Question

Part 1: A jet aircraft is traveling at 220 m/s in horizontal flight. The engine takes in air at a
rate of 123 kg/s and burns fuel at a rate of
1.93 kg/s. The exhaust gases are ejected at
925 m/s relative to the aircraft.
Find the thrust of the jet engine.
Answer in units of N.

Part 2: Find the delivered power.
Answer in units of W.

Answer 1

To find the thrust of the jet engine and the delivered power, we can use the principles of momentum and energy.

### Part 1: Thrust Calculation

**Thrust (F) can be calculated using the formula:**

\[
F = \dot{m} \cdot v_e + (v_e - v_0) \cdot \dot{m}_{\text{air}}
\]

Where:
- \(\dot{m}\) is the mass flow rate of the fuel (1.93 kg/s).
- \(v_e\) is the speed of the exhaust gases relative to the ground.
- \(v_0\) is the speed of the aircraft (220 m/s).
- \(\dot{m}_{\text{air}}\) is the mass flow rate of air (123 kg/s).

1. **Calculate the exhaust velocity in terms of the ground.**

Given that the exhaust is ejected at 925 m/s relative to the aircraft, we can find the absolute speed of the exhaust gases (relative to the ground) as follows:

\[
v_e = v_0 + 925 \, \text{m/s} = 220 \, \text{m/s} + 925 \, \text{m/s} = 1145 \, \text{m/s}
\]

2. **Now we can substitute these values into the thrust equation:**

Since the thrust is mainly due to the mass of the fuel and the generated exhaust velocity, we simplify to:

\[
F = \dot{m}_{\text{fuel}} \cdot v_e + \dot{m}_{\text{air}} \cdot (v_e - v_0)
\]

Plugging in the values:

- \(\dot{m}_{\text{fuel}} = 1.93 \, \text{kg/s}\)
- \(\dot{m}_{\text{air}} = 123 \, \text{kg/s}\)
- \(v_e = 1145 \, \text{m/s}\)
- \(v_0 = 220 \, \text{m/s}\)

Now substituting these into the equation:

\[
F = (1.93 \, \text{kg/s} \cdot 1145 \, \text{m/s}) + (123 \, \text{kg/s} \cdot (1145 \, \text{m/s} - 220 \, \text{m/s}))
\]

Now, calculate each term:

1. Mass flow of fuel contribution:

\[
1.93 \cdot 1145 = 2212.85 \, \text{N}
\]

2. Air contribution:

\[
F_{\text{air}} = 123 \cdot (1145 - 220) = 123 \cdot 925
\]

Calculating it:

\[
123 \cdot 925 = 113175 \, \text{N}
\]

3. Now add the two contributions to find total thrust:

\[
F = 2212.85 + 113175 = 115387.85 \, \text{N}
\]

So, the thrust of the jet engine is approximately **115388 N (rounded to the nearest whole number)**.

### Part 2: Delivered Power Calculation

**The delivered power (P) can be calculated using the formula:**

\[
P = F \cdot v_0
\]

Where:
- \(F\) is the thrust we just calculated (~115388 N).
- \(v_0\) is the velocity of the aircraft (220 m/s).

Substituting the values:

\[
P = 115388 \, \text{N} \cdot 220 \, \text{m/s}
\]

Calculating it:

\[
P = 25383160 \, \text{W}
\]

The delivered power is approximately **25383160 W or 25.4 MW (rounded to two decimal places)**.

### Final Results:
- Thrust: \( F \approx 115388 \, \text{N} \)
- Delivered Power: \( P \approx 25383160 \, \text{W} \)