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Sati [7]
3 years ago
8

Air flows through a device such that the stagnation pressure is 0.4 MPa, the stagnation temperature is 400°C, and the velocity i

s 528 m/s. Determine the static pressure and temperature of the air at this state. The properties of air at an anticipated average temperature of 600 K are cp = 1.051 kJ/kg·K and k = 1.376.
Engineering
1 answer:
RoseWind [281]3 years ago
3 0

To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.

The stagnation temperature can be defined as

T_0 = T+\frac{V^2}{2c_p}

Where

T = Static temperature

V = Velocity of Fluid

c_p = Specific Heat

Re-arrange to find the static temperature we have that

T = T_0 - \frac{V^2}{2c_p}

T = 673.15-(\frac{528}{2*1.005})(\frac{1}{1000})

T = 672.88K

Now the pressure of helium by using the Adiabatic pressure temperature is

P = P_0 (\frac{T}{T_0})^{k/(k-1)}

Where,

P_0= Stagnation pressure of the fluid

k = Specific heat ratio

Replacing we have that

P = 0.4 (\frac{672.88}{673.15})^{1.4/(1.4-1)}

P = 0.399Mpa

Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa

<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>

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Answer:

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Explanation:

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rate of heat remover from water, Qw is given by

Qw = 6607.33[28.89 - 20] * 4.18

Qw = 245529.545kw

For air, inlet condition

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oulet condition,

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check the attached file for complete solution

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3 years ago
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(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. Th
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Answer:

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Explanation:

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\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4}) (1)

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\epsilon - Emissivity, dimensionless.

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\sigma - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

T_{s} - Temperature of the student, measured in Rankine.

T_{b} - Temperature of the bus, measured in Rankine.

If we know that \epsilon = 0.90, A = 16.188\,ft^{2}, \sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}, T_{s} = 554.07\,R and T_{b} = 527.67\,R, then the heat transfer rate due to electromagnetic radiation is:

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Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)

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