N

B. 20, 123 N

D. 15, 267 N

42. The ratio of volume stress to volume strain is called A. Young’s modulus

C. Bulk modulus*

B. Shear modulus

D. Hooke’s Law

43. The valve push rod for an overhead valve engine is ¼ inch in diameter and 14 inches long. Find the diameter and inertia of the rod in inches. A. 1.917 x 10-4*

C. 1.917 x 10 -3

B. 1.917 x 10-7

D. 1.917 x10-1

44. A test specimen is under tension. The load is 20, 000 lb., allowable stress is 10, 00 0 psi, modulus of elasticity is 30 million psi, and original length of specimen is 40 in. what is the required cross section, in sq inches, if the resulting elongation must not be greater than 0.001 inch? A. 2

C. 26.6*

B. 10

D. 62.2

45. A circular aluminum tube of length L=400mm is loaded in compression by forces P as shown in the figure. The outside and inside diameters are 60 mm and 50 mm, respectively. If the compressive stress in the bar is intended to be 40 MPa, what should be the load P?

A. 34.6 kN*

B. 35.6 kN

C. 36.4 kN

D. 37.5 kN

46. A motorist drives north for 35.0 minutes at 85.0 km/hr and then stops for 15.0 minutes. He then continues north, traveling 130 km in 2.00 hours. What is his average velocity? A. 57.5 km/hr

B. 58.4 km/hr

C. 60.6 km/hr

D. 63.6 km/hr*

47. An arrow is shot straight up in the air at an initial speed of 15.0 m/s. After how much time is the arrow heading downward at a speed of 8.00 m/s?

ME 601 A. 2.35 s*

B. 3.22 s

C. 4.28 s

D. 5.27 s

48. Determine the bursting steam pressure of a steel shell with diameter of 10 inches and made of ¼ thick steel plate. The joint efficiency is at 70% and the tensile strength is 60 ksi A. 4200 psi *

C. 42.8 ksi

B. 10.5 ksi

D. 8500 psi

49. A drop hammer of 1 ton dead weight capacity is propelled downward by a 12 inch diameter cylinder. At 100 psi air pressure what is the impact velocity if the stroke is 28 inches? A. 47.4 ft/sec B. 31.6 ft/sec *

C. 15.8 ft/sec D. 63.2 ft/sec

50. Which of the following is the type of stress that differs from compressive stress and it is caused by a contact pressure between separate bodies. A. shearing stress

B. bearing stress*

C. compressive stress D. tensile stress

Engineering Economy:

51. A company needs P60, 000 in five years to buy a new equipment in order to accumulate this sum, a sinking fund consisting of three annual payments is established now. For tax purposes, no further payments will be made after three years. What payments are necessary if money is worth 18% per annum? A. 12, 620

C. 12, 220

B. 12, 062 *

D. 12, 602

52. A man borrowed the amount of P20,000 with an interest of 30%. How much is the interest is he going to pay at the end of 22 months? A. 32, 353

C. 31, 533

B. 12, 353 *

D. 32, 296

53. What is the present worth of a $100 annuity starting at the end of the third year and continuing to the end of the fourth year, if the annual interest rate is 8%? A. $122*

C. $160

B. $153 D. $162 54. An item is purchased for P5,000,000 annual cost are P900,000. Using 8%, what is the capitalized cost of perpetual service? A. P16,250,000.00*

C. P16,890,000.00

ME 601 B. P18,345,243.00

D. P17,435,000.00

55. You borrowed the amount of P10, 000 in a bank with an interest rate of 30% compounded monthly. How much would you have to pay after 2 years? A. P18, 807 *

C. P17, 987

B. P17, 032

D. P17, 825

Recall: F = P(1 + i/p)y = 10000(1 + 0.3/12mos)2 years * 12 mos = 18807.2595

56. A man invests 10,000 today to be repaid in 5 years in 1 lump sum at 12 % compounded annually. If the rate of inflation is 3 % compounded annually, how much profit in present day pesos, is realized over the five years? A. P320

