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Unformatted text preview: Foundation of Technical Education College of Technical/ Basrah Third year 32 Lectures Lectures of Heat Transfer
Heat Transfer Rate Processes
Mode Conduction Convection Radiation Transfer Mechanism Diffusion of energy due to random molecular motion Diffusion of energy due to random molecular motion plus bulk motion Energy transfer by electromagnetic waves Rate of heat transfer (W) q =  kA dT dx q = h A(TsT∞) q = σ ε A(Ts4Tsur4) By Mr. Amjed Ahmed Ali Syllabus of Heat Transfer (English),
(2 hours/ week, Applied 2 hours /week) 1.Heat transfer by conduction, convection and radiation 2.Onedimensional steady state conduction 3.Systems with conductionconvection 4.Radial systems(cylinder and sphere) 4. Overall heat transfer coefficient 5. Critical thickness of the insulator 6. Heat source systems 7. Extended Surface (Fins) 8. Resistance to heat contact 9. Unsteady state conduction • Complete heat capacity system • Limited conditions of convection • Application and Hessler's diagrams 11. Multidimensions systems 12. Principles of heat transfer by convection 13. Boundary layer for laminar and turbulent flow 14. Thermal boundary layer for laminar and turbulent flow 15. Analogy between fluid friction and heat transfer 16. Experimental relations of heat transfer by forced convection inside pipes 17. Flow through cylindrical and spherical bodies 18. Flow through bundle of tubes 19. Heat exchangers Scaling Mean logarithmic difference of temperature NTU method 20. Heat transfer by radiation 21. Properties of radiation 22. Body in thermal radiation 23. Relation between coefficient and the body 24. Heal exchange between nonblack bodies 25. Radiation barriers Hours 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 4 1 1 2 1 2 2 2 2 2 Ch 1: Introdaction 3rd Year College of Technical Chapter Oneِ Introduction
Introduction
A consider the cooling of a hot steal rod which is water Thermodynamics may be used to predict the temperature of the rodwater combination. It will not tell takes to reach this equilibrium condition. Heat Transfer predict the temperature of the rod and the water as a function of time. place in a cold final equilibrium us how long it may be used to 1.1 Definition: ﺗﻌﺎرﻳﻒ ﻣﻬﻤﺔ Heat: is the energy transit as a result of the temperature difference. Heat transfer: is that science which seeks to predict the energy transfer that may take place between material bodes as a result of a temperature difference. Thermodynamics: is the state science of energy, the transformation of energy and the change in the state of matter. (Thermodynamics can be able to determination of heat and work requirements for chemical and physical process and the equilibrium conditions). Heat flux: heat transfer flow in the direction per unit area (q”). Steady state: Temperature is very does not very with time (dT/dt) =0. Unsteady state: temperature is depending on time. 1.2 Modes of Heat Transfer أﻧﻤﺎط اﻧﺘﻘﺎل اﻟﺤﺮارة The engineering area frequently referred to as thermal science includes thermodynamics and heat transfer. The role of heat transfer is to supplement thermodynamic analyses, which consider only systems in equilibrium, with additional laws that allow prediction of time rates of energy transfer. These supplemental laws are based upon the three fundamental modes of heat transfer conduction, convection, and radiation. 1.3 A Conduction Heat Transfer
Conduction may be viewed as the transfer of energy from the more energetic to the less energetic particles of a substance due to interactions between the particles. A temperature gradient within a homogeneous substance results in an energy transfer rate within the medium which can be calculated by Fourier's law dT q = kA (1.1) dx Where q is the heat transfer rate (W or J/s) and k thermal conductivity (W/m K) is an experimental constant for the medium involved, and it may depend upon other properties, such as temperature and pressure. dT dx Is the temperature gradient in the direction normal to the area A. Mr. Amjed Ahmed 1 Ch 1: Introdaction 3rd Year College of Technical dT ∆Τ Τ −Τ = =2 1 dx ∆χ χ 2 − χ1
q T1 T2 T1> T2 X2 X1 L Figure 1.1 Temperature distribution for steady state conduction. Through a plate wall The minus sign in Fourier's Law (1.1) is required by the second law of thermodynamics: thermal energy transfer resulting from a thermal gradient must be from a warmer to a colder region. If the temperature profile within the medium is linear Fig. 1.1 it is permissible to replace the temperature gradient (partial derivative) with (1.2) Τ −Τ2 q = kA 1 L The quantity (L/kA) is equivalent to a thermal resistance Rk (K/W) which is equal to the reciprocal of the conductance. As: Τ − Τ1 L (1.3) q= 2 Rk kA Such linearity always exists in a homogeneous medium of fixed k during steady state heat transfer occurs whenever the temperature at every point within the body, including the surfaces, is independent of time. Rk T1 q T1> T2 T2 Figure 1.2 Association of conduction heat transfer with diffusion of energy due to molecular activity. If the temperature changes with time from the body. This storage rate is dT , energy is either being stored in or removed dt
(1.4) qstored = mc p dT dt Where m is the mass of substance and Cp is specific heat capacity. Mr. Amjed Ahmed 2 Ch 1: Introdaction 3rd Year College of Technical 1.3.1 Thermal Conductivity
