Processes (ideal Gas): P P Ln V P W  

Processes (Ideal Gas) A steady flow compressor handles 113.3 m3/min of nitrogen ( M = 28; k = 1.399) measured at intake where P1= 97 KPa and T1= 27 C. Discharge is at 311 KPa. The changes in KE and PE are negligible. For each of the following cases, determine the final temperature and the work if the process is: a) PVk= C b) PV = C Given: V1 = 113.3/60 = 1.89 m3sec ; P1 = 97 KPa ; T1 = 27 + 273 = 300 K P2 = 311 KPa P  a. T2  T1 2   P1  W

k 1 k

 418 K

kP1V1  T2      1  - 253 KW 1  k  T1  

T1 = T2 = 300K W  P1V1 ln

P1  - 213.6 KW P2

Processes (Ideal Gas) Air is contained in a cylinder fitted with a frictionless piston. Initially the cylinder contains 500 L of air at 150 KPa and 20 C. The air is then compressed in a polytropic process ( PVn = C) until the final pressure is 600 KPa, at which point the temperature is 120 C. Determine the work W and the heat transfer Q. (R = 0.287 KJ/kg-K ; k = 1.4) Given: V1 = 0.500 m3 ; P1 = 150 KPa; T1 = 293K P2 = 600 KPa; T2 = 393K T2 n 1 T1  P2 n ln P1 ln

n  1.27

m

P1 V1  0.892 kg RT1

W

P1V1  T2      1  95 KJ 1  n  T1  

∆U = mCv(T2 – T1) = 64 KJ Q = ∆U + W = 64 – 95 = - 31 KJ Ideal Gas (Paddle Work) A piston cylinder arrangement contains 0.02 m3/min of air at 50 C and 400 KPa. Heat is added in the amount of 50 KJ and work is done by a paddle wheel until the temperature reaches 700 C. If the pressure is held constant, how much paddle wheel work must be added to the air. (R = 0.287 KJ/kg-K ; k = 1.4) m3 min T1  50  273  323 K; T2  993 K V1  0.02

P  400 KPa Q  50 KJ Process : Isobaric PV  mRT m  0.086 kg V2  0.06 m 3 W  P( V2  V1 )  16.6 KJ Q  ΔU  W - WP WP  8.1KJ

Ideal Gas Mixture A 3 m3 drum contains a mixture at 101 KPa and 35C of 60% Methane (CH4) and 40% oxygen (O2) on a volumetric basis. Determine the amount of oxygen that must be added at 35C to change the volumetric analysis to 50% of each component. Determine also the new mixture pressure. For CH 4: M = 16; k = 1.321 O2: R = 32 ; k = 1.395 GIVEN : V  3 m 3 ; P  101KPa ; T  308K ; y CH 4  60% ; y O 2  40% M  0.60(16)  0.40(32)  22.4 8.3143

R

22.4 PV  mRT

 0.371

KJ kg - K

101(3)

m

0.371(308) m  2.65 kg xi 

yiMi M 0.60(16)

x CH 4  xO2 

22.4 0.40(32)

x CH 4 

22.4 m CH 4

 0.429

 0.571

2.65 m CH 4  1.137 kg m O 2  2.65 - 1.137  1.513 kg a) at y CH 4  0.50 ; y O 2  0.50 M  0.50(16)  0.50(32)  24 KJ  0.346 24 kg - K 0.50(16) x CH 4   0.33 24 x O 2  0.67

R

8.3143

x CH 4  0.33 

1.137  m CH 4 2.65  m CH 4

1.137  m CH 4 2.65  m CH 4

m CH $  - 0.39 kg ; therefore some CH4 is removed from the mixture because it is negative. m  2.65 - .39  2.26 kg b) y CH 4  0.50 ; y O 2  0.50 M  0.50(16)  0.50(32)  24 KJ  0.346 24 kg - K 0.50(16) x CH 4   0.33 24 x O 2  0.67

R

8.3143

0.67 

1.513  m O 2 2.65  m O 2

m O 2  0.796 m  2.65  0.796  3.446 kg c) for condition a; P P

mRT V 2.26(0.346)(308)

3 P  80.28 KPa For condition b : P

3.446(0.346)(308)

3 P  122.41KPa

Ideal Gas (Closed System) Air from the discharge of a compressor enters a 1 m 3 storage tank. The initial air pressure in the tank is 500 KPa and the temperature is 600K. The tank cools, and the internal energy decreases 213 KJ/kg. Determine a) the work done b) the heat loss c) the change of enthalpy d) the final temperature For Air k = 1.4 R = 0.287