C. P5202*

B. P5628

D. P 7623

57. An investment of x dollar is made at the end of each year for three years, at an interest rate of 9% per year compounded annually. What will the dollar value of the total investment be upon the deposit of the third payment? A. 0.727x

C. 3.278x*

B. 1.295x D. 3x 58. A mining company invested 25,000,000 to develop an oil well which is estimated to contain 1,000,000 barrels of oil. During a certain year, 200,000 barrels were produced from this well. Compute the depletion charge during the year. A. 10M

C. 8M

B. 5M*

D. 1M

60. A machine has a first cost of ₱60,000 and has an expected salvage value after 10 years of ₱6,000. Find the book value after 5 years using declining balance method. A. ₱10,028.34 B. ₱15,071.32* C. ₱20,083.45

D. ₱25,083.45

Fluid Mechanics:

61. Water, density = 62.4 lbf/ft3, is flowing through a pipe. A pitot static gage registers 3.0 inches of mercury. What is the velocity of water in the pipe? Note: densityhg = 848.6 lbf/ft3. A. 14.24 ft/s *

C. 8.24 ft/s

B. 11.24 ft/s

D. 7.45 ft/s

ME 601 Recall: Static Pressure: V = Cv SQRT(2ghDpressure) = Cv *SQRT(2 * 32.2*3/12 ft (62.4/848.6 - 1)) 62. A cylindrical 1 ft diameter tank, 4 ft high contains 3 ft of water. What rotational speed is required to spin the water out the top? A. 22.7 rad/s *

C. 23.7 rad/s

B. 21.4 rad/s

D. 18.90 rad/s

Recall: Angular Velocity = Change in Angular Position/Time (Plug into bernouli’s) ω=

㌳䁣

= sqrt((2*32.2*4ft/0.5ft)) = 22.698 rad/sec = 22.7 rad/sec

63. A 1 m x 1.5 m cylindrical tank is full of oil with SG = 0.92. Find the force acting at the bottom of the tank in dynes. E. 106.33 x 107 dynes F. 106.33 x 104 dynes G. 106.33 x 105 dynes H. 106.33 x 106 dynes* Recall: Static Fluids: Pressure = Gamma * Height = (S.G*density water)*Height = 0.92 * 9.81 KN/m3 * 1.5 m = 13.5378 kPA Force = Pressure * Area = 13.5378 * (pi/4 * (1m)2 *) = 10.632 kN = 10632 N = 106.32 * 106 dynes Note: 1N = 100 000 dynes : Force has Newton unit; Pressure in Pascal

ME 601 64. Find the pressure at the 100 fathom depth of water in kPag. E. 1,793.96 kPag* F. 1,893.96 kPag G. 1,983.96 kPag H. 1,693.96 kPag Height = 100 fathom depth (6 ft/1fathom depth) = 600 ft Pressure = pgh = 1000 kg/m3 * 9.81 m/sec * 600 ft(1m/3.28ft) = 1794512.195 Pagauge = 1794.51 kPag

65. A large mining company was provided with a 3 m3 of compressed air tank. Air pressure in the tank drops from 700 kPa to 150 kPa while the temperature remains constant at 28oC. What percentage has the mass of air in the tank been reduced? E. 74.00 F. 72.45 G. 78.56 H. 78.57 * Fluid mach % of Reduced Mass = Change in Pressue/Initial Pressure = 700 – 150/150 = 0.7857 = 78.57%

Thermodynamics 1: 66. Air is compressed adiabatically from 30oC to 100oC. If mass of air being compressed is 5 kg. Find the change in entropy. E. 1.039 kJ/kg F. 0.746 kJ/kg G. 0 * H. 1.245 kJ/kg Change in Entropy = Qreversible/Temp = Qrev/O adiabatic or no exchange of heat = 0

