Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
8th Edition
ISBN: 9780073398174
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 9.12, Problem 146P
To determine

The exergy destruction associated with each process of the Brayton cycle and the exergy of the exhaust gases at the exit of the regenerator.

Expert Solution & Answer
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Answer to Problem 146P

The exergy destruction associated with process 1-2 of the given Brayton cycle is 38.42kJ/kg.

The exergy destruction associated with process 3-4 of the given Brayton cycle is 24.23kJ/kg.

The exergy destruction associated with regeneration process of the given Brayton cycle is 4.47kJ/kg.

The exergy destruction associated with process 5-3 of the given Brayton cycle is 24.16kJ/kg.

The exergy destruction associated with process 6-1 of the given Brayton cycle is 148.7kJ/kg.

The exergy of the exhaust gases at the exit of the regenerator is 148.9kJ/kg.

Explanation of Solution

Show the regenerative Brayton cycle with air as the working fluid, on Ts diagram as in Figure (1).

Thermodynamics: An Engineering Approach, Chapter 9.12, Problem 146P

For the given regenerative Brayton cycle with air as the working fluid, let Ti, Pi, hi and Pri be the temperature, pressure, specific enthalpy and relative pressure at ith state respectively.

Write the expression of pressure ratio for the regenerative Brayton cycle (rp).

rp=P2/P1 (I)

Write the pressure ratio and pressure relation for the process 3-4.

Pr4=P4P3Pr3 (II)

Write the expression of efficiency of the turbine (ηT).

ηT=h3h4h3h4s (III)

Write the expression of heat added due to regeneration (qregen).

qregen=ε(h4h2) (IV)

Here, the effectiveness of the regenerator is ε.

Write the expression of net work output of the regenerative Brayton cycle (wnet).

wnet=wT,outwC,in

wnet=(h3h4)(h2h1) (V)

Here, the work output by the turbine is wT,out, and the work input to the compressor is wC,in.

Write the expression of heat input to the regenerative Brayton cycle (qin).

qin=(h3h2)qregen (VI)

Write the expression of heat rejected by the regenerative Brayton cycle (qout).

qout=qinwnet (VII)

Write the expression of specific enthalpy at state 6 (h6), using the heat rejection relation.

h6=h1+qout (VIII)

Write the specific enthalpy relation for the regenerator.

h5h2=h4h6 (IX)

Write the expression of exergy destruction associated with the process 1-2 of the given Brayton cycle (xdestroyed,12).

xdestroyed,12=T0(s2s1)

xdestroyed,12=T0(s2s1RlnP2P1) (X)

Here, the temperature of the surroundings is T0, the gas constant of air is R, entropy of air at state 2 as a function of temperature alone is s2, and entropy of air at state 1 as a function of temperature alone is s1.

Write the expression of exergy destruction associated with the process 3-4 of the given Brayton cycle (xdestroyed,34).

xdestroyed,34=T0(s4s3)

xdestroyed,34=T0(s4s3RlnP4P3) (XI)

Here, entropy of air at state 3 as a function of temperature is s3, and entropy of air at state 4 as a function of temperature is s4.

Write the expression of exergy destruction associated with the regeneration process of the given Brayton cycle (xdestroyed,regen).

xdestroyed,regen=T0[(s5s2)+(s6s4)]

xdestroyed,regen=T0[(s5s2)+(s6s4)] (XII)

Here, entropy of air at state 5 as a function of temperature alone is s5, and entropy of air at state 6 as a function of temperature alone is s6.

Write the expression of exergy destruction associated with the process 5-3 of the given Brayton cycle (xdestroyed,53).

xdestroyed,53=T0(s3s5+qinTH)

xdestroyed,53=T0(s3s5RlnP3P5qinTH) (XIII)

Here, the temperature of the heat source is TH.

Write the expression of exergy destruction associated with the process 6-1 of the given Brayton cycle (xdestroyed,61).

xdestroyed,61=T0(s1s6+qoutTL)

xdestroyed,61=T0(s1s6RlnP1P6+qoutTL) (XIV)

Here, the temperature of the sink is TL.

Write the expression of stream exergy at the exit of the regenerator (state 6) (ϕ6).

ϕ6=(h6h0)T0(s6s0) (XV)

Here, the specific enthalpy of the surroundings is h0.

Write the expression of change entropy for the exit of the regenerator (s6s0).

s6s0=s6s0RlnP6P1 (XVI)

Here, entropy of air at the surroundings as a function of temperature alone is s0, and the pressure of air at the surroundings is P0.

