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MECH4411 Heat Transfer L. Q. Wang Professor 7-3 Haking Wong Building 3917-7908 [email protected] Reference 1. 2. Holman J.P., Heat Transfer, 10th Edition, McGraw-Hill, 2010. Cengel Y. A., Heat Transfer: A Practical Approach, 2nd Edition, McGraw-Hill, 2003. 1 Chapter 1 Introduction Heat In thermodynamics, heat is defined as a form of energy which is transferred from one body (at a high temperature) to another body (at a lower temperature) due to the temperature difference between the bodies. Heat Transfer is to predict and control the rate at which heat is transferred. To find the relation between temperature difference and the heat transfer rate forms one major purpose of heat transfer study. 1-1 Three Ways of Heat Transfer 1. Heat Conduction (Conduction) The transfer of heat from one part of a substance to another part of the same substance, or from one substance to another in physical contact with it, without macroscopic relative motion of the medium. T2 T1 T1 T2 The transfer of heat through solid bodies is usually by conduction alone. 2. Heat Convection The transfer of heat within substances in macroscopic relative motion such as within a fluid by the mixing of one portion of the fluid with another Fluid T1 T2 Natural Convection Movement due to density variation in a force field (Gravitational, Centrifugal etc.) 2 T2 T1 Air g Forced Convection Movement due to external mechanical means Tw Water Air Fan Pump Heat transfer from a solid surface to a liquid or gas takes place partly by conduction and partly by convection. Whenever there is an appreciable movement of the gas or liquid, the heat transfer by conduction in the gas or liquid becomes negligibly small compared with the heat transfer by convection. However there is always a thin boundary layer of fluid on a surface (due to the viscosity of the fluid) and through this thin film the heat is transferred by conduction. 3. Radiation 3 1-2 Heat Conduction (Fourier Law) δq Fourier Law of Conduction δq = −kdA ∇T dA δq - heat transfer rate W(J/s), kW dA – heat transfer area m2 T – temperature k, °C ∇T – temperature gradient k/m, °C/m k – thermal conductivity W/m⋅k, kW/m⋅k can be defined as the heat flow per unit area per unit time when the temperature decreases by one degree (°C, k) in unit distance Material with high thermal conductivity k are good conductors of heat, whereas materials with low thermal conductivity are good thermal insulators. k – thermodynamic state property depending on two independent, intensive state properties for simple compressible materials (T, p etc.). But for engineering analysis, k = f (material) Thermal Conductivities of Some Materials Sbstance Pure copper Pure aluminium Cast iron Lead Rubber Cork board k(W/m⋅k) 386 229 52 34.6 0.15 0.043 k determined by (1) Experiments (Fourier Law) (2) Theoretical Analysis (Microscopic analysis and statistical mechanics) ∇T determined by: (1) (2) Experiments Solving heat-conduction equation (1st Law) either analytically or numerically 4 1-3 Heat Conduction Equation Applying 1st Law of thermodynamics to the system (element) in Fig. 1-3(a) and the process taking in the unit time period yields qx + q y + qz + qgen − qx + dx − q y + dy − qz + dz = dE dτ By Fourier Law of Conduction and Taylor series expansion: ∂T ∂x ∂T ∂ ∂T + k q x + dx = − k dx dydz ∂x ∂x ∂x ∂T q y = − k dxdz ∂y q x = −kdydz ∂T ∂ ∂T + k q y + dy = − k dy dxdz ∂ ∂ y y y ∂ ∂T q z = −k dxdy ∂z ∂T ∂ ∂T + k q z + dz = − k dz dxdy ∂z ∂z ∂z qy+dy y qz x z dy qx Fig. 1-3a qz+dz qx+dx dz dx Also qgen = q dxdy dz qy q gen = q dxdydz q - energy generated per unit volume, W/m3 dE ∂T = ρcdxdydz dτ ∂τ ρ - density, kg/m3 c - specific heat of material, J/kg⋅°C (cv) (cv ≈ cp for solids) 5 dE = dkE + dPE + dU = 0 + 0 + ρdxdydz cv dT + ….. dv (dv = 0 for solids) So that the general three-dimensional heat conduction equation is (in Cartesian Coordinates): ∂ ∂T ∂ ∂T ∂ ∂T ∂T (k ) + (k ) + (k ) + q = ρc ∂x ∂x ∂y ∂y ∂z ∂z ∂τ (1-1) For constant thermal conductivity (Uniform Conductivity) Eq. (1-1) is written ∂ 2T ∂ 2T ∂ 2T q 1 ∂T + + + = ∂x 2 ∂y 2 ∂z 2 k α ∂τ (1-1a) α = k/ρc is called the thermal diffusivity of the material (another thermodynamic state property) In cylindrical coordinates (Fig. 1-3b), Eq. (1-1a) → ∂ 2T 1 ∂T 1 ∂ 2T ∂ 2T q 1 ∂T + + + 2 + = k α ∂τ ∂r 2 r ∂r r 2 ∂φ ∂z (1-1b) z dφ r φ dr dz y x Fig. 1-3b In spherical coordinates: (Fig. 1-3c), Eq. (1-1a) → 1 ∂2 1 ∂ ∂T 1 ∂ 2T q 1 ∂T ( ) rT + (sin ) + θ + = r ∂r 2 ∂θ r 2 sin θ ∂θ r 2 sin 2 θ ∂φ 2 k α ∂τ 6 (1-1c) z φ dφ r r dr dθ φ y x Fig. 1-3c Special Cases Steady-state one-dimensional ( q = 0) d 2T =0 dx 2 (1-2) Steady-state 1-D in cylindrical coordinates ( q = 0) : d 2 T 1 dT + =0 dr 2 r dr (1-3) Steady-state 1-D with heat sources: d 2 T q + =0 dx 2 k (1-4) Two-dimensional steady-state without heat sources ( q = 0) ∂ 2T ∂ 2T + =0 ∂x 2 ∂y 2 (1-5) 7 1-4 Heat Convection (Newton Law) Newton Law of Cooling q = hA(Tw - T∞) T∞ q Tw q – heat transfer rate W, kW A – heat transfer area m2 ∆T – temperature difference between wall and liquid k, °C h – convection heat-transfer coefficient W/m2⋅k, kW/m2⋅k h – determined by temperature fields T T – determined by (1) experiment (2) solving governing Eqs. Analytically or numerically (conservation of mass conservation of momentum conservation of energy) h – can also be determined experimentally (Newton law of cooling) Non-dimensional number Theoretical analysis based on governing Eqs. (or dimensional analysis) ⇒ non-dimensional numbers which control heat transfer process (heat convection) 1. Nusselt number Nu = hL k h – convection heat-transfer coefficient L – characteristic length k – thermal conductivity of fluid Nu characterizes the strength of the convective heat transfer 8 2. Prandtl number Pr = ν/α (a state property) ν - kinematic viscosity of fluid α - thermal diffusivity of fluid Pr characterizes the effect of fluid properties on the heat transfer. It expresses the relative magnitudes of diffusion of momentum and heat in the fluid 3. Reynolds number Re = UL ν U – relative velocity of fluid (characteristic velocity) Re characterizes the effect of macroscopic relative motion on the heat transfer. It represents the ratio of the inertial force over the viscous force in the forcedconvection system. 