2 Heat transfer by conduction: Fourier’s law: “The heat flux is directly proportional to the temperature gradient”.k = thermal conductivity of material [W/m.K]
3 General heat conduction equation: z a) In Cartesian co-ordinates (x,y,z):xyb) In cylindrical polar co-ordinates (r,,z):
4 Plane wall:For one dimensional steady-state heat transfer by conduction in a plane wall without heat generation, the general heat conduction is reduced toBy integrationBy another integrationThe boundary conditions are the known temperatures. That isand,
5 When these boundary conditions are applied to the equation for temperature distribution, we obtain Accordingly, the expression for the temperature distribution (temperature profile) becomes:The temperature distribution is thus linear (straight line) across the plane wall.
6 Rate of heat Transfer: Fourier’s law Evidently for heat conduction in a plane wall, the thermal resistance takes the form
7 Cylindrical pipes:For one dimensional steady-state heat transfer by conduction in a cylindrical pipe without heat generation, the general heat conduction is reduced toBy integrationBy another integration
8 Incorporating the relevant boundary conditions that 1)2)The constants are determined as followsBy subtraction
9 The substitution in the equation of gives When the values of the constants are substituted into the equation of the temperature distribution, one obtains the following expression for temperature distribution in the pipe wall
10 Rate of heat flow: Fourier’s law evaluated at r1or r2 Then, the thermal resistance for heat conduction in a cylindrical pipe takes the form
11 Heat transfer by convection: Equivalent thermal circuit for heat flow by conduction through the walls of a cylindrical pipe is shown in the following figure:Heat transfer by convection:Newton-Rikhman’s law of convection:“The heat flux is directly proportional to the temperature difference between the wall and the fluid”.
12 In case of heat convection from/to a cylindrical pipe, the thermal resistance takes the form The equivalent thermal resistance circuit for heat transfer by convection is shown in the following figure:
13 Overall heat transfer coefficient through a plane wall (U): Consider the plane wall, shown in the figure, exposed to a hot fluid A on one side and a cooler fluid B on the other side. The rate of heat transfer is expressed bythese three expressions1) Heat transfer by convection fromfluid A to wall surface (1):2) Heat transfer by conductionthe plane wall:
14 3) Heat transfer by convection from wall surface (2) to fluid B: The three previous equations can be rewritten as follows:By addition
15 The equivalent thermal resistance circuit for heat transfer through the plane wall with convective boundaries is shown in the following figure:LetU = overall heat transfer coefficient [ W/m²K]Comparing these two equations, one obtains
16 h1 = heat transfer coefficient of surface (1) [W/m2.K] = heat transfer film coefficient of wall surface (1)= individual heat transfer coefficient of wall surface (1)h2 = heat transfer coefficient of surface (2) [W/m2.K]= heat transfer film coefficient of wall surface (2)= individual heat transfer coefficient of wall surface (2)k = thermal conductivity of the wall material [W/m.K]L = wall thickness [m]
17 The wall conduction term may often can be neglected, since a thin wall of large thermal conductivity is generally used in heat exchangers. Also, one of the convection coefficients is often much smaller than the other and hence dominates determination of the overall heat transfer coefficient. For example, if one of the fluids is a gas and the other is a liquid or a liquid-vapor mixture such as boiling or condensation, the gas-side convection coefficient is much smaller.h is small, in case of gases (low viscosity and low specific heat) and in case of laminar flow (low velocity).h is big, in case of liquids (high viscosity and high specific heat) and in case of turbulent flow (high velocity).
18 Example 1:An iron plate of thickness L with thermal conductivity k is subjected to a constant, uniform heat flux qo (W/m²) at the boundary surface at x = 0. From the other boundary surface at x = L, heat is dissipated by convection into a fluid at temperature T∞ with a heat transfer coefficient h. The figure shows the geometry and the nomenclature.Develop an expression for the determination of the surface temperatures T1 and T2 at the surfaces x = 0 and x = L. Also, develop an expression for the overall heat transfer coefficient U.Calculate the surface temperatures T1 and T2 and the overall heat transfer coefficient U for L = 2 cm, k = 20 W/m.K, qo = 105 W/m2 , T∞ = 50oC, and h = 500 W/m2.K.
19 Data: L = 2 cm, k = 20 W/m2 °C, qo = 105 W/m2 , T∞ = 50oC, and h = 500 W/m2 °C. Find: T1, T2, USolution: Applying the thermalresistance concept:
20 By equating the first and the last expression, T1 is found and by equating the first and the third expressions, T2 is found:and since there is no convective heat transfer on surface (1),
21 Introducing The numerical values of various quantities in the above results, we obtains Note if the wall thickness is 2 mm, then T1 = 260 °C, T2 = 250°C and U = W/m2 °C.
22 Overall heat transfer coefficient in pipes: Consider a pipe exposed to a hot fluid on the inner side and a cooler fluid on the outer side, as shown in the figure. The area of convection is not the same for both fluids in this case, these areas depend on the inside pipe diameter and wall thickness.The heat transfer is expressed bythe following relations:1) Heat transfer by convection fromthe hot fluid on the inner side tothe inner wall surface of the pipe:
23 2) Heat transfer by conduction through the pipe wall itself: 3) Heat transfer by convection from the outer wall surface of the pipe to the cold fluid on the outer side:The three previous equations can be rewritten as follows:
24 By additionThe equivalent thermal resistance circuit for heat transfer through the pipe wall with convective boundaries is shown in the following figure:
25 LetUo = overall heat transfer coefficient based on theouter area of pipe.Ui = overall heat transfer coefficient based on theinner area of pipe.Ao = 2ro L, is the outer surface area of the pipe.Ai = 2ri L, is the inner surface area of the pipe.Upon comparing the two equations of , one obtains
27 Example 2:Steam at 120oC flows in an insulated pipe. The pipe is made of mild steel (kp =45 W/m.K) and has an inside radius of 5 cm and an outside radius of 5.6 cm. The pipe is covered with 2.5 cm layer of magnesia insulation (kin = W/m.K). The inside heat transfer coefficient is 85 W/m2.K and the outside heat transfer coefficient is 12.5 W/m2.K. Determine the overall heat transfer coefficients Uo and Ui and the heat transfer rate from the steam per meter of pipe length, if the surrounding air temperature is 20oC.Data: Ti = 120oC, kp = 45 W/m.K, r1 = 0.05 m, r2 = m,r3 = m, kin = W/m.K, hi = 85 W/m2.K,ho = W/m2.K, To = 20oC.Find: Uo , Ui ,
30 Values of the overall heat transfer coefficient U (W/m².K)Fluid combination850 – 1700Water to water heat exchanger110 – 350Water to oil heat exchanger1000 – 6000Steam condensers (water in tubes)800 – 1400Ammonia condenser (water in tubes)250 – 700Alcohol condenser (water in tubes)Finned-tube heat exchanger (water in tubes, air in cross flow)10 – 40Gas to gas heat exchanger