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SAMPLE PROBLEMS 1. An air-standard Diesel Cycle operates with a compression ratio of 15. The maximum cycle temperature is 1,700ºC. At the beginning of compression the air is at 100 KPa abs and 40ºC. What is the cut-off ratio? T2 = T1 rk K-I = (313 915) 0.4 = 924.657 ºK rc = V3 /V2 = T3 /T2 =197.3/924.657 = 2.13 A) 2.13 B) 2.31 C) 1.23 D) 2.87 2. A spark- ignition engine produces brake power of 224 KW while using 0.0169 kg/s of fuel. The fuel has a higher heating value of 44,186 KJ/kg, and the engine has a compression ratio of 8. The frictional power lost is found to be 22.4 KW. Determine the indicated thermal efficiency. A) 11 B) 22 C) 33 D) 44 ei = WKi / QA = WK i / mf x H V = (224 + 22.4) / (0.0169 x 44,186) = 33% 3. The volume in the clearance space of a 152.40 mm x 254.0 mm Otto gas engine is 1.70 liters. Find the ideal thermal efficiency of the engine on the standard air basis, if the exponent of the expansion and compression lines is 1.35. Express in percent. A) 38.55 B) 36.89 C) 26.98 D) 35.95 2 3 VD =[π ] (0.1524) (0.254) = 0.004633m ; C = V2 / VD = 0.0017/0.004633 = 36.69% rk = (1+ C)/C = (1 +0.3669)/0.3669 = 3.726 e = 1 – (1 / rk k-1 ) = 1 – [1/3.726] 0..35 = 36.89% 4. Calculate from the following data, the kg of air per kg of fuel used by an automobile engine: air temperature is 21.50ºC, barometer reads 760 mm Hg, air entering 1.75 m3 /min., measured gasoline 14.25 liters per hr, specific gravity of gasoline 0.735. A)12.02 B) 15.45 C) 14.45 D) 21.45 ma =PV/RT = (101.325 x 1.75) / 0.287)(294.5) = 2.098 Kg/min mf = ρ x v =0.735 (1,000) x (0.01425)/(60) = 0.1746 kg/min ra/f = 2.098 / 0.1746 = 12.02 kga /kgf 5. An engine has an efficiency of 26%. It uses 2 gallons of gasoline per hour. Gasoline has a heating value of 20,500 Btu/lbf and a specific gravity of 0.8. What is the power output of the engine? A. 20.8KW B. 29KW C. 19.8KW D. 18.9KW Solution: mf

2 gal ft 3 0.862.4lb lbs 13.346 3 hr 7.481gal ft hr

Power Output m f Q f eff Power Output

13.346lb hr 20500 Btu 1.055kj 0.26 hr 3600 sec lb Btu

Power Output 20.8Kw

6. A supercharged six-cylinder four stroke cycle diesel engine of 10.48 cm bore and 12.7 cm stroke has a compression ratio of 15. When it is tested on a dynamometer with a

53.34 cm arm at 2,500 rpm, the scale reads 81.65 kg, 2.86 kg of fuel of 45,822.20 KJ/kg heating value are burned during a 6 min test, and air metered to the c ylinders at the rate of 0.182 kg/s. Find the brake thermal efficiency. A. 0.327 B. 0.367 C. 0.307 D. 0.357 Solution: T = load x lever arm = [81.65kg x 9.91N/kg](0.5334m) = 427.25N- m = 0.42725 KN-m Brake Power = 2nT = 2rad/rev[(2,500/60)rev/sec](0.42725 KN-m) = 111.854 KW Mass of fuel, mf = 2.86kg / 6(60)sec = 0.00794 kgf /sec Brake thermal efficiency = brake power/ mf x HVf = 111.854 KW / [0.00794 kg/sec x 45,822.20 KJ/kg] = 0.307 = 30.7% 7. A Carnot engine receives 130 BTU of heat form a hot reservoir at 700°F and rejects 49 BTU of heat. Calculate the temperature of the cold reservoir. (April ’97) A. -21.9ºF B. -24.2 C. -20.8 D. -22.7 Solution: TH = 700 + 460 = 1160°R

Q A QR T TL 130 49 1160 TL = H = = QA TH 1160 130 TL = 437.23°R TL = -22.77°F

Carnot Engine efficiency

=

8. A Carnot engine requires 35 KW from the power and the temperature of the sink is source? (April ’97) A. 245.6ºC B. 210.1 Solution: TC = 26 + 273 = 299°K T TL W Efficiency = = H TH QA TH = 523.25°K

hot source. The engine produces 15 KW of 26°C. What is the temperature of the hot C. 250.2