KJ kg  K

V  1 m3 P  500 KPa T  600 K ΔU  -213

KJ kg

W  0  Cons tan t Volume PV  mRT m  2.9 kg Q  mΔU  2.9(-213)  618.5 KJ (Heat is rejected) Δh ΔU Δh  1.4(-618.5)  866 KJ

k

Δh  mC p (T2  T1 ) T2  303 K

Entropy Change (Ideal Gas) Calculate the change of entropy per kg of air when heated from 300K to 600K while the pressure drops from 400 Kpa to 300 KPa. S = 0.78 KJ/kg-K) Ideal Gas (Open System; Compressor) An air compressor receives 20 m3/min of air at 101.325 KPa and 20C and compresses it to 10 000 KPa in an isentropic process. Calculate the power of the compressor. Ideal Gas (Closed System) A closed gaseous system undergoes a reversible process in which 40 KJ of heat are rejected and the volume changes 0.15 m3 to 0.60 m3. The pressure is constant at 200 KPa. Determine the change of internal energy U in KJ. Given: Q = -40 KJ; V1 = 0.15 m3 V2 = 0.60 m3; P = 200 KPa Solution: Q = U + W ; W = P(V2 – V1) W = 200(0.60 – 0.15) = 90 KJ U = -40 – 90 = -130 KJ

Ideal Gas (Isentropic Process) Air in a piston - cylinder occupies 0.12 m3 at 550 KPa. The air expands isentropically doing work on the piston until the volume is 0.25 m3. Determine the work W in KJ. For Air R = 0.287 KJ/kg-K and k = 1.4. Given: V1 = 0.12 m3; P1 = 550 KPa; V2 = 0.25 m3 k 1  PI V1  V1     42 KJ   W  1 (1  k )  V2    

Ideal Gas (Polytropic Process) Air at 0.07 m3 and 4000 KPa is expanded in an engine cylinder and the pressure at the end of expansion is 320 KPa. If the expansion is polytropic with PVn = C where n = 1.35, find the final volume. Given: V1 = 0.07 m3 ; P1 = 4000 KPa ; P2 = 320 KPa ; PV1.35 = C n

P1V1  P2 V2

n 1

 P n V2  V1 1   0.455  P2 

Ideal Gas Gas at a pressure of 100 KPa, volume 0.20 m3 and temperature 300K, is compressed until the pressure is 320 KPa and the volume is 0.09 cu.m.. Calculate the final temperature in K. Given: P1 = 100 KPa; V1 = 0.20 m3; T1 = 300K; P2 = 320 KPa; V2 = 0.09 m3 P1V1 P2 V2  T1 T2 T 2  432K

Ideal Gas Helium gas ( R=2.077 KJ/kg-K; k= 1.667) at 800 KPa and 300K occupies a volume of 0.30 m3. Determine the mass of helium in kg. PV  mRT

m  0.39kg

Ideal Gas (Isometric Process) A rigid container containing 25 kg of nitrogen gas at 298K receives heat and its temperature is increased to 370K. Determine the amount of heat added in KJ. For N2: R = 0.297 KJ/kgK; k = 1.399 Cv = 0.7444KJ/kg-K Q = mCvT = 1334 KJ Pure Substance (Isometric Process) A 2 kg steam-water mixture at 1000 KPa and 90% quality (U = 2401.41 KJ/kg) is contained in a rigid tank. Heat is added until the pressure rises to 3500 KPa and the temperature is 400C (U = 2926.4 KJ/kg). Determine the heat added in KJ. Q = m(U2 – U1) = 1045 KJ Ideal Gas Calculate the pressure of 2 moles of air at 400K, with a total volume of 0.5 m3.

