MODULE II HEAT TRANSFER BY CONDUCTION Lesson 1: Conduction Rate Equation Lesson 2: Steady-State Conduction of Plane Walls (Composite Walls) and Radial Systems Lesson 3: Conduction with Film Coefficient of Convection THELMA T. OBILLO, PME FACULTY, MECHANICAL ENGINEERING 1ST SEMESTER SY 2021-2022 Module II 28 Lesson 3 Conduction with Convection Film Coefficient When a moving fluid comes into contact with a stationary surface, a thin boundary layer develops adjacent to the wall and in this layer, there is no relative velocity with respect to surface. In a heat exchange process, this layer is called stagnant film and heat flow in the layer is covered both by conduction and convection processes. Since thermal conductivity of fluids is low, the heat flow from the moving fluid of the wall is mainly due to convection. The rate of convective heat transfer between a solid boundary and adjacent fluid is given by the Newton-Rikhman law, q=hA(ts-tf), where tf is the temperature of the hot moving fluid, ts is the temperature of the wall surface, and A is the area exposed to heat transfer. The convective coefficient (film coefficient) h depends upon the thermodynamic and transport properties (e.g., density, viscosity, specific heat and thermal conductivity) of the fluid flow and the prevailing thermal conditions. The resistance of the stagnant boundary layer is included in the convection coefficient. The heat transfer through A. Plane Wall (Film Considered) Equivalent electrical analogy Figure 2-3 Based on the figure above, a plane wall with fluids A and B on both sides are considered where the fluid A is hotter and fluid B is colder. The temperature profile and equivalent thermal network is also shown. As discussed in the previous lesson, a convection film coefficient, h, measures how effectively a fluid transfers heat by convection. It is measured in W/m2-K, and is determined by factors such as the fluid density, viscosity, and velocity. Thus, for the given illustration above, a combination of conduction and convection are considered: conduction heat transfer across the plane wall and convection heat transfer for fluids A and B. Therefore, the corresponding resistance and total heat transfer will be: Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 29 qwall = qFluid A = qFluid B qFluid A = h1 A (TA − T1 ) q = h1 A (TA − T1 ) = q= qwall = kA (T1 − T2 ) x qFluid B = h2 A (T2 − TB ) kA (T1 − T2 ) =h2 A (T2 − TB ) x Overall Temperature Drop Total Resistance q= TA − TB 1 x 1 + + h1 A kA h2 A B. Composite Walls with Convection Film Coefficient Considering Figure 2-4 with fluids A and B on both sides of the composite (series-parallel) walls and hA and hB are the convection coefficients of the two fluids, respectively, the following are the corresponding equations for the resistances and total heat transfer across the composite wall: Figure 2-4 thermal heat transfer: q= Overall Temperature Drop T = Total Resistance Rthoverall Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 30 thermal resistances: For the composite wall : x A xB RA = RB = k A AA k B AB RE = xE k E AE RF = xF k F AF RC = xC kC AC RG = xG kG AG RD = xD k D AD For the two fluids : 1 1 RFluidA = RFluidB = hA A hB A The total thermal resistances will be: Rth overall = RFluidA + R12 + R23 + R34 + R35 + RFluidB Where: R12 = thermal resistance across 1 to 2 or across material A R23 = thermal resistance across 2 to 3 or across materials B, C and D R34 = thermal resistance across 3 to 4 or across material E R45 = thermal resistance across 4 to 5 or across materials F and G RFluidA = thermal resistance of Fluid A (hot fluid) RFluidB = thermal resistance of Fluid B (cold fluid) NOTE: Insulation and R-Value In classifying the performance of insulation, it is a common practice in the building industry to use a term called the R-value, which is defined as R= T q A ---------Equation 2.6 The units for R are ◦C·m2/W or ◦F·ft2·h/Btu. Note that this differs from the thermalresistance concept discussed above in that a heat flow per unit area is used. At this point it is worthwhile to classify insulation materials in terms of their application and allowable temperature ranges. Table 2-1 (Heat Transfer, J.P. Holman 10th Ed.) furnishes such information and may be used as a guide for the selection of insulating materials. Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 31 The Overall Heat Transfer Coefficient, U Equivalent electrical analogy Figure 2-3 q= Overall Temperature Drop Total Resistance q= TA − TB 1 x 1 + + h1 A kA h2 A Observe that the value 1/hA is used to represent the convection resistance. The overall heat transfer by combined conduction and convection is frequently expressed in terms of an overall heat-transfer coefficient U, defined by the relation q = UAToverall ---------Equation 2.7 where A is some suitable area for the heat flow. The overall heat transfer coefficient represents the intensity of heat transfer from one fluid to another through a wall separating them. It is numerically equal to the quantity of heat passing through unit area of wall surface in a unit time at a temperature difference of unit degree. Referring to the figure (Figure 2-3), the overall heat-transfer coefficient would be U= 1 1 x 1 + + h1 k h2 1 x 1 U = + + h1 k h2 Module II Heat Transfer by Conduction −1 THELMA T. OBILLO, PME Faculty, Mechanical Engineering 32 The overall heat-transfer coefficient is also related to the R value U= 1 RValue = U = ( RValue ) 1 RTotal −1 ---------Equation 2.8 Heat Transfer by Conduction through Walls (with convection film) Solved Problems: Problem 10: A composite wall is made up of an external thickness of brickwork 110 mm thick inside which is a layer of fiberglass 75 mm thick. The fiberglass is faced internally by an insulating board 25 mm thick. The coefficient of thermal conductivities are as follows: Brickwork = 1.5 W/m-K Fiberglass = 0.04 W/m-K Insulating board = 0.06 W/m-K The surface transfer coefficient of the inside wall is 3.1 W/m 2-K while that of the outside wall is 2.5 W/m2-K. Determine the (a) overall coefficient of heat transfer and using this coefficient, (b) find the heat lost through such wall 6 m high and 10 m long. Take the internal ambient temperature as 10 oC and external temperature as 27oC. (c) Find also the interface temperature between the brickwork and the fiberglass. Given: Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 33 Required: a. Overall heat transfer coefficient, U b. Heat lost, Q c. Interface temperature between brickwork and fiberglass Solution: a. Overall heat transfer coefficient, U U= 1 x x x 1 1 + 1 + 2 + 3 + ho k BW k FG k IB hi 1 1 0.11m 0.075m 0.025m 1 + + + + 2 2.5W / m − K 1.5W / m − K 0.04W / m − K 0.06W / m − K 3.1W / m 2 − K U = 0.3239W / m2 − K U= b. Heat lost, Q Q = UAT Q = ( 0.3239W / m 2 − K ) (10m 6m )( 27 − 10 ) C o Q = 330.38W c. Interface temperature between brickwork and fiberglass Q1− 2 = A ( t1 − t2 ) x1 k BW ------equation 1 Solving for t1 , Qo −1 = A ( to − t1 ) 1 ho 330.38W = (10m 6m ) ( 27o C − t1 ) 1 2.5W / m 2 − K Substituting t1 in equation 1: 330.38W = (10m 6m ) ( 24.8o C − t2 ) 0.11m 1.5W / m − K t2 = 24.4o C Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 34 Problem 11: Hot gases at 980oC flow past the upper surface of the blade of a gas turbine and the lower surface is cooled by air bled off the compressor. The convective heat transfer coefficient at the upper and lower surfaces are estimated to be 2830 and 1415 W/m 2-oC respectively. The blade material has a thermal conductivity of 11.6 W/m-oC. If metallurgical considerations limit the blade temperature at 870oC, workout the temperature of the cooling air. Consider the blade as a flat plate 0.115 cm thick and presume that steady state conditions have been reached. Given: Required: Temperature of the coolant air Solution: Heat transfer per unit area through the coolant q = hc A ( tl − tc ) q = q " = hc ( tl − tc ) A Heat flow rate per unit area, from the hot gases to the upper surface of the blade is q = hu A ( t g − tu ) ( ) q = 2830W 2 o ( 980 − 870 ) oC m − C A q = q " = 311,300W 2 m A Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 35 Since steady state conditions prevail, this heat would be conducted across