10 m/s 4 m/s 4. Grashof number Gr = gβ ∆T L3 ν2 g – gravitational acceleration β - thermal (volume) expansion coefficient of fluid (a state property) ∆T – characteristic temperature difference Gr characterizes the effect of gravitational buoyancy force on heat transfer. It represents the ratio of the buoyancy force over the viscous force in the free convection system. 9 Forced-Convection Nu = f (Re, Pr, Geometry Parameters) d Nud = 0.023 Red 0.8 Pr n 0.4 for heating of the fluid d n= 0.3 for cooling of the fluid …. ≤ Red ≤ …. …. ≤ Pr ≤ …. Free (Natural) – Convection Nu = f (Gr, Pr, Geometry Parameters) Free convection from horizontal cylinders d Nu d = 0.36 + 0.518(Grd Pr)1 / 4 [1 + (0.559 / Pr) 9 / 16 ] 4 / 9 10-6 < GrdPr < 109 10 Chapter 2 2-1 Steady-State Conduction: One Dimension Introduction In this Chapter we will use Fourier’s law of heat conduction and heat-conduction equation to calculate heat flow in steady-state, one-dimensional systems. By steady-state we mean that every quantities (temperature in particular) are independent of time. By onedimensional we mean that every quantities (temperature in particular) depend on space only through one spatial coordinate. 2-2 The Plane Wall Feature: heat transfer area is constant along heat flux direction y T1 k T2 q x o ∆x z A Fig. 2-1 A. Single-layer Wall Under 1D steady-state condition (dimension along y, z → ∞; T1 and T2 are independent of y, z, k uniform ect.), the heat-conduction equation for a coordinate system in Fig. 2-1 becomes d 2T =0 dx 2 (qgen = 0, k = constant) Integrating w.r.t. x yields dT = c1 dx 1 Integrating w.r.t. x again yields T = c1x + c2 Applying boundary conditions x = 0, x = ∆x, T = T1 T = T2 leads to c2 = T1 T2 = c1∆x + c2 c1 = i.e. T2 − T1 ∆x T2 − T1 x + T1 ∆x dT T2 − T1 = dx ∆x ∴T= 0 ≤ x ≤ ∆x Fourier law of conduction q = −kA ∇T = −kA where R = dT T1 − T2 T1 − T2 = = dx ∆x / kA R (2-1) ∆x - thermal resistance kA q T1 Heat Flow Rate = B. R T2 Thermal Potential Difference Thermal Resistance Multilayer Wall q12 = − k A A T2 − T1 ∆x A q 23 = −k B A T3 − T 2 ∆x B 2 q 34 = −k C A T 4 − T3 ∆x C Temperature profile A q q A Fig. 2-2a 1 B 2 C 3 4 For steady-state heat conduction q12 = q23 = q34 = q (1st law) Three Eqs. for three unknowns T2, T3, q q= T1 − T4 ∆x A / k A A + ∆x B / k B A + ∆x C / k C A (2-2) q T1 ∆x A kAA RC RB RA T2 ∆x B kB A T3 ∆x C kC A T4 Fig. 2-2b 3 B A F E C G D (a) 1 2 3 5 4 RB RF q RC RA T1 T2 (b) RE RD T3 RG T4 T5 Fig. 2-2c Series and parallel one-dimensional heat transfer through a composite wall and electrical analog. q= 2-3 ∆Toverall ∑ Rth (2-3) R values Thermal conductivity is an important parameter in classifying the performance of insulation. It is, however, a common practice to use a term called the R value, which is defined as R= ∆T q/ A (2-4) to classify the performance of insulation. The units for R by Eq. (2-4) are °C⋅m2/W. Note that this differs from the thermal-resistance concept in that a heat flow per unit area is used. 4 It makes sense to talk about the R value rather than thermal conductivity of insulation material, because, if we are only told about thermal conductivity of insulation material, we would also need to know its thickness to find its resistance. 2-4 Radial Systems Feature: Heat transfer area varies along heat flux direction. A. Cylinders Consider a long cylinder of inside radius ri, outside radius ro, and length L, such as the one shown in Fig. 2-3. We expose this cylinder to a temperature difference Ti-To and ask what the heat flow will be. For a cylinder with length very large compared to diameter, it may be assumed that the heat flows only in a radial direction, so that the only space coordinate needed to specify the system is r. q ro ri r dr L q Ti To Rth = ln( ro / ri ) 2πkL Fig. 2-3 One-dimensional heat flow through a hollow cylinder and electrical analog. Heat conduction equation (steady-state 1D in cylindrical coordinates, no heat generation) d 2 T 1 dT + =0 dr 2 r dr which can be rewritten as 5 r2 d 2T dT +r =0 2 dr dr (*) Euler equation (linear ODE with variable coefficients) Let Then r = et or t = ln r dT dT dt 1 dT = = dr dt dr r dt d 2 T d 1 dT = dr 2 dr r dt 1 dT 1 d dT + =− 2 r dt r dr dt 1 dT 1 d 2T 1 d 2 T dT =− 2 + = 2 2 − dt r dt r 2 dt 2 r dt ∴ (*) becomes d 2 T dT dT − + =0 dt dt dt 2 i.e. d 2T =0 dt 2 Integrating twice w.r.t.t yields T = c1t + c2 = c1 ln r + c2 Applying boundary conditions r = ri r = ro T = Ti T = To gives Ti = c1 ln ri + c2 To = c1 ln ro + c2 To − Ti ln ro ri T − Ti c 2 = Ti − o ln ri ln(ro ri ) ∴ c1 = ∴ T= To − Ti T − Ti ln r + Ti − o ln ri ln ro ri ln(ro ri ) 6 To − Ti r ln + Ti ln ro ri ri dT To − Ti 1 = dr ln ro ri r = Fourier law of conduction q r = −kAr T − Ti 1 dT = −k 2πrL o dr ln ro ri r (2-5) i.e. qr = 2πkL(Ti − To ) Ti − To = ln (ro ri ) ln (ro ri ) 2πkL (2-6) and the thermal resistance in this case is ln(ro ri ) Rth = 2πkL q To Ti Rth = ln( ro ri ) 2πkL The thermal-resistance concept may be used for multiple-layer cylindrical walls just as it was used for plane walls. For the three-layer system shown in Fig. 2-4 the solution is q= T1 − T4 ln( r2 r1 ) 2πk A L + ln( r3 r2 ) 2πk B L + ln( r4 r3 ) 2πk C L The thermal circuit is shown in Fig. 2-4b. The derivation of Eq. (2-7) is left as an exercise. 