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THERMODYNAMICS - CASE STUDY SOLUTION


 

Several suggestions are given to modify a power plant. The power output and thermal efficiency of each plan need to be determined.

Assumptions:

  • Cold-Air-standard assumption is valid for this analysis. The constant pressure specific heat of air CP = 1.005 kJ/(kg-K), the specific heat ratio k = 1.4.
  • Model the cycles as a simple ideal Brayton cycle, an ideal Brayton cycle with regeneration, and an ideal Brayton cycle with intercooling, reheating and regeneration respectively.
     

 

The network output from a cycle can be determined by the energy balance of the cycle. It is

      wnet,out = qin - qout

And the thermal efficiency of a cycle can be determined by its general definition.

      ηth,Brayton =1 - qout /qin

     

T-s Diagram of the Simple Ideal Brayton Cycle
 

(1) Determine the net work output and the thermal efficiency of the simple ideal Brayton cycle

The T-s diagram of the simple ideal Brayton cycle is shown on the left.

Under the cold-air-standard assumption, heat input equals the enthalpy difference in the constant-pressure heat addition process. Likewise, heat output equals the enthalpy difference in the constant-pressure heat rejection process.

      qin = h3 - h2 = cP(T3 - T2)

      qout = h4 - h1 = cP(T4 - T1)

Therefore, temperature at each state in the cycle needs to be determined first.

      T1 = 300 K (given)
      P1 = 100 kPa (given)

1-2: Isentropic compression:

      rP = 8 (given)
      P2 = rPP1 = 8(100) = 800 kPa
      T2 = T1 (P2/P1)(k-1)/k = 543.4 K

2-3: Constant pressure heat addition:

      P3 = P2 = 800 kPa
      T3 = 1300 K (given)

4-1: Constant pressure heat rejection:

      P4 = P1 = 100 kPa

3-4: Isentropic expansion:

      T4 = T3 (P4/P3)(k-1)/k = 717.7 K

After determining all the temperatures, the heat input and heat rejection can be calculated.

      qin = cP(T3 - T2)
           = 1.005(1300 - 543.4) = 760.3 kJ/kg

      qout = cP(T4 - T1)
            = 1.005(717.7 - 300) = 419.7 kJ/kg

Net work output and thermal efficiency of this cycle can be determined with qin and qout.

      wnet,out = qin - qout = 760.3 - 419.7 = 340.6 kJ/kg

      ηth,Brayton = wnet,out /qin = 340.6/760.3 = 44.7%

     

T-s Diagram of the Ideal Brayton Cycle with Regeneration
 

(2) Determine the net work output and the thermal efficiency of the ideal Brayton cycle with regeneration

The T-s diagram of the ideal Brayton cycle with regeneration is shown on the left.

 Heat input and heat output in this cycle are:

      qin = h3 - h5 = cP(T3 - T5)

      qout = h6 - h1 = cP(T6- T1)

Temperatures of state 1, 2, 3, and 4 can be determined by the same way as the simple ideal Brayton cycle and they are the same as in (1).

2-5: constant pressure regeneration:

      P5 = P2 = 800 kPa
      ε = 0.8 (given)
      

4-6: constant pressure regeneration:

      P6 = P4 = 100 kPa

The energy balance in the regenerator is:

      h5 - h2 = h4 - h6
      cP(T5 - T2) = cP(T4 - T6)

Under cold-air-standard assumption, the temperature at state 6 is

      T6 = T4 - (T5 - T2) = 578.3 K

Substituting the temperatures into the expression of qinand qout gives

      qin = cP(T3 - T5)
           =1.005(1,300 - 682.8) = 620.3 kJ/kg

      qout = cP(T6- T1)
            = 1.005(578.3 - 300) = 279.6 kJ/kg

Net work output and thermal efficiency of this cycle can be determined as

      wnet,out = qin - qout = 620.3 - 279.6 = 340.6 kJ/kg

      ηth,Brayton = wnet,out /qin = 340.6/620.3 = 54.9%

     

T-s Diagram of the Ideal Brayton Cycle with Intercooling, Reheating and Regeneration
 

(3) Determine the net work output and the thermal efficiency of the ideal Brayton cycle with intercooling, reheating and regeneration

The T-s diagram of the ideal Brayton cycle with intercooling, reheating, and regeneration is shown on the left.