the blade. Then from Fourier’s law of heat conduction, kA ( tu − tl ) x k ( tu − tl ) q = q" = A x q= 11.6W (870 C − t ) ( m − C) = o 311,300W o m 2 l 0.00115m tl = 839.138 C o The heat conducted across the blade would finally be transferred to the coolant by convection, then q = hc A ( tl − tc ) q = q " = hc ( tl − tc ) A 311,300W 2 = 1415W 2 o (839.138 oC − tc ) m m − C tc = 619.138 oC ( Module II Heat Transfer by Conduction ) THELMA T. OBILLO, PME Faculty, Mechanical Engineering 36 Conduction with Film Coefficient of Convection Cylinders: ➢ The figure below shows conduction of heat through a composite cylindrical wall having three layers of different materials and with inside and outside convection films considered. ➢ There is a perfect contact between the layers and so an equal interface temperature for any two neighboring layers. ➢ For example, steam pipe used for conveying high pressure steam in a steam power plant may have cylindrical metal wall, a layer of insulation material and then a layer of protecting plaster. ➢ The arrangement is called lagging of the pipe system. Figure 3.9 Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 37 ➢ If the internal and external heat transfer coefficients for the composite cylinder as shown in Figure 3.9 are ho and ho, respectively, then for a steady state conduction, the heat flow through each layer is same and it can be described by the following set of equations: qi1 = hi Ai ( ti − t1 ) q23 = q= Where: t2 − t3 D ln 3 D2 2 k2 L q4 o = ho Ao ( t4 − to ) q34 = q12 = t1 − t2 D ln 2 D1 2 k1 L t3 − t4 D ln 4 D3 2 k3 L 2 L ( ti − to ) r r r ln 2 ln 3 ln 4 r r r 1 1 + 1+ 2+ 3+ hi Ai k1 k2 k3 ho Ao Ai = 2 ri L = Di L Ao = 2 ro L = Do L ri = r1 Di = D1 ti t1 ro = r4 Do = D4 to t 4 OVERALL HEAT TRANSFER COEFFICIENT, U ➢ The heat transfer equation can be written as q = UAt ➢ Since the flow area varies for a cylindrical pipe, it becomes necessary to specify the area on which U is based. ➢ Thus, depending upon whether the inner or outer area is specified, two different values are defined for U. qo = U o Ao t qi = U i Ai t Uo= Overall heat transfer coefficient based on the outside surface area Ui= Overall heat transfer coefficient based on the inside surface area Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 38 Derivation of Uo: Equating equations of heat transfer, t Rt t U o Ao t = = Rt U o Ao t = U o ( D4 L ) = t D D D ln 4 ln 2 ln 3 D3 D1 D2 1 1 + + + + hi Ai 2 k1 L 2 k2 L 2 k3 L ho Ao 1 D D D2 ln 4 ln ln 3 D3 D1 D2 1 1 + + + + hi ( D1 L ) 2 k1 L 2 k2 L 2 k3 L ho ( D4 L ) divide both sides by D4 L : Uo = 1 D D D D4 ln 2 D4 ln 3 D4 ln 4 D4 D1 + D2 + D3 + 1 + hi D1 2k1 2k 2 2 k3 ho D D D D4 ln 2 D4 ln 3 D4 ln 4 D3 1 D1 D2 D4 Uo = + + + + hi D1 2k1 2k 2 2 k3 ho −1 Do the same procedure for Ui: Ui = 1 D D D D1 ln 2 D1 ln 3 D1 ln 4 1 D1 + D2 + D3 + D1 + hi 2k1 2k 2 2 k3 ho D4 D D D D1 ln 2 D1 ln 3 D1 ln 4 D3 D1 D2 D1 1 Ui = + + + + hi 2k1 2k 2 2 k3 ho D4 Module II Heat Transfer by Conduction −1 THELMA T. OBILLO, PME Faculty, Mechanical Engineering 39 Hollow Sphere: Heat conduction through composite sphere can be obtained similar to heat conduction through composite cylinder. If the convective heat transfer is considered with hi and ho as the inner and outer convection film coefficients, respectively, then the heat conduction through composite sphere will be: q= ti − to 1 1 1 1 − − r r r r 1 1 + 1 2+ 2 3+ hi Ai 4 k1 4 k2 ho Ao where : Ai = 4 ri 2 = 4 r12 Ao = 4 ro 2 = 4 r32 q= q= ti − to r2 − r1 r3 − r2 r1r2 r2 r3 1 1 + + + 2 4 k1 4 k2 hi ( 4 r1 ) ho ( 4 r32 ) 4 ( ti − to ) r2 − r1 r3 − r2 r1r2 r2 r3 1 1 + + + 2 hi r1 k1 k2 ho r32 Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 40 Critical Thickness of Insulation − There is some misunderstanding about that addition of insulating material on a surface always brings about a decrease in the heat transfer rate. − But addition of insulating material to the outside surfaces of cylindrical or spherical walls (geometries which have non-constant cross-sectional areas) may increase the heat transfer rate rather than decrease under the certain circumstances. − To establish