PV  mRT  nMRT  n RT P  13.3MPa Ideal Gas A 100 Liters oxygen tank use in a hospital has a pressure of 1 atmosphere and a temperature of 16C. For a molecular weight of 32 kg/kgmol, determine the mass of oxygen in kilograms. PV = mRT T = 289K ; P = 101.325 KPa m = 0.135 kg

Ideal Gas An unknown gas has a mass of 1.5 kg and occupies 2.5 m3 while at a temperature of 300K and a pressure of 200 KPa. Determine the gas constant for the gas in KJ/kg-K. R

PV = 1.111 KJ/kg-K mT

Open System (Polytropic Process) A gas turbine expands 50 kg/sec of helium (M = 4; k = 1.666) polytropically, PV 1.8 = C, from 1000K and 500 KPa to 350K. Determine; a. The final pressure in KPa b. The power produced in KW c. The heat loss in KW d. The entropy change in KW/K Given : m  50 kg/sec R

8.3143 KJ  2.08 M kg - K

Pr ocess : PV1.8  C (Polytropic) System : Gas Turbine (Open) T1  1000 K; P1  500 KPa T2  350 K T2  P2    T1  P1 

n 1 n

n

P2  T2  n 1   P1  T1  P2  627.14 KPa nmR (T2  T1 )  151,995.80 KW 1 n Q  mC n (T2  T1 )  -16,989.85 KW W  Q  Δh 

k n Cn  C v    0.523  1 n  Δh  mC p (T2  T1 )  -168,985.65 kw

Open System At one point in a pipeline the water speed is 3 m/sec and the gage pressure is 50 KPa. Find the gage pressure at a second point in the line, 11 m lower than the first , if the pipe diameter at the second point is one half the first.

Ideal Gas (Open system; Nozzle) Oxygen expands in a reversible adiabatic manner through a nozzle from an initial pressure and initial temperature and with an initial velocity of 50 m/sec. There is a decrease of 38K in temperature across the nozzle. Determine a. the exit velocity b. for inlet conditions of 410 KPa and 320 K, find the exit pressure.

Q  Δh  ΔKE  ΔPE  W Q  0; W  0; ΔPE  0 ΔKE  Δh  Cp (T2  T1) 2

2

v 2  v1  Cp (T1  T2 ) 2(1000) 2

v 2  2(1000)Cp (T1  T2 )  v1 Cp 

Rk KJ 8.3143 KJ  0.918 ;R   0.26 ; k  1.395 k 1 kg - K 32 kg - K

v 2  268.8 T2  P2    T1  P1 

m sec

k 1 k

k

k

 T  k 1 P2  T2  k 1    ;P2  P1 2  P1  T1   T1  (T1  T2 )  38 (320  T2 )  38 T2  282 K P2  262.14

Ideal Gas A pressure in the cylinder in the figure varies in the following manner with volume, P = C/V2. If the initial pressure is 500 KPa, initial volume is 0.05 m3 and the final pressure is 200 KPa, find the work done by the system. P1  500 KPa; V1  0.05 m3 ; P2  200 KPa PV 2  C 2

2

P1V1  P2 V2  C 1

 P 2 V2   1  V1  0.08 m3  P2 

Hydrostatic Pressure A storage tank contains oil with a specific gravity of 0.88 and depth of 20 m. What is the hydrostatic pressure at the bottom of the tank in kg/cm2. PBottom  0  0.88(9.81)20  172.656 KPa x

1.033 kg

cm2  1.76 kg cm2 101.325 KPa

Variation of Pressure A hiker carrying a barometer that measures 101.3 KPa at the base of the mountain. The barometer reads 85 KPa at the top of the mountain. The average air density is 1.21 kg/m 3. Determine the height of the mountain. ΔP   γΔh γ

ρg 1000

P2  P1  

ρg h2  h1  1000

h  h2  h1  h

1000P1  P2   1373.2 m ρg

Closed System During the execution of a non-flow process by a system, the work per degree temperature increase is dW/dT = 170.9 J/K and the internal energy may be expressed as U = 27 + 0.68T Joules, a function of the temperature T in Kelvin. Determine the heat in KJ if the temperature changes from 10C to 38C. dW  170.9dT U  27  0.68T dU  0.68dT dQ  dU  dW dQ  0.68dT  170.9dT dQ  171.58dT Q  171.5811 dT Q  171.58(T2  T1)  171.58(38  10)  4804.24 Joules  48.0424 KJ

Pure Substance (Nozzle) A nozzle receives 10 kg/sec of steam at 4 MPa and 260C (h = 2836.3 KJ/kg ;  = 51.74 x10-3 m3/kg) and discharges it at 1.4 MPa (h =2634.07 KJ/kg;  = 129.72 x 10-3 m3/kg). If the velocity at inlet is negligible, find the diameter at exit in cm. Q  Δh  ΔKE  ΔPE  W Q  0; W  0; ΔPE  0 ΔKE  Δh 2

2

v 2  v1  Δh 2(1000) 2

v 2  2(1000)( Δh)  v1 v1  0 v 2  2(1000)( Δh)

