Question

A large gas-turbine power plant delivers a net power output of 325 MW to an electric generator. The minimum temperature in the cycle is 300 K, and the maximum temperature is 1500 K. The minimum pressure in the cycle is 100 kPa, and the compressor pressure ratio is 12. Calculate the power output of the turbine, the back work ratio, and the thermal efficiency of the cycle. Also, determine the improved thermal efficiency if a regenerator, with an effectiveness of 85 percent, were installed in the power plant.

Answer 1

A) Power output = 541.67 MW

B) Backwork ratio = 0.4

C) Improved thermal efficiency = 0.598

Step-by-step explanation:

We are given;

Net power output; W'_net = 325 MW

Minimum Temperature; T1 = 300 K

Maximum Temperature; T4 = 1500 K

Compression Pressure ratio;P2/P1 = 12

From online tables, we have the following properties of air;

Specific heat capacity; C_p = 1.004 KJ/Kg

Adiabatic constant; k = 1.4

Temperature at stage 2 will be given by the formula;

T2 = T1(P2/P1)^((k - 1)/k)

Plugging in the relevant values gives;

T2 = 300(12)^((1.4 - 1)/1.4)

T2 = 610.18 K

Similarly, Temperature at stage 3 will be;

T3 = T4(P2/P1)^((k - 1)/k)

Plugging in the relevant values gives;

T3 = 750(12)^((1.4 - 1)/1.4)

T3 = 1525.45 K

Now, let's calculate the specific heat addition given by the formula;

q = C_p(T3 - T2)

q = 1.004(1525.45 - 610.18)

q = 918.93 KJ/Kg

Let's now calculate the specific net work output;

w_net = C_p[(T3 - T4) - (T2 - T1)]

w_net = 1.004[(1525.45 - 750) - (610.18 - 300)]

w_net = 467.13 KJ/Kg

A) Power output is given by;

W_T = (W'_net/W_net) × C_p(T3 - T4)

W_T = (325/467.13) × 1.004(1525.45 - 750)

W_T = 541.67 MW

B) back work ratio is;

f = (T2 - T1)/(T3 - T4)

f = (610.18 - 300)/(1525.45 - 750)

f = 0.4

C) Thermal efficiency is given by;

η = w_net/q

Since we are told that a regenerator, with an effectiveness of 85 percent, were installed in the power plant.

Thus;

η = (w_net/q)÷0.85

η = (467.13/918.94) ÷ 0.85

η = 0.598