Air discharges steadily through a horizontal nozzle and impinges on a stationary vertical plate as shown in figure.

The inlet and outlet areas of the nozzle are 0.1 m² and 0.02 m², respectively. Take air density as constant and equal to 1.2 kg/m³. If the inlet gauge pressure of air is 0.36 kPa, the gauge pressure at point O on the plate (the point where the axis of nozzle touches the plate) is ________ kPa (round off to two decimal places).

The correct answer is 0.375

1. Analysis of the Image and Given Data:
Based on the provided schematic diagram, air discharges steadily from a horizontal converging nozzle and impinges on a stationary vertical plate. The diagram highlights three critical parameters:
- The inlet gauge pressure is labeled as $P_{\text{inlet}}=0.36\text{kPa}$.
- The outlet discharges into the surrounding environment, which is at atmospheric pressure $P_{\text{atm}}$.
- The dashed horizontal centerline represents the nozzle axis, which terminates at point O on the vertical plate. Since the jet impinges normally on the plate, point O acts as a stagnation point where the air velocity is reduced to zero.

We are given the following values:
- Inlet area, $A_1=0.1{\text{m}}^2$
- Outlet area, $A_2=0.02{\text{m}}^2$
- Air density (constant), $\rho =1.2{\text{kg/m}}^3$
- Inlet gauge pressure, $P_1=0.36\text{kPa}=360\text{Pa}$

2. Applying the Continuity Equation:
Since the flow is steady and the density of air is constant, the mass flow rate through the nozzle remains constant. The volumetric flow rate is also conserved:

$A_1{V}_1=A_2{V}_2$
Substituting the given areas:

$0.1\times V_1=0.02\times V_2$
Simplifying this relation gives:

$V_2=5V_1$

3. Applying Bernoulli's Equation to the Nozzle:
Applying Bernoulli's equation between the inlet (section 1) and the outlet (section 2) of the horizontal nozzle (neglecting potential energy differences since the nozzle is horizontal):

$P_1+\frac{1}{2}\rho V_1^2=P_2+\frac{1}{2}\rho V_2^2$
Since the nozzle discharges to the atmosphere, the gauge pressure at the outlet is:

$P_2=0\text{Pa (gauge)}$
Substituting the values of $P_1$, $\rho$, and the velocity relation $V_2=5V_1$ into Bernoulli's equation:

$360+\frac{1}{2}\left(1.2\right)V_1^2=0+\frac{1}{2}\left(1.2\right){\left(5V_1\right)}^2$

$360+0.6V_1^2=0.6\times 25V_1^2$

$360+0.6V_1^2=15V_1^2$

$14.4V_1^2=360$

$V_1^2=\frac{360}{14.4}=25$

$V_1=5\text{m/s}$
Thus, the outlet velocity $V_2$ is:

$V_2=5\times 5=25\text{m/s}$

4. Applying Bernoulli's Equation along the Free Jet Streamline:
Applying Bernoulli's equation along the central streamline from the nozzle exit (section 2) to the stagnation point O on the plate:

$P_2+\frac{1}{2}\rho V_2^2=P_O+\frac{1}{2}\rho V_O^2$
Since the flow stops completely at the stagnation point O, the velocity at point O is:

$V_O=0$
Thus, the gauge pressure $P_O$ is given by:

$P_O=P_2+\frac{1}{2}\rho V_2^2$
Substituting $P_2=0$, $\rho =1.2{\text{kg/m}}^3$, and $V_2=25\text{m/s}$:

$P_O=0+\frac{1}{2}\left(1.2\right){\left(25\right)}^2$

$P_O=0.6\times 625=375\text{Pa}$
Converting the pressure to kilopascals (kPa):

$P_O=\frac{375}{1000}=0.375\text{kPa}$
Therefore, the gauge pressure at point O on the plate is 0.375 kPa.