The thermal conductivity of a material is a measure of the ability of the material to conduct heat. I Thermal Conductivity of Solids: In general, k for a pure metal decreases with temperature; alloying elements tend to reverse this trend. The thermal conductivity of a metal can usually be represented over a wide range of temperature by k = k 0(a + bθ + cθ 2 ) (1.5) Where θ = T − Tref and k0 is the conductivity at the reference temperature Tref The thermal conductivity of a non homogeneous material is usually markedly dependent upon the apparent bulk density, As a general rule, k for a no homogeneous material increases both with increasing temperature and increasing apparent bulk density II Thermal Conductivity of Liquids: Thermal conductivities of most liquids decrease with increasing temperature. But insensitive to pressure the exception is water, which exhibits increasing k up to about 150°C and decreasing k there after. Water has the highest thermal conductivity of all common liquids except the socalled liquid metals. III Thermal Conductivity of Gases: The thermal conductivity of a gas increases with increasing temperature, but is essentially independent of pressure for pressures close to atmospheric. For high pressure (i.e., pressure of the order of the critical pressure or greater), the effect of pressure may be significant. Fig(1.3) The mechanism of heat conduction of different phases of a substance. 1.4 A Convection Heat Transfer
Whenever a solid body is exposed to a moving fluid having a temperature different from that of the body, energy is carried or convected from or to the body by the fluid If the upstream temperature of the fluid is T∞, and the surface temperature of the solid is Ts the heat transfer per unit time is given by Newton’s Law of cooling: (1.6) Where h is Convective Heat transfer coefficient (W/m K) as the constant of proportionality relating the heat transfer per unit time and area to the overall temperature difference. It is important to keep in mind that the fundamental energy exchange at a solidfluid boundary is by conduction, and that this energy is then converted away by the fluid flow. The thermal resistance to convection heat transfer Rc, as:
2 q = h A(Ts  T∞ ) 1 hA T T q= s ∞ Rc Rc =
Mr. Amjed Ahmed (1.6) (1.7) Fig (1.4) Velocity and temperature distribution on flat plate 3 Ch 1: Introdaction 3rd Year College of Technical 1.5 A Radiation Heat Transfer
The third mode of heat transmission is due to electromagnetic wave propagation, which can occur in a total vacuum as well as in a medium. Experimental evidence indicates that radiant heat transfer is proportional to the fourth power of the absolute temperature, where as conduction and convection are proportional to a linear temperature difference. The fundamental StefanBoltzmann Law is q = σ A T4 (1.8) Where T is the absolute temperature, σ is Boltzmann constant independent of surface, medium, and temperature; its value is 5.6697 X 108 W/m2.K4 ., the thermal emission from many surfaces (gray bodies) can be well represented by q = σ ε A(Ts4Tsur4) (1.9) Where ε, the emissivity of the surface, ranges (01). The ideal emitter or blackbody is one, All other surfaces emit some what less than one. Ts and Tsur The temperature of surface and surroundings respectively. Similarly, The thermal resistance to radiation heat transfer Rr, as: Τs − Τsur (1.11) Rr = 4 4 σε Α( Τs − Τsur ) q= Ts  Tsur Rr (1.12) Table 1.1 Summary of heat transfer rate processes Rate of heat transfer Mode Transfer Mechanism (W) dT Diffusion of energy due to q =  kA Conduction random molecular motion dx Diffusion of energy due to Convection random molecular motion q = h A(TsT∞) plus bulk motion Energy transfer by Radiation electromagnetic waves q = σ ε A(Ts4Tsur4) Thermal Resistance (K/W) L Rk = kA 1 Rc = hA
Rr = Τs − Τsur σε Α( Τs 4 − Τsur 4 ) Figure (1.5) Conduction, Convection and Radiation Heat transfer Modes The concept of thermal resistance (analogous to electrical resistance) is introduced as an aid to solving conduction heat transfer problems. Mr. Amjed Ahmed 4 Ch 1: Introdaction 3rd Year College of Technical Example 1.1 Calculate the rate of heat transfer by natural convection between a shed roof of area 20 m x 20 m and ambient air, if the roof surface temperature is 27°C, the air temperature 3°C, and the average convection heat transfer coefficient 10 W/m2 K. Figure 1.7 Schematic Sketch of Shed for Analysis of Roof Temperature. Solution Assume that steady state exists and the direction of heat flow is from the air to the roof. The rate of heat transfer by convection from the air to the roof is then given by Eq: Note we initially assumed that the heat transfer would be from the air to the roof. But since the heat flow under this assumption turns out to be a negative quantity the direction of heat flow is actually from the roof to the air. Example 1.2 Determine the steady state rate of heat transfer per unit area through a 4.0cm thick homogeneous slab with its two faces maintained at uniform temperatures of 38I o C and 21 oC. The thermal conductivity of the material is 0.19 W/m K. Example 1.3 The forced convective heat transfer coefficient for a hot fluid x1 x2 flowing over a cool surface is 225 W/m2.oC for a particular problem. The fluid temperature upstream of the cool surface is 1200C, and the surface is held at 10 0C. Determine the heat transfer rate per unit surface area from the fluid to the surface. q = h A(TsT∞) q/A= 225(12010)=24750 W/m2 Example 1.4 After sunset, radiant energy can be sensed by a person standing near a brick wall. Such walls frequently have surface temperatures around 44 oC, and typical brick emissivity values are on the order of 0.92. What would be the radiant thermal flux per square foot from a brick wall at this temperature?