Δh  2634.07  2836.3  -202.23 v 2  636

m sec

m

A 2v 2 kg  10 υ2 sec

m

πd2 v 2 ρ 2 πd2 v 2  4υ2 4

d

4m x 100 cm  5.10 cm πρ 2 v 2

Properties of fluids A 3 m diameter by 4.5 m height vertical tank is receiving water ( = 978 kg/m3) at the rate of 1.13 m3/min and is discharging through a 150 mm  with a constant velocity of 1.5 m/sec. At a given instant, the tank is half full. Find the water level and the mass change in the tank 15 minutes later. m1  1.13(978)15  16,577.1kg  entering π(0.150)2 (1.5)( 60)(15)(978)  23331.63 kg  leaving 4 π(3)2 ( 4.5)(978) mi   15,554.42 kg  initial mass of water 4(2) Δm  m1  mi  m2  8,799.9 kg m2 

Volume of water in the tank  Height of water 

8,799.9  9. 0 m 3 978

9. 0  1.27 m π 2 (3 ) 4

Zeroth Law An engineering student wants to cool 0.25 kg of Omni Cola (mostly water) initially at 20C by adding ice that is initially at -20C. How much ice should be added so that the final temperature will be 0C with all the ice melted, if the heat capacity of the container neglected. QCola  Qice

0.25( 4.187)(20  0)  mice 2.0935(0  20)  334.9 mice  0.06 kg

Zeroth Law 2.5 kg of brass of specific heat 0.39 KJ/kg-C at a temperature of 176C is dropped into a 1.2 liters of water at 14C. Find the resulting temperature of the mixture. (At 14C density of water is 999 kg/m3) QBrass  Q w ater 2.5(0.39)(176  t ) 

1.2 999(4.187)t  14 1000

t  40.35C

Open system (Turbine) Steam enters a turbine with a velocity of 1.5 m/sec and an enthalpy of 2093 KJ/kg and leaves with an enthalpy of 1977 KJ/kg and a velocity of 91.5 m/sec. Heat losses are 8 KCal/min and the steam flow rate is 27 kg/min. The inlet of the turbine is 3.5 m higher than its outlet. What is the work output of the turbine if the mechanical losses is 15%. (1 KCal = 4.187 KJ) Q  Δh  ΔKE  ΔPE  W  8( 4.187)  0.56 KW  Rejected 60 27 1977  2093  -52.2 KW Δh  60 27  (91.5)2 - (1.5)2  ΔKE     1.9 KW 60  2(1000) 

Q

27(9.81)  3.5  0  15.45 KW 60 W  0.56  52.2 - 1.9  15.45  65.2 KW

ΔPE 

Woutput  (65.2)(1 - .15)  55.41KW (work done by the system)

Open System (Compressor) An air compressor handles 8.5 m3/min of air with a density of 1.26 kg/m 3 and a pressure of 101 KPaa and discharges at 546 KPaa with a density of 4.86 kg/m3. The changes in specific internal energy across the compressor is 82 KJ/kg and the heat loss by cooling is 24 KJ/kg. Neglecting changes in kinetic and potential energies, find the work in KW. System : Compressor (0pen system) 8 .5 1.26  0.18 kg 60 sec Q  Δh  ΔKE  ΔPE  W

m

Δh  Δ(Pυ)  ΔU ΔU  82

KJ kg

ΔPE  0  P  546 101 KJ Δ(Pυ)  Δ     32.19 kg  ρ  4.86 1.26 KJ Δh  32.19  82  114.19 kg Q  24

KJ (heat loss) kg

W  Q - Δh  ΔKE  ΔPE  -138.2

KJ kg

Closed system (Piston in cylinder) Four kilograms of a certain gas is contained within a piston–cylinder assembly. The gas undergoes a process for which the pressure - volume relationship is PV1.5 = C. The initial pressure is 300 KPa, the initial volume is 0.10 m 3, and the final volume is 0.2 m3. The change in specific internal energy of the gas in the process is U = - 4.6 kJ/kg. There are no significant changes in kinetic or potential energy. Determine the net heat transfer for the process, in kJ. System : Piston - Cylinder (Closed) P1  300 KPa; V1  0.10 m3 ; V2  0.20 m3 ; ΔU  4.6

KJ kg

Q  ΔU  W W   PdV  n

P2 V2 - P1V1 1 n

n

P2 V2  P1V1

P2  106.07 KPa W  17.57 KJ ΔU  4.6(4)  18.4 KJ Q  35.97 KJ

Entropy Change Calculate the change of entropy per kg of air (R = 0.287 KJ/kg-K; k = 1.4) when heated from 300K to 600K while the pressure drops from 400 KPa to 300 KPa.

