This question is incomplete, the complete question is;
A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle? Assume the specific heat ratio is 1.2.
Answer:
- stagnation pressure is 274.993 Mpa
- the stagnation temperature Tt is 4048 K
Explanation:
Given the data in the question;
To determine the stagnation pressure and temperature in the exit plane of the nozzle;
we us the expression;
Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1) = ( Tt/T )^(γ/γ -1)
where Pt is stagnant pressure = ?
P is static pressure = 4 MPa = 4 × 10⁶ Pa
Tt is stagnation temperature = ?
T is the static temperature = 2000 K
γ is ratio of specific heats = 1.2
M is Mach number M = 3.2
we substitute
Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)
Pt = P(1 + (γ-1 / 2) M²)^(γ/γ -1)
Pt = 4 × 10⁶(1 + (1.2-1 / 2) 3.2²)^(1.2/1.2 -1)
Pt = 4 × 10⁶ × 68.7484
Pt = 274.993 × 10⁶ Pa
Pt = 274.993 Mpa
Therefore stagnation pressure is 274.993 Mpa
Now, to get our stagnation Temperature
Pt/P = ( Tt/T )^(γ/γ -1)
we substitute
274.993 × 10⁶ Pa / 4 × 10⁶ Pa = ( Tt / 2000 )^(1.2/1.2 -1)
68.7484 = Tt⁶ / 6.4 × 10¹⁹
Tt⁶ = 68.7484 × 6.4 × 10¹⁹
Tt⁶ = 4.3998976 × 10²¹
Tt = ⁶√(4.3998976 × 10²¹)
Tt = 4047.999 ≈ 4048 K
Therefore, the stagnation temperature Tt is 4048 K