Mr. Amjed Ahmed 5 Ch 1: Introdaction 3rd Year College of Technical Example 1.5 In the summer, parked automobile surfaces frequently average 4050 oC. Assuming 45 o C and surface emissivity of 0.9, determine the radiant thermal flux emitted by a car roof Example 1.6 The air inside an electronics package housing has a temperature of 50°C. A "chip" in this housing has internal thermal power generation (heating) rate of 3 X 103 W. This chip is subjected to an air flow resulting in a convective coefficient h of 9 W/m2.oC over its two main surfaces which are 0.5 cm X 1.0 cm. Determine the chip surface temperature neglecting radiation and heat transfer from the edges Example 1.7 Calculate the thermal resistance and the rate of heat transfer through a pane of window glass (k = 0.78 W/m K) 1 m high, 0.5 m wide, and 0.5 cm thick, if the outersurface temperature is 24°C and the innersurface temperature is 24.5°C 24 °C
Figure 1.5 heat transfer by conduction through a window pane. Mr. Amjed Ahmed 6 Ch 1: Introdaction 3rd Year College of Technical Solution Assume that steady state exists and that the temperature is uniform over the inner and outer surfaces. The thermal resistance to conduction Rk is from Eq 0.005m K L = = 0.0128 Rk = kA 0.78w / mk × 1m × 0.5m W The rate of heat loss from the interior to the exterior surface is ∆T 24.5  24 q= = = 39.1 W Rk 0.0128 Example 1.8 A long, cylindrical electrically heated rod, 2 cm in diameter, is installed in a vacuum furnace as shown in Fig.1.8. The surface of the heating rod has an emissivity of 0.9 and is maintained at 1000 K, while the interior walls of the furnace are black and are at 800 K. Calculate the net rate at which heat is lost from the rod per unit length and the radiation heat transfer coefficient. Figure 1.8 Schematic Diagram of Vacuum Furnace with Heating Rod Solution Assume that steady state has been reached. Moreover, note that since the walls of the furnace completely enclose the heating rod, all the radiant energy emitted by the surface of the rod is intercepted by the furnace walls. Thus, for a black enclosure, Eq. (1.9) applies and the net heat loss from the rod of surface A1 is Note that in order for steady state to exist, the heating rod must dissipate electrical energy at the rate of 1893 W and the rate of heat loss through the furnace walls must equal the rate of electric input to the system, that is, to the rod. Mr. Amjed Ahmed 7 Ch 1: Introdaction 3rd Year College of Technical Example 1.9 An instrument used to study the Ozone depletion near the poles is placed on a large 2cmthick duralumin plate. To simplify this analysis the instrument can be thought of as a stainless steel plate 1 cm tall with a 10 cm x 10 cm square base, as shown in Fig. 1.6. The interface roughness of the steel and the duralumin is between 20 and 30 rms (µm) the contact resistance is 0.05 k/w. Four screws at the corners. The top and sides of the instrument are thermally insulated. An integrated circuit placed between the insulation and the upper surface of the stainless steel plate generates heat. If this heat is to be transferred to the lower surface of the duralumin, estimated to be at a temperature of 0°C, determine the maximum allowable dissipation rate from the circuit if its temperature is not to exceed 40°C. Figure 1.6 Schematic Sketch of Instrument for Ozone Measurement. Solution Since the top and the sides of the instrument are insulated, all the heat generated by the circuit must flow downward. The thermal circuit will have three resistances the stainless steel, the contact, and the duralumin. Using thermal conductivities kss = 14.4 W/m K, kM = 164 W/m K the thermal resistances of the metal plates are calculated from Equations: Mr. Amjed Ahmed 8 Ch 1: Introdaction 3rd Year College Of Technical 1.6 The Energy Balance In this special case the control surface includes no mass or volume and appears as shown in Figure 1.8.Accordingly, the generation and storage terms of the Energy expression, Ein –Eout Est + Eg= 0 Consequently, there can be no generation and storage. The conservation requirement then becomes Ein –Eout = 0 In Figure 1.8 three heat transfer terms are shown for the control surface. On a unit area basis they are conduction from the medium to the control surface q"cond convection from the surface to a fluid q"conv, and net radiation exchange from the surface to the surroundings q"rad. The energy balance then takes the Form and we c an express each of the terms according to the a ppropriate rate equations. q"cond = q"conv+ q"rad 1.7 Combined heat transfer systems Summarizes the basic relations for the rate equation of each of the three basic heat transfer mechanisms to aid in setting up the thermal circuits for solving combined heat transfer problems. 