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It is 15C outside and 22C inside. A. 75 m x .5m window is left open, and the wind is blowing in the window at a speed of1.2m/s. What is the thermal power transfer out of the room.

1 Answer

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Here are the step-by-step workings to solve this problem:

Given:

Outside temperature is 15°C

Inside temperature is 22°C

Window area is 75 m x 0.5 m = 37.5 m^2

Wind speed through window is 1.2 m/s

We want to find the thermal power transfer out of the room.

We use the heat transfer equation:

Q = UAΔT

Where:

Q is the heat transfer rate (power)

U is the heat transfer coefficient

A is the area

ΔT is the temperature difference

The heat transfer coefficient for natural convection and an open window is ~ 8 W/(m^2K)

U = 8 W/(m^2K)

The window area is 37.5 m^2

A = 37.5 m^2

The temperature difference is 22°C - 15°C = 7 K

ΔT = 7 K

Plugging into the equation:

Q = UAΔT

Q = (8 W/(m^2K)) * (37.5 m^2) * (7 K)

Q = 2100 W

Therefore, the thermal power transfer out of the room is 2100 W.

The key steps are:

Using the heat transfer equation Q = UAΔT

Finding the relevant values for U, A, and ΔT

Plugging those values into the equation to calculate Q, the heat transfer rate

Hope this helps! Let me know if you have any other questions.

answered
User Lucina
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