A Brayton cycle with regeneration using air as the working fluid has a pressure ratio of 7. The minimum and maximum temperatures in the cycle are 310 and 1150 K. Take an isentropic efficiency of 75 percent for the compressor and 82 percent for the turbine and an effectiveness of 65 percent for the regenerator. Determine the total exergy destruction associated with the cycle, assuming a source temperature of 1500 K and a sink temperature of 290 K. Also, determine the exergy of the exhaust gases at the exit of the regenerator. Use variable specific heats for air.

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

Chapter 9.12, Problem 148P

A Brayton cycle with regeneration using air as the working fluid has a pressure ratio of 7. The minimum and maximum temperatures in the cycle are 310 and 1150 K. Take an isentropic efficiency of 75 percent for the compressor and 82 percent for the turbine and an effectiveness of 65 percent for the regenerator. Determine the total exergy destruction associated with the cycle, assuming a source temperature of 1500 K and a sink temperature of 290 K. Also, determine the exergy of the exhaust gases at the exit of the regenerator. Use variable specific heats for air.

Expert Solution & Answer

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.

Answer to Problem 148P

The exergy destruction associated with process 1-2 of the given Brayton cycle is 38.91 kJ/kg.

The exergy destruction associated with process 3-4 of the given Brayton cycle is 35.83 kJ/kg.

The exergy destruction associated with regeneration process of the given Brayton cycle is 4.37 kJ/kg.

The exergy destruction associated with process 5-3 of the given Brayton cycle is 58.09 kJ/kg.

The exergy destruction associated with process 6-1 of the given Brayton cycle is 142.6 kJ/kg.

The exergy of the exhaust gases at the exit of the regenerator is 143.2 kJ/kg.

Explanation of Solution

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

THERMODYNAMICS LLF W/ CONNECT ACCESS, Chapter 9.12, Problem 148P

Consider, the pressure is Pi , the specific volume is vi, the temperature is Ti, the entropy is si, the enthalpy is hi corresponding to ith state.

Write the pressure and relative pressure relation for the process 1-2.

Pr2=P2P1Pr1 (I)

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

Pr4=P4P3Pr3 (II)

Write the expression of efficiency of the compressor (ηC).

ηC=h2s−h1h2−h1 (III)

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

ηT=h3−h4h3−h4s (IV)

Write the expression of net work output by the gas turbine (wnet).

wnet=wT,out−wC,in

wnet=(h3−h4)−(h2−h1) (V)

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

Write the expression of effectiveness of the regenerator (ε).

ε=h5−h2h4−h2 (VI)

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

qin=h3−h5 (VII)

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

qout=qin−wnetqout=h6−h1 (VIII)

Write the expression of thermal efficiency of the given turbine (ηth).

ηth=wnetqin (IX)

Write the energy balance equation on the heat exchanger.

h5−h2=h4−h6 (X)

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

xdestroyed,12=T0(s2−s1)

xdestroyed,12=T0(s2∘−s1∘−RlnP2P1) (XI)

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(s4−s3)

xdestroyed,34=T0(s4∘−s3∘−RlnP4P3) (XII)

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[(s5−s2)+(s6−s4)]

xdestroyed,regen=T0[(s5∘−s2∘)+(s6∘−s4∘)] (XIII)

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(s3−s5+−qinTH)

xdestroyed,53=T0(s3∘−s5∘−RlnP3P5−qinTH) (XIV)

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(s1−s6+qoutTL)

xdestroyed,61=T0(s1∘−s6∘−RlnP1P6+qoutTL) (XV)

Here, the temperature of the sink is TL.

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

ϕ6=(h6−h0)−T0(s6−s0) (XVI)

Here, the specific enthalpy of the surroundings is h0.

Write the expression of change entropy for the exit of the regenerator (s6−s0).

s6−s0=s6∘−s0∘−RlnP6P1 (XVII)

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 a temperature of 310 K (T1).

h1=310.24 kJ/kgPr1=1.5546s10=1.73498 kJ/kg⋅K

Substitute 7 for P2P1, and 1.5546 for Pr1 in Equation (I).

Pr2=(7)(1.5546)=10.88

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a relative pressure of 10.88 (Pr2).

h2s=541.26 kJ/kg

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a temperature of 1150 K (T3).

h3=1219.25 kJ/kgPr3=200.15s30=3.129 kJ/kg⋅K

Substitute 17 for P4P3, and 200.15 for Pr3 in Equation (II).