15 35

TH = 523.25 – 273

=

D. 260.7

TH 299 TH TH = 250.25°C

9. The compression ratio of an ideal Otto cycle is 6:1. Initial conditions are 101.325 KPa and 20°C. Find the temperature at the end of the adiabatic compression. (October ’94) A. 60ºC B. 600ºK C. 60ºK D. 600ºC Solution: k 1

T2 V1 T1 V2 T2 (6)1.41 20 273

T2 = 599.96°K 10. The maximum thermal efficiency possible for a power cycle operating between 1200°F and 225°F is: (April ’97)

A. 58% B. 58.73 Solution: TH = 1200 + 460 = 1660°R Efficiency =

C. 57.54

D. 57.40

TL = 225 + 460 = 685°R

TH TL 1660 685 = = 58.73% TH 1660

11. A heat engine is operated between temperature limits of 1370 degree C and 260 degree C. Engine supplied with 14,142 per KW-hr. Find the Carnot cycle efficiency in percent. (October ’97) A. 67.56 B. 70.10 C. 65.05 D. 69.32 Solution: T1 = 1370 + 273 = 1643°K Efficiency = 1

T2 = 260 + 273 = 533°K

T2 533 = 1 = 67.56% T1 1643

12. An Otto engine has clearance volume of 7%. It produce s 300 KW power. What is the amount of heat rejected in KW? (October ’97) A. 152 KW B. 95.32 C. 92.16 D. 90.72 Solution: r = compression ratio =

e = 1

e=

1 r

k 1

= 1

W QA

1 c 1 0.07 = c 0.07

1 = 0.664 (15.286)1.41

QA = 452 KW

W = QA – QR

300 = 452 – Q R

QR = 152 kW 13. Calculate the energy transfer rate across 6 in wall of firebricks with a temperature difference across the wall of 50°C. The thermal conductivity of the fireb rick is 0.65 Btu/hr- ft-°F at the temperature interest. (October ’94) A. 285 W/m2 B. 369 C. 112 D. 429 Solution: x = 6 in = 0.50 ft 9 (ta – tb ) = 50 × = 90°F 5 kA( t a t b ) 0.65(1)(90) QC = x 0.50 Btu 1055 J hr (3.28)2 ft 2 QC = 117 = 369 W/m2 Btu 3600 sec hr ft 2 m2

14. The sun generates 1 kW/m2 when used as a source for solar collectors. A collector with an area of 1 m2 heats water. The flow rate is 3.0 liters/min. What is the temperature rise in the water? The specific heat of water is 4,200 J/kg -°C. (April ’97) A. 4.8ºC B. 0.48 C. 0.56 D. 0.84 Solution: 1kW Q (1m 2 ) 1kW 1000 W 2 m m

3 li 1kg min 0.05 kg/sec 2 li 60 sec m

Q = m × cp × Δt

1000 = 0.05 × (4200) × Δt

Δt = 4.76°C

15. An electric current is passed through a wire 1mm diameter and 10 cm long. The wire is submerged in liquid water at atmospheric pressure, and the current is increased until the water boils. For this situation h = 5000W/m2 -K, and the water temperature will be 100 0 C. How much electric power must be supplied to the wire to maintain the wire surface at 1140 C? A. 11.99 B.31.99 C. 41.99 D. 21.99 Solution: convection heat transfer because there is a flow of elec tricity Heat Transfer = electric power supplied Q = hAt = h (πdL) t = 5000 [π(1x10-3 )(1x10-2 )](114 - 100) Q = 21.99 W 16. What is the temperature in degree C of 2 liters of water at 30 degree C after 500 calories of heat have been added to? (April ’98) A. 35.70ºC B. 30.25 C. 38.00 D. 39.75 Solution: Q = m × cp × (t2 – t1 ) 0.500 Kcal (4.187 KJ/Kcal) = 2 liters (1kg/liter)(4.187 KJ/kg-°C)(t2 – 30°C) t2 = 30.25°C 17. At an average temperature of 100°C, hot air flows through a 2.5 m long tube with an inside diameter of 50 mm. The temperature of the tube is 20°C along its entire length. Convective film coefficient is 20.1 W/m2 -°K. Determine the convective heat transfer form air to the tube. (April ‘ 97) A. 900W B. 909 C. 624 D. 632 Solution: A = heat transfer area = π × D × L = π × (0.050) × (2.5) = 0.3927 m2 QC = hC × A × (t2 – t1 ) QC = 20.1 × (0.3927 × (100 – 20) QC = 631.5 W 18. One kg of water initially at 15ºC is to be freeze at –3ºC. Calculate the total amount of heat to be removed during the cooling process? Cpice = 0.5(Cpw)