1.7.1 Plane Walls in Series In Fig. 1.15 for a threelayer system, the temperature gradients in the layers are different. The rate of heat conduction through each layer is qk, and from Eq. (1.1) we get Eliminating the intermediate temperatures T2 and T3 in Eq. qk can be expressed in the form Similarly, for N layers in series we have Mr. Amjed Ahmed 9 Ch 1: Introdaction 3rd Year College Of Technical where T1 is the outersurface temperature of layer 1 and TN+1 is the outersurface temperature of layer N. and ∆T is the overall temperature difference, often called the temperature potential. Figure 1.9 Conduction Through a ThreeLayer System in Series. Example 1.6 Calculate the rate of heat loss from a furnace wall per unit area. The wall is constructed from an inner layer of 0.5 cm thick steel (k : 40 W/m K) and an outer layer of 10 cm zirconium brick (k = 2.5 W/m K) as shown in Fig. The innersurface temperature is 900 K and the outside surface temperature is 460 K. What is the temperature at the interface? Figure 1.10 Schematic Diagram of Furnace Wall. Solution Assumptions: • Assume that steady state exists, • neglect effects at the corners and edges of the wall, • the surface temperatures are uniform. The rate of heat loss per unit area can be calculated from Eq: The interface temperature T2 is obtained from Mr. Amjed Ahmed 10 Ch 1: Introdaction 3rd Year College Of Technical q T1 − T2 = A R1 Solving for T2 gives Note that the temperature drop across the steel interior wall is only 1.4 K because the thermal resistance of the wall is small compared to the resistance of the brick. Example 1.7 Two large aluminum plates (k = 240 W/m K), each 1 cm thick, with 10 µm surface roughness the contact resistance Ri = 2.75 x 104 m2 K/W. The temperatures at the outside surfaces are 395°C and 405°C. Calculate (a) the heat flux (b) the temperature drop due to the contact resistance. Figure 1.11 Schematic Diagram of Interface Between Plates. Solution (a) The rate of heat flow per unit area, q'' through the sandwich wall is the two resistances is equal to (L/k) = (0.01 m)/(240 W/m K) = 4.17 x 105 m2 K/W Hence, the heat flux is (b) The temperature drop in each section. The fraction of the contact resistance is Hence 7.67°C of the total temperature drop of 10°C is the result of the contact resistance. Mr. Amjed Ahmed 11 Ch 1: Introdaction 3rd Year College Of Technical 1.7.2 Plane Walls in Parallel Conduction can occur in a section with two different materials in parallel between the same potential. Fig. 1.18 shows a slab with two different materials of areas AA and AB in parallel. If the temperatures over the left and right faces are uniform at T1 and T2, the total rate of heat flow is the sum of the flows through AA and AB: Figure 1.12 Heat Conduction Through a Wall Section with Two Paths in Parallel. Note that the total heat transfer area is the sum of AA and AB and that the total resistance equals the product of the individual resistances divided by their sum, as in any parallel circuit. A more complex application of the thermal network approach is illustrated in Fig. 1.19, where heat is transferred through a composite structure involving thermal resistances in series and in parallel. For this system the resistance of the middle layer, R2 becomes and the rate of heat flow is Where N is number of layers in series Rn : Thermal resistance of nth layer ∆Toverall : temperature difference across two outer surfaces Figure 1.13 Conduction Through a Wall Consisting of Series and Parallel Thermal Paths. Mr. Amjed Ahmed 12 Ch 1: Introdaction 3rd Year College Of Technical Example 1.8 A layer of 2 in thick firebrick (kb = 1.0 Btu/hr ft °F) is placed between two ¼ in.thick steel plates (ks = 30 Btu/hr ft °F). The faces of the brick adjacent to the plates are rough, having solidtosolid contact over only 30 % of the total area, with the average height of asperities being L2=1/32 in. If the surface temperatures of the steel plates are 200° and 800°F, respectively. the conductivity of air ka is 0.02 Btu/hr ft °F, determine the rate of heat flow per unit area. Figure 1.14 Thermal Circuit for the ParallelSeries Composite Wall. L1 = 1 in.; L2 = 1/32 in.; L3 = 1/4 in.; T1 i s at the center. Solution The overall unit conductance for half the composite wall is then, from an inspection of the thermal circuit Since the air is trapped in very small compartments, the effects of convection are small and it will be assumed that heat flows through the air by conduction. At a temperature of 300°F. Then R5 the thermal resistance of the air trapped between the asperities, is, on the basis of a unit area, equal to Mr. Amjed Ahmed 13 Ch 1: Introdaction 3rd Year College Of Technical The factors 0.3 and 0.7 in R4 and R5, respectively, represent the percent of the total area for the two separate heat flow paths. The total thermal resistance for the two paths, R4 and R5 in parallel, is The thermal resistance of half of the solid brick, Rl is and the overall unit conductance is Inspection of the values for the various thermal resistances shows that the steel offers a negligible resistance Mr. Amjed Ahmed 14 Ch 1: Introdaction 3rd Year College Of Technical 1.5.2 Convection and Conduction in Series
F igure (1.15) shows a situation in which heat is transferred between two fluids separated by a wall, the rate of heat transfer from the hot fluid at temperature Thot to the cold fluid at temperature Tcold is Figure 1.15 Thermal Circuit with Conduction and Convection in Series. Example 1.8 A 0.1 m thick brick wall (k = 0.7 W/m K) is exposed to a cold wind at 270 K through a convection heat transfer coefficient of 40 W/m2 K. On the other side is air at 330 K, with a natural convection heat transfer coefficient of 10 W/m2 K. Calculate the rate of heat transfer per unit area. Solution The three resistances are the rate of heat transfer per unit area is : Mr. Amjed Ahmed 15 Ch 1: Introdaction 3rd Year College Of Technical 1.5.3 Convection and Radiation in Parallel