ME 601 67. A perfect gas has a value of R = 58.8 ft-lb/lb-oR and k = 1.26. if 20 BTU are added to 10 lbs of the gas at constant volume when initial temperature is 90o F. Find the final temperature. E. 97oF * F. 104oF G. 154oF H. 185oF Recall Ideal Gas, Q = mCvdT; Cv = R/k-1 = (58.8ft-lb/lb-R)/(1.26 – 1) * (1/788) = 0.29086BTU/lb-F **788 is conversion of FT-lb to BTU 20 lbs = 10lbs*(0.29086)*(T2 – 90F); T2 = 96.88F = 97F

68. Air enters a nozzle steadily at 1.71 kg/m3 and 35 m/s. what is the mass flow rate through the nozzle if the inlet area of the nozzle is 80 cm2? B. 0.479 kg/s*B. 3.57 kg/s

C. 4.79 kg/s

D. 0.53 kg/s

Recall: Mass Flow Rate = Density * Area * Velocity = (pav) = 1.71 kg/m3 * 80cm2(1m/100cm)2 * 35 m/sec = 0.4788 kg/sec

69. A carnot cycle has a maximum temperature of 580F and minimum temperature of 150F. if the heat added is 4200 BTU/min, find the horsepower output of the engine. B. 33.53

B. 40.94*

C. 44.69

D. 75.40

WorkNET= efficiency * QADDED Recall Efficiency of Carnot = (THIGH – TLOW)/THIGH = (580 – 150)RANKINE/ (580 + 460RANKINE) = 0.4134 Finally, WorkNET = 0.4134(4200)BTU/MIN *(60/HR)(1HP/2545 BTU/HR) = 40.94 HP

70. Steam turbine is receiving 1000 lbm/hr of steam, determine the horsepower output of the turbine if the work done by steam is 250 Btu/lbm. B. 98.23 Hp*

B. 362.7 Hp

Mass Flow Rate = 1000 lbm/hr ; Q = 250 Btu/lbm.

C. 280 Hp

D. 6072.7 Hp

ME 601 Work = mQ = 1000 lbm/hr*(250 Btu/lbm)(1HP/2545 btu/hr) = 98.2318 HP

Thermodynamics 2: 71. Determine th air – standard efficiency of an engine operating on the diesel cycle with clearance of 6% when the suction pressure is 99.97 kPa and the fuel is injected for 7% of the stroke. Assume k = 1.4 E. 62.11% * F. 51.20% G. 73.58% H. 60.02% Take note of DIESEL CYCLE. Recall Eff = 1 – [1/rk^(k-1)]*[(rc^k -1)/k(rc-1)] V3 – V2 = 0.07VD ; V2 = 0.06VD ; V3 = 0.07VD + 0.06VD = 0.13VD Cut-off Ratio, Rc = V3/V2 = 0.13VD/0.06VD = 2.1666 Compression Ratio, Rk = (1 + clearance)/clearance = (1 + 0.06)/0.06 = 17.6667 Then Eff = 1 – [1/rk^(k-1)]*[(rc^k -1)/k(rc-1)] = 1 – [(1/17.6667^(1.4-1))]*[2.1666^1.4 – 1/1.4(2.1666 – 1)] Eff = 0.62109 = 62.11%

72. A 2000 kW Diesel engine unit uses 1 bbl oil per 525 kWh produced. Oil is 25°API. Efficiency of generator 93%, mechanical efficiency of engine 80%. What is the thermal of engine based on indicated power (%)? A.39.3%

B. 39.4%

C. 39.5%

D.39.6% *

This is asking for THERMAL EFFICIENCY of engine! BBL is BARREL. 1 BBL = 42 US gallons = 35 UK gallons 7.33 BBL = 1000kg = 1 metric ton Recall Thermal Eff (Diesel) = Indicated Work/Mfuel(Qh) 25 API = 141.5/SG15.6 – 131.5 SG15.6 = 0.9042; Density = (0.9042 * 1000 kg/m3 )= 0.9042 kg/li Mass of Fuel, Mf = 0.9042 kg/li * (42 us gal)(3.784 li) = 143.7026 kg Qh = 41130 + 139.6(API) = 41130 + 139.6(25) = 44620 kj/kg Work * EffGEN * EffMECH = QOUTPUT