A) 39.831KJ B) 403 Q 1 = mc∆T = 1 (4.187)15 = 62.8 Q 3 = 1(0.5)(4.187)3 = 6.28

C) 3983.1 Q2 = 334 QT = 403KJ

D) 39831

19. Calculate the enthalpy in KJ of 1.50 kg of fluid that occupy a volume of 0.565 m3 if the internal energy is 555.60 Kcal per kg and the pressure is 2 atm. A. 2,603.7 KJ B. 3,603.7 C. 4,603.7 D. 5,603.7 h = u + pv = [555.6 kcal x (1.055/0.252)] + 2(101.325)(0.565/1.5) = 2402.35 KJ/kg H = mh = 1.5(2402.35) = 3,603.5KJ 20. Steam flows into a turbine at the rate of 10 kg/s and 10 KW of heat are lost from the turbine. Ignoring elevation and kinetic energy effects, calculate the power output from the turbine. (October ’94) Given: hin = 2739.0 KJ/kg; hout = 2300.5 KJ/kg A. 4605KW B. 4973 C. 4375 D. 4000 Solution: H1 + PE1 + KE1 = H2 + PE2 + KE2 + QR + W T W T = H1 – H2 + PE1 – PE2 + KE1 – KE2 – QR = m(h 1 – h 2 ) – QR = [10 kg/s (2739 – 2300.5)KJ/kg] – 10KW = 4385 – 10 = 4375 KW

21. The enthalpy of air is increased by 139.586 KJ/kg in a compressor. The rate of air flow is 16.42 kg/min. The power input is 48.2 KW. Which of the following values most nearly equals the heat loss from the compressor in KW? (April ’97) A. 10KW B. 90.95 C. 100 D. 995 Solution: H1 + PE1 + KE1 + W C = H2 + PE2 + KE2 + QR QR = H1 – H2 + PE1 – PE2 + KE1 – KE2 – W C = m(–∆h) + W C = – (16.42/ 60)kg/s (139.586KJ/kg ) + 48.2 KW = –38.2 KW + 48.2 KW = 10 KW

22. A volume of 400 cc of air is measured at a pressure of 740 mmHg abs and a temperature of 18°C. What will be the volume at 760 mmHg and 0°C? A. 376 cc B. 326 C. 356 D. 365 Solution: 760 V2 P1V1 PV 740( 400) = 2 2 = V2 = 365.4 cc T2 0 273 T1 18 273 23. When the pressure of an ideal gas is doubled while the absolute temperature is halved, the volume is: (A) quadrupled (B) doubled (C) halved (D) quarte red. P1 V1 / T1 = P2 V2 / T2

V2 / V1 = P1 T2 / P2 T1 = P1 (0.5T1 ) / 2P1 T1 = (0.5)(0.5) = 0.25

24. An ideal gas at 45 psig and 80°F is heated in a closed container to 130°F. What is the final pressure? (April ’97) A. 54psia B. 65 C. 75 D. 43 Solution: P1 = 45 + 14.7 = 59.7 psia

P1 P2 T1 T1

T1 = 80 + 460 = 540°R

59.7 P2 540 590

T2 = 130 + 460 = 590°R

P2 = 65.23 psia

25. A closed vessel contains air at a pressure of 160 KPa gauge and temperature of 30 degree C. The air is heated at constant volume to 60 degree C with the atmospheric pressure as 759 mmHg. What is the final gauge pressure? (April ’98) A. 186KPag B. 169 C. 167 D. 172 Solution: 101.325 KPa Patm = 759 mmHg = 101.2 KPa 760 mmHg P1 P2 P2 (160 101.2) T1 T2 (30 273) (60 273) P2 = 287 KPa abs

P2 = 287 – 101.325 = 185.675 KPag

26. Three resistors with 4, 5 and 6 ohms respectively are connected in parallel. Calculate the total resistance in the circuit. A. 6.122 B. 1.622 C. 2.661 D. 1.226 1/ RT = 1/R1 + 1/R2 + 1/R3 = ¼ + 1/5 + 1/6 = 37/60 RT = 60/37 = 1.622 ohms 27. What current would flow for three hours through a resistance of 50 ohms to produce enough heat to raise the temperature of ten kg water from the freezing point to boiling point? A) 7.76amp B) 77.6 C) 0.776 D) 776 Q = mc∆T = 10(4.187)100 = 4,187 KJ in 3 hrs Heat rate = 4,187KJ / 3hrs = 1396 KJ/hr = 0.388 KW = 388 W Therefore; I = √ P/R = √388/50 = 7.76 amp 28. How many 100-W lamps connected in parallel can be placed in a 230 volts circuit, which is protected by a 30-amp fuse? A. 69 B. 70 C. 71 D. 72 P = E x I = 230volts x 30amp = 6900 watts NL = 6900 / 100 = 69 29. A 12-V automotive storage battery is charged with a constant current of 10 amp for 3 hours. Determine the amount of energy supplied to the battery in kw- hr. A. 0.33 B. 0.36 C. 0.39 D. 0.42 E = 12V x 10amp x 3hrs = 360watt-hrs = 0.36 KW-hr 30. Electrical resistance of 7 ohms and 11 ohms are connected in parallel and the combination of which is then connected in series with a resistance of 15 ohms and a source of 220 volts. Calculate the total current flowing in the circuit.

