To calculate the thrust generated by the engine under isentropic flow conditions, we can use the following formula for thrust:
\[ F = \dot{m} \cdot v_{e} + (P_e - P_0) \cdot A_e \]
Where:
- \( F \) is the thrust.
- \( \dot{m} \) is the mass flow rate of the exhaust gases.
- \( v_e \) is the effective exhaust velocity.
- \( P_e \) is the pressure at the nozzle exit.
- \( P_0 \) is the atmospheric pressure.
- \( A_e \) is the exit area of the nozzle.
First, we will determine the effective exhaust velocity \( v_e \) using the temperature and pressure in the combustion chamber, the exit pressure, and the specific heat ratio.
- Calculate the specific gas constant \( R \):
For this problem, it is already given as \( R = 378.00 , J/(kg \cdot K) \).
- Calculate the exit temperature \( T_e \):
We can calculate the exit temperature using the isentropic relations. The isentropic temperature ratio is given by the formula:
\[ \frac{T_e}{T} = \left(\frac{P_e}{P}\right)^{\frac{y-1}{y}} \]
Where:
- \( T \) is the combustion chamber temperature (3685 K).
- \( P \) is the combustion chamber pressure (9.7 MPa = 9700 kPa).
- \( P_e \) is the exit pressure (1 atm = 101.325 kPa).
- \( y \) is the specific heat ratio, given as 1.14.
Plugging in the values:
\[ \frac{T_e}{3685} = \left(\frac{101.325}{9700}\right)^{\frac{1.14 - 1}{1.14}} \approx \left(0.01045\right)^{0.09722} \approx 0.774 \]
Now calculating for \( T_e \):
\[ T_e = 3685 \times 0.774 \approx 2850.39 , K \]
- Calculate the effective exhaust velocity \( v_e \):
The effective exhaust velocity can be determined using:
\[ v_e = \sqrt{2 \cdot c_p \cdot (T - T_e)} \]
Where:
- \( c_p \) can be calculated as \( c_p = \frac{y}{y - 1} R \).
Now calculating \( c_p \):
\[ c_p = \frac{1.14}{1.14 - 1} \times 378 \approx \frac{1.14}{0.14} \times 378 \approx 10.14 \times 378 \approx 3839.32 , J/(kg \cdot K) \]
Replacing \( v_e \):
\[ v_e = \sqrt{2 \times 3839.32 \times (3685 - 2850.39)} \approx \sqrt{2 \times 3839.32 \times 834.61} \approx \sqrt{6411028.588} \approx 2535.62 , \text{m/s} \]
- Calculate the mass flow rate \( \dot{m} \):
We can find the mass flow rate using the ideal gas relation at the throat area:
\[ \dot{m} = \frac{P \cdot A_t}{R \cdot T} \]
- \( A_t = 0.05 , m^2 \)
- Using \( P = 9700 , kPa = 9700000 , Pa \)
- Using \( T = 3685 , K \)
Now let's calculate:
\[ \dot{m} = \frac{9700000 \cdot 0.05}{378 \cdot 3685} = \frac{485000}{1390575} \approx 0.348 , kg/s \]
- Calculate the thrust \( F \):
Now we can put everything together:
\[ F = \dot{m} \cdot v_e + (P_e - P_0) \cdot A_e \]
Noting that \( P_e = 101325 , Pa \), \( P_0 = 101325 , Pa \), and substituting into thrust:
\[ F = 0.348 \cdot 2535.62 + (101325 - 101325) \cdot 0.05 \]
The second term cancels out:
\[ F = 0.348 \cdot 2535.62 \]
Finally, calculating:
\[ F \approx 883.107 , N \]
Convert to kN:
\[ F \approx 0.883 , kN \]
Thus, the thrust generated by one engine is approximately 0.883 kN.