In many engineering problems a surface loses or receives thermal energy by convection and radiation simultaneously. Figure 1.23 illustrates the co current heat transfer from a surface to its surroundings by convection and radiation. q= qc + qr q = hcA(T1  T2)+ hrA (T1  T2) q =(hc+ hr)A (T1  T2) where hc is the average convection heat transfer coefficient between area A1 and the surroundings air at T2, the radiation heat transfer coefficient The combined heat transfer coefficient is h = hc + hr Example 1.5 Air at 20C blow over a hot plate 50 x 75 cm and thick 2 cm maintained at 250 oC. the convection heat transfer coefficient is 25 W/m2 C. calculate the inside plate temperature if it is mode of carbon steel and that 300 W is lost from the plate surface by radiation. Where thermal conductivity is 43 w/m C. Solution qconv = h A(TsT∞) qconv = 25 (0.5 *0.75) (250  20) qconv =2.156 KW qcond = qconv + qrad qcond = 2.156 +0. 3=2.456 kW
qcond = kA Τ1 − Τ 2 L 2.456 = 43 (0.5 × 0.75)
T1 = 253.05 oC Τ1 − 250 0.02 Mr. Amjed Ahmed 16 Ch 1: Introdaction 3rd Year College Of Technical Example 1.9 A 0.5 m diameter pipe (ε = 0.9) carrying steam has a surface temperature of 500 K. The pipe is located in a room at 300 K, and the convection heat transfer coefficient between the pipe surface and the air in the room is 20 W/m2 K. Calculate the combined heat transfer coefficient and the rate of heat loss per meter of pipe length. Figure 1.17 Schematic Diagram of Steam Pipe Solution hr = 13.9 W/m2 K The combined heat transfer coefficient is h = hc + hr = 20 + 13.9 = 33.9 W/m2 K and the rate of heat loss per meter is 1.5.4 Overall Heat Transfer Coefficient
We noted previously that a common heat transfer problem is to determine the rate of heat flow between two fluids, gaseous or liquid, separated by a wall. If the wall is plane and heat is transferred only by convection on both sides, the rate of heat transfer in terms of the two fluid temperatures is given by: the rate of heat flow is expressed only in terms of an overall temperature potential and the heat transfer characteristics of individual sections in the heat flow path., the overall transmittance, or the overall coefficient of heat transfer U Writing Eq. (1.29) in terms of an overall coefficient gives An overall heat transfer coefficient U can be based on any chosen area Mr. Amjed Ahmed 17 Ch 1: Introdaction 3rd Year College Of Technical Example 1.10 In the design of a heat exchanger for aircraft application, the maximum wall temperature in steady state is not to exceed 800 k. For the conditions tabulated below, determine the maximum permissible unit thermal resistance per square meter of the metal wall that separates the hot gas Tgh = 1300 K from the cold gas Tgc = 300 K. Combined heat transfer coefficient on hot side h 1 = 200 W/m2 K Combined heat transfer coefficient on cold side h3 = 400 W/m2 K Figure 1.18 Physical System and Thermal Circuit. Solution In the steady state we can write Solving for R2 gives R2 = 0.0025 m2 K/W Mr. Amjed Ahmed 18 Ch 1: Introdaction 3rd Year College Of Technical Example 1.11 The door for an industrial gas furnace is 2 m x 4 m in surface area and is to be insulated to reduce heat loss to no more than 1200 W/m2. The interior surface is a 3/8in.thick Inconel 600 sheet (K= 25 W/m K), and the outer surface is a l/4 in.thick sheet of Stainless steel 316. Between these metal sheets a suitable thickness of insulators material is to be placed. The effective gas temperature inside the furnace is 1200°C, and the overall heat transfer coefficient between the gas and the door is Ui = 20 W/m2 K. The heat transfer coefficient between the outer surface of the door and the surroundings at 20°C is hc= 5 W/m2 K. calculate the thickness of insulated should be use Figure 1.19 Cross section of composite wall of gas furnace door Solution The thermal resistance of the two metal sheets are approximately 25 W/m K the thermal resistance of the two metal sheets are approximately: L1+L2=0.25+0.375=0.625 in These resistances are negligible compared to the other three resistances shown in the simplified thermal circuit below; The temperature drop between the gas and the interior surface of the door at the specified heat flux is: Q=AU ∆T Hence, the temperature of the Inconel will be about (120060)=1140°C. This is acceptable since no appreciable load is applied. The temperature drop at the outer surface is The insulation thickness for k = 0.27 W/m K is: X (1140240 ) Mr. Amjed Ahmed 19 Ch 2: Heat Conduction 3rd Year College of Technical Chapter Two Heat Conduction
2.1 Introduction
A major objective in a conduction analysis is to determine the temperature field in a medium (Temperature Distribution), which represents how temperature varies with position in the medium. knowledge of the temperature distribution: • Determination of thermal stresses, It could be used to ascertain structural integrity through • To determine the optimize thickness of an insulating material • To determine the compatibility of special coatings or adhesives used with the material. 2.2 Conservation of Energy
Applying energy conservation to the control volume. At an instant, these include the rate at which thermal and mechanical energy enter Ein and leave Eout. through the control surface, Is additional to the rate of change of energy generation Eg and stored Est. A general form of the energy conservation requirement may then be expressed on rate basis as: Ein + Eg − Eout = Est 2.1 Eg Ein Est Eout 2.3 The Conduction Equation of Rectangular Coordinate
