Intro to Food Engineering

Question

Refrigerant is flowing though a ¾ inch O.D. copper tube under conditions such that its mixed temperature is 15 °F. Condensation of moisture from the atmosphere on the outside of the tube is considered a nuisance, so it is decided to insulate the tube with glass wool [k = 0.025 Btu/ (h) (ft) (°F)]. Find the minimum thickness of glass wool necessary to prevent condensation on the outer surface of the insulation. Air in the room is at 70 °F but it has a dew-point temperature of 45 °F. The natural convection heat-transfer coefficient between the insulation surface and the air in the room is represented by h = 0.5 (Δt)1/4, in which Δt is the temperature drop in °F between the insulation surface and the air in the room, and h is in Btu/(h)(ft2)(°F). The convective coefficient between the inner wall of the tube and refrigerant is sufficiently large that the resistance to heat transfer in the refrigerant is negligible. The resistance of the copper tube to heat transfer is also negligible.

Answer 1

The heat transfer by conduction (in Btu/h) through the insulation is given by

H = k A ΔT / d,

where k is the thermal conductivity we are given (k = 0.025 Btu/ (h) (ft) (°F)), A the insulation's surface area (in ft²), ΔT the temperature difference (in °F) between the pipe's surface and the insulation's surface, and d the thickness of the insulation we are asked to find. To prevent condensation, the pipe's surface temperature must be at least equal to the dew point of 45°, so that ΔT=45-15=30 °F.

 

The convective heat transfer between the insulation's surface and the air that surrounds it is given by

H = h A = 0.5 (Δt)1/4 A = 0.125 Δt A  (in Btu/(h)(ft)²(°F))

where Δt is the temperature difference between the insulation and the air: Δt=70-45=25 °F.

 

Assuming all of the heat transferred from the air to the insulation by convection is then transferred through the insulation by conduction, we can set the two expressions for H equal, cancel the surface area A, and solve for the thickness d:

 

H =  0.125 Δt A = k A ΔT / d

0.125 Δt = k ΔT / d

d = k ΔT / (0.125 Δt) = 0.025 (30)/(0.125 *25) ft = 0.24 ft = 2.88 in