CP 

Rk KJ  1.0045 k 1 kg - K

ΔS  0.78

KJ kg - K

Law of Conservation of Mass Steam at 1 MPa, 300°C flows through a 30 cm diameter pipe with an average velocity of 10 m/s. Determine mass flow rate of this steam. (Density at 1 MPa and 300C = 3.875 kg/m3) m  ρAv

kg π 2 m  3.875 0.30 10  2.74 sec 4

Closed System (Piston in Cylinder) Air is compressed in a piston-cylinder device. Using constant specific heats and treating the process as internally reversible, determine the amount of work required to compress this air from 100 KPa, 27°C to 2000 KPa, 706°C. System : Piston - Cylinder (closed) Process : PVn  C Q  ΔU  W P2 V2 - P1V1 R(T2  T1)  1 n 1 n

W   PdV  n

P1V1  P2 V2

n

n

 V2  P1   V  P 2  1 n 1

T2  P2  n   T1  P1  U sin g Laws of logarithm P1 P2 n V ln 2 V1 ln

n 1  n

ln

T2 T1

P  ln 2   P1  n  1.652 W  266.8

KJ kg

Open System (Nozzle) Air enters the after burner nozzle of a jet fighter at 427°C with a velocity of 100 m/s. It leaves this adiabatic nozzle at 377°C. Assuming that the air specific heats do not change with temperature, determine the velocity at the nozzle exit. Q  Δh  ΔKE  ΔPE  W Q  0; W  0; ΔPE  0 ΔKE  Δh  Cp (T2  T1 ) 2

2

v 2  v1  Cp (T1  T2 ) 2(1000) 2

v 2  2(1000)Cp (T1  T2 )  v1 Cp  1.0045 T1  427C T2  377C v1  100

m sec

v 2  332.3

m sec

Gas Mixture One mole of a gaseous mixture has the following gravimetric analysis: O2 = 16% ; CO2 = 44% ; N2 = 40%. Find a. the molecular weight of the mixture b. the mass of each constituents c. the moles of each constituent in the mixture

Formulas yi 

xi

Mi xi Σ Mi

M  ΣyiMi

xi 

mi m

M

m n

8.3143 M ni yi  n

R

Open System (Nozzle) Air enter the nozzle as shown at a pressure of 2700 KPa at a velocity of 30 m/sec and with an enthalpy of 923 KJ/kg, and leaves with a pressure of 700 KPa and enthalpy of 660 KJ/kg. If the heat loss is 0.96 KJ/kg, find the exit velocity in m/sec if the mass flow rate is 0.2 kg/sec. System : Nozzle (Open) Q  Δh  ΔKE  ΔPE  W Q  -0.96

KJ kg

W  0; ΔPE  0 ΔKE  Q - Δh 2

2

v 2  v1  Q - Δh 2000 v 2  2000Q - Δh  v1

2

Δh  (660 - 923)  -263 v1  30

KJ kg

m sec

v 2  724.6

m sec

Ideal Gas A certain perfect gas of mass 0.1 kg occupies a volume of 0.03 m3 at a pressure of 700 KPa and a temperature of 131C. The gas is allowed to expand until the pressure is 100 KPa and the final volume is 0.02 m 3. Calculate; a) the molecular weight of the gas b) the final temperature

PV  mRT 700(0.03)  0.1R(131 273) R  0.52 M

KJ kg - K

8.3143 kg  16 0.52 kgmol

100(0.02)  0.1(0.52)T T  38.5 K

Force 10 liters of an incompressible liquid exert a force of 20 N at the earth’s surface. What force would 2.3 Liters of this liquid exert on the surface of the moon? The gravitational acceleration on the surface of the moon is 1.67 m/sec 2. V  0.01m3 m ; m  Vρ V m  0.01ρ ρ

F  ma 20  0.01ρ(9.81) kg m3  0.0023(203.9)1.67  0.80 N

ρ  203.9 FMoon

Manometers A closed tank contains compressed air and oil (S = 0.90) as shown in the figure.a U – tube manometer using mercury (S = 13.6) is connected to the tank as shown. For column heights h1 = 90 cm; h2 = 15 cm and h3 = 22 cm, determine the pressure reading of the gage.

