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(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.

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Answer:

[tex]V_2 = 45.44m/s[/tex]

Explanation:

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

[tex]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[/tex]

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

[tex]\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[/tex]

Solving for V_2

[tex]V_2 = \sqrt{V_1^2+2(h_1-h_2)}[/tex]

From the table of gas properties we calculate for [tex]T_1 = 216K[/tex] and [tex]T_2 = 250K[/tex]

[tex]h_1 = 209.97+(219.97-209.97)(\frac{216-210}{220-210})[/tex]

[tex]h_1 = 215.97kJ/kg[/tex]

For T_2;

[tex]h_2 = 250.05kJ/kg[/tex]

Substituting in equation for V_2

[tex]V_2 = \sqrt{V_1^2+2(h_1-h_2)}[/tex]

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

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