7 (2-7) q r1 r2 T2 T1 A r3 T3 q r4 T4 B T1 RA T2 RB T3 RC T4 C (a) (b) Fig. 2-4. One-dimensional heat flow through multiple cylindrical sections and electrical analog. B. Spheres Spherical systems may also be treated as one dimensional when the temperature is a function of radius only. The heat flow is then q= 4πk (Ti − To ) 1 ri − 1 ro (2-8) (k = constant, steady-state, T = Ti at r = ri, T = To at r = ro) The derivation of Eq. (2-8) is left as an exercise. Example 2-1 Multilayer Cylindrical System A tube of stainless steel (k = 19 W/m⋅°C) with 2-cm inner diameter (ID) and 4-cm outer diameter (OD) is covered with a 3-cm layer of insulation (k =0.2 W/m⋅°C). If the inside wall temperature of the pipe is maintained at 600°C, calculate the heat loss per meter of length. Also calculate the tube-insulation interface temperature (T2 = 100°C) 8 Stainless steel T1 = 600°C r2 r1 r3 Asbestos T2 = 100°C T2 T1 ln( r3 r2 ) 2πk a L ln( r2 r1 ) 2πk s L Fig. Ex. 2-1 Solution: Assume a steady-state 1D process along radial direction. The accompanying figure shows the thermal network for this problem. The heat flow is given by 2π (T1 − T2 ) q = L ln( r2 r1 ) k s + ln( r3 r2 ) k i = 2π (600 − 100) = 680 W/m ln(2) 19 + (ln 5 2) 0.2 This heat flow may be used to calculate the interface temperature between the outside tube wall and the insulation. We have Ti − T2 q = = 680 W/m L ln( r3 r2 ) 2πk i where Ti is the interface temperature, which may be obtained as Ti = 595.8°C The largest thermal resistance clearly results from the insulation, and thus the major portion of the temperature drop is through that material. 9 C. Convection Boundary Conditions Convection heat transfer can be calculated from qconv = hA(Tw - T∞) An electric-resistance analogy can also be drawn for the convection process by rewriting the equation as q conv = T w − T∞ 1 / hA (2-9) where now the 1/hA term becomes the convection resistance. D. Fur Thickness in Small and Large Animals Fur provides a natural thermal insulation for animals. It is interesting to see how the thermal resistance of fur varies with its thickness. Consider different size animals that can be treated as if they were approximately of cylindrical shape. Heat loss per unit surface area of the animal is, by Eq. (2-6), qsurface = Ti − To ln r / r (2πri L) o i 2πkL where Ti is the temperature at the body surface and To is the temperature at the outer surface of the fur. The body surface area of the animal without the fur is 2πri L ( ri is its radius without the fur). The quantity in the denominator is the R value for the layer of fur over a cylindrical shaped animal. This R value can be written as ∆r ∆r ln1 + ln1 + ri ri Rfur = 2πri L = ri k 2πkL (2-10) where ∆r = ro − ri . For small values of ∆r / ri , the term ln (1 + ∆r / ri ) can be approximated as ∆r / ri . Using this approximation, ∆r (2-11) Rfur = k 10 Therefore, the Rfur increases linearly with the thickness of fur ∆r for small values of ∆r relative to the radius of the animal, ri . Eq. (2-10) is plotted in Fig. 2-5 for two values of radius of the animal, ri . It shows that the initial linear increases in Rfur with fur thickness ∆r (as predicted by Eq. (2-11)) continues for a larger value of fur thickness when the animal is bigger (large ri ) . While for a small animal, Rfur quickly levels off as the fur thickness increases. Thus, increasing fur thickness beyond a certain value is not as beneficial for a small animal as that for a large animal. Therefore, the smaller the animal, the more difficult it is to provide the insulation. Indeed, small animals may only survive in cold climates by behavioral responses that avoid severe stress and by having very high rates of metabolism. Fig. 2-5 Plot of Rfur as the function of ∆r for an assumed fur thermal conductivity of 0.05W/m.K. 2-5 The Overall Heat-Transfer Coefficient A. Plane Wall Consider the Plane Wall shown in Fig. 2-6 exposed to a hot fluid A on one side and a cooler fluid B on the other side. The heat transfer is, for 1D case, expressed by qA1 = h1A(TA – T1) kA q12 = (T1 − T2 ) ∆x q2B = h2A(T2 – TB) For steady-state heat transfer qA1 = q12 = q2B = q (1st law) 11 Fluid A The heat-transfer process can be represented by the resistance network in Fig. 2-6b. Three Eqs. for three unknowns T1, T2, q. TA q Fluid B T1 T2 h2 h1 TB (a) q TA 1 h1 A TB T2 T1 ∆x kA 1 h2 A (b) Fig. 2-6 Overall heat transfer through a plane wall. q= T A − TB 1 h1 A + ∆x kA + 1 h2 A (2-12) The overall heat transfer by combined conduction and convection is often expressed in terms of an overall heat transfer coefficient U, defined by the relation q = UA∆Toverall Here (2-13) A – some suitable area for the heat flow ∆Toverall = TA - TB Eqs, (2-12) and (2-13) together yields U= 1 1 h1 + ∆x k + 1 h2 12 B. Hollow Cylinder For a hollow cylinder exposed to a convection environment on its inner and outer surfaces, the electric-resistance analogy would appear as in Fig. 2-7 (1D, steady state) where, again, TA and TB are the two fluid temperatures. Note that the area for convection is not the same for both fluids in this case, these areas depending on the inside tube diameter and wall thickness. In this case the overall heat transfer would be expressed by T A − TB ln( ro ri ) 1 1 hi Ai + + 2πkL ho Ao (2-14) Fluid B q= q Ti TA Fluid A 1 1 hi Ai 2 (a) To ln( ro ri ) 2πkL TB 1 ho Ao (b) Fig. 2-7 Resistance analogy for hollow cylinder with convection boundaries. The overall heat transfer coefficient may be based on either the inside or outside area of the tube. q = UiAi ∆Toverall (a) q = UoAo ∆Toverall (b) Eqs. (a) and (2-14) leads to Ui = 1 A ln( r A 1 1 o ri ) + i + i hi 2πkL Ao ho (2-15) Eqs. (b) and (2-14) yields 13 Uo 1 Ao 1 Ao ln( ro ri ) 1 + + Ai hi ho 2πkL (2-16) Eqs. (a) (b) give U i Ao = ≥1 U o Ai Example 2-2. Over Heat Transfer Coefficient for a Tube Water flows at 50°C inside a 2.5-cm-inside-diameter tube such that hi = 3500 W/m2⋅°C. The tube has a wall thickness of 0.8 mm with a thermal conductivity of 16 W/m2⋅°C. The outside of the tube loses heat by free convection with ho = 7.6 W/m2⋅°C. Calculate the overall heat transfer coefficient and heat loss per unit length to surrounding air at 20°C. Solutions: These are three resistances in series for this problem, as in Eq. (2-14). With L = 1.0 m, di = 0.025 m and do = 0.025 + 2 × 0.008 = 0.0266 m, the resistance may be calculated as Ri = 1 1 = = 0.00364 o C/W hi Ai 3500π × 0.025 × 1 ln(d o d i ) ln(0.0266 / 0.025) = = 0.00062 o C/W 2πkL 2π × 16 × 1 1 1 = = 1.575 o C/W Ro = ho Ao 7.6π × 0.0266 × 1 Rt = Clearly Ro is the largest. It is the controlling resistance for the total heat transfer because the other resistance (in series) are negligible in comparison. We shall base