Consider the energy processes that are relevant to this control volume. If there are temperature gradients, conduction heat transfer will occur across each of the control surfaces at the x, y, and z coordinate. The conduction heat rates at the opposite surfaces can then be expressed as a Taylor series expansion where, neglecting higher order terms, Figure 2.1 Differential control volume, dx dy dz. q x+ dx = q x +
q y + dy q z + dz dq x dx dx dq y = qy + dy dy dq = q z + z dz dz Slope = dq x dx Mr. Amjed Ahmed 20 Ch 2: Heat Conduction 3rd Year College of Technical The rate of change of energy generation Eg and stored Est & & E g = qV = qdxdydz dT dT dT = ρVC p = ρ (dxdydz)C p dt dt dt & where q is the rate at which energy is generated per unit volume (W/m3) and to express conservation of energy using the foregoing rate equation Est = mC p Ein + E g − Eout = E st
and, substituting equations, we obtain dq y dq dq dT & q x + q y + q z − (q x + x dx) − (q y + dy ) − (q z + z dz ) + qdxdydz = ρdxdydzC p dz dt dx dy The conduction heat rates may be evaluated from Fourier's law, dT dT q x = − kA = − kdzdy dx dx dT dT q y = − kA = − kdzdx 2.3 dy dy 2.2 dT dT = − kdzdx dz dz Substituting Equations 2.3 into Equation 2.2 and dividing out the dimensions of the control volume (dx dy dz), we obtain ∂ ∂T ∂ ∂T ∂ ∂T ∂T & (k ) + (k ) + (k ) + q = ρC p ∂x ∂x ∂y ∂y ∂z ∂z ∂t 2.4 It is often possible to work with simplified versions of Heat Equation (k=Const)is & ∂ 2T ∂ 2T ∂ 2T q 1 ∂T + 2+ 2+ = 2 ∂x ∂y ∂z k α ∂t 2.5 q z = − kA
where α = k/ρCp (m2/s) is the thermal diffusivity. 2.3.1 One Dimension Steady State Conduction
A plane wall separates two fluids of different temperatures. Heat transfer occurs by convection from the hot fluid at T∞ ,1 to one surface of the wall at Ts1, by conduction through the wall, and by convection from the other surface of the wall at Ts2 to the cold fluid at T∞ , 2 Figure 2.2 Heat transfer through a plane wall. If the heat transfer one dimensional and under steadystate conditions (there can be no change in the amount of energy storage and generation; hence Heat Equation reduces to ∂ ∂T (k )=0 ∂x ∂x 2.6 If the thermal conductivity is assumed to be constant (k=Const), the equation may be integrated twice to obtain the general solution T(x)=C1 x+C2 Mr. Amjed Ahmed 21 Ch 2: Heat Conduction 3rd Year College of Technical To obtain the constants of integration, C1 and C2 boundary conditions must be introduced. Applying the conditions B.C.1 x=0 at T=Ts1 C2=Ts1 B.C.2 x=L at T=Ts2 T − Ts1 Ts2=C1L+C2 = C1L+Ts1 C1 = s 2 L Substituting into the general solution, the Temperature Distribution is then T(x) = (TS 2 TS 1 ) x + Ts1 L Linearly equation. 2.7 2.3.2 Contact Resistance
The existence of a finite contact resistance is due principally to surface roughness effects. Contact spots are interspersed with gaps that are, in most instances, air filled. Heat transfer is therefore due to conduction across the actual contact area and to conduction and/or radiation across the gaps. The contact resistance may be viewed as two parallel resistances: that due to : (1)the contact spots (2) that due to the gaps (the major contribution to the resistance). The resistance is defined as
T A − TB q ′′ x ′′ Rtc = TB TA Figure 2.3 Temperature drop due to thermal contact resistance. Mr. Amjed Ahmed 22 Ch 2: Heat Conduction 3rd Year College of Technical Example 2.1 The temperature distribution across a wall 1 m thick at a certain instant of time is given as (T(x) = a+ bx + cx2 ) where T is in degrees Celsius and x is in meters, while a = 900° C, b = 300°C/m, and c= 50°C/m2. A uniform heat generation q=1000 W/m3, is present in the wall of area 10 m2 having the properties ρ = 1600 kg/m3, k = 40 W/m K, and Cp = 4 kJ/kg K. 1. Determine the rate of heat transfer entering (x = 0) and leaving the wall (x = 1 m). 2. Determine the rate of change of energy storage in the wall. 3. Determine the time rate of temperature change at x = 0, 0.25 and 0.5 m. Solution Assumptions: 1. Onedimensional conduction in the x direction. 2. Homogeneous medium with constant properties. 3. Uniform internal heat generation, q (W/m3). 1. 2. Ein + Eg − Eout = Est 2.1 3. The time rate of change of the temperature at any point in the medium may be determined from the heat equation, Equation 2.15, as Mr. Amjed Ahmed 23 Ch 2: Heat Conduction 3rd Year College of Technical Example 2.2 The diagram shows a conical section from pyroceram (k = 3.46 W/m K). It is of circular cross section with the diameter D = ax. The small end is at x1 = 50 mm and the large end at x2 = 250 mm. The end temperatures are T1 = 400 K and T2 = 600 K, while the lateral surface is well insulated and a=0.25. 1. Derive an expression for the temperature distribution T(x) in symbolic form, assuming onedimensional conditions. 2. Sketch the temperature distribution. 3. Calculate the heat rate through the cone. Solution Assumptions: 1. Steadystate conditions. 2. Onedimensional conduction in the x direction. 3. No internal heat generation. 4. Constant properties. dT q x = − kA dx With A=лD2/4= лa2x2/4 and separating variables 4q x dx = − kdT πa 2 x 2 Integrating from x1 to any x within the, it follows that x T 4q x dx (k = const ) ∫ = −k T∫1dT πa 2 x1 x 2 Hence
4q x 11 ( − + ) = − k (T − T1 ) 2 x x1 πa and solving for q
qx = 4[((1 / x1 ) − (1 / x))] πa 2 k (T1 − T ) or solving for T T ( x ) = T1 −
B.C.2 T=Ts2