ΔP   γΔh 0  13 .6(9.81)(0.22 )  0.90(9.81)(0.15 )  0.90(9.81)(0.90 )  Pgage Pgage  20.08 KPa

Manometer In the figure pipe A contains carbon tetrachloride (S = 1.60) and the closed storage tank B contains a salt brine (S =1.15). Determine the air pressure in tank B in KPa if the gage pressure in pipe A is 1.75 kg/cm2.

(101.325)  171.65 KPa 1.033 3 ft  0.914 m

PA  1.75

171.65 - (0.914)(1.6)(9.81)  (0.914)(1.15)(9.81) - 1.2(1.15)(9.81)  PB PB  154.08 KPa gage

Ideal Gas (Isothermal Process) Air which is initially at 120 KPa and 320K occupies 0.11 m3. It is compressed isothermally until the volume is halved and then compressed it at constant pressure until the volume decreases to ¼ of the initial volume. Sketch the process on the PV and TS diagrams. Then determine the pressure, the volume and temperature in each state. Pr ocess Isothermal : P1V1  P2 V2 Fluid : Air (R  0.287 KJ/kg - K) k  1.4 P1  120 KPa; T1  320 K; V1  0.11 m3 PV  mRT 120(0.11)  0.144 kg 0.287(320) At 1 to 2 (Isorherma l)

m

0.11  0.055 m3 2 PV P2  1 1  240 KPa;T2  320 K V2 V2 

At 2 to 3 (Isobaric) P2  P3  240 KPa 0.11  0.0275 m3 4 T3 V3  T2 V2

V3 

T3  160 K

Gas MIxture One mole of a gaseous mixture has the following gravimetric analysis; O2 = 16% ; CO2 = 44%; N2 = 40%.Find a) the molecular weight of the mixture b) the mass of each constituent c) the moles of each constituent in the mixture d) the gas constant R e) the partial pressure for P = 207 KPa

Carnot Cycle A Carnot engine operates between temperature levels of 397C and 7C and rejects 20 KJ/min to the environment. The total network output of the engine is used to drive heat pump which is supplied with heat from the environment at 7C and rejects heat to a home at 40C. Determine: a) the network delivered by the engine in KJ/min b) the heat supplied to the heat pump in KJ/min c) the overall COP for the combined devices which is defined as the energy rejected to the home divided by the initial energy supplied to the engine.

Gas Mixture A 3 m3 drum contains a mixture at 101 KPa and 308K of 60% CH4 and 40% O2 on a volumetric basis. Determine: ( For O2: M = 32; k = 1.395 ; For CH4 : M = 16; k = 1.321) a. The amount of CH4 in kg that must be added at 308K to change the volumetric analysis to 50% for each component. b. The amount of O2 in kg that must be added at 308K to change the volumetric analysis to 50% for each component. c. The new mixture pressure in KPa for conditions (a) and (b)

With CH4 added or removed 1.136  mx x CH 4  2.65  mx 1.136  mx .333  2.65  mx 0.333mx  (0.333)( 2.65)  1.136  mx (0.333)( 2.65)  1.136  0.38 kg 1  .333 therfore CH4 is removed from the mixture

mx 

mCH4  1.136  0.38  0.756 m  2.65  .38  2.27 kg With O2 added x O2 

1.515  my  0.667 2.65  my

0.667(2.65)  1.515  0.758 kg 1  0.667 m  2.65  0.758  3.408 kg my 

Diesel Cycle An air standard diesel cycle has a compression ratio of 18 and the heat transferred is 1800 KJ/kg. At the beginning of the compression process the pressure is 100 KPa and the temperature is 15C. For 1kg of air, determine The entropy change during isobaric heat addition The temperature at the start of heat rejection For Air The work in KJ k = 1.4 The thermal efficiency R = 0.287 KJ/kg-K The mean effective pressure in KPa T2  T1 r 

k 1

 915K

P2  P1 r   10018 k

1.4

 5719.8 KPa

Q A  mC p (T3 - T2 ) QA  T2  2707.1K mC p

T3 

S 3  S 2  mC p ln

T3 KJ  1.0894 T2 K

S1  S 4  (S 3  S 2 )  mC v ln T4  1314.61K QR  mC v (T4  T1 )  736.6 KJ e

Q A - QR  59.1% QA

W  Q A - QR  1063.4 KJ Pm 

W VD

VD  V 1 V2 V1 

mRT1 mRT2 ; V2  P1 P2

VD  0.781 m3 Pm  1362.2 KPa

T1 T4

Diesel Cycle In the air standard diesel cycle, the air is compressed isentropically from 26C and 105 KPa to 3700 KPa. The entropy change during heat rejection is -0.6939 KJ/kg-K. Determine a) the heat added per kg b) the thermal efficiency c) the maximum temperature d) the temperature at the start of heat rejection

