MSE 252 Heat and Mass Transfer PDF

MSE 252 Heat and Mass Transfer Prof. A. A. ADJAOTTOR 0243408866, [email protected] 1  MSE 252 Heat and Mass Transfer Reading List: Bird, R.B., Stewart, W.E. and Lightfoot, E.N., Transport Phenomena, John Wiley and Sons, N...

MSE 252 Heat and Mass Transfer Prof. A. A. ADJAOTTOR 0243408866, [email protected] 1  MSE 252 Heat and Mass Transfer Reading List: Bird, R.B., Stewart, W.E. and Lightfoot, E.N., Transport Phenomena, John Wiley and Sons, New York. Sindo Kou, Transport Phenomena and Materials Processing, John Wiley and Sons, New York. Poirier, D.R. and Geiger, G.H, Transport Phenomena in Materials Processing, The Minerals, Metals and Materials Society, Pennsylvania 2  MSE 252 Heat and Mass Transfer Darby R., Chemical Engineering Fluid Mechanics, Second Edition, Marcel Dekker Inc, New York Gaur, R.K., and Gupta, S.L., Engineering Physics, 8th Edition, Dhanpat Rai Publications, New Delhi, India McCabe, W.L., Smith, J.C., Harriot, P., Unit Operations of Chemical Engineering, 7th Edition, McGraw Hill Higher Education Publishers, USA, 2005 Kumar D.S., Heat and Mass Transfer, 7th Revised Edition, Katson Books, Kataria and Sons, New Delhi, India 3 MSE 252 Heat and Mass Transfer 4  Heat Transfer Heat transfer is as a result of a temperature difference. Heat transfer can occur by three different mechanisms. Conduction Convection Radiation Conduction: This refers to heat transfer that occurs across a stationary solid or fluid in which temperature gradient exists. 5  Heat Transfer Convection: It refers to heat transfer that occurs across a moving fluid in which a temperature gradient exists. Radiation: Thermal radiation refers to heat transfer between two surfaces at different temperatures separated by a medium transparent to electromagnetic waves emitted by the surfaces. 6  Heat Transfer Conduction is the transfer of heat by molecular motion which occurs (i) between two parts of the same body (ii) between two bodies which are in physical contact with each other. In solid, heat is conducted by either lattice waves in non-conductors or combination of lattice waves with drift of the conduction electrons in conducting materials. In fluid, heat is conducted by molecular collision. 7  Heat Transfer Fourier’s law is the macroscopic theory of conduction. There are basically two types of heat transfer mechanisms: Conduction and Radiation Convection is rather a process involving mass movement of fluids, than a real mechanism of heat transfer. We can therefore talk of “heat transfer with convection” rather than “heat transfer by convection”. Convection implies fluid motion. 8  Heat Transfer Convection (a) Forced convection or (b)Free (natural) convection. When a pump or other mechanical device causes fluid to move we talk of forced convection 9  Fourier’s Law of Conduction One Dimension Heat flux tends to flow from warmer lower surface to the cooler upper layer 10  Fourier’s Law of Conduction Heat flux, q, is defined as the amount of heat transferred per unit area per unit time Greater the temperature difference between the two layers 𝑑𝑇 is or the steeper the temperature gradient is, the greater 𝑑𝑦 the heat flux qy 𝑑𝑇 𝑞𝑦 = −𝑘 -------- Fourier’s Law of Conduction 𝑑𝑦 q is directly proportional to T2 – T1 but inversely proportional to distance k is the thermal conductivity of the slab (material property) 11  Fourier’s Law of Conduction Units qx, qy, qz ------[W/m2] T --------------[K] x, y, z ----------[m] k ---------------[W/m K] 𝑑𝑇 𝑞𝑦 = −𝑘 Fourier’s Law of Conduction 𝑑𝑦 𝑑𝑉𝑧 𝜏𝑦𝑧 = −𝜇 Newton’s Law of Viscosity 𝑑𝑦 Minus sign shows that heat conduction occurs in the direction of decreasing temperature 12  Fourier’s Law of Conduction Thermal conductivity - reflects the relative ease or difficulty of the transfer of energy through the material Thermal conductivity - depends on the bonding and structure of the material Non – conductors have low thermal conductivities. 13  Three-Dimensional Fourier’s Law For an isotropic material, three – dimensional heat flow can be written as 𝜕𝑇 qx = - k 𝜕𝑥 𝜕𝑇 qy = - k 𝜕𝑦 𝜕𝑇 qz = - k 𝜕𝑧 Assumption of isotropic equalities is valid for fluids and for most homogeneous solids 14  Three-Dimensional Fourier’s Law Fibrous or laminated solids are non-isotropic For Isotropic material heat flux can be written in condensed form as q = - k 𝛻T Expanding into Rectangular coordinate system 𝜕𝑇 𝜕𝑇 𝜕𝑇 qx = - k ; qy = - k ; qz = - k 𝜕𝑥 𝜕𝑦 𝜕𝑧 15  Three-Dimensional Fourier’s Law Cylindrical 𝜕𝑇 1 𝜕𝑇 𝜕𝑇 qr = - k ; qθ = - k ; qz = - k 𝜕𝑟 𝑟 𝜕𝜃 𝜕𝑧 Spherical 𝜕𝑇 1 𝜕𝑇 1 𝜕𝑇 qr = - k ; qθ = - k ; qØ = - k 𝜕𝑟 𝑟 𝜕𝜃 𝑟 sin 𝜃 𝜕∅ 16  Thermal Conductivity of Gases Conduction of energy in a gas phase is primarily by transfer of translational energy (kinetic energy) Energy transfer from faster – moving (higher energy) molecule as it collides with the slower – moving one 17  Thermal Conductivity of Gases λ = Mean free path Mean free path = distance travelled by a molecule between two successive collisions It is deduced that Thermal Conductivity, k is 1Τ 3 2 1 𝑘𝐵 𝑇 k= ----------------------------------- (9) 𝑑2 𝜋3 𝑚 From Eqn (9) thermal conductivity of gases does not depend on pressure but rather on the square root of the temperature 18  Thermal Conductivity of Gases d = center to center distance of two molecules κB = Boltzmann constant T = Temperature in Kelvin m = mass of the molecule. True for pressures up to at least 106 pa (approximately 10 atm) Following figure shows thermal conductivities of several common gases as a function of temperature. 