A) 0.57amp B) 11.41 RT = 15 + 1 / (1/7 + 1/11) = 19.28 ohms

C) 10.41 D) 0.057 I = E / R = 220 / 19.28 = 11.41 amp

31. What retarding force is required to stop a 0.45 caliber bullet of mass 20 grams and speed of 300m/sec as it penetrates a wooden block to a depth of 2 inches? A. 17,716 N B. 19,645 C. 15,500 D. 12,500 x = 2 in = 0.0508 m Energy -- Work relation: Work = KE F(x) = ½ mV² F(0.0508m) = ½(0.02kg)(300m/s)² F = 17,716.53 N 32. A 50 mm dia. pipe is 15 m long extends vertically downward from the bottom of an elevated tank and discharges into air. When the water in the tank is 3 m from the entrance to the pipe, what is the discharge in liters/sec? A. 36.88L/s ___ B. 38.86 C. 68.38 D. 63.88 Vel = √2gh = √2(9.81)18 = 18.783 m / s Q = V x A = (18.783) x 0.7854 x (0.05)2 = 0.03688 m3 / s = 36.88 liters / s 33. An opened topped cylindrical tank with oil (S.G. = 0.8) has a circular base 10 m in diameter when filled to a height of 8 m. Calculate the force exerted in its base. A. 4296kN B. 4926 C. 4629 D. 4962 2 F = pressure x area = hγA = 8 x 0.8(1000)9.81 x 0.7854(10) = 4,926,028.8 N 34. What pressure is a column of water 100 cm high equivalent to? (October ’94) A. 9807 D/cm2 B. 9807 N/m2 C. 0.1 bar D. 98,100 Pa Solution: 1m N Pressure = γ × h = 9807 3 × 100 cm × = 9807 N/m2 100 cm m 35. A pressure of 1 millibar is equivalent to: (April ‘95) A. 1000 D/cm2 B. 1000 cm Hg C. 1000psi

D. 1000kg/cm2

Solution: 1 millibar = 0.001 bar = 0.1 kPa = 0.1

kN m2

kN 1000 N 100,000 dynes m2 kN N m2 10,000 cm 2 dynes 1 millibar 1000 cm2 0.1

36. If the velocity in a 300 mm pipe is 0.50 m/s, determine the discharge in a 75 mm diameter jet attached to the pipe. A. 33.5L/s B. 35.3 C. 53.3 D. 55.3 Q = Vel1 x Area1 = Vel2 x Area2

= 0.50 m /s x 0.7854 x 0.3m2 = 0.03534 m3 / sec = 35.3 L/s 37. A wheel starts from rest and acquires a speed of 500 rad/sec in 15 seconds. Calculate the approximate number of turns made by the wheel. A. 119.4 B. 1194 C. 11940 D. 119400 NT = 500rad/s x 15 sec x 1rev / 2πrad = 1,194 rev 38. A ball dropped from the roof of a building 40 m high will hit the ground with a velocity of: A. 20 m/s B. 28 C. 38 D. 40 V = √2gh = √2(9.81)(40) = 28 39. A water temperature rise of 18°F in the water cooled condenser is equivalent in °C to: (October ’94) A. 7.78ºC B. 10 C. 263.6 D. -9.44 Solution: 5 5 Temperature change, Δ°C = F (18) 10°C 9 9 40.

One kg-mole of oxygen gas has a molecular mass of: A. 16kg B. 32kg C. 16g D. 32g 1 kg- mole = 32kg 1lb- mole = 32lb 1g- mole = 32gram

41. Ten liters of 25% salt solution and 15 liters of 35% salt solution are poured into a drum originally containing 30 liters of 10% salt solution. What is the percent concentration of salt in the mixture? A. 19.55% B. 22.15% C. 27.05% D. 25.72%

Solution: 25% 10

+

35%

+

15

10% 30

=

x% 55

0.25(10) + 0.35(15) + 0.10(30) = x(55) x = 19.55%

42. As2 O 3 + 3 C →3 CO + 2As Atomic weights may be taken as 75 for arsenic, 16 for oxygen, and 12 for ca rbon. According to the equation above, the reaction of 1 gram- mole of As2 O 3 with carbon will result in the formation of

(A) 1 gram- mole of As. (C) 150 grams of As

(B) 28 grams of CO. (D) a greater amount of weight of CO than of As.