ME 601 Work = QOUTPUT/(0.93*0.8) = 525 KWh/0.744 = 705.645 KWh 705.645 KW/hr (3600sec) = 2540323 KWsec = 2540323 kj/sec * sec = 2540323 KJ Finally, ƞTHERMAL = Indicated Work/Mfuel(Qh) = 2540323 KJ/[(143.7026kg)(44620 kJ/kg)] ƞTHERMAL =0.3961 = 39.61% 73. An air-standard Brayton cycle has a pressure ratio of 8. The air properties at the start of compression are 100 kPa and 25°C. The maximum allowable temperature is 1100°C. Determine the net work. A. 373.23 kJ/kg

B. 373.24 kJ/kg *

C. 373.25 kJ/kg

D. 373.26 kJ/kg

When referring to Pressure Ratio, (take note of HI/LOW: Phigh/Plow, BRAYTON CYCLE is SPSP) PHIGH/PLOW = 8 PLOW = 100 kPa PHIGH = 8 * 100 kPa = 800 kPa Recall: Net Work = Work of Turbine – Work of Compressor T1 = 25 C + 273 = 298K T2 = ? T2 = T1(PHIGH/PLOW)^(k-1/k) = 100(8)^(1.4-1/1.4) = 539.811K Work of Compressor = mCpdT = 1(539.811-298) = 241.8113 Max Temp = T3 = 1100C = 1373K T4 = T3/(8)^(1.4-1/1.4)= 757.9574 K Work of Turbine = mCp(dT) = 615.0425 Finally: Work Net = 615.0425 kJ/kg – 241.8113 kJ/kg = 373.2312 kJ/kg

74. Steam leaves an industrial boiler at 827.4 kPa and 171.6C. A portion of the steam is passed through a throttling calorimeter and is exhausted to the atmosphere when the calorimeter pressure is 101.4 kPa. How much moisture does the steam leaving the boiler contain if the temperature of the steam at the calorimeter is 115.6C? At 827.4 kPa (171.6C): hf = 727.25 kJ/kg, hfg = 2043.2 kJ/kg From table 3: at 101.4 kPA and 115.6C: h2 = 20707.6 kJ/kg B. 3.78%

B. 3.08%*

C. 4.56%

Let x = steam entering the throttling calorimeter h1 = h2 2707.6 kj/kg = 727.25 + x(2043.2); x = 0.9692 1-x = 0.0308; 3.08%

D. 2.34%

ME 601

75. A turbine has an available enthalpy of 3300 kJ/kg in a Rankine cycle. The pump work has also 25 kJ/kg. For flow of 3 kg/s, find the system output. A.5960 kW

B. 6080 kW

C. 6343 kW

D. 9825 kW*

Note the cycle (Rankine – common to steam pplant ). Recall: System Output = mass flowrate * (Change in Enthalpy) = 3 kg/sec (3300kj/kg – 25 kj/kg) = 9825 kW This is Work is the ideal Rankine cycle. (Pump – Turbine)

Heat Transfer: 76. A 5 cm diameter spherical ball whose surface is maintained at a temperature of 70 deg. C is suspended in the middle of a room at 20 deg. C. if the convection heat transfer coefficient is 15 W/m^2C and the emissivity of the surface is 0.8, determine the total heat transfer from the ball. A. 23.56 watts B. 32.77 watts* C. 9.22 watts D. 43.45 watts 77. For heat transfer purposes, a standing man can be modeled as a 30 cm diameter, 170 cm long vertical cylinder with bottom both the top and bottom surfaces insulated and with the side surface at an average temperature of 34 deg C. For a convection heat transfer coefficient is 15 W/m2-C, determine the rate of the heat loss from this man by convection in an environment at 20 deg C. A. 316.46 watts

