Question

(eText prob. 4.25 with some values changed) Air enters a diffuser of a jet engine operating at steady state at 2.65 psia, 389◦R, and a velocity of 869 ft/s, all data corresponding to high-altitude flight. The air flows adiabatically through the diffuser and exits the diffuser at 450◦R. Using the ideal gas model for air, determine the velocity [ft/s] of the air at the diffuser exit.

Answer 1

Answer:

$V_2 = 45.44m/s$

Explanation:

We have to many data in different system, so we need transform everything to SI, that is

$P_1 = 2.65 Psi = 18.271 kPa\\T_1= 389\°R = 216 K\\V_1 = 869ft/s = 264m/s\\T_2 = 450\°R = 250K$

When we have all this values in SI apply a Energy Balance Equation,

$\dot{Q}_{cv}-\dot{W}_{cv}+\dot{m}[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0$

Solving for V_2

$V_2 = \sqrt{V_1^2+2(h_1-h_2)}$

From the table of gas properties we calculate for $T_1 = 216K$ and $T_2 = 250K$

$h_1 = 209.97+(219.97-209.97)(\frac{216-210}{220-210})$

$h_1 = 215.97kJ/kg$

For T_2;

$h_2 = 250.05kJ/kg$

Substituting in equation for V_2

$V_2 = \sqrt{V_1^2+2(h_1-h_2)}$

$V_2 = \sqrt{265^2+2(215.97-250.05)*10^3}\\V_2 = 45.44m/s$