Calculate Heat-Transfer Coefficients in Convection
How to calculate heat-transfer coefficients for hot and cold air flow in a duct?
Given the properties of the two fluid streams, how can we determine the overall heat-transfer coefficient and the temperature of the duct wall?
Answer:
a) The heat-transfer coefficient for the hot air flow is 1205.36 W/(m^2 K).
b) The average heat-transfer coefficient for the cold air flow across the duct is 1469.67 W/(m^2 K).
c) The overall heat-transfer coefficient is 0.00175 K/(W/m^2).
d) The temperature of the wall of the duct is 0.878 K.
e) The mode of convection is forced convection, and the heat-transfer coefficients would change depending on the wind speed.
To calculate the heat-transfer coefficient for the hot air flow, we can use the formula Q = mcΔT, where Q is the rate of heat transfer, m is the mass-flow rate, c is the specific heat, and ΔT is the temperature difference. We can rearrange this equation to solve for the heat-transfer coefficient h: h = Q / (A * ΔT), where A is the cross-sectional area of the duct. Plugging in the values, we get h = (0.5 kg/s) * (1007 J/(kg K)) / (1.5 m * 0.28 m * 10 K) = 1205.36 W/(m^2 K).
The average heat-transfer coefficient for the cold air flow across the duct can be calculated using the same formula. Plugging in the values, we get h = (1.235 kg/m^3) * (1006 J/(kg K)) / (1.5 m * 0.28 m * 10 K) = 1469.67 W/(m^2 K).
The overall heat-transfer coefficient can be calculated using the formula 1/h = 1/h1 + 1/h2, where h1 is the heat-transfer coefficient for the hot air flow and h2 is the heat-transfer coefficient for the cold air flow. Plugging in the values, we get 1/h = 1/1205.36 + 1/1469.67 = 0.00175 K/(W/m^2).
To calculate the temperature of the wall of the duct, we can use the formula Q = hAΔT, where Q is the rate of heat transfer, h is the heat-transfer coefficient, A is the surface area of the wall, and ΔT is the temperature difference between the hot air and the wall. Rearranging this equation, we get ΔT = Q / (hA), where Q is the rate of heat transfer from the hot air, h is the heat-transfer coefficient for the hot air flow, and A is the surface area of the wall. Plugging in the values, we get ΔT = (0.5 kg/s) * (1007 J/(kg K)) / (1205.36 W/(m^2 K) * (1.5 m * 3 m) = 0.878 K.
The mode of convection in this scenario is forced convection, as the air flow is caused by an external force (the cross wind). If the wind speed is doubled on a windy day, the heat-transfer coefficients would increase. The heat-transfer coefficient is directly proportional to the velocity of the fluid.
If the wind is not blowing on a calm day, there would be no forced convection, and only natural convection would occur. The heat-transfer coefficients would be lower compared to a windy day.