C. 336.46 watts *

B. 326.46 watts

D. 346.46 watts

Recall: Convective Heat xfer. Q = hA(dT) = 15 W/m2 . °C * pi * 0.3 m * 1.7 m * (34-20)°C = 336.46

78. Consider a person standing in a breezy room at 20 deg C. Determine the total rate transfer from this person of the exposed surface area and the average outer surface temperature of the person are 1.6m2 and 29 deg C, respectively, and the convection heat transfer coefficient is 6 W/m2 with emissivity of 0.95. A. 86.40 watts

C. 198.1 watts

B. 81.70 watts

D. 168.1 watts *

ME 601 Two Modes of Heat Transfer is occuring, as per stated in the problem. Convective and Radiative Heat Transfer. Then Total Heat Transfer = Qconv + Qrad = HAdT + SEAT4 Qconv = 6 W/m2 * 1.6m2 * (29 - 20) = 86.4 Watts Qrad = 5.7 * 10-8 (0.95) * (1.6m2) * ((29+273)4 – (20 + 273)4) = 81.7184 Watts Total Heat = 86.4 + 81.7184 = 168.1184 Watts

79. A tank contains liquid nitrogen at -190 c0 is suspected in a vacuum shell by three stainless steel rods 0.80 cm in diameter and 3 meters long with a thermal conductivity of 16.3 W/m2-C0. If the ambient air outside the vacuum shell is 15 C0, calculate the magnitude of the conductive heat flow in watts along the support rods. A. 0.168 * B. 0.176 C. 0.182 D. 0.0587

Problem States, determine CONDUCTIVE HEAT FLOW. Recall Heat Conduction = HA(dT) Then, Q = 16.3 W/m2 – C * (pi/4 * (0.008 m)2) * (15 – (-190)) = 0.167 Watts

80. How many watts will be radiated from a spherical black body 15 cm in diameter at a temperature of 800 degC? A. 5.34 KW* B. 4.34 KW C. 6.34 KW D. 3.34 KW RECALL: Radiation (SEAT4) Q = 5.6704 * 10-8 * (1 blackbody) * (4*pi*(0.15m/2)2) * (800+273)4 = 5313.063 Watts = 5.31KW Combustion: 81. A gaseous fuel mixture has a molar analysis: H2 = 14%

CH4 = 3%

CO = 27%

O2 = 0.6%

CO2 = 4.5%

N2 = 50.9%

ME 601 Determine the air – fuel ratio for complete combustion on molar basis. E. 2.130 F. 3.230 G. 1.233* H. 1.130 Compute first the Oxygen Chemical Reaction (as this pertains to the product after reacting to Fuel) H2 + O2 = H2O  0.14 H2 + xO2 = 0.14H2O  0.14 H2 + 0.07O2 = 0.14H2O CH4 + O2 = CO2 + H2O  0.03 CH4 + xO2 = 0.03CO2 + 0.06H2O  0.03 CH4 + 0.06O2 = 0.03CO2 + 0.06H2O CO + O2 = CO2  0.27CO + xO2 = 0.27CO2  0.27CO + 0.135O2 = 0.27CO2 Total Oxygen in Chem Reaction = 0.07 + 0.06 + 0.135 = 0.265 Less the 0.6%O2 from fuel = 0.265 – 0.006 = 0.259 O2 from air only. Then, THEO A/F = [0.259 + 0.259(3.76)]/1 = 1.233 AirMOLES/FuelMOLES ***RECALL: the 3.76 is the mole of NITROGEN, from 79%/21% composition of oxygen