19 Thermal Conductivity of Gases 20  Thermal Conductivity of Gases For more accurate treatment of monoatomic gases, we use the Chapman-Enskog formula −4 𝑇Τ𝑀 𝑘 = 1.9891 𝑥10 𝜎 2 Ω𝑘 k [=] cal cm-1 sec-1 (°K)-1 σ [=] Å given on Table B-1 Ωk = Ωμ given on Table B.2 M [=] Molecular weight 21 Thermal Conductivity of Gases, Table B-1 22 Thermal Conductivity of Gases, Table B-1 23 24  Thermal Conductivity of Gases Example Compute the thermal conductivity of neon at 1 atm and 373.2 K. Solution Using the Chapman-Enskog equation given below for monoatomic gases, we need to determine the various parameters from Tables B-1 and B-2. −4 𝑇Τ𝑀 𝑘 = 1.9891 𝑥10 𝜎 2 Ω𝑘 25  Thermal Conductivity of Gases The Leonard-Jones constants for neon, from Table B-1, are σ = 2.789 Å ε/K = 35.7 K M = 20.183 Then at 373.2 K ΚT/ε = 373.2/35.7 = 10.45 From Table B-2 Ωk = Ωμ = 0.821 26 27 28  Thermal Conductivity of Gases Substituting the values into the equation we have 𝑇Τ𝑀 𝑘= 1.9891 𝑥10−4 𝜎 2 Ω𝑘 373.2Τ20.183 𝑘 = 1.9891 𝑥 10−4 = 2.789 2 0.821 1.338 𝑥 10−4 𝑐𝑎𝑙 𝑐𝑚−1 𝑠𝑒𝑐 −1 𝐾 −1 1 cal = 4.184 J 29  Thermal Conductivity of Gases For polyatomic gases, Eucken developed an equation for the thermal conductivity of these gases at normal pressures 1.25𝑅 k = µ 𝐶𝑝 + ----------------------------- 10 𝑀 where M = molecular weight Cp = heat capacity at constant pressure µ = Viscosity of gas 30  Thermal Conductivity of Gases For gas mixtures, we can estimate the thermal conductivity 1ൗ σ𝑖 𝑋𝑖 𝑘𝑖 𝑀𝑖 3 kmix = 1ൗ -------------------------------- 11 σ𝑖 𝑋𝑖 𝑀𝑖 3 Where Xi = mole fraction of component i Mi = Molecular weight of i ki = intrinsic thermal conductivity of i Only 2.7 % error over the temperature range 273 – 353 K 31  Thermal Conductivity of Solids Solids transmit thermal energy by two modes: either or both (1) energy may be transferred by means of elastic vibrations of the lattice moving through the crystal in the form of waves (Nonconductors) (2) notably metals (Conductors), free electrons moving through the lattice also carry energy in a manner similar to thermal conduction in a gas All solids store thermal energy as Vibrational motion - Kinetic energy of their atoms, and Bonding energy between atoms - Potential energy 32  Thermal Conductivity of Solids The waves in a crystal exhibit the attributes of particles and are called phonons The thermal conductivity of a solid (non-conductor) which conducts energy only by phonons, ഥ𝜆 𝐶𝑣 𝑉 𝑘= −−−−− −7 3 𝑉ത = Average speed of the molecules 𝐶𝑣 = Heat capacity, 𝜆 = mean free path Debye showed that 33  Thermal Conductivity of Solids 20𝑇𝑚 𝑑 λph = ----------------------------------- 12 𝛾2𝑇 Tm = melting point T = absolute temperature d = crystal lattice dimension 𝛾 = Gruneisen constant approximately = 2 for most solids at ordinary temperature. High melting point material has a large value of λ at low temperature From eqn (12) a large value of k at room temperature. E.g diamond 34  Thermal Conductivity of Solids Phonons are scattered by Differences in isotropic mass Chemical impurities Dislocations Second phases We can be sure that the less perfect the crystal, the lower the thermal conductivity. 35  Thermal Conductivity of Solids Figure 6.3 below shows the thermal conductivity of oxides and various electrical insulating materials From Eqn 12, 20𝑇𝑚 𝑑 λph = ----------------------------------- 12 𝛾2𝑇 as Temperature increases, λph decreases with a corresponding decrease in k (Eqn 7) ഥ𝜆 𝐶𝑣 𝑉 𝑘= −−−−− −7 3 36 37  Thermal Conductivity of Solids Conductors Conductors are materials with increasing concentration of conduction electrons Electronic contribution to thermal conductivity, kel given as: 2 𝜋2 𝑛𝑒 𝑘𝛽 𝑇𝜆𝑒𝑙 kel = ----------------------------------------- 15 3𝑚𝑒 𝑉𝑓 This predicts that the electronic contribution in metals increases with temperature Provided that 𝜆𝑒𝑙 does not decrease just as strongly with temperature. 38 Thermal Conductivity of Solids (Metals) Figure 6.6 39  Thermal Conductivity of Solids Pure nickel and pure iron show decrease in k with low temperature At higher temperature the electronic contribution presumably overwhelms the phonons contribution and k increases with temperature. 40  Amorphous Solids For amorphous solids such as high polymer and glasses Thermal conduction is mainly via atomic or molecular migration (or radiation at high temperatures) (1) Material is too irregular in structure to support a phonon mechanism and (2) Electron contribution is negligible. Results: Very low values of conductivity, as indicated in Table 6.1 41 Table 6.1 Thermal conductivities of amorphous or molecular solids Substance Temperature (K) k, W/m K Glass 373 0.76 Lead glass 273 0.87 Pyrex glass 373 1.16 Quartz glass 373 1.42 Asphalt 293 0.76 Polystyrene 293 0.12 Polyvinyl chloride 293 0.26 42  Summary Thermal Conductivity of Solids ഥ𝜆 𝐶𝑣 𝑉 Using 𝑘 = to describe k of a solid by phonons, k decreases as 3 temperature increases. Equation applies to electrically insulating substances as oxides Because of the presence of imperfections quantitative prediction is difficult. The less perfect a crystal, the lower the thermal conductivity. 