Question

Consider a converging nozzle with a low velocity at the inlet and sonic velocity at the exit plane. Now the nozzle exit diameter is reduced by half while the nozzle inlet temperature and pressure are maintained the same. The nozzle exit velocity will \((a)\) remain the same \((b)\) double \((c)\) quadruple \((d)\) go down by half \((e)\) go down by one-fourth

Short answer

Answer: The exit velocity of the converging nozzle will remain the same after the exit diameter is reduced by half.

Step-by-step solution

Define the given information

The given information can be summarized as: 1. Converging nozzle with a low velocity at the inlet 2. Sonic velocity at the exit plane 3. The nozzle exit diameter is reduced by half while maintaining the same inlet temperature and pressure. Now, we will determine which of the given options is the correct interpretation of how the exit velocity of the nozzle will change.

Apply the mass flow rate conservation law

First, apply the mass flow rate conservation law, given as: \text{Mass flow rate} = \rho AV Where: - \(\rho\) denotes the fluid density - \(A\) denotes the cross-sectional area - \(V\) denotes fluid velocity Since the nozzle inlet conditions and the mass flow rate through the nozzle remain constant, the same mass flow rate must pass through the reduced nozzle exit diameter. We can write the mass flow rate equation at the entrance and exit of the nozzle as: \(\rho_1 A_1 V_1 = \rho_2 A_2 V_2\)

Calculate the change in cross-sectional area

As the diameter of the nozzle is reduced by half, the new cross-sectional area A at the exit plane is reduced by a factor of 4, because: New cross-sectional exit area = \(\dfrac{\pi (D/2)^2}{4}\) Where: - D is the initial diameter of the nozzle exit Now, since the exit plane is experiencing sonic conditions (Mach number = 1), we can use the isentropic flow relations to relate the density and cross-sectional area changes.

Isentropic flow relation for density and area

For an isentropic flow with Mach number = 1, the isentropic flow relation between the density ratio and the area is given as: \(\dfrac{\rho_2}{\rho_1} = \dfrac{A_1}{A_2}\) Since the cross-sectional area at the exit is reduced by a factor of 4, the new density is increased by a factor of 4: \(\rho_2 = 4\rho_1\)

Calculate the new exit velocity

Using the mass flow rate equation from Step 2 and the density and cross-sectional area relations from Steps 3 and 4, we can find the new exit velocity: \(\rho_1 A_1 V_1 = (4\rho_1) (\dfrac{A_1}{4}) V_2\) Solving for \(V_2\) (new exit velocity): \(V_2 = V_1\) From this analysis, we can conclude that the exit velocity of the converging nozzle will \((a)\) remain the same after the exit diameter is reduced by half.