STEM Problem Detail

Problem Statement

Prompt

Provide the step-by-step solution to example 11.6 from Gas Turbines.

Solution Rendering...

$$\textbf{Example 11.6 : Solution}

\textbf{Given:}
$$

\[
P_1 = 100 \text{ kPa}, \quad T_1 = 300 \text{ K}
\]
\[
P_2 = 600 \text{ kPa}
\]
\[
T_4 = 1000 \text{ K}, \quad T_6 = 1000 \text{ K}
\]
\[
\eta_c = 0.8, \quad \eta_{T,HP} = \eta_{T,LP} = 0.85
\]
\[
\varepsilon = 0.72
\]
\[
CV = 18500 \text{ kJ/kg}
\]
\[
c_{pa} = 1.0 \text{ kJ/kgK}, \quad \gamma_a = 1.4
\]
\[
c_{pg} = 1.15 \text{ kJ/kgK}, \quad \gamma_g = 1.34
\]
\[
P = 6 \text{ MW} = 6000 \text{ kW}
\]

**Temperature after isentropic compression:**

$$

T_{2s} = T_1 \left(\frac{P_2}{P_1}\right)^{\frac{\gamma_a - 1}{\gamma_a}}
= 300 \left(\frac{600}{100}\right)^{\frac{0.4}{1.4}}
= 500.55 \text{ K}

$$

**Actual compressor outlet temperature:**
$$

\eta_c = \frac{T_{2s} - T_1}{T_2 - T_1}

T_2 = 300 + \frac{500.55 - 300}{0.8}
= 550.68 \text{ K}

$$
$$
\textbf{Compressor work:}
$$

\[
W_c = c_{pa}(T_2 - T_1)
= 1.0(550.68 - 300)
= 250.68 \text{ kJ/kg}
\]

$$
\textbf{Intermediate pressure:}
$$

\[
P_3 = \sqrt{P_1 P_2} = \sqrt{100 \times 600} = 245 \text{ kPa}
\]
**Isentropic HP turbine outlet temperature:**
\[
T_{5s} = T_4 \left(\frac{P_3}{P_2}\right)^{\frac{\gamma_g - 1}{\gamma_g}}
= 1000 \left(\frac{245}{600}\right)^{\frac{0.34}{1.34}}
= 796.7 \text{ K}
\]

$$
\textbf{Actual HP turbine outlet temperature:}
$$

\[
\eta_{T,HP} = \frac{T_4 - T_5}{T_4 - T_{5s}}
\]

\[
T_5 = 1000 - 0.85(1000 - 796.7)
= 827.2 \text{ K}
\]

$$
\textbf{Regenerator effectiveness:}
$$

\[
\varepsilon = \frac{T_7 - T_2}{T_5 - T_2}
\]

\[
0.72 = \frac{T_7 - 550.68}{827.2 - 550.68}
\]

\[
T_7 = 749.7 \text{ K}
\]

$$
\textbf{(a) Air–fuel ratio}
$$

Let $$m_{f1}$$ = fuel per kg of air.

Energy balance in main combustor:

\[
m_{f1} \cdot CV = (1 + m_{f1}) c_{pg} (T_4 - T_7)
\]

\[
18500 m_{f1} = (1 + m_{f1}) 1.15 (1000 - 749.7)
\]

\[
18500 m_{f1} = 250.3 + 250.3 m_{f1}
\]

\[
m_{f1} = 0.0137 \text{ kg fuel/kg air}
\]

\[
A/F = \frac{1}{m_{f1}} = 72.9
\]

For reheater, let additional fuel be $$m_{f2}$$.

\[
m_{f2} \cdot CV = (1 + m_{f1} + m_{f2}) c_{pg} (T_6 - T_5)
\]

\[
18500 m_{f2} = (1 + 0.0137 + m_{f2}) 1.15 (1000 - 827.2)
\]

\[
m_{f2} = 0.011 \text{ kg fuel/kg air}
\]

$$
\textbf{Turbine work:}
$$

\[
W_t = (1 + m_{f1}) c_{pg}(T_4 - T_5)
+ (1 + m_{f1} + m_{f2}) c_{pg}(T_6 - T_5)
\]

\[
W_t = (1.0137)(1.15)(1000 - 827.2)
+ (1.0247)(1.15)(1000 - 827.2)
\]

\[
W_t = 201.14 + 203.62 = 404.76 \text{ kJ/kg}
\]

$$
\textbf{Net work:}
$$

\[
W_{net} = W_t - W_c
= 404.76 - 250.68
= 154.08 \text{ kJ/kg}
\]

$$
\textbf{Heat supplied:}
$$

\[
Q_{in} = (m_{f1} + m_{f2}) CV
= (0.0137 + 0.011) \times 18500
= 456.95 \text{ kJ/kg}
\]

$$
\textbf{Thermal efficiency:}
$$

\[
\eta_{th} = \frac{W_{net}}{Q_{in}}
= \frac{154.08}{456.95}
= 0.337 \approx 33\%
\]
$$\textbf{(c) Air supply}

$$
\[
P = \dot{m}_a W_{net}
\]
\[
6000 = \dot{m}_a \times 154.08
\]
\[
\dot{m}_a = 38.94 \text{ kg/s}
\]
$$\textbf{(d) Fuel consumption per hour}
$$

\[
\dot{m}_f = \dot{m}_a (m_{f1} + m_{f2}) \times 3600
\]

\[
\dot{m}_f = 38.94 (0.0137 + 0.011) \times 3600
\]

\[
\dot{m}_f = 3462.62 \text{ kg/h}
\]

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Problem #34