Unit 42 Heat Transfer and Combustion Lesson 5

Unit 42: Heat Transfer and Combustion Lesson 5: Lagging

Aim • LO 1: Understanding Heat Transfer Rates for Composite Systems.

Critical Thickness of Insulation • Let’s consider a layer of insulation which might be installed around a circular pipe. • If the pipe is metal then as has been shown the temperature on the inner surface of the pipe varies little to that of the outer surface and therefore for practical purposes it can be ignored ri h, T∞ ro Ti T∞ To Ln(ro/ri) 2πk. L 1 2πk. L

Critical Thickness of Insulation • The inner temperature of the insulation is fixed at Ti and the outer surface is exposed to a convection environment at T∞. • From the equivalent resistor network the heat transfer is… q = 2πL(Ti – T∞) Ln(ro/ri) + 1 k r oh

Critical Thickness of Insulation • In interesting consideration is the amount of insulation (lagging) required (i. e. radius ro) that will maximise the heat transfer… i. e. dq = 2πL(Ti – T∞)[(1/kro) – (1/hro 2)] = 0 dro Ln(ro/ri) + 1 2 k r oh Thus at a maximum ro = k/h ro being referred to as the critical radius

Critical Thickness of Insulation q This shows that lagging (insulation) when placed onto the outer surface of a pipe will increase the heat flow if it is less that the critical radius ro. Greater that ro, the heat flow will decrease. ro r

Critical Thickness of Insulation • Thus if the outer radius is less than the value given by ro = k/h, then the heat transfer will be increased by adding further insulation. • For radii greater than the critical value an increase insulation thickness will cause a decrease in heat transfer. • The central concept is that for sufficiently small of h the convection heat loss may actually increase with the addition of insulation because of the increased surface area.

Critical Thickness of Insulation • Calculate the critical radius of insulation for asbestos (k=0. 17 W/m. o. C) surrounding a pipe and exposed to room air at 20 o. C with h = 3. 0 W/m 2. o. C. Calculate the heat loss from a 200 o. C, 5. 0 cm diameter pipe when covered with the critical radius of insulation and without insulation.

Critical Thickness of Insulation ro = k/h = 0. 17/3. 0 = 0. 0567 m = 5. 67 cm q= 2π(200 – 20) = 105. 7 W/m L Ln(5. 66/2. 5) + 1 0. 17 (0. 0567)(3. 0) Without insulation the convection from the outer surface of the pipe is… q = h(2πr)(Ti – To) = 3 x 2π x 0. 025 x (200 -20) = 84. 8 W/m L

Critical Thickness of Insulation • So the addition of 3. 17 cm (5. 67 – 2. 5) of insulation actually increases the heat transfer by 25%. • As an alternative, fibreglass having a thermal conductivity of 0. 04 W/m. o. C might be employed as the insulation material. Then the critical radius would be… ro = k/h = 0. 04/3. 0 = 0. 0133 = 1. 33 cm

Critical Thickness of Insulation • Now the critical radius is less than the outside radius of the pipe (2. 5 cm) so addition of any fibreglass insulation would cause a decrease in the heat transfer. • In a practical pipe insulation problem, the total heat loss will also be influenced by radiation as well as convection from the outer surface of the insulation.