Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:An adiabatic gas turbine expands air at
1

3
0
0
kPa and
5
0
0

°
C to
1
0
0
kPa and
1
2

Question

Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:An adiabatic gas turbine expands air at 
1
 
3
0
0
 ﻿kPa and 
5
0
0
 
°
C to 
1
0
0
 ﻿kPa and 
1
2
7
 
°
C
.
 ﻿Air enters the turbine through a 
0
.
2
 ﻿m
^
2
 ﻿opening with an average velocity of 
4
0
 ﻿m
/
s
,
 ﻿and exhausts through a 
1
 ﻿m
^
2
 ﻿opening. Determine 
(
a
)
 ﻿the mass flow rate of air through the turbine and 
(
b
)
 ﻿the power produced by the turbine.

VIPSolutions · Accepted Answer

# Gas Turbine Expansion: Step-by-Step Solution

We are given:

- **Inlet conditions**:  
  - Pressure, $ P_1 = 130 $ kPa  
  - Temperature, $ T_1 = 500^\circ \mathrm{C} = 773.15 $ K  
  - Area, $ A_1 = .2 $ m²  
  - Velocity, $ V_1 = 40 $ m/s

- **Exit conditions**:  
  - Pressure, $ P_2 = 100 $ kPa  
  - Temperature, $ T_2 = 127^\circ \mathrm{C} = 400.15 $ K  
  - Area, $ A_2 = 1 $ m²

- **The process is adiabatic** (no heat transfer).

We are to find:
1. The **mass flow rate** of air ($ \dot{m} $)
2. The **power produced** by the turbine ($ \dot{W}_{turbine} $)

---

## Step 1: Find Mass Flow Rate ($ \dot{m} $)

The mass flow rate is:
$$
\dot{m} = \rho_1 A_1 V_1
$$

First, calculate the **density at inlet** ($ \rho_1 $) using the ideal gas law:

$$
\rho_1 = \frac{P_1}{R T_1}
$$

Where $ R = .287 $ kJ/kg·K $ = 287 $ J/kg·K

Plug in values (convert kPa to Pa):

- $ P_1 = 130 $ kPa $ = 1,300,000 $ Pa
- $ T_1 = 773.15 $ K

$$
\rho_1 = \frac{1,300,000}{287 \times 773.15} \approx \frac{1,300,000}{221,849} \approx 5.86 \text{ kg/m}^3
$$

Now, calculate mass flow rate:

$$
\dot{m} = 5.86 \times .2 \times 40 = 5.86 \times 8 = 46.88 \text{ kg/s}
$$

---

## Step 2: Power Produced by the Turbine ($ \dot{W}_{turbine} $)

Use the **steady-flow energy equation** for an adiabatic turbine:

$$
\dot{W}_{turbine} = \dot{m} \left[ (h_1 - h_2) + \frac{V_1^2 - V_2^2}{2} \right]
$$

### 2.1. Find Exit Velocity ($ V_2 $)

First, find the **exit density** ($ \rho_2 $):

- $ P_2 = 100 $ kPa $ = 100,000 $ Pa
- $ T_2 = 400.15 $ K

$$
\rho_2 = \frac{100,000}{287 \times 400.15} \approx \frac{100,000}{114,843} \approx .87 \text{ kg/m}^3
$$

The exit mass flow rate is also:

$$
\dot{m} = \rho_2 A_2 V_2 \implies V_2 = \frac{\dot{m}}{\rho_2 A_2}
$$

$$
V_2 = \frac{46.88}{.87 \times 1} \approx 53.9 \text{ m/s}
$$

---

### 2.2. Find Enthalpy Change ($ h_1 - h_2 $)

For air,  
$ h = C_p T $,  
where $ C_p = 1.005 $ kJ/kg·K.

$$
h_1 - h_2 = C_p (T_1 - T_2)
$$
$$
= 1.005 \times (773.15 - 400.15) = 1.005 \times 373 = 374.87 \text{ kJ/kg}
$$

---

### 2.3. Find Kinetic Energy Change

$$
\frac{V_1^2 - V_2^2}{2} = \frac{40^2 - 53.9^2}{2}
$$

$$
40^2 = 160,\quad 53.9^2 \approx 2906.5
$$
$$
\Rightarrow \frac{160 - 2906.5}{2} = \frac{-1306.5}{2} = -653.25 \text{ (m}^2\text{/s}^2)
$$

Convert to kJ/kg: $ 1 \text{ kJ/kg} = 100 \text{ m}^2/\text{s}^2 $:

$$
\text{Kinetic energy change} = -.653 \text{ kJ/kg}
$$

---

### 2.4. Total Work Output

$$
\dot{W}_{turbine} = \dot{m} \left[ (h_1 - h_2) + \left( \frac{V_1^2 - V_2^2}{2 \times 100} \right) \right]
$$

$$
= 46.88 \left[ 374.87 - .653 \right] = 46.88 \times 374.22 = 17,554 \text{ kW}
$$

---

## **Final Answers**

**(a) Mass flow rate:**  
$$
\boxed{46.9 \ \text{kg/s}}
$$

**(b) Power produced by the turbine:**  
$$
\boxed{17,550 \ \text{kW}}
$$

---

### **Summary Table**

| Quantity              | Value           |
|-----------------------|----------------|
| Mass flow rate ($\dot{m}$) | 46.9 kg/s      |
| Power output ($\dot{W}_{turbine}$) | 17,550 kW      |

---

**Note:**  
- All values rounded appropriately.  
- If you have access to air property tables, you could use more accurate enthalpy values.  
- The kinetic energy term is small compared to the enthalpy term, but should be included for accuracy.

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