FREEZE PROTECTION DESIGN PROCEDURE

EXAMPLE

• Straight water line (105') to be maintained at 50°F.
• Minimum ambient temperature is -10°F. • Pipe is three-inch diameter steel.
• Insulation is one inch thick mineral fiber.
• Three valves
Calculate Temperature Differential

.T = TF-TA

.T = 50-(-10)°F

.T = 60°F

Heat Loss

Use Table 1 to find heat loss. Where the desired

temperature differential falls between two values,

use interpolation:

From Table 1: @ 50°F Q = 4.4 w/ft.

@ 100°F Q = 9.2 w/ft.

QF = 4.4 w/ft + 10/50 × (9.2 - 4.4 w/ft.)

QF = 4.4 + .96 = 5.4 w/ft.

Adjustment to Heat Loss

Adjust the heat loss for mineral fiber. From
Table 2, the adjustment factor is 1.2.

HSR Self-Regulating Pipe Heating Cable Application/Specification Guide 5

QM = QF × 1.2

QM = 5.4 w/ft. × 1.2

QM = 6.5 w/ft.
Since the piping is outdoors, no adjustment is
necessary for the absence of wind.

Determine Cable Power

Select 8 w/ft cable. Apply single cable straight along
the pipe.

Determine Cable Length

Length = 105 × 1 + 3 × 3 + Slack

Slack = 2 + 0 × 2 + 1 × 2 = 4
Total Length = 114 + 4 = 116'