A water pipe of length L=1 m is surrounded by ambient air at a temperature of T_

Question

A water pipe of length L=1 m is surrounded by ambient air at a temperature of T_infinity = -10 degree C. The water in the pipe is frozen. A thin electric heating tape is wrapped around the pipe in order to melt the ice. The physical nature of the contact between the tape and the surface of the pipe introduces a thermal contact resistance of r_t,c = 0.01 m^2 K/W. The inner radius of pipe is r_i = 25 mm and an outer radius of the pipe is r_o = 75 mm. The pipe material has a thermal conductivity of k = 10 W/m/K. The convection coefficient for heat transfer from the pipe to the surrounding air is h = 100W/m/K. For ice, the latent heat release on melting is H-ts = 334.0 kj/kg, and the density of ice is rho_ice = 0.9167 g/cm^3. What is the minimum power required by the heater tape to melt the ice in the pipe? Once the ice is heated to the melting point, it is desired to melt all the ice in the pipe in 5 hours. To accomplish this, what temperature must the outer surface of the heating tape be held at? What power is required by the heater tape to achieve this surface temperature? Suppose that the heater tape is held on indefinitely. To what temperature will the water in the pipe be heated?

Explanation / Answer

for melt the ice

Q= hls (J/kg) x density (kg/m^3) x Volume (intel) = 334000 x 0.9167 x 1000 x (25x10^-3)^2 x 1 x 3.1416 =601179J

Q in 5 hours, 5 hours= 18000 s

Q_t= 601179/18000=33.3988 J/s

T_in= 100°C

T_out=?

k=10

Q_t= k x (T_in-T_out) x A / z   = 10 x (100-T-out) x 3.1416 x (2 x 25x 10^-3) x 1 / (50 x 10 ^-3) = 33.3988

T_out= 78.7433°C

Q_t= k x (T_in-T_out) x A / z   + h x A_out x ( T_out_T_inf) = 10 x (100-78.74) x 3.1416 x (2 x 25x 10^-3) x 1 / (50 x 10 ^-3) + 100 x ( 78.74 - (-10)) x 3.1416 x (2x25x10^-3)^2 x 1

Q_t= 103.091 W

t=infinity

Q/t = 0

0=0 x (T_in-78.74) x 3.1416 x (2 x 25x 10^-3) x 1 / (50 x 10 ^-3) +

T_in= 78.74°C (the temperature of the outer surface

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