the overall heat transfer coefficient on the outside tube area and write q= ∆T = U o Ao ∆T R ∑ th Uo = 1 Ao ∑ Rth = (a) 1 π × 0.0266 × 1(0.00364 + 0.0062 + 1.575) = 7.577 W/m 2 ⋅ o C The heat transfer is obtained from Eq. (a), with q = Uo Ao ∆T = 7.577π × 0.0266 × 1(50 – 20) = 19W 14 (For 1 m length) Comment: This example shows an important point that many practical heat transfer problems involve multiple modes of heat transfer acting in combination; in this case, as a series of thermal resistances. It is not unusual for one mode of heat transfer to dominate the overall problem. In this example, the total heat transfer could have been computed very nearly by just calculating the free convection heat loss from the outside of the tube maintained at a temperature of 50°C. Because the inside convection and tube wall resistance are so small, there are corresponding small temperature drops, and the outside temperature of the tube will be very nearly that of the liquid inside, or 50°C. 2-6 Critical Thickness of Insulation Let us consider a layer of insulation which might be installed around a circular pipe, as shown in Fig. 2-8. The inner temperature of the insulation is fixed at Ti, and the outer surface is exposed to a convection environment at T∞. From the thermal network the heat transfer is q= 2πL(Ti − T∞ ) ln( ro ri ) 1 + k ro h (2-17) 2πL(Ti − T∞ ) 1 dq 1 =− − 2 2 dro ln( ro ri ) 1 kro hro + k ro h k ro > < 0 h k ro = = rc = = 0 h k ro < > 0 h (2-18) h, T ri Ti ro T∞ Ti ln( ro ri ) 2πkL 1 2πro Lh Fig. 2-8 Critical insulation thickness 15 Example 2-3. Critical Insulation Thickness Calculate the critical radius of insulation for asbestos (k = 0.17 W/m⋅°C) surrounding a pipe and exposed to room air at 20°C with h = 3.0 W/m2⋅°C. Calculate the heat loss from a 200°C, 5.0-cm-diameter pipe when covered with the critical radius of insulation and without insulation. Solution: From Eq. (2-18) we calculate rc as rc = k 0.17 = = 0.0567 m = 5.67 cm h 3.0 The inside radius of the insulation is 5.0/2 = 2.5 cm, so the heat transfer is calculated from Eq. (2-17) as q = L 2π ( 200 − 20) = 105.7 W/m 5.67 ln 1 2.5 + 0.17 0.0567 × 3 Without insulation the convection from the outer surface of the pipe is q = h(2πri)(Ti - T∞) = 3 × 2π × 0.025(200 – 20) L = 84.8 W/m So, the addition of 3.17 cm (5.67 – 2.5) of insulation actually increases the heat transfer by 25 percent. As an alternative, fiberglass having a thermal conductivity of 0.04 W/m⋅°C might be used as the insulation material. Then, the critical radius would be rc = k 0.04 = = 0.0133 m = 1.33 cm h 3.0 Now, the value of rc is less than the outside radius of the pipe (2.5 cm), so addition of any fiberglass 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. 16 2-7 Plane Wall with Heat Sources x=0 Consider the plane wall with uniformly distributed heat sources shown in Fig. 2-9. The thickness of the wall in the x direction is 2L, and it is assumed that the dimensions in the other directions are sufficiently large that heat flow may be considered as one-dimensional. The heat generated per unit volume is q , and we assume that the thermal conductivity does not vary with temperature. This situation might be produced in a practical situation either by passing a current through an electrically conducting material or by bio-chemical reaction in biological systems. q = heat generated per unit volume To Tw L L Tw x Fig. 2-9 Sketch illustrating one-dimensional conduction problem with heat generation. 1D, steady-state heat conduction equation (for coordinates in Fig. 2-9) d 2T q + =0 dx 2 k q (2-19) = constant Boundary condition T = Tw at x = -L (2-20a) T = Tw at x=L (2-20b) Integrating (2-19) twice w.r.t. x yields T =− q 2 x + c1 x + c 2 2k (2-21) 17 Applying B.C. (2-20a) and (2-20b) gives q 2 Tw = − L – c1L + c2 2k Tw = − (a) q 2 L + c1L + c2 2k (b) (a) + (b) gives 2Tw = − i.e. q 2 L + 2c2 k c2 = Tw + q 2 L 2k (c) (a) or (b) then yields c1 = 0 ∴ T = Tw + q 2 2 (L – x ) 2k Fourier Law: q = − kA dT q = −kA (−2 x) = Aqx dx 2k 18 2-8 Cylinder with Heat Sources A. Cylinder Consider a cylinder of radius R with uniformly distributed heat sources and constant thermal conductivity. If the cylinder is sufficiently long that the temperature may be considered a function of radius only. The appropriate heat conduction equation is (1D steady state in cylindrical system). q dr o r R L r d 2 T 1 dT q + + =0 dr 2 r dr k (2-22) The boundary conditions are T=Tw at r=R (2-22a) dT dr (2-22b) q πR 2 L = −k 2πRL r=R we rewrite Eq. (2-22) r i.e. d 2 T dT q r + =− 2 dr k dr d dT r dr dr q r =− k (a) integrating (a) w.r.t. r yields dT qr 2 r =− + c1 dr 2k (b) 19 dT q r c1 =− + dr 2k r i.e. (bb) integrating (bb) w.r.t. r gives q r 2 + c1 ln r + c 2 4k dT q R c1 =− + 2k R dr r = R T =− Applying B.C. (2-22a) and (2-22b) yields Tw = − q R 2 + c1 ln R + c 2 4k qπR 2 L = (c1) c 2πkRL qR − k 2πRL 1 2k R (c2) Then c1 = 0 c2 = Tw + q R 2 4k ∴ T = Tw + q 2 2 (R – r ) 4k (2-23) Fourier Law of conduction q r = −k 2πrL B. dT = πLqr 2 dr Hollow Cylinder For a hollow cylinder with uniformly distributed heat sources, the general solution is still (1D, steady-state) q r 2 T =− + c1 ln r + c 2 4k Applying boundary conditions T = Ti at r = ri (inside surface) T = To at r = ro (outside surface) leads to 20 2 q ri + c1 ln ri + c 2 Ti = − 4k 2 q r To = − o + c1 ln ro + c 2 4k c1 = (a2) Ti − To + q ( ri − ro ) / 4k ln( ri ro ) 2 i.e. (a1) 2 2 q ro c 2 = To + − c1 ln ro 4k (or 2 q ri Ti + − c1 ln ri ) 4k q ro ∴ or ri r dr L q r 2 ( ro − r 2 ) + c1 ln 4k ro q 2 r T = Ti + ( ri − r 2 ) + c1 ln 4k ri T = To + (2-24) (2-24a) Fourier Law: qr = -k2πrL dT = πLq r 2 − 2πkLc1 dr 21 2-9 Conduction-Convection Systems The heat which is conducted through a body must often be removed (or delivered) by some convection process. For example, the heat lost by conduction through a furnace wall must be dissipated to the surroundings through convection. In heat exchanger applications a finned-tube arrangement might be used to remove heat from a hot liquid. The heat transfer from the liquid to the finned-tube is by convection. The