Question

Air enters the diffuser of a jet engine operating at steady state at 18 kPa, 216 K and a velocity of 265 m/s, all data corresponding to high-altitude flight. The air flows adiabatically through the diffuser and achieves a temperature of 250 K at the diffuser exit. Using the ideal gas model for air, determine the velocity of the air at the diffuser exit,

Answer 1

Answer:

45.44m/s

Explanation:

To solve this problem it is necessary to go back to the concepts related to the first law of thermodynamics,

in which it deepens on the conservation of the Energy.

The first law of Thermodynamics is given by the equation:

$0 = \dot{Q}-\dot{W}+\dot{m}(h_1-h_2)+\dot{m}(\frac{V_1^2-V^2_2}{2})+\dot{m}g(z_1-z_2)$

Where,

$\dot{Q}=$ Heat transfer

$\dot{W} =$Work

$\dot{m} =$Flow mass

$V_i =$Velocity

$h_i =$ Specific Enthalpy

$g =$Gravity

$z_i =$Height

From this equation we have that there is not Heat transfer, Work and changes in Height. Then,

Then our equation would be,

$0=\dot{m}(h_1-h_2)+\dot{m}(\frac{V_1^2-V^2_2}{2})$

Solving for $V_2,$

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

From the tables of ideal gas (air) at 216K we have,

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

$h_1 = 215.97kJ/kg$

From the tables at 250K, we have that

$h_2 = 250.05kJ/kg$

The velocity was previously given, then

$V_1 =265m/s$

Replacing in the equation:

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

$V_2 = 45.44m/s$

Therefore the velocity of the air at the diffuser exit is 45.44m/s