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dalvyx [7]
3 years ago
15

A house is maintained at 1 atm and 24°C. Outdoor air at 5.5°C infiltrates into the house through the cracks and warm air inside

the house is forced to leave the house at a rate of 90 m3 /hr. The gas constant of air is R = 0.287 kPa·m3 /kg·K. The constant pressure specific heat of air at room temperature is cp = 1.005 kJ/kg:°C.
a. Create a schematic representation of the system, indicating all the mass and energy interactions across the boundary.
b. Determine the mass flow rate of the warm air exiting the house (in kg/s).
c. Determine the mass flow rate of the cold air entering the house (in kg/s).
d. Determine the rate of net energy loss of the house due to mass transfer (in kW).
e. List the assumptions and approximations that were needed to solve this problem.
Physics
1 answer:
Lunna [17]3 years ago
3 0

Answer:

5454gjb

Explanation:

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Answer:

v_1  = 3.5 \ m/s

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H_1= H_2

mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2

2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2

1637.1 v_1 = 3841.575 \omega_2

\omega_2 = \frac{1637.1 v_1}{3841.575}

\omega _2 = 0.4626 \ v_1

From the conservation of energy as well;we have :

T_2 +V_{2  \to 3} = T_3 \\ \\ \\  \frac{1}{2} I_A \omega_2^2 - mgh =0

[\frac{1}{2} [875+2140(0.765^2+0.895^2)](0.4262 \ v_1)^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0

706.93 \ v_1^2 - 8657.49 =0

706.93 \ v_1^2  = 8657.49

v_1^2  =  \frac{8657.49}{706.93 }

v_1 ^2 =  12.25

v_1  = \sqrt{ 12.25

v_1  = 3.5 \ m/s

6 0
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