14 - Conduction - Critical Radius of Insulation

• A 0.083 inch diameter electrical wire at 115 is covered by 0.02 inch thick plastic insulation with \(k=0.075\) (Btu/h.ft.). The wire is exposed to a medium at 50, with a combined convection and radiation heat transfer coefficient of 2.5 Btu/(h.ft^{2 \( \circ \)}C.). Determine if the plastic insulation on the wire will increase or decrease heat transfer from the wire. (Ans: insulation increases the heat transfer from the wire; as \(r_{oc}>r_i\))

• Calculate the critical radius of insulation for asbestos with \(k=0.17\) W/(m.K) surrounding a pipe and exposed to room air at 20^{\( \circ \)}C with \(h=3\) W/(m²).K). Calculate the heat loss from a 5 cm diameter pipe with a outer surface temperature of 100^{\( \circ \)}C when covered with the critical radius of insulation and without insulation. (Ans: \(r_{oc}=5.67\) cm; \(Q_{\text{with insulation}}=47\) W/m; \(Q_{\text{without insulation}}=37.7\) W/m)

• Modify the previous problem with fiberglass insulating material with \(k=0.04\) W/(m/K), and find the heat loss with 31.7 mm thickness of insulation. (Ans: \(r_{oc}=1.33\) cm; \(Q_{\text{with insulation}}=19.07\) W/m

• Evaluate the thickness of rubber insulation necessary in the case of a 10 mm dia copper conductor to ensure max. heat transfer to the atmosphere, given the thermal conductivity of rubber as 0.155 W/(m.K) and the surface coefficient as 8.5 W/(m²).K). Estimate the max. heat transfer rate per meter length of conductor if the temperature of rubber is not to exceed 65^{\( \circ \)}C. while the atmosphere is at 30^{\( \circ \)}C.(Ans: 13.2 mm; 14.9 W/m)

• A pipe of 20 mm inner diameter and 30 mm outer diameter is insulated with 35 mm thick insulation. The thermal conductivity of insulating material is 0.15 W/m.K and the convective heat transfer coefficient of outside air is 3 W/m².K. The temperature of bare pipe is 200^{\( \circ \)}C and the ambient air temperature is 30^{\( \circ \)}C. The heat transfer resistance of the pipe metal can be neglected.

1. Comment with reasoning about the heat transfer rates with and without insulation. (Ans: Heat transfer rate with insulation is higher than that without insulation. This is because of insulation thickness here corresponds to critical radius of insulation. \(Q_{\text{without insulation}}=48.07\) W/m; \(Q_{\text{with insulation}}=72.7\) W/m).
2. If the same insulating material is used, what is the minimum thickness above which there is a reduction in heat loss as compared to the bare pipe? (Ans: 352 mm; obtain this by equating \(Q_{\text{without insulation}}=Q_{\text{with insulation}}\))
3. For optimum design, what conductivity of insulating material do you suggest for the conditions given in the problem? (Ans: 0.045 W/m.K; obtain this from \(r_{oc}= r_i=k/h \qquad \Longrightarrow \quad k = r_i h\))   (G-1993-21.b)

• A steel pipe of outside diameter 30 cm carries steam and its surface temperature is 220 ^{\( \circ \)}C. It is exposed to surroundings at 25^{\( \circ \)}C. Heat is lost both by convection and radiation. The combined heat transfer coefficient has a value of 22 W/(m².K). Determine the heat loss for 1 m length. Check the economical merits of adding insulation pads of 7.5 cm thickness with thermal conductivity of 0.36 W/(m.K). The cost of heat is Rs. 1000 per GJ. The cost of insulation is Rs. 8000/m length. The unit is in operation for 200 hour/year. The capital cost should should be recovered in 2 years. After adding the insulation also the same convection and radiation prevail over the surface.

(Ans: Critical radius of insulation, \(r_{oc}=k/h=0.36/22=0.0164\) m; As \(r_i>r_{oc}\), any addition of insulation leads to decrease of heat loss. Heat loss without insulation = 4043.8 W; Heat loss with insulation = 919.2 W; Reduction of heat loss due to insulation = \(4043.8-919.2=3124.6\) W; Energy saved due to insulation per year = 3124.6\( \times \)200\( \times \)3600=2.25\( \times \)10⁹\) J = 2.25 GJ; Cost of energy saved per year = Rs.2.25\( \times \) 1000 =R. 2250; Payback period = Investment / savings per year = 8000/2250 =3.56 year; which is higher than the recommended recovery period of 2 year. Hence, there is no economic incentive because of insulation. The pipe can be left bare.)