Please Help a Calculate the heat flux in Wm2 through a sheet

Please Help

(a) Calculate the heat flux (in W/m2) through a sheet of a metal 13 mm thick if the temperatures at the two faces are 320 and 150 C. Assume steady-state heat flow and that the thermal conductivity of this metal is 52.1 W/m·K.
(b) What is the heat loss per hour if the area of the sheet is 0.43 m2?
(c) What will be the heat loss per hour if a material with a thermal conductivity of 1.9 W/m·K is used?
(d) Calculate the heat loss per hour if the first metal is used and the thickness is increased to 24 mm.

Solution

>> According to Fourier Law of Heat Consuction,

Q = k*A*dT/dx

, where,

k = Thermal Conductivity

A = Area

dT = Temperature Difference

dx = Thickness

>> Part (a).

>> Now, k = 52.1 W/m-K

dT = 320 - 150 = 170

dx = 13 mm = 0.013 m

So, Q/A = Heat Flux = k*dT/dx = 52.1*170/0.013 = 0.681 MW/m2

=> Heat Flux = 0.681 MW/m2

>> Part (b).

Now, Area, A = 0.43 m2

So, Heat Loss, Q = Heat Flux*A = 0.681*0.43 = 0.293 MW

So, Total heat Loss per hour = 0.293*3600 = 1054.66 MJ ...........ANSWER.....

>> Part (c).

Now, k = 1.9 W/m-K

So, Q = 1.9*0.43*170/0.013 = 10.684 KW

So, total heat transfer per hour = 10.684*3600 = 38.462 MJ .......ANSWER.....

>> Part (d).

Now, dx = 24 mm = 0.024 m

and, k = 52.1 W/m-K

So, Q = 52.1*0.43*170/0.024 = 0.159 MW

So, total heat transfer per hour = 0.159*3600 = 571.277 MJ .......ANSWER.....

Please Help (a) Calculate the heat flux (in W/m2) through a sheet of a metal 13 mm thick if the temperatures at the two faces are 320 and 150 C. Assume steady-s

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