Conclusion:

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at 310 K (T1), 650K (T2) and 1400 K (T3), which gives 310.24kJ/kg for h1, 659.84kJ/kg for h2, 1,515.42kJ/kg for h3 and 450.5 for Pr3 respectively.

Substitute 900 kPa for P2, and 100 kPa for P1 in Equation (I).

rp=900kPa/100kPa=9

Substitute 19 for P4P3, and 450.5 for Pr3 in Equation (II).

Pr4=(19)(450.5)=50.06

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at 50.06 (Pr4), which gives 832.44kJ/kg for h4s.

Rearrange Equation (III) and substitute 1,515.42kJ/kg for h3, 832.44kJ/kg for h4s, and 0.90 for ηT.

h4=h3ηT(h3h4s)=1,515.42kJ/kg+(0.90)(1,515.42832.44)kJ/kg=900.74kJ/kg

Substitute 0.80 for ε, 900.74kJ/kg for h4, and 659.84kJ/kg for h2 in Equation (IV).

qregen=(0.80)(900.74659.84)kJ/kg=192.7kJ/kg

Substitute 1,515.42 kJ/kg for h3, 900.74 kJ/kg for h4, 659.84 kJ/kg for h2, and 310.24 kJ/kg for h1 in Equation (V).

wnet=(1,515.42900.74)kJ/kg(659.84310.24)kJ/kg=265.08kJ/kg

Substitute 192.7kJ/kg for qregen, 1,515.42kJ/kg for h3, and 659.84kJ/kg for h2 in Equation (VI).

qin=[(1,515.42659.84)192.7]kJ/kg=662.88kJ/kg

Substitute 662.88kJ/kg for qin, and 265.08kJ/kg for wnet in Equation (VII).

qout=(662.88265.08)kJ/kg=397.80kJ/kg

Substitute 310.24 kJ/kg for h1, and 397.80 kJ/kg for qout in Equation (VIII).

h6=(310.24+397.80)kJ/kg=708.04kJ/kg

Rearrange Equation (IX), and substitute 659.84 kJ/kg for h2, 900.74 kJ/kg for h4, and 708.04 kJ/kg for h6.

h5=(h4h6)+h2=(900.74708.04)kJ/kg+659.84kJ/kg=852.54kJ/kg

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at 310 K (T1), 659.84kJ/kg (h2), 1400 K (T3), 900.74kJ/kg (h4), 852.54kJ/kg (h5) and 708.04kJ/kg (h6) which gives 1.73498kJ/kgK for s10, 2.49364kJ/kgK for s20, 3.36200kJ/kgK for s30, 2.81216kJ/kgK for s40,2.75537kJ/kgK for s50, and 2.56525kJ/kgK for s60.

Substitute 300 K for T0, 1.73498kJ/kgK for s1, 2.49364kJ/kgK for s2, 0.287kJ/kgK for R, and 9 for P2P1 in Equation (X).

xdestroyed,12=(300K)[(2.493641.73498)kJ/kgK(0.287kJ/kgK)ln(9)]=38.42kJ/kg

Thus, the exergy destruction associated with process 1-2 of the given Brayton cycle is 38.42kJ/kg.

Substitute 300 K for T0, 3.36200kJ/kgK for s3, 2.81216kJ/kgK for s4, 0.287kJ/kgK for R, and 19 for P4P3 in Equation (XI).

xdestroyed,34=(300K)[(2.812163.36200)kJ/kgK(0.287kJ/kgK)ln(19)]=24.23kJ/kg

Thus, the exergy destruction associated with process 3-4 of the given Brayton cycle is 24.23kJ/kg.

Substitute 300 K for T0, 2.75537kJ/kgK for s5, 2.81216kJ/kgK for s4, 2.49364kJ/kgK for s2, and 2.56525kJ/kgK for s6 in Equation (XII).

xdestroyed,regen=(300K)[(2.755372.49364)kJ/kgK+(2.565252.81216)kJ/kgK]=4.47kJ/kg

Thus, the exergy destruction associated with regeneration process of the given Brayton cycle is 4.47kJ/kg.

Substitute 300 K for T0, 3.36200kJ/kgK for s3, 2.75537kJ/kgK for s5, 0.287kJ/kgK for R, 1260 K for TH, and 662.88kJ/kg for qin in Equation (XIII). For the process 5-3, pressure remains constant, hence substitute 0 for lnP3P5.

xdestroyed,53=(300K)[(3.362002.75537)kJ/kgK(0.287kJ/kgK)(0)662.88kJ/kg1260K]=24.16kJ/kg

Thus, the exergy destruction associated with process 5-3 of the given Brayton cycle is 24.16kJ/kg.

