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A plane wall of thermal conductivity of 45\(\frac{W}{mK}\) was initially maintained at a temperature of 35°C. It is subjected to an ambient temperature of 45°C at one surface. If the heat transfer coefficient at the surface of the wall is 9\(\frac{W}{m^2K}\), then what is the temperature gradient developed at the surface?

(a) 1

(b) 2

(c) 3

(d) 4

This question was posed to me during a job interview.

The query is from Eigen Value and Time Dependent Problems in portion Single Variable Problems of Finite Element Method

1 Answer

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Best answer
Right answer is (b) 2

The best I can explain: Let Tx be temperature gradient developed at the surface. If the heat transfer coefficient at the surface of a wall is is 9\(\frac{W}{m^2K}\) then the heat interaction at the surface of the wall is evaluated by equating the conduction heat transfer to the convection heat transfer, i.e.,

45Tx = 9(35-45)

Tx = \(\frac{9}{-45}\) (35-45)

=\(\frac{1}{-5}\) (-10)

=2.

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