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 C_P = 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

      w_net,out = q_in - q_out

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

      η_th,Brayton =1 - q_out /q_in

(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.

      q_in = h₃ - h₂ = c_P(T₃ - T₂)

      q_out = h₄ - h₁ = c_P(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:

      r_P = 8 (given)
      P₂ = r_PP₁ = 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.

      q_in = c_P(T₃ - T₂)
           = 1.005(1300 - 543.4) = 760.3 kJ/kg

      q_out = c_P(T₄ - T₁)
            = 1.005(717.7 - 300) = 419.7 kJ/kg

Net work output and thermal efficiency of this cycle can be determined with q_in and q_out.

      w_net,out = q_in - q_out = 760.3 - 419.7 = 340.6 kJ/kg

      η_th,Brayton = w_net,out /q_in = 340.6/760.3 = 44.7%

(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:

      q_in = h₃ - h₅ = c_P(T₃ - T₅)

      q_out = h₆ - h₁ = c_P(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
      c_P(T₅ - T₂) = c_P(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 q_inand q_out gives

      q_in = c_P(T₃ - T₅)
           =1.005(1,300 - 682.8) = 620.3 kJ/kg

      q_out = c_P(T₆- T₁)
            = 1.005(578.3 - 300) = 279.6 kJ/kg

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

      w_net,out = q_in - q_out = 620.3 - 279.6 = 340.6 kJ/kg

      η_th,Brayton = w_net,out /q_in = 340.6/620.3 = 54.9%

(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.

      q_in = (h₆ - h₅) + (h₈ - h₇)
           = c_P(T₆- T₅) + c_P(T₈- T₇)

      q_out = (h₁₀ - h₁) + (h₂ - h₃)
             = c_P(T₁₀- T₁) + c_P(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 r_P = 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
      c_P(T₅ - T₄) = c_P(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 q_in
and q_out gives

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

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

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

      w_net,out = q_in - q_out = 784.5 - 321.6 = 463.0 kJ/kg

      η_th,Brayton = w_net,out /q_in = 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.