heat is conducted through the material and finally dissipated to the surroundings by convection. Obviously, an analysis of combined conduction-convection systems is very important from a practical standpoint. We shall defer part of our analysis of conduction-convection systems to Chap. 3 on heat exchangers. For the present we wish to examine some simple extended surface problems. Consider 1D fin exposed to a surrounding fluid at a temperature T∞ as shown in Fig. 2-10. The temperature of the base of the fine is To. We approach the problem by making an 1st law analysis on an element of the fin of thickness dx as shown in the figure and a process taking in the unit time period. Thus Energy in left face = energy out right face + energy lost by convection (for steady-state process) dqconv = h Pdx(T - T∞) t A qx qx+dx Z dx Base L x Fig. 2-10. Sketch illustrating one-dimensional conduction and convection through a rectangular fin. 22 Energy in left face = qx = -kA dT dx Energy out right face = q x + dx = −kA dT dx x + dx dT d 2T = −kA + 2 dx dx dx Energy lost by convection = h Pdx(T - T∞) Here ∴ A – cross-sectional area of the fin, m2 P – the perimeter of the fin, m d 2T − m 2 (T − T∞ ) = 0 2 dx where m2 = (2-25a) hP ≥0 kA Let θ = T - T∞ (T∞ = constant) Then Eq. (2-25a) becomes d 2θ − m 2θ = 0 dx 2 (2-25b) This is a second-order, linear, homogenous ODE with constant coefficient. The two roots of the auxiliary equation r2 – m2 = 0 are r1 = -m (m = hP - constant) kA r2 = +m The general solution for Eq. (2-25b) is thus, θ = c1 e-mx + c2 emx (2-26) c1 and c2 are to be determined by two boundary conditions. One boundary condition is θ = θo = To - T∞ at x = 0 (*) 23 The other boundary condition depends on the physical situation. Several cases may be considered: Case 1: The fin is very long, and the temperature at the end of the fin is essentially that of the surrounding fluid. To T(x) T x →∞ = T∞ T∞ ∞ x 0 θ=0 at x = ∞ (a) Case 2: The fin is of finite length and loses heat by convection from its end −k To dT = hL (TL − T∞ ) dx x = L T∞ L x −k dT dx = h L (T x = L − T∞ ) (b) x=L 24 Case 3: The end of the fin is insulated so that dT/dx = 0 at x = L. T(x) Ts dT dx =0 x =L T∞ L x dT = 0 at x = L dx (c) Case 1: Applying (*) and (a) in (2-26) ∴ ∴ c1 + c2 = θo (a1) c1 e-m∞ + c2 em∞ = 0 (a2) c2 = 0 c1 = θo T − T∞ θ = = e − mx θ o To − T∞ (2-27) All of the heat lost by the fin for steady-state process, must be conducted into the base at x = 0. The heat loss (Fourier Law of heat conduction): q = −kA dT dx = −kA x =0 = kAθ o m e − mx x =0 dθ dx x =0 = kAθ o (2-28) hP = hP kA θ o kA Case 2: Applying (*) and (b) in (2-26) 25 c1 + c2 = θo (b1) − k ( −c1 me − mx + c 2 me + mx ) x=L = h L (T x = L − T∞ ) i.e. c1 e − mL − c 2 e mL = ∴ hL ( c1 e − mL + c 2 e mL ) km (b2) c1 = ….. c2 = ….. (left as an exercise) T − T∞ θ = θ o To − T∞ cosh m( L − x) + (hL / mk ) sinh m( L − x) = cosh mL + (hL / mk ) sinh mL (2-29) Fourier Law of heat conduction q = − kA dT dx = −kA x =0 dθ dx x =0 sinh mL + (hL / mk ) cosh mL = hP kA (To − T∞ ) cosh mL + (hL / mk ) sinh mL (2-30) Case 3: Using (*) and (c) in (2-26) ∴ c1 + c2 = θo (c1) 0 = m(-c1 e-mL + c2 emL) (c2) θo c1 = 1 + e − 2 mL c2 = θ o − θ= i.e. θo 1+ e θo 1+ e − 2 mL − 2 mL = e − mx + θo 1 + e 2 mL θo 1+ e 2 mL e mx e − mx e mx θ = + θ o 1 + e − 2 mL 1 + e 2 mL = (2-31a) cosh[m( L − x)] cosh mL (2-32b) 26 Fourier Law of heat conduction q = − kA dT dx = − kA x =0 dθ dx x =0 1 1 = −kA θ o m − − 2 mL 2 mL 1+ e 1+ e (2-33) = hP kA θ o tanh mL e mL − e − mL mL = tanh(mL) − mL e +e Note: In the above development it has been assumed that the temperature gradients occur only in the x-direction. This assumption will be satisfied if the fin is sufficiently thin. For most fins of practical interest the error introduced by this assumption is less than 1 percent. The overall accuracy of practical fin calculations will normally be limited by uncertainness in values of h. The convection coefficient is seldom uniform over the entire surface, as has been assumed above. If severe nonuniform behavior is encountered, numerical techniques must be used to solve the problem. 2-10 Fins A. Different Types of Finned Surface In 2-9 we derived relations for the heat transfer from a rod or fin of uniform cross-sectional area from a flat wall. In practical applications, fins may have varying cross-sectional areas and may be attached to circular surface. Fig. 2-11 and Fig. 2-12 show some fins used in practical engineering. Fig. 2-11. Different types of finned surfaces. (a) Straight fin of rectangular profile on plane wall, (b) straight fin of rectangular profile on circular tube, (c) cylindrical tube with radial fin of rectangular profile, (d) cylindrical-spine or circular-rod fin. 27 Fig. 2-12. Some fin arrangements used in electronic cooling applications. Wakefield Engineering Inc., Wakefield, Mass.) (Courtesy As the cross-sectional area varies, solution of the basic differential equation and the mathematical techniques become more tedious. Please refer to: 1. Schneider P.J. Conduction Heat Transfer, Addison-Wesley, Reading, Mass., 1955. 2. Kern D.Q. and Kraus A.D. Extended Surface Heat Transfer, 2nd Ed., McGraw-Hill, New York, 1972. 28 B. Fins and Bio-Heat Transfer Warm blooded animals appear to have adapted to new or changing environments by varying the size and shape of their bodies and extremities. These physical changes facilitate heat conservation or dissipation as dictated by ambient conditions. Larger animals need to develop means of dealing with great amounts of heat that they produce. The African elephant (Fig. 2-14), the largest land mammal, has accordingly developed the largest thermoregulatory organ known in any animal, the pinna or external ear. The pinna is considered to be the main external organ responsible for the temperature regulation of the body. Fig. 2-14 An African elephant with a large external ear or pinna that is important for thermoregulation. The combined surface area of both sides of both ears of an African elephant is about 20% of its total surface area. The high surface to volume ratio (see discussion following Eq. (2-35)), large surface area, and extensive vascular network of subcutaneous vessels in the medial side of the ear make it behave like a fin and pay a role in temperature regulation. Temperature distribution measured in a pinna is shown in Fig. 2-15. The temperature changes from where the ear attaches to the head (base of the fin) to the outer edges show the characteristics of a fin. Movement of the pinna (flapping) also increases the heat loss due to the increased air flow. Fig. 2-15 Temperature distribution in the right pinna at two different ambient temperatures of 18 0 C (left) and 32.1 0 C . The change in pattern indicates that a change in blood flow occurs at higher temperatures. 