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

Question

Want to see more full solutions like this?

Answer 1

Question

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

Answer to Problem 146P

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

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

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

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

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

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

Explanation of Solution

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

EBK THERMODYNAMICS: AN ENGINEERING APPR, 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=h3−h4h3−h4s (III)

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

qregen=ε(h4−h2) (IV)

Here, the effectiveness of the regenerator is ε.

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

wnet=wT,out−wC,in

wnet=(h3−h4)−(h2−h1) (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=(h3−h2)−qregen (VI)

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

qout=qin−wnet (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.

h5−h2=h4−h6 (IX)

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) (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(s4−s3)

xdestroyed,34=T0(s4∘−s3∘−RlnP4P3) (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[(s5−s2)+(s6−s4)]

xdestroyed,regen=T0[(s5∘−s2∘)+(s6∘−s4∘)] (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(s3−s5+−qinTH)

xdestroyed,53=T0(s3∘−s5∘−RlnP3P5−qinTH) (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(s1−s6+qoutTL)

xdestroyed,61=T0(s1∘−s6∘−RlnP1P6+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=(h6−h0)−T0(s6−s0) (XV)

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 (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 310.24 kJ/kg for h1, 659.84 kJ/kg for h2, 1,515.42 kJ/kg for h3 and 450.5 for Pr3 respectively.

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

rp=900 kPa/100 kPa=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 832.44 kJ/kg for h4s.

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

h4=h3−ηT(h3−h4s)=1,515.42 kJ/kg+(0.90)(1,515.42−832.44) kJ/kg=900.74 kJ/kg

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

qregen=(0.80)(900.74−659.84) kJ/kg=192.7 kJ/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=(1,515.42−900.74) kJ/kg−(659.84−310.24) kJ/kg=265.08 kJ/kg

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

qin=[(1,515.42−659.84)−192.7] kJ/kg=662.88 kJ/kg

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

qout=(662.88−265.08) kJ/kg=397.80 kJ/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=708.04 kJ/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=(h4−h6)+h2=(900.74−708.04) kJ/kg+659.84 kJ/kg=852.54 kJ/kg

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

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

xdestroyed,12=(300 K)[(2.49364−1.73498)kJ/kg⋅K−(0.287kJ/kg⋅K)ln(9)]=38.42 kJ/kg

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

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

xdestroyed,34=(300 K)[(2.81216−3.36200)kJ/kg⋅K−(0.287kJ/kg⋅K)ln(19)]=24.23 kJ/kg

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

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

xdestroyed,regen=(300 K)[(2.75537−2.49364)kJ/kg⋅K+(2.56525−2.81216) kJ/kg⋅K]=4.47 kJ/kg

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

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

xdestroyed,53=(300 K)[(3.36200−2.75537)kJ/kg⋅K−(0.287kJ/kg⋅K)(0)−662.88 kJ/kg1260 K]=24.16 kJ/kg

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

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

xdestroyed,61=(300 K)[(1.73498−2.56525)kJ/kg⋅K−(0.287kJ/kg⋅K)(0)+397.80 kJ/kg300 K]= 148.7 kJ/kg

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

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

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

s6−s0=(2.56525−1.70203) kJ/kg⋅K=0.86322 kJ/kg⋅K

Substitute 708.04 kJ/kg for h6, 300.19 kJ/kg for h0, 300 K for T0, and 0.86322 kJ/kg⋅K for (s6−s0) in Equation (XV).

ϕ6=(h6−h0)−T0(s6−s0)=(708.04−300.19) kJ/kg−(300 K)(0.86322 kJ/kg⋅K )=148.9 kJ/kg

Thus, the exergy of the exhaust gases at the exit of the regenerator is 148.9 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

Q4/ A compressor is driven motor by mean of a flat belt of thickness 10 mm and a width of 250 mm. The motor pulley is 300 mm diameter and run at 900 rpm and the compressor pulley is 1500 mm diameter. The shaft center distance is 1.5 m. The angle of contact of the smaller pulley is 220° and on the larger pulley is 270°. The coefficient of friction between the belt and the small pulley is 0.3, and between the belt and the large pulley is 0.25. The maximum allowable belt stress is 2 MPa and the belt density is 970 kg/m³. (a) What is the power capacity of the drive and (b) If the small pulley replaced by V-grooved pulley of diameter 300 mm, grooved angle of 34° and the coefficient of friction between belt and grooved pulley is 0.35. What will be the power capacity in this case, assuming that the diameter of the large pulley remain the same of 1500 mm.

You are tasked with designing a power drive system to transmit power between a motor and a conveyor belt in a manufacturing facility as illustrated in figure. The design must ensure efficient power transmission, reliability, and safety. Given the following specifications and constraints, design drive system for this application: Specifications: Motor Power: The electric motor provides 10 kW of power at 1,500 RPM. Output Speed: The output shaft should rotate at 150 rpm. Design Decisions: Transmission ratio: Determine the necessary drive ratio for the system. Shaft Diameter: Design the shafts for both the motor and the conveyor end. Material Selection: Choose appropriate materials for the gears, shafts. Bearings: Select suitable rolling element bearings. Constraints: Space Limitation: The available space for the gear drive system is limited to a 1-meter-long section. Attribute 4 of CEP Depth of knowledge required Fundamentals-based, first principles analytical approach…

- | العنوان In non-continuous dieless drawing process for copper tube as shown in Fig. (1), take the following data: Do-20mm, to=3mm, D=12mm, ti/to=0.6 and v.-15mm/s. Calculate: (1) area reduction RA, (2) drawing velocity v. Knowing that: ti: final thickness V. Fig. (1) ofthre

Answer 2