qx = 4[((1 / x1 ) − (1 / x 2 ))] 4q x (T1 − T2 ) = 2 πa k [((1 / x1 ) − (1 / x 2 ))] πa 2 k (T1 − T2 ) 4q x 11 (− + ) 2 πa k x x1 at x=x2 Substituting for q into the expression for T(x), the temperature distribution becomes ⎡ (1 / x) − (1 / x1 )) ⎤ T ( x ) = T1 + (T1 − T2 ) ⎢ ⎥ ⎣ (1 / x1 ) − (1 / x2 ) ⎦ Substituting numerical values into the foregoing result for the heat transfer rate Mr. Amjed Ahmed 24 Ch 2: Heat Conduction 3rd Year College of Technical 2.4 The Conduction Equation of Cylindrical Coordinate
A common example is the hollow cylinder, whose inner and outer surfaces are exposed to fluids at different temperatures. Figure 2.4 Hollow cylinder with convective surface conditions. For a general transient threedimensional in the cylindrical coordinates T= T(r, φ ,z, t), the general form of the conduction equation in cylindrical coordinates becomes & 1 ∂ ∂T 1 ∂ 2T ∂ 2T q 1 ∂T + 2+ = 2.8 (r )+ 2 2 r ∂r ∂r r ∂φ ∂z k α ∂t If the heat flow in a cylindrical shape is only in the radial direction and for steadystate conditions with no heat generation, the conduction equation reduces to 1 ∂ ∂T (r )=0 r ∂r ∂r Integrating once with respect to radius gives ∂T ∂T C1 = C1 and = r ∂r ∂r r T = C1 ln r + C2. 2.9 A second integration gives To obtain the constants (C1 and C2), we introduce the following boundary conditions B.C.1 T=Ti at r=ri Ti = C1 ln ri+ C2. B.C.2 T=To at r=ro To = C1 ln ro + C2. Solving for C1 and C2 and substituting into the general solution, we then obtain r To − Ti = C1 ln o ri T − Ti T − Ti C2 = To − o ln ro C1 = o ln(ro / ri ) ln(ro / ri ) T − Ti r T (r ) = o ln( ) + Ti 2.10 ln(ro / ri ) ri we obtain the following expression for the heat transfer rate 2πLk (Ti − To ) C dT qr = − kA = −(2πrLk ) 1 = 2.11 dr r ln(ro / ri ) (T − To ) ln(ro / ri ) qr = i R= 2.12 R 2πLk Mr. Amjed Ahmed 25 Ch 2: Heat Conduction 3rd Year College of Technical 2.4.1 Overall Heat Transfer Coefficient
A hot fluid flows through a tube that is covered by an insulating material. The system loses heat to the surrounding air through an average heat transfer coefficient hc,o. the thermal resistance of the two cylinders at the inside of the tube and the outside of the insulation gives the thermal network shown below the physical system where Th∞ hot fluid temperature and
Tc,∞ the environmental air temperature the rate of heat flow is
2.13 it is often convenient to define an overall heat transfer coefficient by the equation q = UAo (ThotTcold) The area varies with radial distance. Thus, the numerical value of U will depend on the area selected. Since the outermost diameter is the easiest to measure in practice, Ao= 2л r3L is usually chosen as the base area. Comparing between above Equations. we see that Note that UA=UiAi=UoAo Ao = 2πro L and the overall coefficient becomes
3 2.14 2.15 Mr. Amjed Ahmed 26 Ch 2: Heat Conduction 3rd Year College of Technical Example 2.3 Compare the heat loss from an insulated and an uninsulated copper pipe (k = 400 W/m K) has an internal diameter of 10 cm and an external diameter of 12 cm. Saturated steam flows inside the pipe at 110°C ( hci = 10,000 W/m2 K). The pipe is located in a space at 30°C and the heat transfer coefficient on its outer surface is estimated to be 15 W/m2 K. The insulation available to reduce heat losses is 5 cm thick and its thermal conductivity is 0.20 W/m K Solution Figure 2.5 Schematic Diagram and Thermal Circuit for a Hollow Cylinder with Convection Surface Conditions The heat loss per unit length is Ts − T∞ q = L R1 + R2 + R3 Hence we get Since R1 and R2 are negligibly small compared to R3 For the uninsulated pipe. q/L = 80/0.177 = 452 W/m
For the insulated pipe, we must add a fourth resistance between r1 and r3. ln(ri / ro ) ln(11 / 6) R4 = = = 0.482mK / W 2πk 2π (0.2W / mK ) Also, the outer convection resistance changes to
1 = 0.096mK / W 2π (0.11 × 15) The total thermal resistance per meter length (RTotal=R4+Ro= 0.578 m K/W) Ro = q/L = 80/0.578 = 138 W/m. Adding insulation will reduce the heat loss from the steam by 70%. Mr. Amjed Ahmed 27 Ch 2: Heat Conduction 3rd Year College of Technical Example 2.4 A hot fluid at an average temperature of 200oC flows through a plastic pipe of 4 cm OD and 3 cm ID. The thermal conductivity of the plastic is 0.5 W/m K, and the heat transfer coefficient at the inside is 300 W/m2 K. The pipe is located in a room at 30°C, and the heat transfer coefficient at the outer surface is 10 W/m2 K, Calculate the overall heat transfer coefficient and the heat loss per unit length of pipe. Solution The overall heat transfer coefficient is based on the outside area of the pipe The heat toss per unit length is 2.4.2 Critical Radius of Insulation
Although the conduction resistance increases with the addition of insulation, the convection resistance decreases due to increasing outer surface area. Hence there may exist an insulation thickness that minimizes heat loss by maximizing the total resistance to heat transfer. Air T∞ ln( r / ri ) 1 + 2π k 2π rh An optimum insulation thickness would be associated with the value of r that minimized qr or maximized RTotal. Such a value could be obtained from the requirement that dq =0 at r=r Critical drc RTotal =
dq r − 2πL(Ti − To )((1 / Krc ) − (1 / hrc )) =0 = 2 drc ⎡ ln(rc / ri ) 1 ⎤ +⎥ ⎢ k rh ⎦ ⎣
2 1 1 − 2 =0 krc rc h For spherical shape: 2k rc = h rc = k h 2.16 Mr. Amjed Ahmed 28 Ch 2: Heat Conduction 3rd Year College of Technical Example 2.5 Calculate the total thermal resistance per unit length of tube for a 10 mm diameter tube having the following insulation thicknesses: 0, 2, 5, 10, 20 and 40 mm. The insulation is composed of Cellular Glass (k=0.055 w/m K), and the outer surface convection coefficient is 5 W/m2 K. k 0.055 rc = = = 0.011m Solution h 5 Hence rc > r, and heat transfer will increase with the addition of insulation up to a thickness of rcri =(0.0110.005)=0.006m The thermal resistances corresponding to the prescribed insulation thicknesses may be calculated and are summarized as follows. Mr. Amjed Ahmed 29 Ch 2: Heat Conduction 3rd Year College of Technical 2.5 The Conduction Equation of Spherical Coordinate
