Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:
Consider
a
n
aircraft powered
b
y
a turbojet engine that has a pressure ratio
o
f

9
.

Question

Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:
Consider 
a
n
 ﻿aircraft powered 
b
y
 ﻿a turbojet engine that has a pressure ratio 
o
f
 
9
.
 ﻿The aircraft 
i
s
 ﻿stationary 
o
n
 ﻿the ground, held 
i
n
 ﻿position 
b
y
 ﻿its brakes. The ambient air 
i
s
 
a
t
 
1
7
°
C
 ﻿and 
9
5
 ﻿kPa and enters the engine 
a
t
 ﻿a rate 
o
f
 
1
5
 
k
g
s
.
 ﻿The jet fuel has a heating value 
o
f
 
4
2
,
7
0
0
 
k
J
k
g
,
 ﻿and 
i
t
 
i
s
 ﻿burned completely 
a
t
 ﻿a rate 
o
f
 
0
.
5
 
k
g
s
.
 ﻿Neglecting the effect 
o
f
 ﻿the diffuser and disregarding the slight increase 
i
n
 ﻿mass 
a
t
 ﻿the engine exit 
a
s
 ﻿well 
a
s
 ﻿the inefficiencies 
o
f
 ﻿engine components, determine the force that must 
b
e
 ﻿applied 
o
n
 ﻿the brakes 
t
o
 ﻿hold the plane stationary. Use data from the tables containing the ideal gas properties 
o
f
 ﻿air.
The force that must 
b
e
 ﻿applied 
o
n
 ﻿the brakes 
i
s
 
N
.

VIPSolutions · Accepted Answer

# Turbojet Engine Brake Force Calculation

Let's break down the problem:

- **Pressure Ratio ($r_p$):** 9  
- **Ambient Conditions:** $T_ = 17^\circ C = 290.15 \, \mathrm{K}$, $P_ = 95\, \mathrm{kPa}$  
- **Air Mass Flow Rate ($\dot{m}_{air}$):** $15\, \mathrm{kg/s}$  
- **Fuel Mass Flow Rate ($\dot{m}_{fuel}$):** $.5\, \mathrm{kg/s}$  
- **Fuel Heating Value ($h_{fuel}$):** $42,700\, \mathrm{kJ/kg}$  
- **Neglect diffuser, mass gain, and inefficiencies**

We are to **find the force applied by the brakes** to hold the aircraft stationary.

---

## 1. **Determine the Thrust Equation**

For a stationary aircraft, **velocity in ($V_$) = .**

$$
F = \dot{m}_{exit} V_{exit} - \dot{m}_{air} V_
$$
Since $V_ = $:

$$
F = \dot{m}_{exit} V_{exit}
$$
But $\dot{m}_{exit} \approx \dot{m}_{air} + \dot{m}_{fuel}$, as mass of combustion products is very close to sum of air and fuel.

---

## 2. **Find Exit Velocity ($V_{exit}$)**

### **Energy Balance (Steady-Flow Jet Engine):**
Ignoring potential and kinetic energy at inlet,

$$
\text{Enthalpy rise} = \text{Input energy from fuel}
$$
$$
\dot{m}_{air} (h_{exit} - h_{}) = \dot{m}_{fuel} h_{fuel}
$$
$$
h_{exit} = h_{} + \frac{\dot{m}_{fuel}}{\dot{m}_{air}} h_{fuel}
$$

---

## 3. **Find $ h_ $ and $ h_{exit} $ from Air Tables**

- $T_ = 290.15\,\mathrm{K}$
- $P_ = 95\,\mathrm{kPa}$
- **Assume cp = $1.005\,\mathrm{kJ/kg\,K}$ (approx for air)**

$$
h_ = c_p T_ = 1.005 \times 290.15 \approx 291.6\, \mathrm{kJ/kg}
$$
$$
h_{exit} = h_ + \frac{.5}{15} \times 42,700 = 291.6 + .0333 \times 42,700
$$
$$
.0333 \times 42,700 \approx 1,423.9
$$
$$
h_{exit} = 291.6 + 1,423.9 = 1,715.5\, \mathrm{kJ/kg}
$$

---

## 4. **Find Exit Temperature ($T_{exit}$)**

$$
h_{exit} = c_p T_{exit}
$$
$$
T_{exit} = \frac{h_{exit}}{c_p} = \frac{1,715.5}{1.005} \approx 1,707\, \mathrm{K}
$$

---

## 5. **Find Exit Pressure**

$$
P_{exit} = r_p \times P_ = 9 \times 95 = 855\, \mathrm{kPa}
$$

But for an ideal turbojet, the nozzle expands the gas back to ambient pressure ($P_$), so the exit static pressure is $P_$.

---

## 6. **Find Exit Velocity ($V_{exit}$)**

The gas at the nozzle exit is expanded isentropically to $P_$:

$$
V_{exit} = \sqrt{2 (h_{exit} - h_)}
$$
$$
V_{exit} = \sqrt{2 \times (1,715.5 - 291.6) \times 1,000}
$$
$$
= \sqrt{2 \times 1,423.9 \times 1,000}
$$
$$
= \sqrt{2,847,800}
$$
$$
V_{exit} \approx 1,688\, \mathrm{m/s}
$$

---

## 7. **Find Total Mass Flow Rate at Exit**

$$
\dot{m}_{exit} = \dot{m}_{air} + \dot{m}_{fuel} = 15 + .5 = 15.5\, \mathrm{kg/s}
$$

---

## 8. **Calculate Thrust (Brake Force Needed)**

$$
F = \dot{m}_{exit} V_{exit}
$$
$$
F = 15.5 \times 1,688 \approx 26,164 \, \mathrm{N}
$$

---

## 9. **Final Answer**

**The force that must be applied on the brakes is:**

$$
\boxed{26,200 \;\text{N}}
$$

(rounding to three significant digits)

---

### **Summary Table**

| Parameter            | Value         |
|----------------------|--------------|
| $T_$              | 290.15 K     |
| $P_$              | 95 kPa       |
| $\dot{m}_{air}$    | 15 kg/s      |
| $\dot{m}_{fuel}$   | .5 kg/s     |
| $h_{fuel}$         | 42,700 kJ/kg |
| $h_$              | 291.6 kJ/kg  |
| $h_{exit}$         | 1,715.5 kJ/kg|
| $V_{exit}$         | 1,688 m/s    |
| $F$ (brakes)       | 26,200 N     |

---

## **Key Steps Recap**

1. Used energy balance to find exit enthalpy.
2. Converted to exit temperature.
3. Used nozzle expansion to get exit velocity.
4. Multiplied by total mass flow to find thrust (brake force).

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