43  Thermal Conductivity of Liquids. Lack of knowledge of their structure – problem Bird, Stewart and Lightfoot proposed the thermal conductivity of liquids at densities away from critical value as 2Τ 𝑁𝑜 3 𝑘 = 2.8𝑘𝛽 𝑉𝑠 ------------------------------------ 18 𝑉 44  Thermal Conductivity of Liquids. Where V = molar volume NO = Avogadro number Vs = speed of sound through the liquid 1Τ 𝐶𝑝 1 2 Vs = ---------------------------- 19 𝐶𝑣 𝜌𝛽 𝑇 β = compressibility 45  Thermal Conductivity of Liquids. Assumption: (1) Molecules in the liquid are arranged in a cubic lattice (2) Energy transfer is via collisions between molecules. The thermal conductivity of ordinary liquids near room temperature (Table below) 46  Table 6.3 Thermal conductivities of various Liquids Substance Temperature, K k, Wm-1K-1 Water 289 0.552 Water 311 0.415 Light oil 289 0.13 Light oil 311 0.14 Benzene 354 0.14 Fluoride salts 755 5.5 Slag 1865 4.0 47 Thermal Boundary Layer - External 48  Thermal Boundary Layer At the solid/fluid interface the fluid temperature is TS Fluid temperature T in the region near the plate, varies from Ts at the plate surface to T∞ in the stream: – heat transfer is by Conduction Region is called - Thermal Boundary Layer Thickness δT is typically – distance from the plate surface at which the dimensionless temperature 𝑇− 𝑇𝑠 𝑇𝑠 − 𝑇 or = 0.99 --------------------------------- 6 𝑇∞ − 𝑇𝑠 𝑇𝑠 − 𝑇∞ 49  Thermal Boundary Layer In practice, it is usually specified that T = Ts @ y = 0----------------------------- 7 𝜕𝑇 =0 @ y = δT --------------------------- 8 𝜕𝑦 This effect of conduction is significant only in the boundary layer 50  Thermal Boundary Layer – Internal Consider a fluid of uniform temperature T∞ entering a circular tube of inner diameter D and uniform wall temperature Ts as shown in Figure below 51  Thermal Boundary Layer Thermal boundary layers develop from opposite sides until they approach the centreline at 𝑧 𝜌𝑉∞ 𝐷 𝐶𝑣 𝜇 ≅ 0.05 ---------------------------------- 9 𝐷 𝜇 𝑘 𝑧 ≅ 0.05 𝑅𝑒𝐷 𝑃𝑟 --------------------------------------- 10 𝐷 Where, 𝑉∞ 𝐷 𝜌𝑉∞ 𝐷 𝒊𝒏𝒆𝒓𝒕𝒊𝒂 𝒇𝒐𝒓𝒄𝒆 𝑅𝑒𝐷 = = = −−− −𝟏𝟏 𝜈 𝜇 𝒗𝒊𝒔𝒄𝒐𝒖𝒔 𝒇𝒐𝒓𝒄𝒆 𝜈 𝐶𝑣 𝜇 𝐦𝐨𝐦𝐞𝐧𝐭𝐮𝐦 𝐝𝐢𝐟𝐟𝐮𝐬𝐢𝐯𝐢𝐭𝐲 𝑃𝑟 = = = − −𝑃𝑟𝑎𝑛𝑑𝑡𝑙 𝑁𝑢𝑚𝑏𝑒𝑟 −− −12 𝛼 𝑘 𝐭𝐡𝐞𝐫𝐦𝐚𝐥 𝐝𝐢𝐟𝐟𝐮𝐬𝐢𝐯𝐢𝐭𝐲 52  Prandtl number 𝜈 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦 𝑃𝑟 = 𝛼 = 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦 = 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦 Peclet number 𝐿𝜈 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑝𝑜𝑟𝑡 𝜌𝐶𝑣 𝑉 𝑇1 −𝑇0 𝑃𝑒𝑇 = 𝑅𝑒 𝑃𝑟 = = = 𝛼 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑝𝑜𝑟𝑡 𝑘 𝑇1 −𝑇0 /𝐿 Froude number 𝜈2 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒 𝜌𝑣 2 Τ𝐿 𝐹𝑟 = = = 𝑔𝐿 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 𝑓𝑜𝑟𝑐𝑒 𝜌𝑔 53  Thermal Boundary Layer V∞ = velocity of fluid D = diameter of tube 𝜇 v = kinematic viscosity = (m2/s) 𝜌 µ = dynamic viscosity 𝜌 = density of fluid 𝑘 α = thermal diffusivity = = (m2/s) 𝜌𝐶𝑣 𝐶𝑣 = specific heat capacity of material k = thermal conductivity 54  Thermal Boundary Layer Equation 9 differs from its equivalent expression in momentum (fluid) transfer by a factor of the Prandtl number 𝑧 𝜌𝑉∞ 𝐷 𝐶𝑣 𝜇 ≅ 0.05 ---------------------------------- 9 𝐷 𝜇 𝑘 For flow through a cross – sectional area A, such as that of a tube, with constant heat capacity, the average temperature is defined as ‫𝐴𝑑𝑉𝑇𝜌 𝐴׭‬ 𝑇𝑎𝑣 = -----------------(14) 𝑚ሶ 55  Thermal Boundary Layer Thermally fully developed temperature profile in a tube is one with a dimensionless temperature as follows 𝑇− 𝑇𝑠 𝑇𝑠 − 𝑇 or which is independent of the axial position 𝑇𝑎𝑣 − 𝑇𝑠 𝑇𝑠 − 𝑇𝑎𝑣 That is 𝜕 𝑇𝑠 − 𝑇 = 0 ------------------------------------------------- 15 𝜕𝑧 𝑇𝑠 − 𝑇𝑎𝑣 56  Heat Transfer Coefficient Consider the thermal boundary layer At the solid/fluid interface heat transfer occurs by conduction since there is no fluid motion. Heat flux across the solid/fluid interface is given by 𝜕𝑇 𝑞𝑦 ห = −𝑘 ቚ ------------------------------------------- 16 𝑦=0 𝜕𝑦 𝑦=0 Temperature gradient is unknown 57  Heat Transfer Coefficient Thus a convenient way to avoid this problem is to introduce the Heat Transfer Coefficient (h), defined as follows 𝜕𝑇 𝑞𝑦 ห −k ቚ 𝑦=0 𝜕𝑦 𝑦=0 h= = ------------------------------------------- 17 𝑇𝑠 − 𝑇∞ 𝑇− 𝑇∞ The absolute values are used to keep h always positive. 58  Heat Transfer Coefficient From equation 17𝜕𝑇 𝑞𝑦 ห − k 𝜕𝑦 ቚ 𝑦=0 𝑦=0 h= = ---------------------------------------- 17 𝑇𝑠 − 𝑇∞ 𝑇− 𝑇∞ 𝑞𝑦 ห = ℎ 𝑇𝑠 − 𝑇∞ ----------------------------------------- 18 𝑦=0 Equation 18 is called Newton’s Law of Cooling. For fluid flow through a tube of inner radius R and wall temperature Ts, a similar equation can be used: 𝜕𝑇 𝑞𝑟 ȁ𝑟=𝑅 − k 𝜕𝑟 ቚ h= = 𝑟=𝑅 ------------------------------------------- 19 𝑇𝑠 − 𝑇𝑎𝑣 𝑇𝑠 − 𝑇𝑎𝑣 59  Heat Transfer Coefficient where Tav is the average fluid temperature over the cross – sectional area πR2. Considering the thermally fully developed region as shown 60  Heat Transfer Coefficient For the case of a constant heat flux 𝑞𝑟 ȁ𝑟=𝑅 , From equation 19 we see that (Ts - Tav) is also constant. 𝜕𝑇 𝑞𝑟 ȁ𝑟=𝑅 −k ቚ 𝜕𝑟 𝑟=𝑅 h= = --------------------------------- 19 𝑇𝑠 − 𝑇𝑎𝑣 𝑇𝑠 − 𝑇𝑎𝑣 Hence heat transfer coefficient h is constant in the thermally fully developed region 61  Heat Transfer Coefficient From equation 19 and equation 15 (thermally fully developed) 𝜕𝑇 𝑞𝑟 ȁ𝑟=𝑅 −k ቚ 𝜕𝑟 𝑟=𝑅 h= = --------------------------------- 19 𝑇𝑠 − 𝑇𝑎𝑣 𝑇𝑠 − 𝑇𝑎𝑣 𝜕 𝑇𝑠 − 𝑇 = 0 ------------------------------------------------- 15 𝜕𝑧 𝑇𝑠 − 𝑇𝑎𝑣 𝜕𝑇𝑠 𝜕𝑇 𝜕𝑇𝑎𝑣 we see that = = ------------------------------- 20 𝜕𝑧 𝜕𝑧 𝜕𝑧 Since TS and Tav are independent of r, 𝜕𝑇 is also independent of r. 𝜕𝑧 62  Heat Transfer Coefficient For a case of constant wall temperature Ts, equation 15 𝜕 𝑇𝑠 − 𝑇 = 0 ------------------------------------------------- 15 𝜕𝑧 𝑇𝑠 − 𝑇𝑎𝑣 𝜕𝑇 can be expanded and solved for to give 𝜕𝑧 𝜕𝑇 𝑇𝑠 − 𝑇 𝜕𝑇𝑎𝑣 = ----------------------------------------------- 21 𝜕𝑧 𝑇𝑠 − 𝑇𝑎𝑣 𝜕𝑧 𝜕𝑇 Since T is dependent on r, is also dependent on r. 