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2 months ago

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

voir at 5 MPa. Assuming one-dimensional flow, calculate the following: a. Maximum back pressure to choke the nozzle. b. Range of back pressures over which a normal shock will appear in the nozzle. c. Back pressure for the nozzle to be perfectly expanded to the design Mach number. d. Range of back pressures for supersonic flow at the nozzle exit plane.

1 answer:

mote1985 [299]2 months ago

8 0

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 $_{exit}$ = 1.75 MPa

At 1.6 MPa, A $_{exit}$ /A* = 1.2502

At 1.8 MPa, A $_{exit}$ /A* = 1.4390

Thus, through interpolation at M $_{exit}$ = 1.75 MPa, we determine A

For this derivation, we will utilize M $_{exit}$ = 1.75 MPa and A

Similarly,

P $_{exit}$ /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 $_{exit}$ /A* = 1.387, by interpolation M

It follows that P $_{exit}$ = 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

= 4.279 MPa

Thus, the value of back pressure ranges from 3.198 MPa to 4.279 MPa

$P_b = 3.406\times 0.939 = 3.198 MPa$

c) At M $_{exit}$ = 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.

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Answer:

a) The phase before eutectoid is commonly referred to as cementite, with the chemical formula Fe₃C.

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Explanation:

Provided:

1 kg of austenite

a carbon content of 1.15 wt%

Cooled below 727°C

Questions:

a) Identify the proeutectoid phase.

b) Calculate the mass of total ferrite and cementite, Wf =?, Wc =?

c) Determine the mass of both pearlite and the proeutectoid phase, Wp =?

d) Create a schematic to illustrate the resulting microstructure.

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b) To find the total mass of formed ferrite:

$W_{f} =\frac{C_{cementite}-C_{2} }{C_{cementite}-C_{1} }$

With:

Ccementite = composition of cementite = 6.7 wt%

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C₂ = overall composition = 1.15 wt%

Inserting the values yields:

$W_{f} =\frac{6.7-1.15}{6.7-0.022} =0.8311kg$

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Answer:

The velocity at exit U_2 is 578.359 m/s

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Explanation:

Provided data includes:

Length of the nozzle L = 25 cm

Inlet diameter d_i = 5 cm

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(H_2 - H_1) + (U²_2 - U²_1) / 2 = 0From which we can derive U_2 = sqrt((2* (H_1 - H_2 )) + U²_1) with the calculations leading to U_2 = sqrt ( 2 * 10^3 * (3112.5 -2945.7) + 900) yielding U_2 = 578.359 m/s

b.

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Answer:

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Explanation:

Given data:

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