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Condensation on a Chilled-Water Pipe: More Insulation or Better Vapor Barrier?

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Condensation on a Chilled-Water Pipe: More Insulation or Better Vapor Barrier?

A 4-in nominal chilled-water pipe (4.5-in OD) carries 40°F water through a 95°F, 85% RH mechanical room. It is insulated with 1 in of fiberglass (k = 0.25 Btu·in/h·ft²·°F) with an aluminum jacket. The outer surface film coefficient is 1.5 Btu/h·ft²·°F. The jacket is dented and has several unsealed seams. The maintenance supervisor proposes adding a second inch of the same insulation because the pipe is sweating. The room dew point is about 91.5°F.

1. Estimate the current outer-surface temperature and compare it with the dew point.

Read the worked answer

Work per foot of pipe. Inner radius r_i = 2.25/12 = 0.1875 ft; outer radius r_o = 3.25/12 = 0.2708 ft.

Convert k: 0.25 Btu·in/h·ft²·°F = 0.02083 Btu/h·ft·°F.

Insulation resistance:

R_ins = ln(r_o/r_i)/(2πk) = ln(0.2708/0.1875)/(2π × 0.02083)

R_ins = 0.367/0.1309 = 2.80 h·ft·°F/Btu

Film resistance:

R_film = 1/(h × 2πr_o) = 1/(1.5 × 2π × 0.2708) = 0.392 h·ft·°F/Btu

Total R = 3.19. Heat gain per foot:

q = (95 - 40)/3.19 = 17.2 Btu/h·ft

Surface temperature:

T_s = 95 - 17.2(0.392) = 88.3°F

The surface is about 88°F, below the 91.5°F dew point, so condensation is expected. The sweating is a surface-temperature/vapor-barrier problem, not simply a thickness problem.

2. Evaluate the proposal to add a second inch of insulation.

Read the worked answer

With 2 in, r_o = 4.25/12 = 0.3542 ft.

R_ins = ln(0.3542/0.1875)/(0.1309) = ln(1.889)/0.1309 = 4.86

R_film = 1/(1.5 × 2π × 0.3542) = 0.300

Total R = 5.16

q = 55/5.16 = 10.66 Btu/h·ft

T_s = 95 - 10.66(0.300) = 91.8°F

The calculated surface temperature rises only to about 91.8°F, barely above the 91.5°F dew point. That margin is not reliable. If the jacket seams leak and the insulation absorbs moisture, the effective k rises, the surface temperature drops, and condensation returns. Adding thickness without repairing the vapor barrier does not solve the problem.

3. What is the correct repair, and why?

Read the worked answer

Repair or replace the damaged aluminum jacket and seal all seams, joints, and penetrations with a vapor-retarder mastic or tape. Remove and replace any wet insulation. Maintain a continuous vapor retarder on the warm side of the insulation, including at supports and hangers. In a chronically damp room, consider closed-cell insulation or cellular glass. The vapor barrier keeps room moisture from entering the insulation; without it, added thickness only stores more water and lowers the surface temperature. After repair, recheck the surface temperature and dew point margin.