𝜕𝑧 63 Example: Flow through a cooling tube Cooling water runs through a copper tubing 0.4 cm in diameter and 20 m long at a mass flow rate of 20 g/s. Calculate the heat transfer coefficient, assuming the inner surface is smooth. µ = 1 x 10-2 g cm-1 s-1 ρ = 1 g/cm3 Cp = 4.2 J g-1 C-1 k = 6 x 10-3 W cm-1 C-1 64 Example: Flow through a cooling tube Solution ℎ𝐷 𝑁𝑢𝐷 = The Nusselt number is the ratio of convective to 𝑘 conductive heat transfer across a boundary. 𝐷𝜌𝑣∞ 𝜋𝐷2 𝜌𝑣𝑎𝑣 4 ReD = = 𝑥 𝜇 4 𝜋𝐷𝜇 4𝑚ሶ = --------------------------- 1 𝜋𝐷𝜇 Where 𝑚ሶ = mass flow rate. As such 4 𝑥 20 𝑔/𝑠 ReD = = 6366 --------------- 2 𝜋 𝑥 0.4 𝑐𝑚 𝑥 1 𝑥 10−2 𝑔𝑐𝑚−1 𝑠 −1 Substituting into equation (73) 𝑓 = 0.79 ln 𝑅𝑒𝐷 − 1.64 −2 65 Example: Flow through a cooling tube 𝑓 = 0.79 ln 6366 − 1.64 −2 𝑓 = 0.036 -------------------------- 3 The Prandtl number 𝐶𝑝 𝜇 4.2 𝐽𝑔−1 𝐶 −1 𝑥 10−2 𝑔𝑐𝑚−1 𝑠 −1 Pr = = = 7 -------------------------- 4 𝑘 6 𝑥 10−3 𝑊𝑐𝑚−1 𝐶 −1𝑓 𝑅𝑒𝐷 −1000 𝑃𝑟 From equation (72) - 𝑁𝑢𝐷 = 8 1 𝑓 ൗ2 2ൗ 1+12.7 8 𝑃𝑟 3 −1 66 Example: Flow through a cooling tube 0.036 6366−1000 𝑥 7 𝑁𝑢𝐷 = 8 1 = 𝟓𝟏. 𝟖 0.036 ൗ2 2ൗ 1+12.7 8 7 3 −1 ℎ𝐷 𝑁𝑢𝐷 𝑘 𝑁𝑢𝐷 = ; ℎ= 𝑘 𝐷 51.8 𝑥 6 𝑥 10−3 𝑊𝑐𝑚−1 𝐶 −1 ℎ= = 𝟎. 𝟕𝟖 𝑾𝒄𝒎−𝟐 𝑪−𝟏 0.4 𝑐𝑚 The Nusselt number is the ratio of convective to conductive heat transfer across a boundary. 67  Series Composite Wall Consider a simple series wall made up of two different materials whose thermal conductivities are k1 and k2 68  Series Composite Wall There is a flow of heat from the gas at temperature Ti through its boundary layer, the composite wall, and the boundary layer of the gas at To. The unidirectional heat rate through the four parts of the entire circuit is constant because steady state prevails. Thus 𝑘1 𝐴 𝑘2 𝐴 Q = 𝐴ℎ𝑖 𝑇𝑖 − 𝑇1 = 𝑇1 − 𝑇2 = 𝑇2 − 𝑇3 = Aℎ0 𝑇3 − 𝑇0 -- 𝐿1 𝐿2 1 A solution based on the four equalities in equation 1 – using the resistance concept 69  Series Composite Wall The flow of heat Q through material, subject to a temperature difference Tj – Tk, is analogous to the flow of current I, as a result of a potential difference Ej – Ek through an electrical conductor. Q Current (I) T j – T k, Potential difference (Ej – Ek) Voltage (V) V=IR R=V/I 70  Series Composite Wall From Ohm’s law for electricity, the thermal resistance Rt for heat flow is 𝑇𝑗 −𝑇𝑘 Rt = ----------------------------- 2 𝑄 Thus, for the composite wall, the four thermal resistances are from (1) 𝑘1 𝐴 𝑘2 𝐴 Q = 𝐴ℎ𝑖 𝑇𝑖 − 𝑇1 = 𝑇1 − 𝑇2 = 𝑇2 − 𝑇3 = Aℎ0 𝑇3 − 𝑇0 -1 𝐿1 𝐿2 1 𝐿1 𝐿2 1 , , , 𝑎𝑛𝑑 𝐴ℎ𝑖 𝑘1 𝐴 𝑘2 𝐴 𝐴ℎ0 71  Series Composite Wall The total resistance for the whole circuit is simply their sum, so that the heat flow is 𝑇𝑖 −𝑇0 Q= 1 𝐿1 𝐿2 1 ------------------------------------ 3 + + + 𝐴ℎ𝑖 𝑘1 𝐴 𝑘2 𝐴 𝐴ℎ0 𝑇𝑖 −𝑇𝑜 𝑄= 1 𝐿𝑖 1 −−− −(3) + σ𝑛 + 𝑖=1𝑘 𝐴 𝐴ℎ𝑜 𝐴ℎ𝑖 𝑖 With the total temperature drop one can calculate the heat flux We can use expression to determine the temperature at any position within the composite wall 72  Composite Wall, Example 1 A furnace wall is constructed of 230 mm of fire brick (k = 1.04 W/mK), 150 mm of insulating brick (k = 0.70 W/mK), 50 mm of glass- wool insulation (k = 0.07 W/mK) and 3 mm thick steel plate (k = 45 W/mK) on the outside. The heat transfer coefficients on the inside and outside surfaces are 28 and 5.7 W/m2K, respectively. The gas temperature inside the furnace is 1365 K and the outside air temperature is 305 K. 1) Calculate the heat flux through the wall. 2) Determine the temperatures at all interfaces 73  Composite Wall, Example 1 Calculate the heat flux through the wall. Determine the temperatures at all interfaces 74  Composite Wall, Example 1 Solution 𝑇𝑗 −𝑇𝑘 𝑅𝑡 = −− −(2) 𝑄 𝑇𝑗 −𝑇𝑘 𝑄= 𝑅𝑡 𝑇𝑗 −𝑇𝑘 𝑞= 𝐴𝑅𝑡 1.𝐴 𝐿1.𝐴 𝐿2.𝐴 𝐿3.𝐴 𝐿4.𝐴 1.𝐴 𝐴𝑅𝑡 = + + + + + 𝐴ℎ𝑖 𝑘1 𝐴 𝑘2 𝐴 𝑘3 𝐴 𝑘4 𝐴 𝐴ℎ0 75  Composite Wall, Example 1 1 0.230 0.15 0.05 0.003 1 G𝑖𝑣𝑖𝑛𝑔 𝐴𝑅𝑡 = + + + + + = 28 1.04 0.70 0.07 45 5.7 𝟏. 𝟑𝟔𝟏 𝑾−𝟏 𝒎𝟐 𝑲 Therefore 1365−305 𝑞= = 778.8 𝑊/𝑚2 1.361 𝑞 778.8 𝑇𝑖 − 𝑇1 = = = 27.8 𝐾 ℎ𝑖 28 T1 = 1365 – 27.8 = 1337.2 K 76  Composite Wall, Example 1 Similarly 𝑞 778.8 0.23 𝑇1 − 𝑇2 = 𝑘 = = 172.2 𝐾 1.04 𝐿1 T2 = 1165.0 K The remaining temperatures are determined in the same manner, yielding T3 = 998.2 K and 𝑇4 − 𝑇5 = 441.8 𝐾 77 Example 2: Conduction through cylindrical composite wall The cylindrical composite wall shown below is made of three different materials, A, B and C, each having its own thermal conductivity that is kA, kB, and kc respectively. The temperatures of the bulk fluids inside and outside the composite walls are Ta and Tb, respectively, And the heat transfer coefficients are h1 and h4, respectively. The overall heat transfer coefficients U1 based on the inner surface and U4 based on the outer surface are defined as 78 Example 2: Conduction through cylindrical composite wall 79 Example 2: Conduction through cylindrical composite wall Qr = 2𝜋𝑟1 𝐿 𝑈1 𝑇𝑎 − 𝑇𝑏 --------------------------------------- (A) Qr = 2𝜋𝑟4 𝐿 𝑈4 𝑇𝑎 − 𝑇𝑏 --------------------------------------- (B) Where Qr is the rate of heat flow through the composite wall L = length of the wall/pipe 𝑇𝑎 − 𝑇𝑏 = overall temperature difference 80 Example 2: Conduction through cylindrical composite wall 81 Example 2: Conduction through cylindrical composite wall 1) Determine at the steady state, U1 as a function of the thermal conductivities and heat transfer coefficient. 2) Then calculate Qr from U1 and (𝑇𝑎 − 𝑇𝑏 ). 