Consider a flow through a nozzle, as shown in the figure below: The air flow is steady, incompressible and inviscid. The density of air is 1.23kg/m³. The pressure difference (p₁ – p_atm) is _______ kPa (round off to the nearest integer).

The correct answer is 1.52.

1. Parameter Identification from the Schematic:
Based on the nozzle flow diagram provided, we identify the following given parameters at Section 1 and Section 2:
• Cross-sectional area at Section 1: A₁ = 0.2 m²
• Cross-sectional area at Section 2: A₂ = 0.02 m²
• Air flow velocity at Section 2: v₂ = 50 m/s
• Air pressure at Section 2: p₂ = p_atm
• Density of air: ρ = 1.23 kg/m³

2. Continuity Equation:
Since the air flow is steady and incompressible, mass conservation dictates that the volumetric flow rate remains constant throughout the nozzle:

$$A_1{v}_1=A_2{v}_2$$

Substituting the known values to solve for the velocity at Section 1 (v₁):

$$0.2\times v_1=0.02\times 50$$

$$v_1=\frac{0.02\times 50}{0.2}=5\text{m/s}$$

3. Bernoulli's Equation:
Since the flow is steady, incompressible, and inviscid, we apply Bernoulli's Equation along the central streamline between Section 1 and Section 2:

$$p_1+\frac{1}{2}\rho v_1^2=p_2+\frac{1}{2}\rho v_2^2$$

Given that the exit pressure p₂ is equal to the atmospheric pressure p_atm, we can rewrite the equation to find the pressure difference (p₁ - p_atm):

$$p_1-p_{\text{atm}}=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)$$

4. Numerical Calculation:
Substitute the values into the equation:

$$p_1-p_{\text{atm}}=\frac{1}{2}\times 1.23\times \left({50}^2-5^2\right)$$

$$p_1-p_{\text{atm}}=0.615\times \left(2500-25\right)$$

$$p_1-p_{\text{atm}}=0.615\times 2475=1522.125\text{Pa}$$

Converting the pressure difference from Pascals (Pa) to kilopascals (kPa):

$$p_1-p_{\text{atm}}=\frac{1522.125}{1000}=1.522125\text{kPa}$$

Rounding to two decimal places gives 1.52 kPa.