Substitute 300 K for T0, 1.73498kJ/kgK for s1, 2.56525kJ/kgK for s6, 0.287kJ/kgK for R, 300 K for TL, and 397.80kJ/kg for qout in Equation (XIV). For the process 6-1, pressure remains constant, hence substitute 0 for lnP1P6.

xdestroyed,61=(300K)[(1.734982.56525)kJ/kgK(0.287kJ/kgK)(0)+397.80kJ/kg300K]=148.7kJ/kg

Thus, the exergy destruction associated with process 6-1 of the given Brayton cycle is 148.7kJ/kg.

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at 300 K (T0) which gives 1.70203kJ/kgK for s00 and 300.19kJ/kg for h0.

Substitute 2.56525kJ/kgK for s6, and 1.70203kJ/kgK for s0 in Equation (XVI). Also, at the exit of the regenerator, pressure remains constant, hence substitute 0 for lnP6P1.

s6s0=(2.565251.70203)kJ/kgK=0.86322kJ/kgK

Substitute 708.04kJ/kg for h6, 300.19kJ/kg for h0, 300 K for T0, and 0.86322kJ/kgK for (s6s0) in Equation (XV).

ϕ6=(h6h0)T0(s6s0)=(708.04300.19)kJ/kg(300K)(0.86322kJ/kgK)=148.9kJ/kg

Thus, the exergy of the exhaust gases at the exit of the regenerator is 148.9kJ/kg.

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Chapter 9 Solutions

Thermodynamics: An Engineering Approach

Ch. 9.12 - Prob. 11PCh. 9.12 - Prob. 12PCh. 9.12 - Prob. 13PCh. 9.12 - Prob. 15PCh. 9.12 - Prob. 16PCh. 9.12 - Prob. 17PCh. 9.12 - Prob. 18PCh. 9.12 - Repeat Prob. 919 using helium as the working...Ch. 9.12 - Consider a Carnot cycle executed in a closed...Ch. 9.12 - Prob. 21PCh. 9.12 - Prob. 22PCh. 9.12 - What four processes make up the ideal Otto cycle?Ch. 9.12 - Are the processes that make up the Otto cycle...Ch. 9.12 - How do the efficiencies of the ideal Otto cycle...Ch. 9.12 - How does the thermal efficiency of an ideal Otto...Ch. 9.12 - Prob. 27PCh. 9.12 - Why are high compression ratios not used in...Ch. 9.12 - An ideal Otto cycle with a specified compression...Ch. 9.12 - Prob. 30PCh. 9.12 - Prob. 31PCh. 9.12 - Prob. 32PCh. 9.12 - An ideal Otto cycle has a compression ratio of 8....Ch. 9.12 - Prob. 35PCh. 9.12 - Prob. 36PCh. 9.12 - Prob. 37PCh. 9.12 - An ideal Otto cycle with air as the working fluid...Ch. 9.12 - Repeat Prob. 940E using argon as the working...Ch. 9.12 - Prob. 40PCh. 9.12 - Prob. 41PCh. 9.12 - Prob. 42PCh. 9.12 - Prob. 43PCh. 9.12 - Prob. 44PCh. 9.12 - Prob. 45PCh. 9.12 - Prob. 46PCh. 9.12 - Prob. 47PCh. 9.12 - Prob. 48PCh. 9.12 - Prob. 49PCh. 9.12 - Prob. 50PCh. 9.12 - Prob. 51PCh. 9.12 - Prob. 52PCh. 9.12 - Prob. 53PCh. 9.12 - Prob. 54PCh. 9.12 - Repeat Prob. 957, but replace the isentropic...Ch. 9.12 - Prob. 57PCh. 9.12 - Prob. 58PCh. 9.12 - Prob. 59PCh. 9.12 - The compression ratio of an ideal dual cycle is...Ch. 9.12 - Repeat Prob. 962 using constant specific heats at...Ch. 9.12 - Prob. 63PCh. 9.12 - An air-standard cycle, called the dual cycle, with...Ch. 9.12 - Prob. 65PCh. 9.12 - Prob. 66PCh. 9.12 - Consider the ideal Otto, Stirling, and Carnot...Ch. 9.12 - Consider the ideal Diesel, Ericsson, and Carnot...Ch. 9.12 - An ideal Ericsson engine using helium as the...Ch. 9.12 - An ideal Stirling engine using helium as the...Ch. 9.12 - Prob. 71PCh. 9.12 - Prob. 72PCh. 9.12 - Prob. 73PCh. 9.12 - Prob. 74PCh. 9.12 - Prob. 75PCh. 9.12 - For fixed maximum and minimum temperatures, what...Ch. 9.12 - What is the back work ratio? 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