82 Example 2: Conduction through cylindrical composite wall Solution Considering material “A” as the control volume Ω with no mass flow through nor heat generation within material A. At steady state equation 29 (overall energy balance eqn) below becomes 𝑑𝐸𝑡 = 𝑚𝐶𝑣 𝑇𝑎𝑣 𝑖𝑛 − 𝑚𝐶𝑣 𝑇𝑎𝑣 𝑜𝑢𝑡 + 𝑄 + 𝑆 ---------------------------- 29 𝑑𝑡 0 = Q = 2𝜋𝑟𝐿𝑞𝑟 𝑟1 − 2𝜋𝑟𝐿𝑞𝑟 𝑟2 --------------------------------- 40 83 Example 2: Conduction through cylindrical composite wall Meaning that 2𝜋𝑟𝐿𝑞𝑟 𝑟1 = 2𝜋𝑟𝐿𝑞𝑟 𝑟2 = 2𝜋𝑟𝐿𝑞𝑟 𝐴 = Q ---------------------- 41 𝑄𝑟 Or rqr = (constant) ------------------------------------ 42 2𝜋𝐿 Repeating for the other two materials, we obtain 𝑄𝑟 𝑟𝑞𝑟 𝐴 = 𝑟𝑞𝑟 𝐵 = 𝑟𝑞𝑟 𝑐 = 𝑟𝑞𝑟 = -------------------------- 43 2𝜋𝐿 84 Example 2: Conduction through cylindrical composite wall From Fourier’s Law of conductivity 𝑑𝑇 𝑑𝑇 𝑑𝑇 𝑄𝑟 −𝑘𝑟 = −𝑘𝑟 = −𝑘𝑟 = ---------------------------- 44 𝑑𝑟 𝐴 𝑑𝑟 𝐵 𝑑𝑟 𝑐 2𝜋𝐿 Integrating over each individual material, (Verify Equations) 𝑄𝑟 𝑟2 𝑇1 − 𝑇2 = 𝑙𝑛 ----------------------------------- 45 2𝜋𝐿𝑘𝐴 𝑟1 𝑄𝑟 𝑟3 𝑇2 − 𝑇3 = 𝑙𝑛 ----------------------------------- 46 2𝜋𝐿𝑘𝐵 𝑟2 85 Example 2: Conduction through cylindrical composite wall 86 Example 2: Conduction through cylindrical composite wall 𝑄𝑟 𝑟4 𝑇3 − 𝑇4 = 𝑙𝑛 ----------------------------------- 47 2𝜋𝐿𝑘𝐶 𝑟3 At the two fluid/solid interfaces, from Newton’s Law of cooling we obtain 𝑞𝑟1 𝑄𝑟 𝑇𝑎 − 𝑇1 = = --------------------------------- 48 ℎ1 2𝜋𝐿𝑟1 ℎ1 𝑄𝑟 Knowing that 𝑟𝑞𝑟 = 2𝜋𝐿 𝑞𝑟4 𝑄𝑟 And 𝑇4 − 𝑇𝑏 = = --------------------------- 49 ℎ 4 2𝜋𝐿𝑟 ℎ 4 4 87 Example 2: Conduction through cylindrical composite wall Adding equation 45 through 49 𝑟 𝑟 𝑟 𝑄 1 𝑙𝑛 𝑟2 𝑙𝑛 𝑟3 𝑙𝑛 𝑟4 1 𝑇𝑎 − 𝑇𝑏 = + 1 + 2 + 3 + --------- 50 2𝜋𝐿 𝑟1 ℎ1 𝑘𝐴 𝑘𝐵 𝑘𝐶 𝑟4 ℎ4 By definition in the problem Qr = 2𝜋𝑟1 𝐿 𝑈1 𝑇𝑎 − 𝑇𝑏 --------(A) 𝑄 𝑇𝑎 − 𝑇𝑏 = ----(A1) 2𝜋𝑟1 𝐿𝑈1 88 Example 2: Conduction through cylindrical composite wall 𝑄 𝑇𝑎 − 𝑇𝑏 = ---(A1) 2𝜋𝑟1 𝐿𝑈1 𝑟 𝑟 𝑟 𝑄 1 𝑙𝑛 𝑟2 𝑙𝑛 3 𝑙𝑛 4 1 𝑟2 𝑟3 𝑇𝑎 − 𝑇𝑏 = + 1 + + + ---(50) 2𝜋𝐿 𝑟1 ℎ1 𝑘𝐴 𝑘𝐵 𝑘𝐶 𝑟4 ℎ4 Equating Eqn (A1) to (50) 1 𝑈1 𝑟1 = 𝑟 𝑟 𝑟 1 𝑙𝑛 𝑟2 𝑙𝑛 𝑟3 𝑙𝑛 𝑟4 1 + 1 + 𝑘 2 + 𝑘 3 +𝑟 ℎ 𝑟1 ℎ1 𝑘𝐴 𝐵 𝐶 4 4 1 1 𝑈1 = 𝑟2 𝑟 𝑟 𝑟1 1 𝑙𝑛 𝑙𝑛 𝑟3 𝑙𝑛 𝑟4 1 𝑟1 2 3 + + + + 𝑟1 ℎ1 𝑘𝐴 𝑘𝐵 𝑘𝐶 𝑟4 ℎ4 89 Example 2: Conduction through cylindrical composite wall Qr = 2𝜋𝑟1 𝐿 𝑈1 𝑇𝑎 − 𝑇𝑏 = 2𝜋𝑟4 𝐿 𝑈4 𝑇𝑎 − 𝑇𝑏 Divide through by 2𝜋𝐿 𝑇𝑎 − 𝑇𝑏 to give 𝑈1 𝑟1 = 𝑈4 𝑟4 𝑈1 𝑟1 𝑈4 = 𝑟4 1 1 1 𝑈4 = = 𝑟2 𝑟 𝑟 𝑟4 𝑟4 1 𝑙𝑛 𝑙𝑛 𝑟3 𝑙𝑛 𝑟4 1 𝑟1 𝑟1 ℎ1 + 𝑘𝐴 + 𝑘 2 + 𝑘 3 +𝑟 ℎ 𝐵 𝐶 4 4 90 Example 2: Conduction through cylindrical composite wall 1 𝑈4 = 𝑟4 Substituting equation “A” into equation 50 𝑟2 𝑟3 𝑟4 −1 1 𝑙𝑛 𝑙𝑛 𝑙𝑛 1 𝑟1 𝑟2 𝑟3 𝑈1 = 𝑟1−1 + + + + 𝑟1 ℎ1 𝑘𝐴 𝑘𝐵 𝑘𝐶 𝑟4 ℎ4 From equation “B”, U4 can be found since 𝑈1 𝑟1 𝑈4 = 𝑟4 91 Example 3: Insulation Selection Example 3 As part of a continuous annealing process, a rod passes through a cylindrical furnace chamber 101 mm inside diameter and 15.2 m long. The inside surface temperature of the furnace wall under operating conditions is predicted to be about 920 K and the outside surface temperature at about 310 K. If it is decided that a heat loss of 73 kW is an acceptable figure, then which of the following insulations would you use? 92 Example 3: Insulation Selection k, Wm-1K-1 Cost, ($) per m3 Insulation A 0.70 350 Insulation B 0.35 880 Solution 93 Example 3: Insulation Selection Heat flow in radial direction in cylindrical coordinates 𝜕𝑇 Heat flux, qr = - k 𝜕𝑟 𝑄𝑟 Heat flow, Q = 𝑞𝑟 2𝜋𝑟𝐿 𝑟𝑞𝑟 = 2𝜋𝐿 From equation 44 and 45 as having integrated over a controlled volume From Fourier’s Law of conductivity 𝑑𝑇 𝑑𝑇 𝑑𝑇 𝑄𝑟 −𝑘𝑟 = −𝑘𝑟 = −𝑘𝑟 = ---------------------------- 44 𝑑𝑟 𝐴 𝑑𝑟 𝐵 𝑑𝑟 𝑐 2𝜋𝐿 94 Example 3: Insulation Selection 𝑄𝑟 𝑟2 𝑇1 − 𝑇2 = 𝑙𝑛 ----------------------------------- 45 2𝜋𝐿𝑘𝐴 𝑟1 2𝜋𝐿𝑘 Qr = 𝑟 𝑇1 − 𝑇2 𝑙𝑛𝑟2 1 𝑟2 2𝜋𝐿𝑘 𝑙𝑛 = (𝑇1 − 𝑇2 ) ----------------------------- A 𝑟1 𝑄 For Material “A” 𝑟2 2𝜋 0.70 15.2 920−310 𝑙𝑛 = = 0.559 𝑟1 73 𝑥 103 95 Example 3: Insulation Selection So that with r1 = 50.5 mm r2 = 88.3 mm Similarly for Material “B” using ratio of conductivities: 𝑟2 2𝜋𝐿 𝑇1 −𝑇2 𝑙𝑛 = 𝑘𝐴 𝑟1 𝐴 𝑄 𝑟2 𝑟2 𝑙𝑛 = 𝐶 𝑘𝐴 Implying 𝑙𝑛 ∝ 𝑘𝐴 𝑟1 𝐴 𝑟1 𝐴 96 Example 3: Insulation Selection Hence ratio of conductivities gives 𝑟 𝑙𝑛 2 𝑟1 𝐴 𝑘𝐴 𝑟2 = 𝑙𝑛 𝑘𝐵 𝑟1 𝐵 Thus 𝑟 𝑘𝐵 𝑙𝑛 𝑟2 𝑟2 1 𝐴 𝑙𝑛 = 𝑟1 𝐵 𝑘𝐴 97 Example 3: Insulation Selection 𝑟2 0.35 𝑙𝑛 = 0.559 = 0.28 𝑟1 𝐵 0.70 So that r2 = 66.8 mm Calculating the volume of insulation and cost (πr2L x unit cost) 98 Example 3: Insulation Selection 1𝑚2 Cost A = 𝜋 88.32 − 50.52 𝑚𝑚2 10002 𝑚𝑚2 350$ = 15.2 𝑚 𝑚3 = $ 87.69 Cost B = $ 80.34 Choice is B. 99  DERIVATION OF OVERALL ENERGY BALANCE EQUATION Consider an arbitrary stationary control volume Ω bounded by surface A as shown in Fig 6A below 100  DERIVATION OF OVERALL ENERGY BALANCE EQUATION The control surface A can be considered to consist of three different regions: Ain for the region where the fluid enters the control volume Aout, where the fluid leaves Awall, where the fluid is in contact with a wall In other words A = Ain + Aout + Awall -------------------------------- 22 101  DERIVATION OF OVERALL ENERGY BALANCE EQUATION Consider the outward heat transfer, for example, conduction rate through dA = q.n dA shown in Figure 6B Since n points outward 102 DERIVATION OF OVERALL ENERGY BALANCE EQUATION Inward heat transfer rate through area dA, dQ = - q.n dA ------------ 23 Applying the following energy conservation law to the control volume (Equation here, PTO) This equation is in fact, the first law of thermodynamics written for an open system under the unsteady state condition. In Terms 1 through 3 – the energy includes the thermal, kinetic and potential energies. 