Consider an isentropic flow of air (ratio of specific heats = 1.4) through a duct as shown in the figure.

The variations in the flow across the cross-section are negligible. The flow conditions at Location 1 are given as follows:

                                                  𝑃₁ = 100 kPa, 𝜌₁ = 1.2 kg/m³ , 𝑢1= 400 m/s

The duct cross-sectional area at Location 2 is given by A₂ = 2A₁, where A₁ denotes the duct cross-sectional area at Location 1. Which one of the given statements about the velocity 𝑢₂ and pressure 𝑃₂ at Location 2 is TRUE?

The correct option is:
𝑢₂ > 𝑢₁ , 𝑃₂ < 𝑃₁

Step-by-Step Explanation:

Step 1: Identify the flow conditions at Location 1
We are given the following conditions at Location 1 (as seen in the diagram):
Pressure, 𝑃₁ = 100 kPa = 100,000 Pa
Density, 𝜌₁ = 1.2 kg/m³
Velocity, 𝑢₁ = 400 m/s
Ratio of specific heats, 𝛾 = 1.4
Specific gas constant for air, 𝑅 = 287 J/(kg·K)

Step 2: Determine the temperature at Location 1
Using the ideal gas equation of state:

$$P_1={\rho }_{1}R{T}_1$$

Solving for 𝑇₁:

$$T_1=\frac{P_1}{{\rho }_{1}R}=\frac{100000}{1.2\times 287}\approx 290.36\text{K}$$

Step 3: Calculate the local speed of sound at Location 1

$$c_1=\sqrt{\gamma R{T}_1}=\sqrt{1.4\times 287\times 290.36}\approx 341.56\text{m/s}$$

Step 4: Find the Mach number at Location 1

$${\mathrm{Ma}}_1=\frac{u_1}{c_1}=\frac{400}{341.56}\approx 1.17$$

Since 𝑀𝑎₁ > 1, the flow at Location 1 is supersonic.

Step 5: Analyze the effect of the diverging duct
The relation between cross-sectional area 𝐴 and flow velocity 𝑢 in a 1D isentropic flow is given by the area-velocity relation:

$$\frac{dA}{A}=({\mathrm{Ma}}^2-1)\frac{du}{u}$$

From the diagram and problem description, the area increases downstream from Location 1 to Location 2 (𝐴₂ = 2𝐴₁), meaning:
𝑑𝐴 > 0
Since the flow is supersonic (𝑀𝑎 > 1), we have:
(𝑀𝑎² - 1) > 0
To satisfy the relation, the velocity gradient must be positive:
𝑑𝑢 > 0 ⇒ 𝑢₂ > 𝑢₁

For a supersonic flow, a diverging duct acts as a nozzle, causing the fluid velocity to increase.

Step 6: Determine the pressure change
According to the 1D momentum equation (Euler's equation) for inviscid flow:

$$dP=-\rho udu$$

Since 𝑑𝑢 > 0, the change in pressure must be negative:
𝑑𝑃 < 0 ⇒ 𝑃₂ < 𝑃₁

Therefore, as the supersonic flow expands through the diverging duct, its velocity increases and its pressure decreases, verifying that:
𝑢₂ > 𝑢₁ , 𝑃₂ < 𝑃₁