Under the cold-air-standard assumption, heat input is the sum of the enthalpy differences in the constant-pressure heat addition process and the reheating process. Heat output equals the sum of the enthalpy differences in the constant-pressure heat rejection process and the intercooling process.

      qin = (h6 - h5) + (h8 - h7)
           = cP(T6- T5) + cP(T8- T7)

      qout = (h10 - h1) + (h2 - h3)
             = cP(T10- T1) + cP(T2- T3)

The temperature at each state can be determined by the same way as in (1) and (2).

      T1 = 300 K (given)
      P1 = 100 kPa (given)

For optimum operation, equal pressure ratios are maintained across each state.

      

The total pressure ratio is given as rP = 8

      

      P2 = 2.828P1= 282.8 kPa

1-2: isentropic compression:

      T2 = T1(P2/P1)(k-1)/k = 403.8 K

2-3: constant pressure intercooling. For optimum operation, air is cooled to a temperature same as the temperature at the compressor inlet. Hence,

      T3 = T1 = 300 K
      P3 = P2 = 282.8 kPa

3-4: isentropic compression:

      P4 = 2.828P3 = 800 kPa
      T4 = T3(P4/P3)(k-1)/k = 403.8 K

Temperature at state 6 reaches the maximum value. The temperature is given as

      T6 = 1,300 K

4-5-6: constant pressure process:

      P6 = P4 = 800 kPa

6-7: isentropic expansion:

      P7 = P6 /2.828 = 282.8 kPa
      T7 = T6(P7/P6)(k-1)/k = 965.9 K

7-8: constant pressure reheating process:

      P8 = P7 = 282.8 kPa

For optimum operation, air is heated to the same temperature at each inlet of the turbine. Hence,

      T8 = T6 = 1,300 K

9-10-1: constant pressure process:

      P9 = P1 = 100 kPa

8-9: isentropic expansion:

      T9 = T8(P9/P8)(k-1)/k = 965.9 K

4-5 and 9-10: constant pressure regeneration:

The regeneration has an effectiveness ε = 0.8


      P5 = P4 = 800 kPa
      P10 = P9 = 100 kPa
      

The energy balance in the regenerator is:

      h5 - h4 = h9 - h10
      cP(T5 - T4) = cP(T9 - T10)

Under cold-air-standard assumption, the temperature at state 10 is

      T10 = T9 - (T5 - T4) = 516.2 K

Substitute the temperatures into the expression of qin
and qout gives

      qin = cP(T6- T5) + cP(T8- T7)
           = 1.005( 1300 - 853.5) + 1.005(1300 - 965.9)
           = 784.5 kJ/kg

      qout = cP(T10- T1) + cP(T2- T3)
             = 1.005(516.2- 403.8) + 1.005(403.8- 300)
             = 321.6 kJ/kg

After determining qin and qout, the net work output and the thermal efficiency of this cycle can be calculated.

      wnet,out = qin - qout = 784.5 - 321.6 = 463.0 kJ/kg

      ηth,Brayton = wnet,out /qin = 463.0/784.5 = 59.0%

     
   

(4) Results analysis

The heat input, heat output, net work output, and thermal efficiency of all three cycles are summarized in the following table.

     
  qin
(kJ/kg)
qout
(kJ/kg)
wnet,out
(kJ/kg)
ηth,Brayton
(%)
Simple Brayton Cycle 760.3 419.7 340.6 44.8
Brayton Cycle with Regeneration 620.3 279.7 340.6 54.9
Two-stage Brayton Cycle
with Regeneration
784.5 321.6 463.0 59.0
     
    Among the three cycles discussed above, the simple Brayton cycle has the lowest thermal efficiency and the smaller work output. Compared with the simple Brayton cycle, Brayton cycle with regeneration can produce the same work with less heat input. Brayton cycle with intercooling, reheating, and regeneration has the highest thermal efficiency and larger work output. Also, it consumes more fuel due to reheating compared with the other two cycles.