103 DERIVATION OF OVERALL ENERGY BALANCE EQUATION 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛 𝑏𝑦 𝑚𝑎𝑠𝑠 𝑜𝑢𝑡 𝑏𝑦 𝑚𝑎𝑠𝑠 = - + 𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑛𝑓𝑙𝑜𝑤 𝑜𝑢𝑓𝑙𝑜𝑤 1 2 3 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑜𝑡ℎ𝑒𝑟 ℎ𝑒𝑎𝑡 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑡𝑜 𝑠𝑦𝑠𝑡𝑒𝑚 𝑏𝑦 𝑠𝑦𝑠𝑡𝑒𝑚 𝑜𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 - + 𝑓𝑟𝑜𝑚 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 𝑖𝑛 𝑠𝑦𝑠𝑡𝑒𝑚 4 5 6 ---------------------------------------- 24 104 DERIVATION OF OVERALL ENERGY BALANCE EQUATION The thermal, kinetic and potential energy unit per unit mass of the fluid are: Thermal : CvT 𝑉2 Kinetic : 2 Potential : ∅ Cv = specific heat T = temperature The Total energy per unit mass of the fluid is given as 1 eT = CvT + 𝑉 2 + ∅ ----------------------------------------- 25 2 105  DERIVATION OF OVERALL ENERGY BALANCE EQUATION 𝜕 ‫𝑑 𝑡𝑒𝜌 ׮‬Ω = − ‫𝑉 𝑡𝑒𝜌 ׭‬. 𝑛𝑑𝐴 − ‫𝑞 𝐴׭‬. 𝑛𝑑𝐴 − 𝜕𝑡 ‫𝑣𝑝 𝐴׭‬. 𝑛𝑑𝐴 − ‫𝜏 𝐴׭‬. 𝑣 𝑛𝑑𝐴 + ‫׮‬Ω 𝑠𝑑Ω − 𝑊𝑠 ----------26 In materials processing problems, the kinetic and potential energies are negligible as compared to the thermal energy. Furthermore, the pressure, viscous and shaft work are usually negligible or even absent. As such equation 26 reduces to 106  DERIVATION OF OVERALL ENERGY BALANCE EQUATION 𝜕 ‫𝑑𝑇 𝑉𝐶𝜌 ׮‬Ω = − ‫𝑉𝑇 𝑉𝐶𝜌 ׭‬. 𝑛𝑑𝐴 − ‫𝑞 𝐴׭‬. 𝑛𝑑𝐴 + ‫׮‬Ω 𝑠𝑑Ω ----- 27 𝜕𝑡 Integral energy balance equation For the overall form of terms 1 through 6 into equation 24 (Rate equation) and neglecting Kinetic energy Pressure work Viscous work Potential energy Shaft work 107  DERIVATION OF OVERALL ENERGY BALANCE EQUATION 𝑑𝐸𝑡 = ‫𝐴𝑑𝑉𝑇 𝑉𝐶𝜌 ׭‬ − ‫𝐴𝑑𝑉𝑇 𝑉𝐶𝜌 ׭‬ + 𝑄 + 𝑆 −−−−− 28 𝑑𝑡 𝑖𝑛 𝑜𝑢𝑡 Where ET is the thermal energy in the control volume, ‫׮‬Ω 𝜌𝐶𝑉 𝑇𝑑Ω. Substituting equation (14) ‫𝐴𝑑𝑉𝑇𝜌 𝐴׭‬ 𝑇𝑎𝑣 = -----------------(14) 𝑚ሶ into equation (28) And assuming Cv is constant, we obtain: 108  DERIVATION OF OVERALL ENERGY BALANCE EQUATION 𝑑𝐸𝑡 = 𝑚𝐶𝑣 𝑇𝑎𝑣 𝑖𝑛 − 𝑚𝐶𝑣 𝑇𝑎𝑣 𝑜𝑢𝑡 + 𝑄 + 𝑆 ------------ 29 𝑑𝑡 Overall Energy Balance Equation Where ET = thermal energy in the control volume = 𝜌𝐶𝑉 𝑇Ω ET = 𝑚𝐶𝑣 𝑇 if 𝜌𝐶𝑉 𝑇 is uniform in Ω 109  DERIVATION OF OVERALL ENERGY BALANCE EQUATION 𝑚ሶ = mass flow rate at inlet or outlet = 𝜌𝑉𝑎𝑣 𝐴 Q = heat transfer rate into control volume from surrounding (other than the two 𝑚𝐶𝑣 𝑇𝑎𝑣 terms), that is, by conduction S = heat generation rate in the control volume = s Ω if uniform s. 110  BERNOULLI EQUATION Considering a steady – state isothermal flow of an inviscid incompressible fluid Term (1) = 0 Without heat generation Term (6) = 0 No heat conduction Term (4) = 0 No shaft work Term (5a) = 0 No viscous work Term (5c) = 0 1 Substituting eT = CvT + 𝑉 2 + ∅ into equation 26 and assuming 2 uniform properties over the cross-sectional area A, we obtain 111  BERNOULLI EQUATION 𝜕 ‫𝑑 𝑡𝑒𝜌 ׮‬Ω = − ‫𝑉 𝑡𝑒𝜌 ׭‬. 𝑛𝑑𝐴 − ‫𝑞 𝐴׭‬. 𝑛𝑑𝐴 − ‫𝑣𝑝 𝐴׭‬. 𝑛𝑑𝐴 − 𝜕𝑡 ‫𝜏 𝐴׭‬. 𝑣 𝑛𝑑𝐴 + ‫׮‬Ω 𝑠𝑑Ω − 𝑊𝑠 ---------------------------26 𝑃 0 = −𝜌 ‫𝐴׭‬ 𝑒𝑡 + 𝑉. 𝑛𝑑𝐴 𝜌 1 2 𝑃 1 2 𝑃 0=𝜌 𝐶𝑣 𝑇 + 𝑉 +∅+ 𝑉𝐴 − 𝜌 𝐶𝑣 𝑇 + 𝑉 +∅+ 𝑉𝐴 - 2 𝜌 1 2 𝜌 2 ---------------------- 30 112  BERNOULLI EQUATION Since T1 = T2 and (ρvA)1 = (ρvA)2 Equation 30 reduces to 1 2 𝑃1 1 2 𝑃2 𝑉1 + ∅1 + = 𝑉2 + ∅2 + ------------------------- 31 2 𝜌 2 𝜌 If the z direction is taken vertically upwards, ∅ = 𝑔𝑧 Where g is the gravitational acceleration. As such equation 31 on multiplying by ρ becomes 113  BERNOULLI EQUATION 1 1 𝜌𝑉12 + 𝜌𝑔𝑧1 + 𝑝1 = 𝜌𝑉22 + 𝜌𝑔𝑧2 + 𝑝2 ----------- 32 2 2 Or simply 1 𝜌𝑉 2 + 𝜌𝑔𝑧 + 𝑝 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ---------------------------- 33 2 This is the Bernoulli equation. 114  Example (1) Fluid Temperature in a Mixing Tank A mixing tank receives fluid from two inlets and discharges it through one outlet. The mass of the fluid in the tank is M and its temperature T is uniform as a result of stirring. Both the tank wall and the stirrer are very light in mass and a very good insulator. The initial fluid mass and temperature in the tank are M0 and T0 respectively. 115 Example (1) Fluid Temperature in a Mixing Tank The mass flow rates and temperatures are respectively m1 and T1 for inlet 1, m2 and T2 for inlet 2, and m3 and T for the outlet. 116 Example (1) Fluid Temperature in a Mixing Tank Determine the fluid temperature in the tank as a function of time Solution Assumptions (1) Since (a) stirrer is very light in mass (b) tank is a very good thermal insulator We can ignore them in the heat flow analysis (2) (a) Kinetic energy, (b) Potential energy, (c) Pressure work of the fluid, (d) work done on the fluid by the stirrer are negligible 117 Example (1) Fluid Temperature in a Mixing Tank Compared with the thermal energy of the fluid. (3) Heat transfer due to conduction is negligible as compared to that due to convection at the inlet and outlet. Since temperature is uniform, the thermal energy in the control volume is 𝐸𝑇 = 𝑀𝐶𝑣 𝑇 ---------------------------------------------- 34 118 Example (1) Fluid Temperature in a Mixing Tank From overall energy balance equation (Eqn 29) with Q = S = 0 𝑑𝐸𝑡 = 𝑚𝐶 ሶ 𝑣 𝑇𝑎𝑣 𝑖𝑛 − 𝑚𝐶 ሶ 𝑣 𝑇𝑎𝑣 𝑜𝑢𝑡 + 𝑄 + 𝑆 ------------ 29 𝑑𝑡 𝑑 𝑀𝐶𝑣 𝑇 = 𝑚ሶ 1 𝐶𝑣 𝑇1 + 𝑚ሶ 2 𝐶𝑣 𝑇2 − 𝑚ሶ 3 𝐶𝑣 𝑇 -----------------35 𝑑𝑡 Taking a material balance around the control volume 𝑑𝑀 = 𝑚ሶ 1 + 𝑚ሶ 2 − 𝑚ሶ 3 @ t = 0 , M = Mo 𝑑𝑡 𝑀 𝑡 As such ‫𝑀𝑑 𝑀׬‬ = ‫׬‬0 𝑚ሶ 1 + 𝑚ሶ 2 − 𝑚ሶ 3 𝑑𝑡 0 119 Example (1) Fluid Temperature in a Mixing Tank Since d(xy) = xdy + ydx 𝑑𝑇 𝑚𝑡 + 𝑀0 + 𝑚𝑇 = 𝑚1 𝑇1 + 𝑚2 𝑇2 − 𝑚3 𝑇 ------------ 38 𝑑𝑡 So 𝑑𝑇 − 