43. Ethane gas burns according to the equation, 2C 2 H6 + 7 O2 → 4CO2 + 6H2 O. What volume of CO 2 , measured at standard temperature and pressure, is formed for each gram- mole of C2 H6 burned? (A) 22.4 liters (B) 44.8 liters (C) 88.0 liters (D) 89.6 liters 44. Express X = 0.351351351…. into fractional number A.) 351/999 B.) 351/100 C.) 99/351

D.) 351/98

45. Express X = 0.353535 into fractional number A.) 35/99 B.) 35/100 C.) 99/35

D.) 35/98

46. Express X = 0.77777777 into fractional number : X = _______ A.) X = 6/9 B.) X = 7/ 10 C.) X = 7/9 D.) X = 6/10

47. Evaluate: x 2 2 2 ....... A.) X= 1 B.) X = 2

C.) X= 3

D.) X= 4

2 gal ft 3 0.862.4lb lbs 13.346 3 hr 7.481gal ft hr

Power Output m f Q f eff Power Output

13.346lb hr 20500 Btu 1.055kj 0.26 hr 3600 sec lb Btu

Power Output 20.8Kw

6. A supercharged six-cylinder four stroke cycle diesel engine of 10.48 cm bore and 12.7 cm stroke has a compression ratio of 15. When it is tested on a dynamometer with a

53.34 cm arm at 2,500 rpm, the scale reads 81.65 kg, 2.86 kg of fuel of 45,822.20 KJ/kg heating value are burned during a 6 min test, and air metered to the c ylinders at the rate of 0.182 kg/s. Find the brake thermal efficiency. A. 0.327 B. 0.367 C. 0.307 D. 0.357 Solution: T = load x lever arm = [81.65kg x 9.91N/kg](0.5334m) = 427.25N- m = 0.42725 KN-m Brake Power = 2nT = 2rad/rev[(2,500/60)rev/sec](0.42725 KN-m) = 111.854 KW Mass of fuel, mf = 2.86kg / 6(60)sec = 0.00794 kgf /sec Brake thermal efficiency = brake power/ mf x HVf = 111.854 KW / [0.00794 kg/sec x 45,822.20 KJ/kg] = 0.307 = 30.7% 7. A Carnot engine receives 130 BTU of heat form a hot reservoir at 700°F and rejects 49 BTU of heat. Calculate the temperature of the cold reservoir. (April ’97) A. -21.9ºF B. -24.2 C. -20.8 D. -22.7 Solution: TH = 700 + 460 = 1160°R

Q A QR T TL 130 49 1160 TL = H = = QA TH 1160 130 TL = 437.23°R TL = -22.77°F

Carnot Engine efficiency

=

8. A Carnot engine requires 35 KW from the power and the temperature of the sink is source? (April ’97) A. 245.6ºC B. 210.1 Solution: TC = 26 + 273 = 299°K T TL W Efficiency = = H TH QA TH = 523.25°K

hot source. The engine produces 15 KW of 26°C. What is the temperature of the hot C. 250.2

15 35

TH = 523.25 – 273

=

D. 260.7

TH 299 TH TH = 250.25°C

9. The compression ratio of an ideal Otto cycle is 6:1. Initial conditions are 101.325 KPa and 20°C. Find the temperature at the end of the adiabatic compression. (October ’94) A. 60ºC B. 600ºK C. 60ºK D. 600ºC Solution: k 1

T2 V1 T1 V2 T2 (6)1.41 20 273

T2 = 599.96°K 10. The maximum thermal efficiency possible for a power cycle operating between 1200°F and 225°F is: (April ’97)

A. 58% B. 58.73 Solution: TH = 1200 + 460 = 1660°R Efficiency =

C. 57.54

D. 57.40

TL = 225 + 460 = 685°R

TH TL 1660 685 = = 58.73% TH 1660

11. A heat engine is operated between temperature limits of 1370 degree C and 260 degree C. Engine supplied with 14,142 per KW-hr. Find the Carnot cycle efficiency in percent. (October ’97) A. 67.56 B. 70.10 C. 65.05 D. 69.32 Solution: T1 = 1370 + 273 = 1643°K Efficiency = 1