29 A 4000 kg elephant needs to maintain a heat loss of 4.65 kw or more while moving and feeding. This large amount of heat cannot be dissipated by surface evaporation since elephants do not have sweat glands. Thus, the pinna plays a important role in the heat dissipation and by some estimates, up to 100% of an African elephant’s heat loss can be met by movement of its pinna and by vasodilatation. Use of pinna for heat loss by convection and radiation is not unique to elephants. New Zealand white rabbits are known to do the same. 30 Chapter 3 Heat Exchangers In this Chapter, we will discuss the methods of predicting heat-exchanger performance, along with methods which may be used to estimate the heat-exchanger size and type necessary to accomplish a particular task. We limit our discussion to heat exchangers where the primary modes of heat transfer are conduction and convection. 3-1 Some Types of Heat Exchangers Heat Exchangers are devices where two moving fluid streams exchange heat without mixing (TA ≠ TB). Double-Pipe Heat Exchangers (c) (d) Fig. 3-1. Double-pipe heat exchange: (a) schematic; (b) thermal-resistance network for overall heat transfer; (c) miniature coiled tube-in-tube exchanger; (d) detail of inlet-outlet fluid connections for miniature tube-in-tube exchanger. 1 Either hot or cold fluid occupying the annular space and the other fluid occupying the inside of the inner pipe. Either counterflow or parallel flow may be used in this type of exchanger. Shell-and-Tube Heat Exchangers Fig. 3-2. Photos of commercial heat exchanges. (a) shell-and-tube heat exchanger with one tube pass; (b) head arrangement for exchanger with two tube passes; (c) miniature shell and tube exchanger with one shell pass and one tube pass; (d) internal construction of miniature exchanger. 2 One fluid flows on the inside of the tubes, while the other fluid is forced through the shell and over the outside of the tubes. One or more tube passes may be used depending on the head arrangement at the ends of the exchanger (one tube pass in Fig.3-2a; two tube passes in Fig.3-2b). Cross-flow Heat Exchanger Commonly used in air or gas heating and cooling (Figs. 3-3 and 3-4). Fig. 3-3. Cross-flow heat exchanger: one fluid mixed and one unmixed. Fig. 3-4. Cross-flow heat exchanger: both fluids unmixed. 3 3-2 Overall Heat Transfer Coefficient The overall heat transfer coefficient U is one important parameter to characterize the performance of heat exchangers.We have already discussed the overall heat-transfer coefficient in Sec. 2-5 with the heat transfer through the plane wall of Fig. 3-5. T TA Fluid A q q T1 TA k T1 1 h1 A T2 h1 A A ∆x h2 TB T2 ∆x kA 1 h2 A TB Fluid B Fig. 3-5 Definition of U q= UA ∆Toverall (∆Toverall = TA – TB) as q= T A − TB 1 ∆x 1 + + h1 A kA h2 A 1 (How to increase U?) 1 ∆x 1 + + h1 k h2 To calculate U, we need to know h1, k, h2 and ∆x. ∴ U= ******************************************************************** q1 = h1A (TA – T1) q 2 = kA T1 − T2 ∆x q3 = h2A(T2 – TB) (Newton law of cooling) (Fourier law of heat conduction: one-D, constant k) (Newton law of cooling) Steady heat transfer process: q1 = q2 = q3 = q 4 Solving for q by eliminating T1 and T2 to obtain q= T A − TB 1 ∆x 1 + + h1 A kA h2 A For double-pipe heat exchangers (Fig. 3-1), q = Ui Ai(TA – TB) q = Uo Ao(TA - TB) Heat transfer area between Fluid A and Fluid B: Ai or Ao q= ∴ Ui = TA − TB 1 ln(ro / ri ) 1 + + hi Ai 2πkL ho Ao 1 1 Ai ln(ro / ri ) Ai 1 + + hi 2πkL Ao ho 1 Uo = Ao 1 Ao ln(ro / ri ) 1 + + Ai hi 2πkL ho In most practical problems the conduction resistance is small compared with the convection resistance. Then, if one value of h is much lower than the other values, it will tend to dominate the equation for U. (We usually want large U, ?). 5 Table 3-1 Approximate Values of Overall Heat-transfer Coefficients. Heat Exchangers Brick exterior wall, plaster interior, uninsulated Frame exterior wall, plaster interior: Uninsulated With rock-wool insulation Plate-glass window Double plate-glass window Steam condenser Feedwater heater Freon-12 condenser with water coolant Water-to-water heat exchanger Finned-tube heat exchanger, water in tubes, air Across tubes Water-to-oil heat exchanger Steam to light fuel oil Steam to heavy fuel oil Steam to kerosone or gasoline Finned-tube heat exchanger, steam in tubes, Air over tubes Ammonia condenser, water in tubes Alcohol condenser, water in tubes Gas-to-gas heat exchanger 6 U W/m 2 K 2.55 1.42 0.4 6.2 2.3 1100-5600 1100-8500 280-850 850-1700 25-55 110-350 170-340 56-170 280-1140 28-280 850-1400 255-680 10-40 3-3 The Log Mean Temperature Difference For heat exchangers, we propose to relate q (heat transfer) to temperature difference by introducing the notation of overall heat-transfer coefficient U. q = U A ∆Tm (3-1) where: U = overall heat-transfer coefficient A = surface area for heat transfer consistent with definition of U ∆Tm = suitable mean temperature difference across heat exchanger (between hot and cold fluids) This plays the same role as Fourier law of heat conduction for heat conduction and Newton law of cooling for heat convection. In last section, we assume that temperature of two fluids is constant to relate U to the other parameters which we already know. However, in real heat exchangers, temperature of two fluids changes from inlet to outlet. Why? ∆Tm? Parallel