P

T 2

3

QA

3

P=C S=C

2

4

S=C

V=C

4

QR

1

1

S

V Given: T 1 = 26 + 273 = 299 K P1 = 105 KPa P2 = P3 = 3700 KPa S1 – S4 = -0.6939 KJ/kg-K S3 – S2 = 0.6939 KJ/kg-K k-1

T2  P2  k   T1  P1  T 2  827.3 K

T3 T2 T 0.6939  1.0045ln 3 827.3 0.6939 1.0045 

T3 827.3 T3  1650.71 K

Q A  Cp T3 - T2   827.1 KJ/kg S 4 - S1  C v ln e

0.6939 0.7175 

T4 299

T4  786.47 K

Q R  349.76 KJ/kg W  Q A - Q R  477.34 KJ/kg e

S 3  S2  S  mCp ln

e

Q R  C V T4  T1 

T4 T1

W x 100%  57.71 % QA

Otto Cycle For an ideal Otto engine with 17% clearance and an initial pressure of 93 Kpa, determine the pressure at the end of compression. If the pressure at the end of constant volume heating is 3448 KPa, what is the mean effective pressure.

P

T

QA

3

3 V=C S=C

2 4

2

V=C S=C

4

1 1 QR V

S

Given: c= 0.17 P1 = 93 KPa P3 = 3448 KPa

W QA W  eQ A  eq.9

e

1  c 1.17  c .17 r  6.88

Q A  C v (T3  T2 )

V V r 1  4 V2 V3 P2 k r P1

Q A  2.315T1  eq.10

r

P2  936.881.4  1383.91 KPa 1   e  1  k 1   0.54  54%  r  T2  T1 r k 1  2.163T1  eq.1 P  T3  T2  3   5.39T1  eq.2  P2  T T4  k31  T3 r 1k  2.49T1  eq.3 r  VD  1  2  eq.4 1 2 1  r2  eq.5

r

VD  r2  2

VD  2 r  1  eq.6

RT2 0.2872.163T1    0.00045T1  eq.7 P2 1383.91 VD  0.00045T1 6.88  1 2 

VD  0.0026T1  eq.8

Q A  0.71755.39T1  2.163T1  Q A  0.7175T15.39  2.163 W  0.542.315T1 

W  1.25T1  eq.11 W 1.25T1  VD 0.0026T1 Pm  480.9 KPa Pm 

Carnot Heat Pump A Carnot engine operates between temperature levels of 397C and 7C and rejects 20 KJ/min to the environment. The total network output of the engine is used to drive heat pump which is supplied with heat from the environment at 7C and rejects heat to a home at 40C. Determine: a) the network delivered by the engine in KJ/min b) the heat supplied to the heat pump in KJ/min c) the overall COP for the combined devices which is defined as the energy rejected to the home divided by the initial energy supplied to the engine.

Carnot Reverse Cycle A reversed Carnot cycle is used for refrigeration and rejects 1000 KW of heat at 340 K while receiving heat at 250 K. Determine: a) the COP b) the power required in KW c) the refrigerating capacity in Tons

Gas Mixture One mole of a gaseous mixture has the following gravimetric analysis; O 2 = 16% ; CO2 = 44%; N2 = 40%.Find a) the molecular weight of the mixture b) the mass of each constituent c) the moles of each constituent in the mixture d) the gas constant R e) the partial pressure for P = 207 KPa

xi Mi yi  xi  Mi xi 0.16 .44 .40      0.0293 Mi 32 44 28 yO2  0.171 yCO2  0.341 yN 2  0.488