T2 = 260 + 273 = 533°K

T2 533 = 1 = 67.56% T1 1643

12. An Otto engine has clearance volume of 7%. It produce s 300 KW power. What is the amount of heat rejected in KW? (October ’97) A. 152 KW B. 95.32 C. 92.16 D. 90.72 Solution: r = compression ratio =

e = 1

e=

1 r

k 1

= 1

W QA

1 c 1 0.07 = c 0.07

1 = 0.664 (15.286)1.41

QA = 452 KW

W = QA – QR

300 = 452 – Q R

QR = 152 kW 13. Calculate the energy transfer rate across 6 in wall of firebricks with a temperature difference across the wall of 50°C. The thermal conductivity of the fireb rick is 0.65 Btu/hr- ft-°F at the temperature interest. (October ’94) A. 285 W/m2 B. 369 C. 112 D. 429 Solution: x = 6 in = 0.50 ft 9 (ta – tb ) = 50 × = 90°F 5 kA( t a t b ) 0.65(1)(90) QC = x 0.50 Btu 1055 J hr (3.28)2 ft 2 QC = 117 = 369 W/m2 Btu 3600 sec hr ft 2 m2

14. The sun generates 1 kW/m2 when used as a source for solar collectors. A collector with an area of 1 m2 heats water. The flow rate is 3.0 liters/min. What is the temperature rise in the water? The specific heat of water is 4,200 J/kg -°C. (April ’97) A. 4.8ºC B. 0.48 C. 0.56 D. 0.84 Solution: 1kW Q (1m 2 ) 1kW 1000 W 2 m m

3 li 1kg min 0.05 kg/sec 2 li 60 sec m

Q = m × cp × Δt

1000 = 0.05 × (4200) × Δt

Δt = 4.76°C

15. An electric current is passed through a wire 1mm diameter and 10 cm long. The wire is submerged in liquid water at atmospheric pressure, and the current is increased until the water boils. For this situation h = 5000W/m2 -K, and the water temperature will be 100 0 C. How much electric power must be supplied to the wire to maintain the wire surface at 1140 C? A. 11.99 B.31.99 C. 41.99 D. 21.99 Solution: convection heat transfer because there is a flow of elec tricity Heat Transfer = electric power supplied Q = hAt = h (πdL) t = 5000 [π(1x10-3 )(1x10-2 )](114 - 100) Q = 21.99 W 16. What is the temperature in degree C of 2 liters of water at 30 degree C after 500 calories of heat have been added to? (April ’98) A. 35.70ºC B. 30.25 C. 38.00 D. 39.75 Solution: Q = m × cp × (t2 – t1 ) 0.500 Kcal (4.187 KJ/Kcal) = 2 liters (1kg/liter)(4.187 KJ/kg-°C)(t2 – 30°C) t2 = 30.25°C 17. At an average temperature of 100°C, hot air flows through a 2.5 m long tube with an inside diameter of 50 mm. The temperature of the tube is 20°C along its entire length. Convective film coefficient is 20.1 W/m2 -°K. Determine the convective heat transfer form air to the tube. (April ‘ 97) A. 900W B. 909 C. 624 D. 632 Solution: A = heat transfer area = π × D × L = π × (0.050) × (2.5) = 0.3927 m2 QC = hC × A × (t2 – t1 ) QC = 20.1 × (0.3927 × (100 – 20) QC = 631.5 W 18. One kg of water initially at 15ºC is to be freeze at –3ºC. Calculate the total amount of heat to be removed during the cooling process? Cpice = 0.5(Cpw)

A) 39.831KJ B) 403 Q 1 = mc∆T = 1 (4.187)15 = 62.8 Q 3 = 1(0.5)(4.187)3 = 6.28

C) 3983.1 Q2 = 334 QT = 403KJ

D) 39831

19. Calculate the enthalpy in KJ of 1.50 kg of fluid that occupy a volume of 0.565 m3 if the internal energy is 555.60 Kcal per kg and the pressure is 2 atm. A. 2,603.7 KJ B. 3,603.7 C. 4,603.7 D. 5,603.7 h = u + pv = [555.6 kcal x (1.055/0.252)] + 2(101.325)(0.565/1.5) = 2402.35 KJ/kg H = mh = 1.5(2402.35) = 3,603.5KJ 20. Steam flows into a turbine at the rate of 10 kg/s and 10 KW of heat are lost from the turbine. Ignoring elevation and kinetic energy effects, calculate the power output from the turbine. (October ’94) Given: hin = 2739.0 KJ/kg; hout = 2300.5 KJ/kg A. 4605KW B. 4973 C. 4375 D. 4000 Solution: H1 + PE1 + KE1 = H2 + PE2 + KE2 + QR + W T W T = H1 – H2 + PE1 – PE2 + KE1 – KE2 – QR = m(h 1 – h 2 ) – QR = [10 kg/s (2739 – 2300.5)KJ/kg] – 10KW = 4385 – 10 = 4375 KW