flow For Double-pipe heat exchanger counterflow ∆Tm = (Th 2 − Tc 2 ) − (Th1 − Tc1 ) ln[(Th 2 − Tc 2 ) (Th1 − Tc1 )] Log mean temperature difference (LMTD) (3-2) Under assumptions: (1) (2) (3) (4) steady flow and heat transfer fluid specific heats do not vary with T and P heat exchanger is well-insulated convective heat-transfer coefficients h are constant throughout the heat exchanger Fig. 3-6. Temperature profiles for (a) parallel flow and (b) counterflow in double-pipe heat exchanger 7 Proof: Parallel flow δq = −m h c h dTh = m c c c dTc (3-3) (Assumptions: heat exchanger is well insulated, steady process, neglect the 2nd term in dh) (1st law of thermodynamics for two fluids) Also δq = U(Th – Tc)dA (3-4) By (3-3), dTh = dTc = Thus − δq m h c h δq m c c c 1 1 dTh - dTc = d(Th – Tc) = -δq + m h c h m c c c (3-5) Eqs. (3-4) and (3-5) yield 1 d (Th − Tc ) 1 dA = −U + Th − Tc m h ch m c cc (3-6) Integration with respect to A from inlet 1 to outlet 2: ln 1 T h 2 − Tc 2 1 = −UA + Th1 − Tc1 m h c h m c c c (3-7) (U, m h , ch, m c , cc are constant w.r.t.A; d( ) = ln( ) + c) ) ∫ () Integrate (3-3) w.r.t. Th or Tc from 1 to 2 m h c h = q Th1 − Th 2 m c c c = q Tc 2 − Tc1 8 ( m h , ch , m c , cc are constant w.r.t. Th or Tc) Thus (3-7) gives ln Th 2 − Tc 2 UA =− (Th1 − Th 2 + Tc 2 − Tc1 ) Th1 − Tc1 q (3-8) = UA[(Th 2 − Tc 2 ) − (Th1 − Tc1 )] / q i.e. q = UA(LMTD) LMTD = (Th 2 − Tc 2 ) − (Th1 − Tc1 ) ln[(Th 2 − Tc 2 ) /(Th1 − Tc 2 )] Comparing Eq. (3-8) with Eq. (3-1) (definition of U) q = UA∆Tm we have ∆Tm = LMTD = (Th 2 − Tc 2 ) − (Th1 − Tc1 ) ln[(Th 2 − Tc 2 ) /(Th1 − Tc1 )] Counterflow: Still (3-9), left as an exercise. (under the same notation in Fig. 3-6) Other heat Exchangers ∆Tm = F(LMTD)counterflow F – correction factor (LMTD)counterflow – LMTD for counterflow double – pipe heat exchangers (LMTD)counterflow – f(t1, t2, T1, T2) t – tube side fluid T – shell side fluid 1 – inlet 2 – outlet F = f1(P, R) 9 (3-9) P= t 2 − t1 T1 − t1 T1 − T2 t 2 − t1 f1 depends on type of heat exchangers, see Figs. 3-7 and 3-8 for example. R= Fig. 3-7. Correction-factor plot for exchanger with one shell pass and two, four, or any multiple of tube passes. Fig. 3-8. Correction-factor plot for single-pass cross flow exchanger: one fluid mixed, the other unmixed. 10 Example 3-1. Water at the rate of 68 kg/min is heated from 35 to 75°C by an oil having a specific heat of 1.9 kJ/kg⋅°C. The fluids are used in a counterflow double-pipe heat exchanger, and the oil enters the exchanger at 110° and leaves at 75°C. The overall heat-transfer coefficient is 320 W/m2⋅°C. Calculate the heat-exchanger area. Solution. The total heat transfer is determined from the energy absorbed by the water. q = m w c w ∆Tw = (68)( 4180)(75 − 35) = 11.37 MJ/ min = 189.5 kW (a) Since all the fluid temperatures are known, the LMTD can be calculate by using the temperature scheme in Fig. 3-6: ∆Tm = (110 − 75) − (75 − 35) = 37.44°C ln[(110 − 75) /(75 − 35)] (b) Then since q = UA∆Tm, 1.895 ×105 = A = 15.82 m 2 (320)(37.44) Example 3-2. Instead of the double-pipe heat exchanger of Example 3-1, it is desired to use a shelland-tube exchanger with the water making one shell pass and the oil making two tube passes. Calculate the area required for this exchanger, assuming that the overall heat-transfer coefficient remains at 320 W/m2⋅°C. Solution To solve this problem, we determine a correction factor from Fig. 3-7 to be used with the LMTD calculated on the basis of a counterflow exchanger. The parameters according to the nomenclature of Fig. 3-7 are T1 = 35°C T2 = 75°C t1 = 110°C P= t 2 − t1 75 − 110 = = 0.467 T1 − t1 35 − 110 R= T1 − T2 35 − 75 = = 1.143 t 2 − t1 75 − 110 11 t2 = 75°C so the correction factor is F = 0.81 And the heat transfer is q = UAF(LMTD) so that A = 1.895 ×105 = 19.53 m 2 (320)(0.81)(37.44) 12 3-4 Effectiveness – NTU Method LMTD approach to heat exchanger analysis is useful when Tc1, Tc2, Th1, Th2 are known or are easily determined. When the inlet or exit temperatures are to be evaluated for a given heat exchanger, effectiveness – NTU method is more easily performed. 1. Capacity ratio C, NTU and effectiveness ε When m c cc ≤ m h ch Cmin = m c cc Cmax = m h ch C= NTU= ε= Cmin m c cc = ≤ 1 (Capacity ratio) Cmax m h ch UA UA (Number of transfer units) = Cmin m c cc Tc 2 − Tc1 m c cc (Tc 2 − Tc1 ) q = = ≤ 1 (Effectiveness) Th1 − Tc1 m c cc (Th1 − Tc1 ) maximum possible q When m h ch ≤ m c cc Cmin = m h ch Cmax = m c cc C= NTU= ε= Cmin m h ch = ≤ 1 (Capacity ratio) Cmax m c cc UA UA (Number of transfer units) = Cmin m h ch Th1 − Th 2 m h ch (Th1 − Th 2 ) q = = ≤ 1 (Effectiveness) Th1 − Tc1 m h ch (Th1 − Tc1 ) maximum possible q 13 2. ε-C-NTU relation for Double-pipe Heat Exchanger Parallel Flow T Th1 dTh δq Th2 dTc Tc2 dA Tc1 2 1 δq = - m h ch dTh = m c cc dTc dTh = dTc = − δq m h c h δq m c c c 1 1 + dTh − dTc = −δq m h c h m c c c d (Th − Tc ) δq = U (Th − Tc )dA 1 1 + ∴ d (Th − Tc ) = −U (Th − Tc )dA m h c h m c cc i.e. 1 d (Th − Tc ) 1 = −UdA + Th − Tc m h c h m c cc Integrate with respect to A from 1 to 2 ln Th 2 − Tc 2 1 1 = −UA + Th1 − Tc1 m h c h m c cc 14 A m c cc ≤ m h c h , Assuming ln Th 2 − Tc 2 − UA m c 1 + c c = Th1 − Tc1 m c cc m h c h Th 2 − Tc 2 Th1 − Tc1 or − UA = Exp (1 + c) m c cc c min m h c h (Th1 − Th 2 ) = m c cc (Tc 2 − Tc1 ) Th 2 = Th1 + C (Tc1 − Tc 2 ) T − Tc 2 Th1 + C (Tc1 − Tc 2 ) − Tc 2 ∴ h2 = Th1 − Tc1 Th1 − Tc1 = (Th1 − Tc1 ) + C (Tc1 − Tc 2 ) + (Tc1 − Tc 2 ) Th1 − Tc1 = 1− ∴ε = Tc 2 − Tc1 (1 + C ) = 1 − ε (1 + C ) Th1 − Tc1 1 + Exp[(− UA C min )(1 + C )] 1+ C Or ε= 1 − Exp[− NTU (1 + C )] 1+ C (a) Assuming m c cc ≥ m h c h ln 1 Th 2 − Tc 2 1 = −UA + Th1 − Tc1 m h c h mc cc = − UA (1 + C ) m h c h or Th 2 − Tc 2 = Exp[− NTU (1 + c)] Th1 − Tc1 m c cc (Tc 2 − Tc1 ) = m h c h (Th1 − Th 2 ) 15 Tc2 = Tc1 + C(Th1 – Th2) ∴ Th 2 − Tc 2 Th 2 − Tc1 − C (Th1 − Th 2 ) = Th1 − Tc1 Th1 − Tc1 = (Th1 − Tc1 ) − (Th1 − Th 2 ) − C (Th1 − Th 2 ) Th1 − Tc1 = 1− ∴ε = Th1 − Th 2 (1 + c) = 1 − ε (1 + c) Th1 − Tc1 1 − Exp[− NTU (1 + c)] 1+ C (b) same as (a) Counterflow A similar analysis leads to ∴ε = 3. 