this fact, consider a thin walled metallic cylinder of length l, radius ri and transporting a fluid at temperature ti which is higher than the ambient temperature. − Insulation of thickness (r - ri) and conductivity k is provided on the surface of the cylinder. − With assumption a. Steady state heat conduction b. One-dimensional heat flow only in radial direction c. Negligible thermal resistance due to cylinder wall d. Negligible radiation exchange between outer surface of insulation and surrounding e. The heat transfer can be expressed as ➢ Where hi and ho are the convection coefficients at inner and outer surface respectively. Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 41 ➢ The denominator represents the sum of thermal resistance to heat flow. ➢ The value of k, ri, hi and ho constant; therefore, the total thermal resistance will depend upon thickness of insulation which depends upon the outer radius of the arrangement. ➢ It is clear from the equation above that with increase of radius r (i.e. thickness of insulation), the conduction resistance of insulation increases but the convection resistance of the outer surface decreases. ➢ Therefore, addition of insulation can either increase or decrease the rate of heat flow depending upon a change in total resistance with outer radius r. ➢ To determine the effect of insulation on total heat flow, differentiate the total resistance Rt with respect to r and equating to zero. ➢ To determine whether the foregoing result maximizes or minimizes the total resistance, the second derivative need to be calculated ➢ which is positive, so r = k ⁄ ho represent the condition for minimum resistance and consequently maximum heat flow rate. ➢ The insulation radius at which resistance to heat flow is minimum is called critical radius. ➢ The critical radius, designated by rc is dependent only on thermal quantities k and ho. Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 42 ➢ From the above equation it is clear that with increase of radius of insulation heat transfer rate increases and reaches the maximum at r=rc and then it will decrease. ➢ Two cases of practical interest are: When ri < rc: It is clear from equation 2.14a below, that with addition of insulation to bare pipe increases the heat transfer rate until the outer radius of insulation becomes equal to the critical radius. Because with addition of insulation decrease the convection resistance of surface of insulation which is greater than increase in conduction resistance of insulation. ➢ Any further increase in insulation thickness decreases the heat transfer from the peak value but it is still greater than that of for the bare pipe until a certain amount of insulation r*. ➢ So, insulation greater than (r* - ri) must be added to reduce the heat loss below the bare pipe. ➢ This may happen when insulating material of poor quality is applied to pipes and wires of small radius. ➢ This condition is used for electric wire to increase the heat dissipation from the wire which helps to increase the current carrying capacity of the cable. Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 43 When ri < rc ➢ It is clear from the figure 2.14b that increase in insulation thickness always decrease the heat loss from the pipe. ➢ This condition is used to decrease the heat loss from steam and refrigeration pipes. ➢ Critical radius of insulation for the sphere can be obtain in the similar way: Problem 13: A hot fluid is being conveyed through a long pipe of 4 cm outer dia. And covered with 2 cm thick insulation. It is proposed to reduce the conduction heat loss to the surroundings to one-third of the present rate by further covering with some insulation. Calculate the additional thickness of insulation. Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 44 Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 45 Problem 14: A pipe carrying the liquid at -20oC is 10 mm in outer diameter and is exposed to ambient at 25oC with convective heat transfer coefficient of 50 W/m-K. It is proposed to apply the insulation of material having thermal conductivity of 0.50 W/m-K. Determine the thickness of insulation beyond which the heat gain will be reduced. Also calculate the heat loss for 2.5 mm, 7.5 mm and 15 mm thickness of insulation over 1m length. Which one is more effective thickness of insulation? Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 46 Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 47 Problem 15: An insulated steam pipe located where the ambient temperature is 32 oC has an inside diameter of 50 mm with 10 mm thick wall. The outside diameter of the corrugated asbestos insulation is 125 mm and the surface coefficient of still air, ho=12 W/m2-oC. Inside the pipe is steam having a temperature of 150 W/m2-oC with film coefficient, hi=6000 W/m2-oC. Thermal conductivity of pipe and asbestos insulation are 45 and 0.12 W/m-K respectively. Determine the heat loss per unit length of pipe and the interface temperature. Given: Required: a) heat loss per unit length of pipe, q/L b) interface temperature, t2 Solution: a) heat loss per unit length of pipe, q/L q= ti − to D D2 ln 3 D1 D2 1 1 + + + hi ( D1 L ) 2 k1L 2 k2 L ho ( D3 L ) ln Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 48 q = L ti − to D D ln 2 ln 3 D1 D2 1 1 + + + hi ( D1 ) 2 k1 2 k2 ho ( D3 ) 150 − 32 ) oC ( q = L 70 125 ln ln 1 1 1 50 70 + + + ( 6000 )( 0.05) 2 ( 45) 2 ( 0.12 ) (12 )( 0.125) W o m − C q = ________ W m L b) interface temperature, t2 q= ti − t2 D2 D1 1 + hi ( D1 L ) 2 k1 L = ln substitute q L t 2 − to D3 D2 1 + 2 k2 L ho ( D3 L ) ln from (a) 150 oC − t2 q = L 70 ln 1 1 50 + ( 6000 )( 0.05 ) 2 ( 45 ) W o m − C t2 = ________ oC Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 49 MODULE SUMMARY The rate of heat conduction in a specified direction is proportional to the temperature gradient, which is the rate of change in temperature with distance in that direction. One dimensional steady state heat conduction through homogenous material is given by Fourier Law of heat conduction. The Fourier law is essentially based on the following assumptions: 1. Steady state heat conduction, i.e. temperature at fixed point does not change with respect to time. 2. One dimensional heat flow. 3. Material is homogenous and isotropic, i.e. thermal conductivity has a constant value in all the directions. 4. Constant temperature gradient and a linear temperature profile. 5. No internal heat generation. The Fourier law helps to define thermal conductivity of the material. Hence thermal conductivity may be defined as the amount of heat conducted per unit time across unit area and through unit thickness, when a temperature difference of unit degree is maintained across the bounding surface. When a moving fluid comes into contact with a stationary surface, a thin boundary layer develops adjacent to the wall and in this layer, there is no relative velocity with respect to surface. In a heat exchange process, this layer is called stagnant film and heat flow in the layer is covered both by conduction and convection processes. Since thermal conductivity of fluids is low, the heat flow from the moving fluid of the wall is mainly due to convection. The rate of convective heat transfer between a solid boundary and adjacent fluid is given by the Newton-Rikhman law or otherwise known as Newton’s law of cooling. The overall heat transfer coefficient represents the intensity of heat transfer from one fluid to another through a wall separating them. It is numerically equal to the quantity of heat passing through unit area of wall surface in a unit time at a temperature difference of unit degree. Congratulations! You have just studied Module 2. Now you are ready to evaluate how much you have benefited from your reading by answering the summative test. Good Luck!!! Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 50 Prepared by: Recommending Approval: Approved: THELMA T. OBILLO Professor ROY N. LAQUIDAN Program Chair LORENZO L. BACANI Dean Module II Heat Transfer by Conduction THELMA T. OBILLO, PME Faculty, Mechanical Engineering 51 August 3, 2021 August 10, 2021 Module II Heat Transfer by Conduction August 10, 2021 THELMA T. OBILLO, PME Faculty, Mechanical Engineering