Some characteristics of this engine (Note that, because the exhaust gas is of course not normal air, γ and R have different values! They are also given in the table):

Variable Value
Pressure in combustion chamber 9.7 MPa
Temperature in combustion chamber 3685 K
Pressure at nozzle exit 1 atm (= 101325.0 Pa)
Throat area 0.05
y 1.14
R 378.00
Calculate the thrust generated by one engine in kN.
Assume isentropic flow.

1 answer

To calculate the thrust generated by the engine, we will use the following formula for thrust in an isentropic flow through a nozzle:

\[ F = \dot{m} \cdot V_e + (P_e - P_0) \cdot A_e \]

Where:

  • \( F \) is the thrust,
  • \( \dot{m} \) is the mass flow rate,
  • \( V_e \) is the exit velocity of exhaust gases,
  • \( P_e \) is the exit pressure,
  • \( P_0 \) is the ambient pressure,
  • \( A_e \) is the exit area.

Step 1: Calculate the mass flow rate (\( \dot{m} \))

The mass flow rate can be calculated using the equation:

\[ \dot{m} = P_0 \cdot A_t \cdot \frac{y}{R \cdot T} \]

Where:

  • \( A_t \) is the throat area,
  • \( y \) is the specific heat ratio,
  • \( R \) is the specific gas constant,
  • \( T \) is the temperature in the combustion chamber.

Assuming that the throat area is \( A_t = 0.05 , m^2 \) (since it was not provided, we will use this as an approximation).

First, we need to calculate the exit velocity \( V_e \) using the isentropic relation:

Step 2: Calculate the exit velocity (\( V_e \))

Using the isentropic flow equations, we can find the exit velocity. The exit velocity can be expressed as:

\[ V_e = \sqrt{2 \cdot R \cdot T \cdot \left(1 - \left(\frac{P_e}{P_c}\right)^{\frac{y-1}{y}}\right)} \]

Given values:

  • \( P_c = 9.7 , \text{MPa} = 9700000 , \text{Pa} \)
  • \( T = 3685 , \text{K} \)
  • \( P_e = P_0 = 1 , \text{atm} = 101325 , \text{Pa} \)

Calculate the ratio:

\[ \frac{P_e}{P_c} = \frac{101325}{9700000} \approx 0.01045 \]

Now, substitute these into the equation for exit velocity:

\[ V_e = \sqrt{2 \cdot 378 \cdot 3685 \cdot \left(1 - \left(0.01045\right)^{\frac{1.14-1}{1.14}}\right)} \]

First, calculate \( \left(0.01045\right)^{\frac{0.14}{1.14}} \):

\[ = 0.01045^{0.1228} \approx 0.4185 \]

Now substitute this value in:

\[ V_e = \sqrt{2 \cdot 378 \cdot 3685 \cdot (1 - 0.4185)} \approx \sqrt{2 \cdot 378 \cdot 3685 \cdot 0.5815} \]

Calculate the inner term:

\[ = 2 \cdot 378 \cdot 3685 \cdot 0.5815 \approx 1.5904 \times 10^6 \]

Finally, calculate \( V_e \):

\[ V_e = \sqrt{1590400} \approx 1261.12 , \text{m/s} \]

Step 3: Calculate the mass flow rate (\( \dot{m} \))

Using the mass flow equation:

\[ \dot{m} = \frac{P_c \cdot A_t}{R \cdot T} = \frac{9700000 \cdot 0.05}{378 \cdot 3685} \]

Calculating this gives:

\[ \dot{m} \approx \frac{485000}{1383790} \approx 0.350 , \text{kg/s} \]

Step 4: Calculate the thrust (\( F \))

Using the thrust equation:

\[ F = \dot{m} \cdot V_e + (P_e - P_0) \cdot A_e \]

Substituting the values calculated into this equation:

\[ F = 0.350 \cdot 1261.12 + (101325 - 101325) \cdot A_e \]

The second term becomes zero, leading to:

\[ F = 0.350 \cdot 1261.12 = 442.39 , \text{N} = 0.442 , \text{kN} \]

Final Result

Thus, the thrust generated by one engine is approximately 0.442 kN.