1 − Exp[− NTU (1 − C )] 1 − C Exp[− NTU (1 − C )] ε-C-NTU relation for the other heat exchangers Table 3-2 ε = f(NTU, C) Table 3-3 NTU = f1(ε, C) 16 Table 3-2. Heat-exchanger Effectiveness Relations. N = NTU = UA C min C= Flow geometry Double pipe: Parallel flow Relation 1 − exp[− N (1 + C )] 1+ C 1 − exp[− N (1 − C )] ε= 1 − C exp[− N (1 − C )] N ε= N +1 ε= Counterflow Counterflow, C = 1 Cross flow: Both fluids unmixed Both fluids mixed Cmax mixed, Cmin unmixed Cmax unmixed, Cmin mixed Shell and tube: One shell pass, 2, 4, 6, tube passes C min C max exp(− NCn) − 1 Cn − 0.22 where n = N ε = 1 − exp C 1 1 + − ε = 1 − exp(− N ) 1 − exp(− NC ) N ε = (1/C){1 – exp[-C(1 – e-N)]} ε = 1 – exp{-(1/C)[1 – exp(-NC)]} ε = 21 + C + (1 + C 2 )1 / 2 1 + exp[− N (1 + C 2 )1 / 2 ] × 1 − exp[− N (1 + C 2 )1 / 2 ] Multiple shell passes, 2n, 4n, 6n tube passes (εp = effectiveness of each shell pass, n = number of shell passes) Special case for C = 1 All exchangers with C = 0 ε= ε= [(1 − ε p C ) /(1 − ε p )]n − 1 [(1 − ε p C ) /(1 − ε p )]n − C nε p 1 + (n − 1)ε p ε = 1 – e-N 17 −1 −1 Table 3-3. NTU Relations for Heat Exchangers C = Cmin/Cmax ε = effectiveness Flow geometry Double pipe: Parallel flow N = NTU = UA/Cmin Relation − ln[1 − (1 + C )ε ] 1+ C 1 ε −1 N= ln C − 1 Cε − 1 N= Counterflow Counterflow, C = 1 N= Cross flow: Cmax mixed, Cmin unmixed ε 1− ε 1 N = − ln 1 + ln(1 − Cε ) C −1 N= ln[1 + C ln(1 − ε )] C Cmax unmixed, Cmin mixed Shell and tube: One shell pass, 2, 4, 6, tube passes N = −(1 + C 2 ) −1 / 2 All exchangers, C = 0 2 / ε − 1 − C − (1 + C 2 )1 / 2 × ln 2 1/ 2 2 / ε − 1 − C + (1 + C ) N = -ln(1 - ε) Evaporators and Condensers In a boiling or condensation process, the fluid temperature stays essentially constant, or the fluid acts as if it had infinite specific heat. In these cases Cmin/Cmax → 0 and all the heatexchanger effectiveness relations approach a single simple equation. ε = 1 – e-NTU N = -ln(1 - ε) Example 3-3. Hot oil at 100°C is used to heat air in a shell-and-tube heat exchanger. The oil makes six tube passes and the air makes one shell pass; 2.0 kg/s of air are to be heated from 20 to 80°C. The specific heat of the oil is 2100 J/kg K, and its flow rate is 3.0 kg/s. Calculate the area required for the heat exchanger for U = 200 W/m2 K. Solution The basic energy balance is 18 m o co ∆To = m a c pa ∆Ta or (3.0)(2100)(100 – Toe) = (2.0)(1009)(80 – 20) Toe = 80.78°C m h c h = (3.0)(2100) = 6300 W/K m c cc = (2.0)(1009) = 2018 W/K We have so the air is the minimum fluid and C= C min 2018 = = 0.3203 C max 6300 The effectiveness is ε= ∆Tc 80 − 20 = = 0.75 ∆Tmax 100 − 20 Now, we may use the analytical relation from Table 3-3 to obtain NTU, 2 −1 / 2 N = −(1 + 0.3203 ) 2 / 0.75 − 1 − 0.3203 − (1 + 0.3203 2 )1 / 2 ln 2 1/ 2 2 / 0.75 − 1 − 0.3203 + (1 + 0.3203 ) = 1.99 Now, with U = 200 we calculate the area as A = NTU C min (1.99)(2018) = = 20.09 m 2 U 200 Example 3-4. A counterflow double-pipe heat exchanger is used to heat 1.25 kg/s of water from 35 to 80°C by cooling an oil [cp = 2.0 kJ/kg K] from 150 to 85°C. The overall heat-transfer coefficient is 850 W/m2 K. A similar arrangement is to be built at another plant location, but it is desired to compare the performance of the single counterflow heat exchanger with two smaller counterflow heat exchangers connected in series on the water side and in parallel on the oil side, as shown in the sketch. The oil flow is split equally between the two exchangers, and it may be assumed that the overall heat-transfer coefficient for the smaller exchangers is the same as for the large exchanger. If the smaller exchangers cost 20 percent more per unit heat-transfer area, which would be the most economical arrangement – the one large exchanger or two equal-sized small exchangers? 19 Exchanger 1 = 150°C Tw3 = 80°C Exchanger 2 Tw1 = 35°C Toi Tw2 Toe1 Toe Toe2 = 85°C Fig. Ex. 3-4 Solution We calculate the surface area required for both alternatives and then compare costs. For the one large exchanger q = m c cc ∆Tc = m h c h ∆Th = (1.25)(4180)(80 − 35) = m h c h (150 − 85) = 2.31 × 10 5 W m c cc = 5225 W/K, m h c h = 3617 W/K so that the oil is the minimum fluid: εh = ∆Th 150 − 85 = = 0.565 150 − 35 150 − 35 C min 3617 = = 0.692 C max 5225 From Table 3-3, NTUlarge = 1.09, so that 20 A = NTU large C min (1.09)(3617) = = 4.638 m 2 U 850 We now wish to calculate the surface-area requirement for the two small exchangers shown in the sketch. We have 3617 = 1809 W/K 2 m c cc = 5225 W/K m h c h = C min 1809 = = 0.347 C max 5225 The number of transfer units is the same for each heat exchanger because UA and Cmin are the same for each exchanger. This requires that the effectiveness be the same for each exchanger. Thus, ε1 = ε1 = Toi − Toe ,1 Toi − Tw,1 150 − Toe ,1 150 − 35 = ε2 = Toi − Toe , 2 = ε2 = Toi − Tw, 2 (a) 150 − Toe , 2 150 − Tw, 2 where the nomenclature for the temperatures is indicated in the sketch. Because the oil flow is the same in each exchanger and the average exit oil temperature must be 85°C, we may write Toe,1 + Toe , 2 2 = 85 (b) An energy balance on the second heat exchanger gives (5225)(Tw3 – Tw2) = (1809)(Toi – Toe,2) (5225)(80 – Tw2) = (1809)(150 – Toe,2) (c) We now have the three equations (a), (b), and (c) which may be solved for the three unknowns Toe,1, Toe,2, and Tw2. The solutions are Toe,1 = 76.98°C Toe,2 = 93.02°C Tw2 = 60.26°C The effectiveness can then be calculated as ε1 = ε 2 = 150 − 76.98 = 0.635 150 − 35 From Table 3-3, we obtain NTUsmall = 1.16, so that 21 A = NTU small C min (1.16)(1809) = = 2.47 m 2 U 850 We thus find that 2.47 m2 of area is required for each of the small exchangers, of a total of 4.94 m2. This is greater than the 4.638 m2 required in the one larger exchanger; in addition, the cost per unit area is greater so that the most economical choice would be the single larger exchanger. It may be noted, however, that the pumping costs for the oil would probably be less with the two smaller exchangers, so that this could precipitate a decision in favor of the smaller exchangers if pumping costs represented a sizable economic factor. 22