a thin-walled, uninsulated 0.4-m-diameter duct is used to route chilled air at 0.07 kg/s through the attic of a large commercial building. the attic air is at 37°c, and natural circulation provides a convection coefficient of 4 W m2K at the outer surface of the duct. if chilled air enters a 16-m-long duct at 7°c, what is its exit temperature (°c) and the rate of heat gain (w)? properties of the chilled air may be evaluated at an assumed average temperature of 300 k.

Answers

Answer 1

The exit temperature of the chilled air is 37.7°C and the rate of heat gain is 3040 W.

To determine the exit temperature and rate of heat gain for the chilled air, we need to use the heat transfer equation. The heat transfer equation is:

Q = U × A × DT...(i)

First, we need to calculate the surface area of the duct. Since the duct is a cylinder, the surface area is:

A = 2 × pi × r × l

Where r is the radius of the duct and l is the length of the duct. In this case, the radius of the duct is 0.4 m and the length is 16 m, so the surface area is:

A = 2 × pi × 0.4 × 16

A = 25.1 m2

Next, we need to calculate the temperature difference between the air inside the duct and the air outside the duct. The temperature of the air inside the duct is 7°C, and the temperature of the air outside the duct is 37°C, so the temperature difference is:

DT = 37 - 7

DT = 30

Now, we can plug the values we calculated into the heat transfer equation to find the rate of heat transfer:

Q = U × A × DT

Q = 4 × 25.1 × 30

Q = 3040 W

Finally, we can use the conservation of energy equation to determine the exit temperature of the air:

Q = m × c × DT

In this case, the mass flow rate of the air is 0.07 kg/s, and the specific heat capacity of air at 300 K is 1.005 kJ/kgK.

So, the equation to solve for the exit temperature:

T exit = 7 + Q / (m c)

T exit = 7 + 3040 / (0.07 x 1.005)

T exit = 37.7°C

Therefore, the exit temperature of the chilled air is 37.7°C and the rate of heat gain is 3040 W.

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