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 balanc****e** can be represented as:

\[ E_\in \space – \space E_{out} \space = \space E_{system} \space\]

## Expert Answer

**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 $40.7 \frac{m}{s}$.

## Numerical Answer

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** $256.9 \frac{m}{s}$.