E5.3 Critical Radius of Insulation

Example 5.3: Find the critical radius of insulation for a pipe surrounded by asbestos and exposed to air. Determine the heat loss from the pipe at 527 °F for the following cases: (1) pipe without insulation; (2) pipe with insulation thickness same as the critical thickness; (3) with insulation thickness 0.049 ft more than the critical thickness; and (4) with insulation thickness 0.049 ft less than the critical thickness. Given the following data:

Outside radius of the pipe, r_o = 0.082 ft,

Ambient temperature, T_a = 50 °F,

Convective heat transfer coefficient, h_c = 0.616 Btu/(hr·ft²·°F);

Thermal conductivity of insulation, k = 0.105 Btu/(hr·ft·°F);

Solution: The critical radius of insulation, r_c = k/h = 0.17 ft. Surface temperature, T_s, is 527 °F, whereas the ambient temperature, T_a, is 50 °F. The heat flow rate, q, is given as

where r₂ is the insulation radius and L is the pipe length.

(1) Without insulation, r₂ = r₁ = 0.082 ft:

Resistance from convection term

(2) With pipe having insulation thickness represented by the critical insulation, r₂ = 0.17 ft:

Total resistance is a combination of conductive and convective terms.

Heat loss, q₂ = 181.72 Btu/hr.

(3) With pipe having insulation thickness 0.049 ft greater than critical radius:

r₃ = r_c + 0.049' = 0.219 ft

Resistance, R₃ = 2.67 (hr·°F)/Btu,

Heat loss, q₃ = 178.7 Btu/hr.

(4) With pipe having insulation thickness 0.049 ft less than critical radius:

r₄ = r_c - 0.049' = 0.121 ft

Resistance, R₄ = 2.72 (hr·°F)/Btu,

Heat loss, q₄ = 175.2 Btu/hr.

It can be seen that if outside radius of the pipe is smaller than the critical radius of the insulation, one does not have to insulate to save heat loss. But we do not want to leave hot pipes bare for safety reasons.