Pr4=(17)(200.15)=28.59

Refer Table A-17, “Ideal gas properties of air”, obtain the property of enthalpy (h4s) at a relative pressure of 28.59 (Pr4).

h4s=711.8 kJ/kg

Rearrange Equation (III), and substitute 310.24 kJ/kg for h1, 541.26 kJ/kg for h2s, and 0.75 for ηC.

h2=h1+h2s−h1ηC=310.24 kJ/kg+(541.26−310.24) kJ/kg0.75=618.26 kJ/kg

Refer Table A-17, “Ideal gas properties of air”, obtain the property of entropy (s20) at a enthalpy of 618.26 kJ/kg (h2).

s20=2.42763 kJ/kg⋅K

Rearrange Equation (IV), and substitute 1219.25 kJ/kg for h3, 711.8 kJ/kg for h4s, and 0.82 for ηT.

h4=h3−ηT(h3−h4s)=1219.25 kJ/kg−(0.82)(1219.25−711.8)kJ/kg=803.14 kJ/kg

Refer Table A-17, “Ideal gas properties of air”, obtain the property of entropy (s40) at a enthalpy of 803.14 kJ/kg (h4).

s40=2.69407 kJ/kg⋅K

Substitute 1219.25 kJ/kg for h3, 618.26 kJ/kg for h2, 803.14 kJ/kg for h4, and 310.24 kJ/kg for h1 in Equation (V).

wnet=(1219.25−803.14)kJ/kg−(618.26−310.24)kJ/kg=108.09 kJ/kg

Substitute 0.65 for ε, 803.14 kJ/kg for h4, and 618.26 kJ/kg for h2 in Equation (VI).

0.65=(h5−618.26 kJ/kg)(803.14−618.26) kJ/kg h5=738.43 kJ/kg

Refer Table A-17, “Ideal gas properties of air”, obtain the property of entropy (s50) at a enthalpy of 738.43 kJ/kg (h5).

s50=2.60815 kJ/kg⋅K

Substitute 1219.25 kJ/kg for h3, and 738.43 kJ/kg for h5 in Equation (VII).

qin=(1219.25−738.43) kJ/kg=480.82 kJ/kg

Substitute 738.43 kJ/kg for h5, 803.14 kJ/kg for h4, and 618.26 kJ/kg for h2 in Equation (X).

738.43 kJ/kg−618.26 kJ/kg=803.14 kJ/kg−h6h6=682.97 kJ/kg

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a enthalpy of 682.97 kJ/kg (h6).

s60=2.52861 kJ/kg⋅K

Substitute 682.97 kJ/kg for h6, and 310.24 kJ/kg for h1 in Equation (VIII).

qout=682.97 kJ/kg−310.24 kJ/kg=372.73 kJ/kg

Substitute 108.09 kJ/kg for wnet, and 480.82 kJ/kg for qin in Equation (IX).

ηth=108.09 kJ/kg480.82 kJ/kg=0.225=22.5 %

Substitute 290 K for T0, 1.73498 kJ/kg⋅K for s1∘, 2.42763 kJ/kg⋅K for s2∘, 0.287 kJ/kg⋅K for R, and 7 for P2P1 in Equation (XI).

xdestroyed,12=(290 K)[(2.42763−1.73498)kJ/kg⋅K−(0.287kJ/kg⋅K)ln(7)]=38.91 kJ/kg

Thus, the exergy destruction associated with process 1-2 of the given Brayton cycle is 38.91 kJ/kg.

Substitute 290 K for T0, 3.129 kJ/kg⋅K for s3∘, 2.69407 kJ/kg⋅K for s4∘, 0.287 kJ/kg⋅K for R, and 17 for P4P3 in Equation (XII).

xdestroyed,34=(290 K)[(2.69407−3.129)kJ/kg⋅K−(0.287kJ/kg⋅K)ln(17)]=35.83 kJ/kg

Thus, the exergy destruction associated with process 3-4 of the given Brayton cycle is 35.83 kJ/kg.

Substitute 290 K for T0, 2.60815 kJ/kg⋅K for s5∘, 2.69407 kJ/kg⋅K for s4∘, 2.42763 kJ/kg⋅K for s2∘, and 2.52861 kJ/kg⋅K for s6∘ in Equation (XIII).

xdestroyed,regen=(290 K)[(2.60815−2.42763)kJ/kg⋅K+(2.52861−2.69407) kJ/kg⋅K]=4.37 kJ/kg

Thus, the exergy destruction associated with regeneration process of the given Brayton cycle is 4.37 kJ/kg.

