A converging-diverging nozzle is designed to operate with an exit Mach number of 1.75 . The nozzle is supplied from an air reser

Response:

a. 4.279 MPa

b. Range of back pressures from 3.198 MPa to 4.279 MPa

c. 0.939 MPa

d. Below 3.198 MPa

Clarification:

Based on the mentioned factors

M = 1.75 MPa  

At 1.6 MPa, A/A* = 1.2502

At 1.8 MPa, A/A* = 1.4390

Thus, through interpolation at M = 1.75 MPa, we determine A

For this derivation, we will utilize M = 1.75 MPa and A

Similarly,

P/P₀ = 0.1878

a) At the point of choking at the throat, subsonic flow persists in the subsequent diverging region of the nozzle. From tables, we obtain

A/A* = 1.387, by interpolation M

It follows that P = P₀ × P

This indicates that the nozzle will choke for back pressures less than 4.279 MPa

b) In the case of a normal shock occurring at the nozzle's exit, where M₁ = 1.75 MPa, we find; P₁ = 0.1878 × 5 = 0.939 MPa

When the normal shock takes place at M₁ = 1.75 MPa, then P₂/P₁ = 3.406

When the shock occurs right prior to the exit of the nozzle from the throat, the back pressure computed is

c) At M = 1.75 MPa and P

d) When back pressure is lower than 3.198 MPa, based on isentropic flow relations, supersonic flow will be present at the nozzle exit plane.