 # The diffuser in a jet engine is designed to decrease the kinetic energy of the air entering the engine compressor without any work or heat interactions. Calculate the velocity at the exit of a diffuser when air at 100 kPa and 30 C enters it with a velocity of 355m/s and the exit state is 200 kPa and 90C.

The main objective of this question is to calculate the velocity of the diffuser at the exit.

This question uses the concept of energy balance. The energy balance of the system states that the energy entering the system is equal to the energy leaving the system. Mathematically, the energy balance can be represented as:

$E_\in \space – \space E_{out} \space = \space E_{system} \space$

Given that:

The air at the inlet have the following values:

Pressure $P_1$ =  $100KPa$

Temperature $T_1$ = $30^{\circ}$

Velocity $V_1$ = $355 m/s$

While the air at the outlet has the following values:

Pressure $P_1$ =  $200KPa$

Temperature $T_1$ = $90^{\circ}$

We have to determine the velocity of the diffuser at the exit.

Now we have to use the Energy balance equation which is as follows:

$E_\in \space – \space E_{out} \space = \space E_{system} \space$

$E_\in \space – = \space E_{out} \space$

$m \space (\space h \space + \space \frac{vi^2}{2}\space ) \space = \space m \space (\space h_2 \space + \space \frac{vi_2^2}{2}\space )$

Therefore the velocity at exit is:

$V_2 \space = \space [V_1^2 \ space + \space 2(h_1-h_2)]^{0.5} \space = \space [V_1^2 \space + \space 2c_p \space (T_1 \space – \space T_2)]^{0.5}$

We know that $c_p$ = $1.007 \frac{KJ}{Kg.K}$

By putting the values in the equation, this results in:

$V_2\space = \space [(350\frac{m}{s})^2 + \space 2(1.007 \frac{KJ}{Kg.K}) \space ( 30 \space – \space 90) K \space (\frac{1000}{1}) \space ]^{0.5}$

$V_2\space = \space [(350\frac{m}{s})^2 + \space 2(1.007 \frac{KJ}{Kg.K}) \space ( -60) K \space (\frac{1000}{1}) \space ]^{0.5}$

$V_2\space = 40.7 \frac{m}{s}$

Therefore, the velocity $V_2$ is  $40.7 \frac{m}{s}$.

The velocity of the diffuser at the exit with given values is $40.7 \frac{m}{s}$.

## Example

Find the velocity of the diffuser which has the air at the inlet with the values of pressure of $100KPa$, the temperature of  $30^{\circ}$  and velocity of $455 m/s$. Furthermore, the air at the outlet has a value of pressure is  $200KPa$, and the temperature is $100^{\circ}$.

Given that:

The air at the inlet have the following values:

Pressure $P_1$ b=  $100KPa$

Temperature $T_1$ = $30^{\circ}$

Velocity $V_1$ = $455 m/s$

While the air at the outlet has the following values:

Pressure $P_2$ =  $200KPa$

Temperature $T_2$ = $100^{\circ}$

We have to determine the velocity of the diffuser at the exit.

Energy balance equation  is as follows:

$E_\in \space – \space E_{out} \space = \space E_{system} \space$

$E_\in \space – = \space E_{out} \space$

$m \space (\space h \space + \space \frac{vi^2}{2}\space=\space m \space (\space h_2 \space + \space \frac{vi_2^2}{2}\space )$

Therefore, the velocity at exit is:

$V_2\space = \space [V_1^2 \ space +\space 2(h_1-h_2)]^{0.5} \space = \space [V_1^2 \space + \space 2c_p \space (T_1 \space – \space T_2)]^{0.5}$

We know that $c_p$ = $1.007 \frac{KJ}{Kg.K}$

By putting the values in the equation, this results in:

$V_2\space = \space [(455\frac{m}{s})^2 + 2(1.007 \frac{KJ}{Kg.K}) \space( 30 \space – \space 100) K \space(\frac{1000}{1}) \space]^{0.5}$

$V_2\space = 256.9 \frac{m}{s}$

Hence, the velocity $V_2$ of diffuser at exit is $256.9 \frac{m}{s}$.