Heat Conduction in a cooling fin
ADVANCED TRANSPORT PROCESSES /
TRANSPORT PHENOMENACCB/CBB 30335.Energy TransportLesson 17. Shell Energy Balance Lesson 18. Heat Conduction through a SlabLesson 19. Heat Conduction Through a composite wall2At the end of the lesson the student should be able to 17. Explain shell energy balance18. Solve problems concerning heat transfer through a slabwhen wall temperature is givenwhen the bulk temperature of the fluid before and after the wall is given19. Solve problems about heat transfer through composite walls
Lesson outcomes3Rate of Energy in by conv. transportRate of Energy out by conv. transport0=-Rate of Energy in by molec. transport+Rate of Energy out by molec. transport--Rate of Energy ProductionRate of work done by system by molecular transport+SHELL ENERGY BALANCE FOR STEADY STATERate of work done by External Force Energy Balance Equation for Steady State4Note the following points when you solve Energy Transport Problems:
The convective transport , molecular transport (conduction) and the molecular work done can be added together to give the combined energy flux e.
In non flow systems ( for which v is zero) the e vector simplifies to q vector, which is given by Fouriers Law.
The energy production term includes (i) the degradation of electrical energy in to heat (ii) the heat produced by viscous dissipation, (iii) heat produced in chemical reaction.
In non flow systems v=0 therefore
Energy Balance Equation for Steady State5Common types of boundary conditions in energy transport problems:Temperature may be specified at a surfaceThe heat flux normal to a surface may be givenAt interfaces the continuity of temperature and of the heat flux normal to the interface are required.At solid-fluid interface, the normal heat flux component may be given by Newtons Law of Cooling as follows:
h = heat transfer coefficient Common Types of Boundary ConditionsT0 = The solid surface temperatureTb = The bulk temperature of the fluid6Conduction through a Slab : The temperatures of the two surfaces of q slab are T1 and T2 and T1 > T2 . There is a steady flow of heat because of the temperature difference. Determine the temperature distribution, the heat flux and the heat flow at the surface.HWx0TT0T1 Heat Conduction through a Slab: Case 1x167Assumptionnon-flow system, v =0No energy generation( production)No work done on the system by external force
SolutionB.C.1 T=T0 at x = x0
B.C.1 T=T1 at x = x1 Heat Conduction through a Slab: Case 1Boundary ConditionsEnergy Transport Mechanism8Shell Energy BalanceTqx|xqx|x+xxx
(1)(2)Rate of Energy out by molec. transportRate of Energy in by molec. transport-= 09
Taking the limit as x approaches zero(3)Integrating (3)Applying Fouriers Law of heat conductionRearranging and integrating with the boundary condition , T=T at x=x and T=T1 at x=x1(4)(5)
(6) Heat Conduction through a Slab: Case 110
(7) Heat Conduction through a SlabSimilarly integrating with the boundary conditions B.C.1, T=T0 at x=x0 and B.C.2 T=T1 at x=x1
Rearranging (7) we get the heat flux at the surface at x=x1(8)By dividing (6) with (7) we get the temperature distributionHeat flux10
(9)Temperature distributionThe heat flow is found by multiplying q by the normal area (HW)
(10)11Conduction through a Slab-Case 2 : The Slab shown in the Figure below is exposed to a fluid at bulk temperatures of Ta and Tb where Ta > Tb . There is a steady flow of heat because of the temperature difference. Determine the heat flux and the heat flow at the surface.x0x1xTbTaFluidFluid Heat Conduction through a Slab: Case 2HWTT1T01112Assumptionnon-flow system, v =0No energy generation( production)No work done on the system by external force
Solution Heat Conduction through a Slab: Case 1Boundary ConditionsEnergy Transport Mechanism
Newtons Law of cooling13Shell Energy BalanceTqx|xqx|x+xxx
(1)(2)Rate of Energy out by molec. transportRate of Energy in by molec. transport-= 014
Taking the limit as x is close to zero(3)Integrating (3)Applying Fouriers Law of heat conductionRearranging and integrating ( specific integral) with the boundary condition T=T0 at x=x0 and T=T1 at x=x1(4)(5)
(6) Heat Conduction through a Slab: Case 115Using Newtons Law of cooling
Adding (6) , (7) and (8) together(7)(8)
(9)(10)Rearranging (9) we get Heat Conduction through a Slab: Case 216
The heat flow can then be calculated asIn terms of the overall heat transfer coefficient given by (13)(11) Heat Conduction through a Slab: Case 2
(12)Using (10) in (11) we get Where the overall heat transfer coefficient U is defined by (14)17Conduction through a Composite Walls: The composite wall shown in the Figure below is exposed to a fluid at bulk temperatures of Ta and Tb where Ta > Tb . There is a steady flow of heat because of the temperature difference. Determine the heat flux and the heat flow at the surface. Heat Conduction through Composite Wall
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Schematic diagramSolution19Assumptionnon-flow system, v =0No energy generation( production)No work done on the system by external force
B.C.1 T=T0 at x = x0
B.C.1 T=T1 at x = x1 Heat Conduction through a Slab: Case 1Boundary ConditionsEnergy Transport Mechanism20Shell Energy BalanceTWHqx|xWH(qx|x+x)xx
(1)(2)Rate of Energy out by molec. transportRate of Energy in by molec. transport-= 021
Taking the limit as x is close to zero(3)Integrating (3)Applying Fouriers Law of heat conduction(4)(5a)
(5b)
(5c) Heat Conduction through Composite Wall22Rearranging and integrating (5a) (5b) and (5c) with the boundary condition T=T0 at x=x0, T=T1 at x=x1 and T=T2 at x=x2
(6a)
(6b)
(6c)Using Newtons Law of cooling
(7)(8) Heat Conduction through Composite Wall23Adding (6a), (6b), (6c), (7) and (8) together
(9)(10)Rearranging (9) we get the heat flux asThe heat flow can then be calculated as(11)
Heat Conduction through Composite Wall24
In terms of the overall heat transfer coefficient(13)
(12)Using (10) in (11) we get From (12) and (3) we get overall heat transfer coefficient U as(15)
In general, for systems with n slabs(14) Heat Conduction through Composite Wall2516. Explain shell energy balance17. Solve problems concerning heat transfer through a slabwhen wall temperature is givenwhen the bulk temperature of the fluid before and after the wall is given18. Solve problems about heat transfer through composite walls
Lesson outcomes
