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 > 𝑢_{1 ,} 𝑃₂ < 𝑃₁

Step-by-Step Explanation:

1. Analysis of the Duct Geometry
From the schematic diagram provided in the image, we can observe that the duct has a diverging profile from Location 1 to Location 2. The cross-sectional areas at these locations are labeled as A₁ and A₂ respectively. We are given:
$A_2=2A_1$
Since $A_2>A_1$, the cross-sectional area of the duct increases in the flow direction, confirming it is a diverging duct.

2. Determining the Inlet Mach Number (Location 1)
To understand how the flow behaves in a diverging duct, we must first determine if the incoming flow is subsonic ($M<1$) or supersonic ($M>1$).
We are given the following properties of air at Location 1:
• Pressure, P₁ = 100 kPa = 100 × 10³ Pa
• Density, ρ₁ = 1.2 kg/m³
• Velocity, u₁ = 400 m/s
• Ratio of specific heats, γ = 1.4
• Gas constant for air, R = 287 J/(kg·K)

We first find the temperature T₁ at Location 1 using the ideal gas equation:
$P_1={\rho }_{1}R{T}_1$
Rearranging for T₁:
$T_1=\frac{P_1}{{\rho }_{1}R}$
Substituting the given values:
$T_1=\frac{100\times {10}^3}{1.2\times 287}\approx 290.36\text{ K}$

Next, we calculate the local speed of sound c₁ at Location 1:
$c_1=\sqrt{\gamma R{T}_1}$
Substituting the values:
$c_1=\sqrt{1.4\times 287\times 290.36}\approx 341.56\text{ m/s}$

Now, we find the Mach number M₁ at Location 1:
$M_1=\frac{u_1}{c_1}$
Substituting the velocity and speed of sound:
$M_1=\frac{400}{341.56}$
Since $M_1>1$, the flow entering the duct is supersonic.

3. Flow Behavior in the Diverging Duct
For isentropic, compressible flow, the relation between cross-sectional area changes and velocity changes is given by the area-velocity relation:
$\frac{dA}{A}=(M^2-1)\frac{du}{u}$
Let us analyze this equation for our conditions:
• Since the duct is diverging, area increases: $dA>0$.
• Since the flow is supersonic, $M>1$, which means $(M^2-1)>0$.

To satisfy the area-velocity relation under these conditions, the change in velocity must also be positive:
$du>0$
Therefore, the velocity increases as the flow moves from Location 1 to Location 2, meaning:
𝑢₂ > 𝑢₁

4. Pressure Variation
For isentropic flow, Euler's momentum equation describes the relation between pressure changes and velocity changes:
$dP=-\rho udu$
Since density ρ, velocity u, and velocity change du are all positive, the change in pressure must be negative:
$dP<0$
This means pressure decreases as the flow expands and accelerates, resulting in:
𝑃₂ < 𝑃₁

Conclusion:
In supersonic flow through a diverging duct, the flow continues to expand and accelerate, yielding:
𝑢₂ > 𝑢₁ and 𝑃₂ < 𝑃₁