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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:

512212njun][[]

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A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car ent
Paha777 [63]

Answer:

0.8712 m/s²

Explanation:

We are given;

Velocity of first car; v1 = 33 m/s

Distance; d = 2.5 km = 2500 m

Acceleration of first car; a1 = 0 m/s² (constant acceleration)

Velocity of second car; v2 = 0 m/s (since the second car starts from rest)

From Newton's equation of motion, we know that;

d = ut + ½at²

Thus,for first car, we have;

d = v1•t + ½(a1)t²

Plugging in the relevant values, we have;

d = 33t + 0

d = 33t

For second car, we have;

d = v2•t + ½(a2)•t²

Plugging in the relevant values, we have;

d = 0 + ½(a2)t²

d = ½(a2)t²

Since they meet at the next exit, then;

33t = ½(a2)t²

simplifying to get;

33 = ½(a2)t

Now, we also know that;

t = distance/speed = d/v1 = 2500/33

Thus;

33 = ½ × (a2) × (2500/33)

Rearranging, we have;

a2 = (33 × 33 × 2)/2500

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