Question

Heat Removal of a Cooling Coil. A cooling coil of 1.0 ft of 304 stainless-steel tubing having an inside diameter of 0.25 in. and an outside diameter of 0.40 in. is being used to remove heat from a bath. The temperature at the inside surface of the tube is 40 degrees fahrenheit and is 80 degrees fahrenheit on the outside. The thermal conductivity of 304 stainless steel is a function of temperature:

k=7.75+7.78 x 10^-3T

where k is in btu/h*ft*fahrenheit and T is in fahrenheit. Calculate the heat removal in btu/s and watts.

Answer 1

Inside dia D1 = 0.25 in

Outside dia D2 = 0.40 in x 1ft/12in = 0.033 ft

Inside surface temperature T1 = 40 °F

Outside surface temperature T2 = 80 °F

thermal conductivity

k = 7.75 + 7.78 x 10^-3T

Where k is in Btu/h-ft-F

At T1

k1 = 7.75 + 7.78 x 10^-3 x T1

= 7.75 + 7.78 x 10^-3 x 40 = 8.06 Btu/h-ft-F

At T2

K2 = 7.75 + 7.78 x 10^-3 x 80 = 8.37 Btu/h-ft-F

Mean thermal conductivity

k = (k1 + k2)/2 = (8.06 + 8.37)/2 = 8.215 Btu/h-ft-F

Area A = 3.14 x D2 x L = 3.14 x 0.40 in x (1ft/12in) x 1ft

= 0.2094 ft2

Heat removal rate

q = (Ak/D2) (T1 - T2) /ln(D2/D1)

= (0.2094 ft2 x 8.215 Btu/h-ft-F/ 0.033 ft) (40 - 80)F / ln(0.40/0.25)

= - 2066.33 / 0.470

= - 4396.41 Btu/h x 1h/3600s

= - 1.22 Btu/s

Heat removal in W

q = - 1.22 Btu/s x (1W / 9.486*10^-4Btu/s)

= - 1287.4 W

Negative sign shows that the temperature decreases as the heat flows move on.