Problem 5.1.3

Determine the heat loss from a pipe, per unit length, whose external surface is 527 degrees Fahrenheit for the following situations:

1. Bare pipe (outer radius, r_o is equal to 0.082 foot).
2. Pipe with asbestos insulation of the thickness same as critical radius for the system.
3. Pipe with insulation thickness 0.049 foot greater than for Case 2.
4. Pipe with insulation thickness 0.049 foot less than for Case No. 2.

Outside film coefficient is given as 0.616 BTU per hour per square-foot per degree Fahrenheit. Thermal conductivity of asbestos is 0.105 BTUs per hour per foot per degrees Fahrenheit.

Solution:

(Please check Problem 5.1.2 for additional theoretical discussion.)

Outside radius of the pipe is 0.082 foot.

It would seem at first that the thicker the insulation the less the total heat loss. This is always true for flat insulation but not for curved insulation. As the thickness of the insulation is increased, the surface area from which heat may be removed by air increases and the total heat loss may also increase if the area increases more rapidly than the resistance. The resistance of the insulation of pipe is one over (2 pi L k) times log (r over r_o). The resistance of the air, although a function of both the surface and air temperatures, is given by expression (one over 2 pi r L h_o). The resistance is a minimum and the heat loss a maximum when the derivative of the sum of these resistances with respect to the radius r is set equal to zero. At the maximum heat loss r is equal to r_c, the critical radius, or is equal to the ratio of k to h. In other word, the maximum heat loss occurs when the critical radius equals the ratio of the thermal conductivity of the insulation to the surface coefficient of heat transfer. The ratio has the dimensions of feet. It is desirable to keep the critical radius as small as possible so that the application of insulation will result in a reduction and not an increase in the heat loss from a pipe. This is obviously accomplished by using an insulation of small conductivity so that the critical radius is less than the radius of the pipe.

In our case the value of critical-radius of insulation is 0.105 over 0.616 equal 0.17 feet.

For cases 1 though 4, we get r₁, r₂, r₃, and r₄ as 0.082, 0.17, 0.219, and 0.121 feet, respectively.

Film resistance can be obtained by calculating the value of one over (2 pi r L h_o).

However, the conductive resistance needs to be calculated by using the following expression, that is, one over (2 pi L k) times log (r over r_o). These resistances would be added for the above cases.

We obtain the values of R₁, R₂, R₃, and R₄ as 3.151, 2.625, 2.67, and 2.72, respectively. Total driving force is 527 minus 50 equal 477 degrees Fahrenheit. So, the heat flowing through the system is delta T over R and is found to be 151.39, 181.72, 178.7, and 175.2 BTUs per hour.