Suggested Response - Q2

** Calculate the heat loss from a pipe with hot wort flowing through it. The wort is at 45°C and the air around the pipe is 20°C. The pipe is 12 m long and made of stainless steel with OD of 25 mm and ID of 21 mm and a thermal conductivity of 20 Wm−1·K−1. The film heat transfer coefficients inside and outside the pipe are, respectively, 350 and 7 Wm-2 K-1.

Response:

An illustration of the described scenario is below.

To determine the overall heat loss, we will use the equation for heat transfer through a multi-layer system.

q space equals space U A capital delta T

Where

q is the heat transfer rate in W

U is the overall heat transfer coefficient in Wm-2 K-1 

A is the heat transfer area in m2

∆T is the temperature difference T2 − T1 in K

We first need to determine the equation for the overall heat transfer coefficient, U. The equation needs to have a term for each layer of resistance to heat transfer. In this scenario we have convective heat loss through the liquid, conductive heat loss through the pipe, and convective again into the air. Therefore, the equation for U is:

1 over U equals 1 over h subscript 1 plus x over k plus 1 over h subscript 2

Where h1 is the film transfer coefficient on the inside on the inside of the pipe, h2 is the film transfer coefficient on the outside of the pipe, x is the thickness of the pipe, and k is the thermal conductivity of the pipe. Thickness of the pipe is the difference in the OD and ID divided by 2. Entering these values, we have: 

1 over U equals 1 over 350 plus fraction numerator 0.002 over denominator 20 end fraction plus 1 over 7

1 over U equals 0.003 plus 0.0004 plus 0.143

U equals 6.83 W m to the power of negative 2 end exponent K to the power of negative 1 end exponent

Note that it is the convection film coefficients for the outside of the pipe that has the greatest influence on the value of the overall heat transfer coefficient and hence the rate of transfer.

The surface area of the pipe can be calculated as:

A space equals space 2 pi r L

A space equals space 2 space cross times space pi space cross times space 0.0125 space cross times space 12

A space equals space 0.943 space m squared

Now we can go back and enter our values to find the overall heat loss.

q space equals space 6.83 space cross times space 0.943 space cross times space left parenthesis 20 space minus space 45 right parenthesis

q space equals space minus 161.02 space W

So, the pipe is losing 161.02 J of heat per second. It is suggested that the pipe be insulated. 

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