Isentropic Process A blower compresses 0.02 m 3/sec of gaseous mixture that has the following volumetric analysis: CO2 = 11% ; O2 = 8% ; CO = 1% ; N2 = 80% ; from 101 Kpa and 27 C to 180 KPa. Assuming the compression to be isentropic, find the work in KW. For CO : For O : For CO: For N : M =44 M = 32 M = 28 M = 28 K = 1.288 K = 1.295 K = 1.399 K = 1.399 Gas Mixture A 0.23 m3 drum contains a gaseous mixture of CO2 and CH4 each 50% by mass at P = 689 KPa, 38C; 1 kg are added to the drum with the mixture temperature remaining at 38C. For the final mixture, find; a) the gravimetric analysis b) the volumetric analysis c) the Cp (For CO2: k = 1.288; CH4: k = 1.321; O2: k = 1.395)

of O2

d) the total pressure Gas Mixture A gaseous mixture composed of 25 kg of N2, 3.6 kg of H2, and 60 kg of CO2 is at 200 KPa, 50C. Find the respective partial pressures and compute the volume of each component at its own partial pressure and 50C. Given: mN2 = 25 kg ; mH2 = 3.6 kg ; mCO2 = 60 kg m = 25 + 3.6 + 60 = 88.6 kg xN2 = 0.282 ; xH2 = 0.041 ; xCO2 = 0.678 P = 200 KPa ; T = 323 K xi Mi yi  xi  Mi xi 0.282 .041 .678      0.046 Mi 28 2 44 y N2  0.219 y H2  0.446 y CO2  0.335 Pi P PN2 = .219(200) = 43.8 KPa PH2 = .446(200) = 89.2 KPa PCO2 = 0.335(200) = 67 KPa yi 

Pi Vi  mi R i Ti 25( 0.297( 323)  54.76 m 3 43.8 3.6( 4.16 )( 323) VH2   54.23 m 3 89.2 60( 0.189 )( 323) VCO2   54.67 m 3 67 VN2 

Gas Mixture Assume 2 kg of O2 are mixed with 3 kg of an unknown gas. The resulting mixture occupies a volume of 1.2 m3 at 276 KPa and 65C. Determine a) R and M of the unknown gas constituent b) the volumetric analysis c) the partial pressures Given; mO2 = 2 kg; mx = 3 kg V = 1.2 m3 ; P = 276 KPa; T = 338 K

a) m = 5 kg xO2 = 0.40 ; xx = 0.60 PV  mRT R = 0.1361 KJ/kg-K R = .40(0.26) + 0.60(Rx) Rx = 0.535 KJ/kg-K Mx = 15.54 kg/kgm x .40 .60  i    0.0511 b) Mi 32 15.54 yO2 = 0.245 ; yx = 0.755 c) PO2 = .245(276) = 67.62 KPa ; Px = 0.755(276) = 208.38 KPa

Gas Mixture A 3 m3 drum contains a mixture at 101 KPa and 308K of 60% CH4 and 40% O2 on a volumetric basis. Determine: ( For O2: M = 32; k = 1.395 ; For CH4 : M = 16; k = 1.321) a) The amount of CH4 in kg that must be added at 308K to change the volumetric analysis to 50% for each component. b) The amount of O2 in kg that must be added at 308K to change the volumetric analysis to 50% for each component. c) The new mixture pressure in KPa for conditions (a) and (b) GIVEN : V  3 m 3 ; P  101KPa ; T  308K ; y CH 4  60% ; y O 2  40% M  0.60(16)  0.40(32)  22.4 8.3143

R

22.4 PV  mRT

 0.371

KJ kg - K

101(3)

m

0.371(308) m  2.65 kg xi 

yiMi

M 0.60(16) x CH 4   0.429 22.4 0.40(32) xO2   0.571 22.4 m CH 4 x CH 4  2.65 m CH 4  1.137 kg m O 2  2.65 - 1.137  1.513 kg a) at y CH 4  0.50 ; y O 2  0.50 M  0.50(16)  0.50(32)  24 KJ  0.346 24 kg - K 0.50(16) x CH 4   0.33 24 x O 2  0.67

R

8.3143

x CH 4  0.33 

1.137  m CH 4 2.65  m CH 4

1.137  m CH 4 2.65  m CH 4

m CH $  - 0.39 kg ; therefore some CH4 is removed from the mixture because it is negative. m  2.65 - .39  2.26 kg b) y CH 4  0.50 ; y O 2  0.50 M  0.50(16)  0.50(32)  24 KJ  0.346 24 kg - K 0.50(16) x CH 4   0.33 24 x O 2  0.67

R

8.3143

0.67 

1.513  m O 2 2.65  m O 2

m O 2  0.796 m  2.65  0.796  3.446 kg c) for condition a; P P

mRT V 2.26(0.346)(308)

3 P  80.28 KPa For condition b : P

3.446(0.346)(308)

3 P  122.41KPa

By: ENGR. YURI G. MELLIZA