Water at the rate of 0.8 kg/s flows in a smooth steel pipe (k=53 W/m°C) with 2.5

Question

Water at the rate of 0.8 kg/s flows in a smooth steel
pipe (k=53 W/m°C) with 2.5-cm ID and 3-cm OD at an average bulk temperature of 190°F.The temperature of the surrounding air is 20°C. The room pressure is 1 atm, and the pipe is 15 m long. How much heat is lost by free convection to the surrounding air? Ans:1652.67

Hint: The outer wall temperature is not known.So one must first calculate the resistance (per meter pipe length) of the pipe wall and the resistance due to convective heat transfer of inside wall. Then by an iterative procedure, to find out the resistance due to convective heat transfer outside of the wall.

Explanation / Answer

Rate of heat transfer varies with radius of cylindrical pipe as follows:

Qr = -k*dT/dr*2**r*l

where r is the intermediate radius and l is length of the pipe.

By integrating this with respect to the boundary conditions of Temperture and radius we get the rate of heat transfer as:

Q= 2**l*k*(Twall - Tbulk)/ln(rout/rin)

This is the rate of heat transfer through conduction of walls of pipe.

Now, Rate of heat transfer through convection,Q = h *A*(Tair-Twall) , A=2**rout*l

Rate of heat transfer through conduction = Rate of heat transfer through convection.

By equating the two Qs, we get Twall.

From that rate of heat transfer and total amount of heat transferred can be calculated.

I could not give numerical answer as you gave me the value of k but not the value of h, convective heat transfer coefficient.

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