82. A bituminous coal has the following composition: C = 71.5%

H = 5.0%

S = 3% Ash = 7.6%

O = 7.0%

N = 1.3%

W = 3.4%

Determine the theoretical weight of Nitrogen in lb/lb of coal. E. 2.870 F. 7.526 * G. 2.274 H. 6.233 From Problem, coal stated as fuel (solid fuel). Known ultimate analysis (percentage of each elements), therefore use EMPERICAL FORMULA for COMBUSTION OF SOLID FUELS Theo A/F = 11.5C + 34.5(H – O/8) + 4.3S Theo A/F = 11.5(0.715) + 34.5(0.05 + 0.07/8) + 4.3(0.03) = 10.378 lbair/lbfuel

ME 601 Recall Mass Percentage for Oxygen and Nitrogen: %Mass of O2 = 23.2% * Mass of Air %Mass of N2 = 76.8% * Mass of of Air, in this case = 0.768 * 10.378 = 7.97 lb/lb. (Nearest ans 7.526)

83. A diesel power plant uses fuel with heating vlue of 43,000 kJ/kg. What is the density of the fuel at 25C? B. 840 kg/m3

B. 873 kg/m3

C. 970 kg/m3*

D. 940 kg/m3

RECALL: API transposes to Density of Fuel Qh relates to API with empirical formula of: Qh = 41130 + 139.6 API 43,000 = 41130 + 139.6 of API API = 13.395 13.395 = 141.5/SG15.6 – 131.5; SG15.6 = 0.9765 For correction factor applied to 25C: SG25 = 0.9765(1-0.00072(25-15.6) = 0.96995 Finally, Density = 0.96995 * 1000 kg/m3 = 969.95 approx to 970 kg/m3

84. The heating value of fuel supplied in a boiler is 43,000 kJ/kg. If the factor of evaporation is 1.10 and the actual specific evaporation is 10, what is the efficiency of the boiler? B. 62.07%

B. 53.08%

C. 78.05%

D. 57.77%*

Recall: Boiler Eff: ƞBOILER = MSTEAM(HSTEAM – HFUEL)/MFUEL(QHEATING VALUE) Note FE or Factor of Evaporation = 1.10 FE = (HSTEAM – HFUEL)/2257 Also, Note ASE or Actual Specific Evaporation = 10 ASE = MSTEAM/MFUEL Simplify: ƞBOILER = ASE(FE*2257) /(QHEATING VALUE) = [10(1.10*2257)]/43000 = 0.5773 = 57.73 %

ME 601

85. Find the density of fuel at 50C if fuel used is 27 API. B. 763.556 kg/m3

B. 834.56 kg/m3 C. 853.45 kg/m3 D. 871.30 kg/m3*

Recall: API transposes Specific Gravity 27 API = 141.5/SG15C – 131.5 SG15C = 0.892 Using Correction Factor SG50 = 0.892(1-0.00072(50C – 15.6C)) = 0.8706 Density = Mass/Volume SG*Density = 0.8706 * 1000 kg/m3 = 870.6 kg/m3 = Approx to 871.3

ME Laws: 86. The Mechanical Engineering Law (BASAG) is also known as A. RA 8459 *

C. RA 5984

B. RA 3449

D. RA 6561

87. The Article V of the the Mechanical Engineering Law A. TITLE, STATEMENT OF POLICY, AND DEFINITION OF TERMS B. PENAL AND CONCLUDING PROVISIONS* C. PRACTICE OF THE PROFESSION D. CONCLUSION

88. Article V, Sec. 43 talks about A. Implementing Rules and Regulations* B. Penalties C. Transitory Provisions D. Funding Provisions

ME 601

89. The Article I, Sec. 1 of the Mechanical Engineering Law A. statement of policy

C. Title *

B. Definition of terms

D. Creation and Composition of the Board

90. According to Sec 42, t any person who violates any of the provisions of this Act and its rules and regulations shall, upon conviction be penalized by a fine of not less than A. P 50, 000 *

C. P40, 000

B. 30, 000

D. P100, 000

AC/DC

91. An ideal step-up transformer with 100 turns in the primary and 2500 turns in the secondary carries a load of 2A in the secondary windings. What is the current in the primary side? C. 50A

