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MaRussiya [10]
3 years ago
12

Air flows from a large reservoir in which the pressure and temperature are 1 MPa and 30°C, respectively, through a convergent–di

vergent nozzle and into a constant area duct. The ratio of the nozzle exit area to its throat area is 3.0 and the length– diameter ratio of the duct is 15. Assuming that the flow in the nozzle is isentropic, that the flow in the duct is adiabatic, and that the average friction factor is 0.005, find the back-pressure for a normal shock to appear at the exit to the duct.
Engineering
2 answers:
docker41 [41]3 years ago
7 0

Answer:

The back pressure for a normal shock to appear at the exit to the duct is  375.959 kPa  

Explanation:

Here we have

A₁/A* = 3 =   \frac{(1+0.2Ma^2_1)^3}{1.728Ma_1}}  Which gives

Mₐ₁ = 2.637

P₀₁ =P_1(1+0.2(2.637}^2)^{3.5} = 1 MPa

∴ P₁ = 47298.69 Pa

P₂/P₁ shock = \frac{2.8(2.637)^2-0.4 }{2.4}   = 7.949

∴ P₂ = 47298.69 Pa× 7.949 =  375959.457 Pa          

Therefore, for a normal shock to appear at the exit to the duct we have the back pressure p_b =  375959.457 Pa  = 375.959 kPa      

SSSSS [86.1K]3 years ago
3 0

Answer:

The solution is attached in the attachment.

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By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

<h3>How to determine the differential of a one-variable function</h3>

Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:

dy = y'(x) · dx     (1)

If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:

y' = -\frac{1}{x^{2}}\cdot \sin 2x + \frac{2}{x}\cdot \cos 2x

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By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

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Explanation:

Given that the critical resolved shear stress for a metal is 39 MPa, the maximum possible yield strength for a single crystal of this metal is twice the critical resolved shear stress for the metal. The maximum yield yield strength for a single crystal of this metal that is pulled in tension (\sigma_y) is given as:

\sigma_y=2*critical\ resolved\ shear\ stress(\tau_{css})\\\\\sigma_y=2*\tau_{css}\\\\\sigma_y=2*39\\\\\sigma_y=78\ MPa

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