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):

Question

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.

Answer 1

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:

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

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:

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

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.