𝑚+𝑚3 𝑇− 𝑚1 𝑇1 +𝑚2 𝑇2 / 𝑚+𝑚3 = 𝑚 𝑑𝑡 𝑚 𝑡+ 𝑚0 120 Example (1) Fluid Temperature in a Mixing Tank And so M = 𝑚ሶ 1 + 𝑚ሶ 2 − 𝑚ሶ 3 𝑡 + 𝑀0 𝑀 = 𝑚𝑡 ሶ + 𝑀0 -------------------------------------- 36 Where 𝑚ሶ = 𝑚ሶ 1 + 𝑚ሶ 2 − 𝑚ሶ 3 Substituting equation 36 into 35 and dividing by Cv, we obtain 𝑑 𝑚𝑡+𝑀0 𝑇 = 𝑚ሶ 1 𝑇1 + 𝑚ሶ 2 𝑇2 − 𝑚ሶ 3 𝑇 ------------------------------------- 37 𝑑𝑡 121 Example (1) Fluid Temperature in a Mixing Tank Initial condition At t = 0 , T = T0 𝑑∅ 𝑎 ∅−𝑏 From = (Appendix A: Case B, p619) 𝑑𝑈 𝑐 𝑢−𝑑 With boundary (or initial) condition ∅ = ∅0 @ 𝑈 = 𝑈0 Solution is ∅−𝑏 𝑐 𝑈−𝑑 𝑎 = ∅0 −𝑏 𝑈0 −𝑑 122 Example (1) Fluid Temperature in a Mixing Tank Thus, we have 𝑇− 𝑚1 𝑇1 + 𝑚2 𝑇2 / 𝑚+𝑚3 𝑚 𝑡+ 𝑚0 Τ𝑚 −(𝑚+𝑚3 ) = 𝑇0 − 𝑚1 𝑇1 + 𝑚2 𝑇2 / 𝑚+𝑚3 𝑡0 + 𝑚0 Τ𝑚 123  Differential Energy – Balance Equation In materials processing the kinetic and potential energy are negligible as compared to the thermal energy, Total energy per unit mass 𝑒𝑡 = 𝐶𝑣 𝑇. Pressure, viscosity and shaft work are usually negligible. The surface integrals in the integral energy balance (equation 27) can be converted into volume integrals. From Gauss’ divergence theorem ‫𝑣 𝐴׭‬. 𝑛𝑑𝐴 = ‫𝛻 𝛺׮‬. 𝑣𝑑𝛺 124 Differential Energy – Balance Equation 𝜕 ‫𝑑𝑇 𝑉𝐶𝜌 ׮‬Ω = − ‫𝑣𝑇 𝑉𝐶𝜌 ׭‬. 𝑛𝑑𝐴 − ‫𝑞 𝐴׭‬. 𝑛𝑑𝐴 + ‫׮‬Ω 𝑠𝑑Ω ------ 27 𝜕𝑡 (1) (2) (3) (4) From the integral balance equation (27) and Gauss theorem Term (2) becomes ‫𝑣𝑇 𝑣𝐶𝜌 𝐴׭‬. 𝑛𝑑𝐴 = ‫𝛻 𝛺׮‬. 𝜌𝐶𝑣 𝑇𝑣 𝑑𝛺 ------------------------------ 51 125 Differential Energy – Balance Equation Term (3) also becomes And ‫𝑞 𝐴׭‬. 𝑛𝑑𝐴 = ‫𝛻 𝛺׮‬. 𝑞𝑑𝛺 ---------------------------- 52 Substituting equations (51) and (52) into (27) we have 𝜕 ‫׮‬ 𝜌𝐶𝑉 𝑇𝑑Ω + ‫𝛻 𝛺׮‬. 𝜌𝐶𝑣 𝑇𝑣 𝑑𝛺 + ‫𝛻 𝛺׮‬. 𝑞𝑑𝛺 − ‫׮‬Ω 𝑠𝑑Ω = 0 𝜕𝑡 ----------------- 53 126 Differential Energy – Balance Equation 𝜕 If the control volume Ω does not change with time, in equation (53) 𝜕𝑡 can be moved inside the integration sign: 𝜕 ‫׮‬ 𝜌𝐶𝑉 𝑇 + 𝛻. 𝜌𝐶𝑣 𝑇𝑣 + 𝛻. 𝑞 − 𝑠 𝑑𝛺 = 0 --------------------- 54 𝜕𝑡 The integrand, which is continuous, must be zero everywhere since the equation must hold for any arbitrary region Ω 127 Differential Energy – Balance Equation 𝜕 𝜌𝐶𝑉 𝑇 + 𝛻. 𝜌𝐶𝑣 𝑇𝑣 + 𝛻. 𝑞 − 𝑠 = 0 -------------------------------- 55 𝜕𝑡 (1) (2) (3) (4) The first two terms in the equation can be expanded In the case of the second term by using the idea of the divergence of products we have, from appendix A, Terms (1) and (2) become 𝜕 𝜕𝜌 𝜌 𝐶𝑣 𝑇 + 𝐶𝑣 𝑇 + 𝐶𝑣 𝑇 𝛻. 𝜌𝑣 + 𝜌𝑣. 𝛻 𝐶𝑣 𝑇 ----(A) 𝜕𝑡 𝜕𝑡 128 Differential Energy – Balance Equation Hence Eqn (A) becomes 𝜕 𝜕𝜌 𝜌 𝐶𝑣 𝑇 + 𝐶𝑣 𝑇 + 𝛻. 𝜌𝑣 + 𝜌𝑣. 𝛻 𝐶𝑣 𝑇 -----(B) 𝜕𝑡 𝜕𝑡 Which reduces to 𝜕 𝜌 𝐶𝑣 𝑇 + 𝜌𝑣. 𝛻 𝐶𝑣 𝑇 --------------------------------- 56 𝜕𝑡 Since from the continuity equation 𝜕𝜌 + 𝛻. 𝜌𝑣 = 0 𝜕𝑡 129 Differential Energy – Balance Equation Substituting 𝑞 = −𝑘𝛻𝑇 ----------------- (5) and Substituting equation (56) into (55), we have Eqn (57) 𝜕 𝜌 𝐶𝑣 𝑇 + 𝜌𝑣. 𝛻 𝐶𝑣 𝑇 --------------------------------- 56 𝜕𝑡 𝜕 𝜌𝐶𝑉 𝑇 + 𝛻. 𝜌𝐶𝑣 𝑇𝑣 + 𝛻. 𝑞 − 𝑠 = 0 ------------ 55 𝜕𝑡 𝜕 𝜌 𝐶𝑣 𝑇 + 𝜌𝑣. 𝛻 𝐶𝑣 𝑇 = 𝛻. 𝑘𝛻𝑇 + 𝑠 ------------ 57 𝜕𝑡 130 Differential Energy – Balance Equation Assuming constant Cv and k we have from 57 𝝏 𝝆 𝑪𝒗 𝑻 + 𝝆𝒗. 𝛁 𝑪𝒗 𝑻 = 𝜵. 𝒌𝜵𝑻 + 𝒔 ------------ 57 𝝏𝒕 𝝏𝑻 𝝆𝑪𝒗 + 𝒗. 𝜵𝑻 = 𝒌𝜵𝟐 𝑻 + 𝑺 --------------------------- 58 𝝏𝒕 Equation 57 or 58 is the Differential Energy Balance Equation or the Equation of Energy. 131 Differential Energy – Balance Equation Rectangular Coordinates 𝜕𝑇 𝜕𝑇 𝜕𝑇 𝜕𝑇 𝜕2 𝑇 𝜕2 𝑇 𝜕2 𝑇 𝜌𝐶𝑣 + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 =𝑘 + + + 𝑆 ------ A 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 Cylindrical Coordinates 𝜕𝑇 𝜕𝑇 𝑣𝜃 𝜕𝑇 𝜕𝑇 1 𝜕 𝜕𝑇 1 𝜕2 𝑇 𝜕2 𝑇 𝜌𝐶𝑣 + 𝑣𝑟 + + 𝑣𝑧 =𝑘 𝑟 + + + 𝜕𝑡 𝜕𝑟 𝑟 𝜕𝜃 𝜕𝑧 𝑟 𝜕𝑟 𝜕𝑟 𝑟 2 𝜕𝜃 2 𝜕𝑧 2 𝑆 ---------------- B 132 Differential Energy – Balance Equation Spherical Coordinates 𝜕𝑇 𝜕𝑇 𝑣𝜃 𝜕𝑇 𝑣∅ 𝜕𝑇 1 𝜕 𝜕𝑇 𝜌𝐶𝑣 + 𝑣𝑟 + + = 𝑘ቂ 𝑟 + 𝜕𝑡 𝜕𝑟 𝑟 𝜕𝜃 𝑟 sin 𝜃 𝜕∅ 𝑟 𝜕𝑟 𝜕𝑟 1 𝜕 𝜕𝑇 1 𝜕2 𝑇 2 sin 𝜃 + 2 2 2 ቃ+ 𝑆 ------------------- C 𝑟 sin 𝜃 𝜕𝜃 𝜕𝜃 𝑟 sin 𝜃 𝜕∅ 133  Dimensionless Form Dimensionless forms are used to make solutions more general Reynolds number 𝜌𝑣2 𝐿𝑉 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒 Re = = = 𝐿 𝜇𝑣 ---------------------------------- 59 𝒱 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒 𝐿2 Froude number 𝑣2 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒 𝜌𝑣 2 Τ𝐿 Fr = = = ----------------------------- 60 𝑔𝐿 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 𝑓𝑜𝑟𝑐𝑒 𝜌𝑔 134 Dimensionless Form Prandtl number 𝑣 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦 Pr = = ------------------------------- 61 ∝ 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦 Peclet number 𝐿𝑣 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑝𝑜𝑟𝑡 𝜌𝐶𝑣 𝑉 𝑇1 −𝑇0 PeT = Re Pr = = = -------- 62 ∝ 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑝𝑜𝑟𝑡 𝑘 𝑇1 −𝑇0 Τ𝐿 135  Solution Procedure The purpose of the equation of energy, is to determine the temperature distribution. The following steps can be followed: (1) Choose a coordinate system that best describes the physical system geometrically. (2) Choose the equation of energy from a Table for the coordinate system. (3) Eliminate the zero terms from the equation of energy. 136 Solution Procedure (4) Substitute the velocity distribution, if it is available and temperature – independent, into the equation. (5) Set up the boundary and/or initial conditions (6) Solve the equation of energy, subject to the boundary and/or initial conditions in step 5, for the temperature distribution. Standard solutions to the differential equations are available. 