STEM Problem Detail

Problem Statement

Prompt

Generate the solution of Question 30.20

Solution Rendering...

**Solution:**

**Given:**

\[
T_1 = 20 + 273 = 293 \ \text{K}; \quad P_1 = 1.1 \ \text{bar}
\]

\[
P_2 = 3.3 \ \text{bar}; \quad T_3 = 27 + 273 = 300 \ \text{K}
\]

\[
P_3 = 3.3 - 0.15 = 3.15 \ \text{bar}; \quad P_4 = P_6 = 9.45 \ \text{bar}
\]

\[
T_6 = 973 \ \text{K}
\]

\[
T_8 = 670 + 273 = 943 \ \text{K}; \quad P_8 = 3.5 \ \text{bar}
\]

\[
\eta_{\text{compressor}} = 82\%
\]

\[
\eta_{\text{turbine}} = 85\%; \quad \text{Power generated} = 6000 \ \text{kW}
\]

Effectiveness,

\[
\varepsilon = 0.65; \quad c_{pa} = 1.005 \ \text{kJ/kg K}
\]

\[
\gamma_{air} = 1.44; \quad c_{pg} = 1.15 \ \text{kJ/kg K}; \quad \gamma_{gases} = 1.33
\]

Refer to Fig. 30.11.

---

\[
\frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma - 1}{\gamma}}
\]

\[
= \left( \frac{3.3}{1.1} \right)^{\frac{1.4 - 1}{1.4}} = 1.369
\]

\[
T_2 = 293 \times 1.369 = 401 \ \text{K}
\]

\[
\eta_{\text{compressor}} = 0.82 = \frac{T_2 - T_1}{T_2' - T_1}
\]

\[
0.82 = \frac{401 - 293}{T_2' - 293}
\]

\[
T_2' = \frac{401 - 293}{0.82} + 293 = 425 \ \text{K}
\]

Again,

\[
\frac{T_4}{T_3} = \left( \frac{P_4}{P_3} \right)^{\frac{\gamma - 1}{\gamma}}
\]

\[
= \left( \frac{9.45}{3.15} \right)^{\frac{1.4 - 1}{1.4}} = 1.369
\]

\[
T_4 = 300 \times 1.369 = 411 \ \text{K}
\]

Now,

\[
\eta_{\text{compressor (H.P.)}} = \frac{T_4 - T_3}{T_4' - T_3}
\]

\[
0.82 = \frac{411 - 300}{T_4' - 300}
\]

\[
T_4' = \frac{411 - 300}{0.82} + 300 = 435 \ \text{K}
\]

Similarly,

\[
\frac{T_6}{T_7} = \left( \frac{P_6}{P_7} \right)^{\frac{\gamma}{\gamma - 1}}
\]

\[
= \left( \frac{9.45}{1.33} \right)^{\frac{1.33}{1.33}} = 1.268
\]

\[
T_7 = \frac{973}{1.268} = 767 \ \text{K}
\]

Also,

\[
\eta_{\text{turbine (H.P.)}} = \frac{T_6 - T_7'}{T_6 - T_7}
\]

\[
0.85 = \frac{973 - T_7'}{973 - 767}
\]

\[
T_7' = 973 - 0.85(973 - 767) = 798 \ \text{K}
\]

Again,

\[
\frac{T_8}{T_9} = \left( \frac{P_8}{P_9} \right)^{\frac{\gamma}{\gamma - 1}}
\]

\[
= \left( \frac{3.5}{1.1} \right)^{\frac{1.33}{1.33}} = 1.332
\]

\[
T_9 = \frac{943}{1.332} = 708 \ \text{K}
\]

\[
\eta_{\text{turbine (L.P.)}} = \frac{T_8 - T_9'}{T_8 - T_9}
\]

\[
0.85 = \frac{943 - T_9'}{943 - 708}
\]

\[
T_9' = 943 - 0.85(943 - 708) = 743 \ \text{K}
\]

Effectiveness of heat exchanger,

\[
\varepsilon = 0.65 = \frac{T_5 - T_4'}{T_9' - T_4'}
\]

\[
0.65 = \frac{T_5 - 435}{743 - 435}
\]

\[
T_5 = 0.65(743 - 435) + 435 = 635 \ \text{K}
\]

---

\[
W_{\text{turbine (H.P.)}} = c_{pg}(T_6 - T_7')
\]

\[
= 1.15(973 - 798) = 201.25 \ \text{kJ/kg of gas}
\]

\[
W_{\text{turbine (L.P.)}} = c_{pg}(T_8 - T_9')
\]

\[
= 1.15(943 - 743) = 230 \ \text{kJ/kg of gas}
\]

\[
W_{\text{compressor (L.P.)}} = c_{pa}(T_2' - T_1)
\]

\[
= 1.005(425 - 293) = 132.66 \ \text{kJ/kg of air}
\]

\[
W_{\text{compressor (H.P.)}} = c_{pa}(T_4' - T_3)
\]

\[
= 1.005(435 - 300) = 135.67 \ \text{kJ/kg of air}
\]

Heat supplied,

\[
= c_{pg}(T_6 - T_5) + c_{pg}(T_8 - T_7')
\]

\[
= 1.15(973 - 635) + 1.15(943 - 798)
\]

\[
= 555.45 \ \text{kJ/kg of gas}
\]

---

**(i) Overall efficiency,** $$\eta_{\text{overall}}$$

\[
\eta_{\text{overall}} = \frac{\text{Network done}}{\text{Heat supplied}}
\]

\[
= \frac{(W_{\text{turbine (H.P.)}} + W_{\text{turbine (L.P.)}})
- (W_{\text{comp (L.P.)}} + W_{\text{comp (H.P.)}})}{\text{Heat supplied}}
\]

\[
= \frac{(201.25 + 230) - (132.66 + 135.67)}{555.45}
\]

\[
= \frac{162.92}{555.45}
\]

\[
= 0.293 \ \text{or} \ 29.3\% \quad \text{(Ans.)}
\]

---

**(ii) Work ratio**

\[
\text{Work ratio} = \frac{\text{Network done}}{\text{Turbine work}}
\]

\[
= \frac{(W_{\text{turbine (H.P.)}} + W_{\text{turbine (L.P.)}})
- (W_{\text{comp (L.P.)}} + W_{\text{comp (H.P.)}})}
{(W_{\text{turbine (H.P.)}} + W_{\text{turbine (L.P.)}})}
\]

\[
= \frac{162.92}{(201.25 + 230)}
\]

\[
= \frac{162.92}{431.25} = 0.377
\]

\[
\text{Work ratio} = 0.377 \quad \text{(Ans.)}
\]

---

**(iii) Mass flow rate,** $$\dot{m}$$

\[
\text{Network done} = 162.92 \ \text{kJ/kg}
\]

Since mass of fuel is neglected, for 6000 kW, mass flow rate,

\[
\dot{m} = \frac{6000}{162.92} = 36.83 \ \text{kg/s}
\]

\[
\text{i.e., Mass flow rate} = 36.83 \ \text{kg/s} \quad \text{(Ans.)}
\]

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Problem #33