21. The enthalpy of air is increased by 139.586 KJ/kg in a compressor. The rate of air flow is 16.42 kg/min. The power input is 48.2 KW. Which of the following values most nearly equals the heat loss from the compressor in KW? (April ’97) A. 10KW B. 90.95 C. 100 D. 995 Solution: H1 + PE1 + KE1 + W C = H2 + PE2 + KE2 + QR QR = H1 – H2 + PE1 – PE2 + KE1 – KE2 – W C = m(–∆h) + W C = – (16.42/ 60)kg/s (139.586KJ/kg ) + 48.2 KW = –38.2 KW + 48.2 KW = 10 KW

22. A volume of 400 cc of air is measured at a pressure of 740 mmHg abs and a temperature of 18°C. What will be the volume at 760 mmHg and 0°C? A. 376 cc B. 326 C. 356 D. 365 Solution: 760 V2 P1V1 PV 740( 400) = 2 2 = V2 = 365.4 cc T2 0 273 T1 18 273 23. When the pressure of an ideal gas is doubled while the absolute temperature is halved, the volume is: (A) quadrupled (B) doubled (C) halved (D) quarte red. P1 V1 / T1 = P2 V2 / T2

V2 / V1 = P1 T2 / P2 T1 = P1 (0.5T1 ) / 2P1 T1 = (0.5)(0.5) = 0.25

24. An ideal gas at 45 psig and 80°F is heated in a closed container to 130°F. What is the final pressure? (April ’97) A. 54psia B. 65 C. 75 D. 43 Solution: P1 = 45 + 14.7 = 59.7 psia

P1 P2 T1 T1

T1 = 80 + 460 = 540°R

59.7 P2 540 590

T2 = 130 + 460 = 590°R

P2 = 65.23 psia

25. A closed vessel contains air at a pressure of 160 KPa gauge and temperature of 30 degree C. The air is heated at constant volume to 60 degree C with the atmospheric pressure as 759 mmHg. What is the final gauge pressure? (April ’98) A. 186KPag B. 169 C. 167 D. 172 Solution: 101.325 KPa Patm = 759 mmHg = 101.2 KPa 760 mmHg P1 P2 P2 (160 101.2) T1 T2 (30 273) (60 273) P2 = 287 KPa abs

P2 = 287 – 101.325 = 185.675 KPag

26. Three resistors with 4, 5 and 6 ohms respectively are connected in parallel. Calculate the total resistance in the circuit. A. 6.122 B. 1.622 C. 2.661 D. 1.226 1/ RT = 1/R1 + 1/R2 + 1/R3 = ¼ + 1/5 + 1/6 = 37/60 RT = 60/37 = 1.622 ohms 27. What current would flow for three hours through a resistance of 50 ohms to produce enough heat to raise the temperature of ten kg water from the freezing point to boiling point? A) 7.76amp B) 77.6 C) 0.776 D) 776 Q = mc∆T = 10(4.187)100 = 4,187 KJ in 3 hrs Heat rate = 4,187KJ / 3hrs = 1396 KJ/hr = 0.388 KW = 388 W Therefore; I = √ P/R = √388/50 = 7.76 amp 28. How many 100-W lamps connected in parallel can be placed in a 230 volts circuit, which is protected by a 30-amp fuse? A. 69 B. 70 C. 71 D. 72 P = E x I = 230volts x 30amp = 6900 watts NL = 6900 / 100 = 69 29. A 12-V automotive storage battery is charged with a constant current of 10 amp for 3 hours. Determine the amount of energy supplied to the battery in kw- hr. A. 0.33 B. 0.36 C. 0.39 D. 0.42 E = 12V x 10amp x 3hrs = 360watt-hrs = 0.36 KW-hr 30. Electrical resistance of 7 ohms and 11 ohms are connected in parallel and the combination of which is then connected in series with a resistance of 15 ohms and a source of 220 volts. Calculate the total current flowing in the circuit.

