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Sergio [31]
4 years ago
10

A converging-diverging nozzle with a rectangular cross-section and exit area ratio AAee⁄AAtt = 4.0 operates in an under-expanded

condition. The working gas is air, the back pressure away from the nozzle exit is ppaa = 10 kPa, and the stagnation pressure is pp0 = 500 kPa. The exit pressure is greater than the back pressure (ppee > ppaa), so expansion waves form outside the exit. For the given conditions, what is the Mach number on the downstream side of the first set of expansion waves, and what is the turning angle?

Engineering
1 answer:
Contact [7]4 years ago
8 0

Answer:

3.21 and 84.98 degrees

Explanation:

In this question, we are asked to calculate the Mach number on the downstream side of the first set of expansion wave and its turning angle.

Please check attachment for complete solution and step by step explanation

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Find the percent change in cutting speed required to give an 80% reduction in tool life when the value of n is 0.12.
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Explanation:

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80 % reduction in tool life

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VT^{0.12}=V'\left ( 0.2T\right )^{0.12}

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A power washer is being used to clean the siding of a house. Water enters at 20 C, 1 atm, with a velocity of 0.2 m/s .A jet of w
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A power washer is being used to clean the siding of a house. Water at the rate of 0.1 kg/s enters at 20°c and 1 atm, with the velocity 0.2m/s. The jet of water exits at 23°c, 1 atm with a velocity 20m/s at an elevation of 5m. At steady state, the magnitude of the heat transfer rate from power unit to the surroundings is 10% of the power input. Determine the power input to the motor in kW.

Answer:

Net power of 1.2 KW is being extracted

Explanation:

We are given;

Mass flow rate; m' = 0.1kg/s

Inlet temperature; T1 = 20°C = 293K

Inlet pressure; P1 = 1 atm = 10^(5) pa

Inlet velocity; v1 = 0.2 m/s

Exit Pressure; P2 = 1 atm = 10^(5) pa

Exit Temperature; T2 = 1 atm = 296K

Exit velocity; V2 = 20m/s

Change in elevation; h = Z2 - Z1 = 5m

We are told that the magnitude of the heat transfer rate from the power unit to the surroundings is 10% of the power input.

Thus;

Q = -0.1W

From Bernoulli equation;

Q - W = ∆Potential energy + ∆Kinetic energy + ∆Pressure energy

Where;

∆Potential energy = mg(z2 - z1)

∆Kinetic energy = ½m(v2² - v1²)

∆Pressure energy = mc_p(T2 - T1)

Thus;

-0.1W - W = [m'g(z2 - z1)] + [½m'(v2² - v1²)] + [m'c_p(T2 - T1)]

Where C_p is specific heat capacity of water = 4200 J/Kg.k

Plugging in the relevant values, we have;

-1.1W = (0.1 × 9.81 × 5) + (½ × 0.1(20² - 0.2²)) + (0.1 × 4200 × (296 - 293))

-1.1W = 4.905 + 19.998 + 1260

-1.1W = 1284.903

W = -1284.903/1.1

W ≈ -1168 J/s ≈ -1.2 KW

The negative sign means that work is extracted from the system.

7 0
3 years ago
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