Question 7.DE.4: Calculate the nozzle throat area for the same data as in the......
Calculate the nozzle throat area for the same data as in the precious question, assuming nozzle loss coefficient, T_N = 0.05. Take γ = 1.333, and C_{pg} = 1.147 kJ/kgK.
Nozzle throat area, A = m/ρ_2Ca_{2}
and ρ_2 = \frac{p_2} {RT_2}
T_2 = T_{02} – \frac{C_2^2}{2C_p} = 1100 – \frac{(550.67)^2} {(2)(1.147)(1000)} (T_{01} = T_{02})
i.e., T_2 = 967.81 K
From nozzle loss coefficient
T^′_2 = T_2 – λ_N\frac{C_2^2}{2C_p} = 967.81 – \frac{0.05 \times (550.67)^2}{(2)(1.147)(1000)} = 961.2 K
Using isentropic p–T relation for nozzle expansion
p_2 = p_{01}/(T_{01}/T^′_2)^{γ/(γ – 1)} = 5/(1100/961.2)^4 = 2.915 bar
Critical pressure ratio
p_{01}/p_c = \left(\frac{γ + 1}{2}\right)^{γ/(γ – 1)} = \left(\frac{2.333}{2}\right)^{4} = 1.852
or p_{01}/p_2 = 5/2.915 = 1.715
Since \frac{p_{01}} {p_2} < \frac{p_{01}} {p_c}, and therefore nozzle is unchoked.
Hence nozzle gas velocity at nozzle exit
C_2 = \sqrt{[2C_{pg}(T_{01} – T_{2})]}
= \sqrt{[(2)(1.147)(1000)(1100 – 967.81)]} = 550.68 m/s
Therefore, nozzle throat area
A = \frac{m} {ρ_2C_2}, \text{and} ρ_2 = \frac{p_2} {RT_2} = \frac{(2.915)(10^2)}{0.287)(967.81)} = 1.05 kg/m³
Thus
A = \frac{15} {(1.05)(550.68)} = 0.026 m²