*

D. 0.08A

C. 25A D. 1,250A

Turn Ratio: 100/2500 = 1/25 = 0.04 Primary Turn/Secondary Turn = Current Secondary/Current Primary 0.04 = 2Amp/Current Primary Current Primary = 2/0.04 = 50A

92.Gearmotors are selected based on which of the following? C. Speed requirement

C. both A and B*

D. Torque reuirement

D. neither A or B

Recall: Gearmotors requires speed and torque (gear ratio) Ans: Both A and B 93. If N1/N2 = 2, and the primary voltage is 120 V, what is the secondary voltage?

ME 601 A. 0 V B. 36 V

C. 60 V * D. 240 V

V1/V2 = N1/N2 120V/V2 = 2 V2 = 120/2 = 60V 94.A transformer has a turns ratio of 4: 1. What is the peak secondary voltage if 115 V rms is applied to the primary winding? A. 40.7 V * B. 64.6 V

C. 163 V D. 650 V

V1/V2 = 4/1 Vrms = Vpeak/sqrt(2) 115V rms = Vpeak Primary/sqrt(2) V peak Primary = 115V * sqrt(2) = 162.6345 V 162.6345/V2 = 4/1 V2 or V secondary = 162.6345/4 = 40.6586 = 40.7V 95.Line voltage may be from 105 V rms to 125 rms in a half-wave rectifier. With a 5:1 step-down transformer, the maximum peak load voltage of an ideal approximation is closest to A. 21 V B. 25 V

C. 29.6 V D. 35.4 V *

Max RMS = 125V 125V = Vpeak/sqrt(2) Vpeak = 125V(sqrt(2)) = 176.77 V considered the primary voltage (greater) Using Ratio of 5:1 V1/V2 = 5/1 176.77/5 = V2/1 V2 = 35.355 V =35.4 V Basic Electronics:

96. The purpose of the ballast in a fluorescent lamp assembly is E. To regulate the voltage across the lamp F. To improve the overall power factor

ME 601 G. To limit the current through the lamp* H. To regulate the lumens output

97. If N1/N2 = 2, and the primary voltage is 120 V, what is the secondary voltage? A. 0 V B. 36 V C. 60 V * D. 240 V V1/V2 = N1/N2 120V/V2 = 2 V2 = 120/2 = 60V

98.A transformer has a turns ratio of 4: 1. What is the peak secondary voltage if 115 V rms is applied to the primary winding? A. 40.7 V * B. 64.6 V C. 163 V D. 650 V V1/V2 = 4/1 Vrms = Vpeak/sqrt(2) 115V rms = Vpeak Primary/sqrt(2) V peak Primary = 115V * sqrt(2) = 162.6345 V 162.6345/V2 = 4/1 V2 or V secondary = 162.6345/4 = 40.6586 = 40.7V 99.Line voltage may be from 105 V rms to 125 rms in a half-wave rectifier. With a 5:1 step-down transformer, the maximum peak load voltage of an ideal approximation is closest to A. 21 V B 25 V C. 29.6 V D. 35.4 V* Max RMS = 125V 125V = Vpeak/sqrt(2) Vpeak = 125V(sqrt(2)) = 176.77 V considered the primary voltage (greater) Using Ratio of 5:1

ME 601 V1/V2 = 5/1 176.77/5 = V2/1 V2 = 35.355 V =35.4 V 100.What is the peak load voltage out of a bridge rectifier for a secondary voltage of 15 V rms? (Use second approximation.) A. 9.2 V B. 15 V C. 19.8 V * D. 24.3 V

Approximate computation: 15V = Vpeaksecondary / sqrt(2) Vpeak Secondary = 21.21 Using second approximation: Diode is silicon with 0.7 voltage drop that doubles, considering it’s a full bridge (4 diodes) Finally, output voltage will be 21.21 volts – 2(0.7)volts = 19.81 Volts