137 Solution Procedure (7) Check to see if the temperature distribution satisfies the boundary and/or initial conditions in step 5. 138 Commonly Encountered Heat Flow Boundary Conditions Rectangular Coordinates (1) Plane of symmetry 139 Commonly Encountered Heat Flow Boundary Conditions (2) Constant surface Temperature 140 Commonly Encountered Heat Flow Boundary Conditions (3) Adiabatic or Insulated Surface 141 Commonly Encountered Heat Flow Boundary Conditions (4) Constant Surface Heat Flux 142 Commonly Encountered Heat Flow Boundary Conditions (5) Convection Exchange 143 Commonly Encountered Heat Flow Boundary Conditions The free surface of a fluid may be exposed to a gas of bulk temperature Tf Or the surface of a solid may be exposed to a gas or liquid of bulk temperature Tf. From Newton’s law of cooling 𝜕𝑇 𝑞 = −𝑘 = ℎ 𝑇 − 𝑇𝑓 ----------------------------- 63 𝜕𝑦 144 Commonly Encountered Heat Flow Boundary Conditions (6) Interface I/II :S/S, L/S, G/S, L/L, G/L 145 Commonly Encountered Heat Flow Boundary Conditions The temperature and the heat flux are the same on both sides of the interface. This is for a case when temperature distribution is to be found in both phases. 146 Commonly Encountered Heat Flow Boundary Conditions (7) Solid/solid contact 147 Commonly Encountered Heat Flow Boundary Conditions If there is a small gap between two solids in contact with each other Heat flux across the gap can be expressed as 𝑞 = ℎ 𝑇𝐼 − 𝑇𝐼𝐼 Where h is the heat transfer coefficient, TI = temperature in solid I at the gap TII = the temperature of solid II at the gap. This is for a case where the temperature fields in both I and II are being determined. Eg: gas-filled window pane 148 Scenarios in Heat Conduction 3 Scenarios Scenario 1 Infinite thermal conductivity in solids, k→∞ (Lumped Parameter Analysis) Uniform temperature in the solid body (No temperature gradient in solid) Biot number < 0.1 Convection resistance > Conduction resistance 149 Scenarios in Heat Conduction Typical examples I. Cooling of a small metal casting II. Cooling of a billet in quenching bath after removal from the furnace III. Heating or cooling of a fine thermocouple wire due to change in ambient temperature. 150 Scenarios in Heat Conduction Scenario 2 Transient Heat Conduction In Solids with Finite Conduction and Convective Resistance (0.1 < Bi < 100) Conduction and convection resistances are almost of equal importance Use Heisler charts 151 Scenarios in Heat Conduction Scenario 3 Transient Heat Conduction in Infinite (Semi Infinite) Thick Solids Bi →∞ Use Error Function solution 152 Lumped Parameter Analysis (Bi < 0.1) For the lumped parameter analysis. ℎ𝐿 𝐵𝑖𝑜𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝐵𝑖 = < 0.1 𝑘 Presumed the solid has infinitely large thermal conductivity. Internal conduction resistance is so small that heat flow to or from the solid is controlled by the convective resistance Transient response can be done by relating the rate of change of internal energy with the convective heat exchange at the surface 153 Lumped Parameter Analysis (Bi < 0.1) For a body with Surface area = A Volume = V Density = ρ Thermal conductivity = k Specific heat capacity = Cv Initial temperature = Ti Ambient temperature = Ta 154 Lumped Parameter Analysis (Bi < 0.1) Heat exchange is given as 𝑑𝑇 −𝜌𝑉𝐶𝑣 = ℎ𝐴 𝑇 − 𝑇𝑎 −−− −(1) 𝑑𝑡 Separating the variables 𝑑𝑇 ℎ𝐴 ‫׬‬ = − ‫׬‬ 𝑑𝑡 𝑇−𝑇𝑎 𝜌𝑉𝐶𝑣 Evaluating we have 𝑇−𝑇𝑎 ℎ𝐴𝑡 𝑙𝑛 = − 𝑇𝑖 −𝑇𝑎 𝜌𝑉𝐶𝑣 155 Lumped Parameter Analysis (Bi < 0.1) Or 𝑻−𝑻𝒂 𝒉𝑨𝒕 = 𝒆𝒙𝒑 − 𝑻𝒊 −𝑻𝒂 𝝆𝑽𝑪𝒗 The dimensionless argument of the exponential can be rewritten as ℎ𝐴𝑡 ℎ𝑉 𝐴2 𝑘𝑡 = 𝜌𝑉𝐶𝑣 𝑘𝐴 𝜌𝑉 2 𝐶𝑣 ℎ𝐴𝑡 ℎ𝑙 𝛼𝑡 = 𝜌𝑉𝐶𝑣 𝑘 𝑙2 ℎ𝐴𝑡 = 𝐵𝑖 𝐹𝑜 = 𝐵𝑖𝑜𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 (𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝑛𝑢𝑚𝑏𝑒𝑟) 𝜌𝑉𝐶𝑣 156 Lumped Parameter Analysis (Bi < 0.1) l is a characteristic length given by the ratio of the volume of the solid to its surface area. For simple geometrical shapes, characteristic lengths are given as Sphere 4 3 𝜋𝑅 𝑅 𝑙= 3 = 4𝜋𝑅 2 3 Cylinder 𝜋𝑅 2 𝐿 𝑅 𝑙= = 2𝜋𝑅𝐿 2 157 Lumped Parameter Analysis (Bi < 0.1) Cube 𝐿3 𝐿 𝑙= = 6𝐿2 6 For a flat plate (thickness δ, breadth b, height h) heat exchange occurs from both sides. Area exposed for heat transfer is 2bh. Characteristic length therefore is 𝛿𝑏ℎ 𝛿 𝑙= = = ℎ𝑎𝑙𝑓 𝑡ℎ𝑒 𝑝𝑙𝑎𝑡𝑒 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 2𝑏ℎ 2 158 Lumped Parameter Analysis (Bi < 0.1) Fourier number, Fo 𝛼𝑡 𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝑛𝑢𝑚𝑏𝑒𝑟 = 𝑙2 It signifies the extent of heating or cooling effect through a solid. For instance, a large time t is required to obtain a significant temperature change for small values of (α/l2) 𝑘 𝛼= 𝜌𝐶𝑣 ℎ𝐿 𝐵𝑖𝑜𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝐵𝑖 = 𝑘 159 Lumped Parameter Analysis (Bi < 0.1) Biot number, Bi ℎ𝑙 𝐵𝑖 = 𝑘 It gives an indication of the ratio of internal (conduction) resistance to the surface (convection) resistance. Small value of Bi implies the system has small conduction (internal) resistance. This implies a small temperature gradient or existence of practically uniform temperature within the system. 160 Lumped Parameter Analysis (Bi < 0.1) Convective resistance then predominates, and the convective heat exchange controls the transient phenomenon. Small Biot number can hold with (i) thin plates and (ii) solid with large thermal conductivity k and (iii) situation with small heat transfer coefficient h. Acceptable value of Biot number for lumped parameter analysis is Bi