ii. Constant pressure ycle In a gas turbine unit, air is drawn at.I bar and $$ 20^{\circ} \mathrm{C} $$ and compressed to 6 bar. maximum cycle temperature is $$ 850^{\circ} \mathrm{C} $$. Determine each of the following for an constant pressure cycle (2marks) (1. Thermal efficiency ii. Work ratio (10marks) (iii. Entropy change per kg iv. Heat rejected per kg

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To analyze the constant pressure (Brayton) cycle for the given gas turbine unit, we'll determine the following parameters:

1. **Thermal Efficiency (η)**
2. **Work Ratio**
3. **Entropy Change per kg**
4. **Heat Rejected per kg**

### **Given Data:**
- **Inlet Pressure, $$ P_1 = 1 $$ bar**
- **Inlet Temperature, $$ T_1 = 20^\circ \mathrm{C} = 293 $$ K**
- **Compression Pressure, $$ P_2 = 6 $$ bar**
- **Maximum Cycle Temperature, $$ T_3 = 850^\circ \mathrm{C} = 1123 $$ K**
- **Assumptions:**
  - Air behaves as an ideal diatomic gas.
  - Specific Heat at Constant Pressure, $$ C_p = 1.005 $$ kJ/kg·K
  - Specific Heat at Constant Volume, $$ C_v = 0.718 $$ kJ/kg·K
  - Adiabatic Index, $$ \gamma = \frac{C_p}{C_v} = 1.4 $$

### **1. Thermal Efficiency (η):**
The thermal efficiency of the Brayton cycle is given by:
$$
\eta = \frac{W_{\text{net}}}{Q_{\text{in}}} = 1 - \frac{T_1}{T_2} \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}}
$$

**Step-by-Step Calculation:**

**a. Isentropic Compression (1 → 2):**
$$
T_2 = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} = 293 \times 6^{\frac{0.4}{1.4}} \approx 293 \times 1.667 \approx 488 \, \text{K}
$$

**b. Isentropic Expansion (3 → 4):**
$$
T_4 = T_3 \left( \frac{P_4}{P_3} \right)^{\frac{\gamma-1}{\gamma}} = 1123 \times \left( \frac{1}{6} \right)^{\frac{0.4}{1.4}} \approx 1123 \div 1.667 \approx 674 \, \text{K}
$$

**c. Heat Added (2 → 3):**
$$
Q_{\text{in}} = C_p (T_3 - T_2) = 1.005 \times (1123 - 488) \approx 639 \, \text{kJ/kg}
$$

**d. Heat Rejected (4 → 1):**
$$
Q_{\text{out}} = C_p (T_4 - T_1) = 1.005 \times (674 - 293) \approx 383 \, \text{kJ/kg}
$$

**e. Net Work Done:**
$$
W_{\text{net}} = Q_{\text{in}} - Q_{\text{out}} = 639 - 383 \approx 256 \, \text{kJ/kg}
$$

**f. Thermal Efficiency:**
$$
\eta = \frac{256}{639} \approx 0.40 \text{ or } 40\%
$$

### **2. Work Ratio:**
The work ratio is defined as the ratio of the compressor work to the turbine work.

$$
\text{Work Ratio} = \frac{W_{\text{compressor}}}{W_{\text{turbine}}}
$$

**a. Compressor Work (1 → 2):**
$$
W_{\text{compressor}} = C_p (T_2 - T_1) = 1.005 \times (488 - 293) \approx 196 \, \text{kJ/kg}
$$

**b. Turbine Work (3 → 4):**
$$
W_{\text{turbine}} = C_p (T_3 - T_4) = 1.005 \times (1123 - 674) \approx 451 \, \text{kJ/kg}
$$

**c. Work Ratio:**
$$
\text{Work Ratio} = \frac{196}{451} \approx 0.43
$$

### **3. Entropy Change per kg:**
For a complete cycle, the net entropy change is zero. However, we can calculate the entropy changes for individual processes.

**a. Process 2 → 3 (Heat Addition at Constant Pressure):**
$$
\Delta S_{2-3} = C_p \ln\left(\frac{T_3}{T_2}\right) = 1.005 \times \ln\left(\frac{1123}{488}\right) \approx 1.005 \times 0.832 \approx 0.837 \, \text{kJ/kg·K}
$$

**b. Process 4 → 1 (Heat Rejection at Constant Pressure):**
$$
\Delta S_{4-1} = C_p \ln\left(\frac{T_1}{T_4}\right) = 1.005 \times \ln\left(\frac{293}{674}\right) \approx 1.005 \times (-0.832) \approx -0.837 \, \text{kJ/kg·K}
$$

**Net Entropy Change:**
$$
\Delta S_{\text{cycle}} = 0 \, \text{kJ/kg·K} \quad (\text{since it's a cyclic process})
$$

### **4. Heat Rejected per kg:**
As calculated earlier:
$$
Q_{\text{out}} \approx 383 \, \text{kJ/kg}
$$

### **Summary of Results:**
1. **Thermal Efficiency (η):** ≈ **40%**
2. **Work Ratio:** ≈ **0.43**
3. **Entropy Change per kg:**
   - **Process 2 → 3:** +0.837 kJ/kg·K
   - **Process 4 → 1:** -0.837 kJ/kg·K
4. **Heat Rejected per kg (Q_out):** ≈ **383 kJ/kg**
**Thermal Efficiency:** 40% **Work Ratio:** 0.43 **Entropy Change per kg:** - Process 2 → 3: +0.837 kJ/kg·K - Process 4 → 1: -0.837 kJ/kg·K **Heat Rejected per kg:** 383 kJ/kg

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