A) 0.57amp B) 11.41 RT = 15 + 1 / (1/7 + 1/11) = 19.28 ohms

C) 10.41 D) 0.057 I = E / R = 220 / 19.28 = 11.41 amp

31. What retarding force is required to stop a 0.45 caliber bullet of mass 20 grams and speed of 300m/sec as it penetrates a wooden block to a depth of 2 inches? A. 17,716 N B. 19,645 C. 15,500 D. 12,500 x = 2 in = 0.0508 m Energy -- Work relation: Work = KE F(x) = ½ mV² F(0.0508m) = ½(0.02kg)(300m/s)² F = 17,716.53 N 32. A 50 mm dia. pipe is 15 m long extends vertically downward from the bottom of an elevated tank and discharges into air. When the water in the tank is 3 m from the entrance to the pipe, what is the discharge in liters/sec? A. 36.88L/s ___ B. 38.86 C. 68.38 D. 63.88 Vel = √2gh = √2(9.81)18 = 18.783 m / s Q = V x A = (18.783) x 0.7854 x (0.05)2 = 0.03688 m3 / s = 36.88 liters / s 33. An opened topped cylindrical tank with oil (S.G. = 0.8) has a circular base 10 m in diameter when filled to a height of 8 m. Calculate the force exerted in its base. A. 4296kN B. 4926 C. 4629 D. 4962 2 F = pressure x area = hγA = 8 x 0.8(1000)9.81 x 0.7854(10) = 4,926,028.8 N 34. What pressure is a column of water 100 cm high equivalent to? (October ’94) A. 9807 D/cm2 B. 9807 N/m2 C. 0.1 bar D. 98,100 Pa Solution: 1m N Pressure = γ × h = 9807 3 × 100 cm × = 9807 N/m2 100 cm m 35. A pressure of 1 millibar is equivalent to: (April ‘95) A. 1000 D/cm2 B. 1000 cm Hg C. 1000psi

D. 1000kg/cm2

Solution: 1 millibar = 0.001 bar = 0.1 kPa = 0.1

kN m2

kN 1000 N 100,000 dynes m2 kN N m2 10,000 cm 2 dynes 1 millibar 1000 cm2 0.1

36. If the velocity in a 300 mm pipe is 0.50 m/s, determine the discharge in a 75 mm diameter jet attached to the pipe. A. 33.5L/s B. 35.3 C. 53.3 D. 55.3 Q = Vel1 x Area1 = Vel2 x Area2

= 0.50 m /s x 0.7854 x 0.3m2 = 0.03534 m3 / sec = 35.3 L/s 37. A wheel starts from rest and acquires a speed of 500 rad/sec in 15 seconds. Calculate the approximate number of turns made by the wheel. A. 119.4 B. 1194 C. 11940 D. 119400 NT = 500rad/s x 15 sec x 1rev / 2πrad = 1,194 rev 38. A ball dropped from the roof of a building 40 m high will hit the ground with a velocity of: A. 20 m/s B. 28 C. 38 D. 40 V = √2gh = √2(9.81)(40) = 28 39. A water temperature rise of 18°F in the water cooled condenser is equivalent in °C to: (October ’94) A. 7.78ºC B. 10 C. 263.6 D. -9.44 Solution: 5 5 Temperature change, Δ°C = F (18) 10°C 9 9 40.

One kg-mole of oxygen gas has a molecular mass of: A. 16kg B. 32kg C. 16g D. 32g 1 kg- mole = 32kg 1lb- mole = 32lb 1g- mole = 32gram

41. Ten liters of 25% salt solution and 15 liters of 35% salt solution are poured into a drum originally containing 30 liters of 10% salt solution. What is the percent concentration of salt in the mixture? A. 19.55% B. 22.15% C. 27.05% D. 25.72%

Solution: 25% 10

+

35%

+

15

10% 30

=

x% 55

0.25(10) + 0.35(15) + 0.10(30) = x(55) x = 19.55%

42. As2 O 3 + 3 C →3 CO + 2As Atomic weights may be taken as 75 for arsenic, 16 for oxygen, and 12 for ca rbon. According to the equation above, the reaction of 1 gram- mole of As2 O 3 with carbon will result in the formation of

(A) 1 gram- mole of As. (C) 150 grams of As

(B) 28 grams of CO. (D) a greater amount of weight of CO than of As.

43. Ethane gas burns according to the equation, 2C 2 H6 + 7 O2 → 4CO2 + 6H2 O. What volume of CO 2 , measured at standard temperature and pressure, is formed for each gram- mole of C2 H6 burned? (A) 22.4 liters (B) 44.8 liters (C) 88.0 liters (D) 89.6 liters 44. Express X = 0.351351351…. into fractional number A.) 351/999 B.) 351/100 C.) 99/351

D.) 351/98

45. Express X = 0.353535 into fractional number A.) 35/99 B.) 35/100 C.) 99/35

D.) 35/98

46. Express X = 0.77777777 into fractional number : X = _______ A.) X = 6/9 B.) X = 7/ 10 C.) X = 7/9 D.) X = 6/10

47. Evaluate: x 2 2 2 ....... A.) X= 1 B.) X = 2

C.) X= 3

D.) X= 4