In the operation of the Heat Transfer Service Unit, one of the variables you can control is the rate of cooling water flowing against the cold side of the metal under study. If the temperature of the hot side of the metal is kept constant, what would be the expected effect of an increase in the rate of flow of cooling water against the cold side? 0 A. The rate of heat transfer through the material would increase. O B. The temperature gradient across the specimen would decrease. ° C. The rate of heat transfer through the material would decrease. ° C. The temperature at the hot side would increase.

Question

Chemistry

Damari Huang

4 July, 16:54

In the operation of the Heat Transfer Service Unit, one of the variables you can control is the rate of cooling water flowing against the cold side of the metal under study. If the temperature of the hot side of the metal is kept constant, what would be the expected effect of an increase in the rate of flow of cooling water against the cold side? 0 A. The rate of heat transfer through the material would increase. O B. The temperature gradient across the specimen would decrease. ° C. The rate of heat transfer through the material would decrease. ° C. The temperature at the hot side would increase.

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Answers (2)

Know the Answer?

Heidy Irwin · Answer 1

A. The rate of heat transfer through the material would increase.

Explanation:

Remember that when doing heat transfer problems you will always have a hotter side than the other, and that is the one that you'd like to cool off, ig you increase the rate of flow of cooling water agains the cold side you will be having more cold fluid flowing thru that pipe, meaning that you can transfer more heat to that side of the heat transfer. So the answer would be A. The rate of heat transfer through the material would increase.

Clay Peck · Answer 2

A. The rate of heat transfer through the material would increase.

Explanation:

To calculate the heat transfer in a heat exchanger you decide that there is not heat leakage to the surroundings, that means that magnitude of the two transfer rates will be equal. Any heat lost by the hot fluid, is gained by the cold fluid. The equation that describes this is Q = m*Cp*dT

Where:

heat = mass flow * specific heat capacity * temperature difference

So if we increase the rate of flow of cooling water and the other variables that ypu can control remain the same, the result is that the rate of heat transfer through the material would increase, as it is stated in option a.