Substitute 290 K for T0, 3.129 kJ/kg⋅K for s3∘, 2.60815 kJ/kg⋅K for s5∘, 0.287 kJ/kg⋅K for R, 1500 K for TH, and 480.82 kJ/kg for qin in Equation (XIV). For the process 5-3, pressure remains constant, hence substitute 0 for lnP3P5.

xdestroyed,53=(290 K)[(3.129−2.60815)kJ/kg⋅K−(0.287kJ/kg⋅K)(0)−480.82 kJ/kg1500 K]=58.09 kJ/kg

Thus, the exergy destruction associated with process 5-3 of the given Brayton cycle is 58.09 kJ/kg.

Substitute 290 K for T0, 1.73498kJ/kg⋅K for s1∘, 2.52861 kJ/kg⋅K for s6∘, 0.287 kJ/kg⋅K for R, 290 K for TL, and 372.73 kJ/kg for qout in Equation (XV). For the process 6-1, pressure remains constant, hence substitute 0 for lnP1P6.

xdestroyed,61=(290 K)[(1.73498−2.52861)kJ/kg⋅K−(0.287kJ/kg⋅K)(0)+372.73 kJ/kg290 K]= 142.6 kJ/kg

Thus, the exergy destruction associated with process 6-1 of the given Brayton cycle is 142.6 kJ/kg.

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a temperature of 290 K (T0).

h0=290.16 kJ/kgs00=1.66802 kJ/kg⋅K

At the exit of the regenerator, pressure remains constant, (P6=P1).

Substitute P6 for P1 , 2.52861 kJ/kg⋅K for s6∘, and 1.66802 kJ/kg⋅K for s0∘ in Equation (XVII).

s6−s0=(2.52861−1.66802) kJ/kg⋅K−RlnP6P6=(2.52861−1.66802) kJ/kg⋅K−Rln(1)=0.86059 kJ/kg⋅K

Substitute 682.97 kJ/kg for h6, 290.16 kJ/kg for h0, 290 K for T0, and 0.86059 kJ/kg⋅K for (s6−s0) in Equation (XVI).

ϕ6=(682.97−290.16) kJ/kg−(290 K)(0.86059 kJ/kg⋅K )=143.2 kJ/kg

Thus, the exergy of the exhaust gases at the exit of the regenerator is 143.2 kJ/kg.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

An undamped single-degree-of-freedom system is subjected to dynamic excitation as shown in Figure 1.• System properties: m = 1, c = 0, k = (6π)2.• Force excitation: p(t) = posin(ωt) where po = 9 and ω = 2π.• Initial conditions: u(t = 0) = 0 and ̇u(t = 0) = 0.Please, complete Parts (a) through (d) using any computational tool of your preference. The preferred toolis MATLAB. Print and turn in a single pdf file that will include your code/calculations and your plots.(a) Generate the solution using a linear interpolation of the load over each time step (note that hereyou can use the undamped coefficients). Plot the displacement response for the first 4 seconds andcompare to the exact closed form solution. Repeat using the following time step sizes, ∆t = 0.01,0.05, 0.15, 0.20 seconds. Include the closed form solution and the solutions for different ∆t values in asingle plot. Please, provide your observations by comparing the closed form solution with the solutionsderived using the four…

Assume multiple single degree of freedom systems with natural periods T ∈ [0.05, 2.00] seconds with in-crement of period dT = 0.05 seconds. Assume three cases of damping ratio: Case (A) ξ = 0%; Case (B)ξ = 2%; Case (C) ξ = 5%. The systems are initially at rest. Thus, the initial conditions are u(t = 0) = 0 anḋu(t = 0) = 0. The systems are subjected to the base acceleration that was provided in the ElCentro.txt file(i.e., first column). For the systems in Case (A), Case (B), and Case (C) and for each natural period computethe peak acceleration, peak velocity, and peak displacement responses to the given base excitation. Please,use the Newmark method for β = 1/4 (average acceleration) to compute the responses. Create threeplots with three lines in each plot. The first plot will have the peak accelerations in y-axis and the naturalperiod of the system in x-axis. The second plot will have the peak velocities in y-axis and the natural periodof the system in x-axis. The third plot will have…

Both portions of the rod ABC are made of an aluminum for which E = 70 GPa. Based on the given information find: 1- deformation at A 2- stress in BC 3- Total strain 4- If v (Poisson ratio is 0.25, find the lateral deformation of AB Last 3 student ID+ 300 mm=L2 724 A P=Last 2 student ID+ 300 KN 24 24 Diameter Last 2 student ID+ 15 mm Last 3 student ID+ 500 mm=L1 724 C B 24 Q=Last 2 student ID+ 100 KN 24 Diameter Last 2 student ID+ 40 mm

Answer 2