Page 263 - Facility Piping Systems Handbook for Industrial, Commercial, and Healthcare Facilities

HEAT TRANSFER, INSULATION, AND FREEZE PROTECTION

HEAT TRANSFER, INSULATION, AND FREEZE PROTECTION 5.47

The electronic, or solid-state, sensor is extremely accurate, far beyond the normal
requirements of a freeze protection system, but has the advantage of maintaining the
programmed accuracy for a distance of longer than a mile, if necessary.
11. Very often, the line being heat traced is a critical one and must be assured of continu-
ous operation. The options available are an emergency power supply or two separate
supplies each connected to a different circuit.

System Design Procedure
The following example will describe the design procedure for an electric heat tracing system
to provide freeze protection for an exterior potable water pipe. The installation will be in the
city of Chicago. The pipe will be installed on an exposed roof, run for 100 ft, be 3 in in size,
be made of copper, and have 1-in mineral fiber insulation. There is 1 valve and 10 supports.
1. Determine the true heat loss from the pipe and insulation. Referring to Fig. 5.3, the low
temperature on the map is –15°F (−5°C). The water temperature is to be maintained
at 40°F. This is a difference, or ΔT, of 55°F (−26°C). Referring to Table 5.3, using the
insulation type and thickness, pipe size, and temperature difference of 50°F (30°C)
(closest to 55), read 4.4 watts per foot (W/ft) of pipe. Because of the additional 5°, add
another 0.6 W to make the value 5.0 W/ft. Since cellular glass is being used and not
fiberglass (which the chart is based on), refer to the insulation correction factor in Table
5.3. For mineral fiber, read 1.20. Therefore, to find the true heat loss, multiply 5.0 × l.20
to calculate the true loss of heat from the pipe and insulation together of 6.0 W/ft of pipe.
This is the minimum amount of heat the cable must be capable of producing.
2. Select the cable type. With the true heat loss calculated, the type of environment where
it will be installed, and the maximum temperature that the cable might attain, consult a
manufacturer’s catalog and select a cable type.
3. If the proposed cable is not capable of providing the necessary heat, for example, the
cable selected is capable of producing only 4.0 Btu/ft, there are three options available:
a. Select a thicker or different insulation.
b. Install two cables to give the required heat.
c. Spiral the heating cable around the pipe.
The spiral option will be selected. To design for the spiral condition, the following steps
are required:
1. Determine the required length of cable, since some cables have a maximum permitted length.
2. Starting with a run of 100 ft, add the additional cable required because of spiraling,
valves, and supports. To find the spiraling factor, divide the required heat by the available
cable heat, 6 ÷ 4 = 1.5. Thus, for a cable capable of 4 W/ft if installed straight to now
give 6 W/ft, more than 1 ft of cable must be placed on 1 ft of pipe. This is done by
spiraling the cable around the outside of the pipe. The figure just calculated (1.5) is the
actual number of feet of cable per foot of pipe that will give the required amount of heat.
Therefore, we must add 50 ft for spiraling.
3. For installation purposes, use Table 5.14 to find the spiral pitch factor, which is the actual
installed distance between cable spirals. Enter with the calculated spiraling factor (1.5) and
the pipe size (3 in), read 9.8 at the intersection. That is the separation, in inches, that the
cable should use when wound around the pipe.
4. If spiraling is not necessary, refer to Table 5.15 for the additional valve losses, in Btu.
Multiply the factor found by the Btu heat loss from the pipe.
5. To find the additional length of cable needed for valves, fittings, and supports, refer to
Table 5.16. Entering with the pipe size, find 3 ft for the valve and 3 times the diameter

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