For spherical coordinates, the temperature is a function of the three space coordinates T(r ,θ , φ , t). The general form of the conduction equation is then
2.17 Figure 2.6 Spherical Coordinate System For a hollow sphere with uniform temperatures at the inner and outer surfaces, the temperature distribution without heat generation in the steady state can be obtained by simplifying Eq 2.17. Under these boundary conditions the temperature is only a function of the radius r, and the conduction equation is
1 ∂ 2 ∂T (r )=0 ∂r r 2 ∂r ∂T r2 = C1 ∂r C T (r ) = C 2 − 1 r ∂T = C1 ∂r r2 B.C.1 B.C.2 T=Ti T=To
C1 = at r=ri at r=ro C1 ri C To = C 2 − 1 ro Ti = C 2 − Ti − To 1 1 ( )−( ) ro ri C 2 = To + ( Ti − To 1 ) ((1 / ro ) − (1 / r i )) ro The temperature distribution is
T (r ) = ( Ti − To 1 1 )( − ) + Ti 1 1 r ro − ri ro 2.18 The rate of heat transfer through the spherical shell is
qr = −kA dT dT = − k (4πr 2 ) dr dr A=4πr2 A=πD2 ﻣﺴﺎﺣﺔ اﻟﻜﺮة may be expressed in the integral form ro To 1 qr dr = − ∫ kdT 4π ∫ r 2 ri Ti Assuming constant k and qr, we obtain V=4πr2/3 ﺣﺠﻢ اﻟﻜﺮة Mr. Amjed Ahmed 30 Ch 2: Heat Conduction 3rd Year College of Technical qr = 4πkro ri (Ti − To ) ro − ri R= (ro − ri ) 4πkro ri (2.19)(2.20) Example 2.6 The spherical, thinwalled metallic container is used to store liquid nitrogen at 77 K. The container has a diameter of 0.5 and is covered with an evacuated insulation system composed of silica powder (k = 0.0017 W/m K). The insulation is 25 mm thick, and its outer surface is exposed to ambient air at 300 K. The latent heat of vaporization hfg of liquid nitrogen is 2 × 105 J/kg. If the convection coefficient is 20 W/m2 K over the outer surface, 1. Determine the rate of liquid boiloff of nitrogen per hour? 2. Show expiration of critical radius of insulation? Ans: rc= 2h/k Solution 1. The rate of heat transfer from the ambient air to the nitrogen in the container can be obtained from the thermal circuit. We can neglect the thermal resistances of the metal wall and between the boiling nitrogen and the inner wall because that heat transfer coefficient is large. Hence Figure 2.7 Schematic Diagram of Spherical Container To determine the rate of boiloff we perform an energy balance & Ein=Eout mh fg = q Solving for m gives
& m=
q (13.06 J/s)(3600 s/hr) = = 0.235kg / hr h fg 2 x 105 J/kg Mr. Amjed Ahmed 31 Ch 2: Heat Conduction 3rd Year College Of Technical 2.6 Heat Generation
• • • • A common thermal energy generation process involves The conversion from electrical to thermal energy in a currentcarrying medium Eg=I2R. The deceleration and absorption of neutrons in the fuel element of a nuclear reactor Exothermic chemical reactions occurring within a medium. Endothermic reactions would, of course, have the inverse effect A conversion from electromagnetic to thermal energy may occur due to the absorption of radiation within the medium. Note: Remember not to confuse energy generation with energy storage. 2.6.1 Plane Wall with Heat Generation (a) Asymmetrical plane wall (b)Symmetrical plane wall (c) Adiabatic surface at midline Figure 2.8 Conduction in a with uniform heat generation Assumptions • Uniform heat generation per unit volume q =Const. • For constant thermal conductivity k=Const. • One dimension and steady state heat transfer. The appropriate form of the heat equation, is & ∂ 2T ∂ 2T ∂ 2T q 1 ∂T + 2+ 2+ = 2 ∂x ∂y ∂z k α ∂t 2.5 the equation may be integrated twice to obtain the general solution & q2 T(x) = x + C1 x + C 2 2.21 2k To obtain the constants of integration, C1 and C2 boundary conditions must be introduced. & q2 B.C.1 T=Ts1 at x=L Ts1 = L + C1 L + C2 2k & q2 B.C.2 T=Ts2 at x= L Ts 2 = − L − C1 L + C 2 2k & T − Ts 2 qL2 Ts1 + Ts 2 C1 = s1 C2 = + 2L 2k 2 In which case the Temperature distribution is & T − Ts1 x Ts1 + Ts 2 qL2 x2 (1 − 2 ) + s 2 2.22 T ( x) = + 2k 2 2 L L
Mr. Amjed Ahmed 32 Ch 2: Heat Conduction 3rd Year College Of Technical The Symmetrical Plane Wall when both surfaces are maintained at a common temperature, Ts1= Ts2= Ts. The temperature distribution is given by
T(x) = & qL2 x2 ( 1 − 2 ) + Ts 2k L 2.23 The maximum temperature (T=To) exists at the midline (x=0). & & qL2 qL2 To = + Ts or 2.24 To − Ts = 2k 2k which case the temperature distribution, after substitution eq 2.24 into eq 2.23 T ( x) − Ts x2 = 1− 2 2.25 To − Ts L Consider the surface at x = L for (Fig. 2.8b) or the insulated plane wall (Fig. 2.8c). The energy balance given by E g = Eout & qV = Ah(Ts − T∞ ) & qAL = Ah(Ts − T∞ ) The surface temperature is Neglecting radiation Ts = T∞ + & qL h 2.26 Note :A heat generation cannot be represented by a thermal circuit element Example 2.7 A long electrical heating element made of iron has a cross section of 10 cm x 1.0 cm. It is immersed in a heat transfer oil at 80°C. If heat is generated uniformly at a rate of 106 W/m3 by an electric current, determine the heat transfer coefficient necessary to keep the temperature of the heater below 200°C. The thermal conductivity for iron is 64 W/m K. Solution Tmax − T1 = & qL2 10 6 × (0.01) 2 = = 0.2o C 8k 8 × 64 & qV = Ah(Ts − T∞ ) h= & qL = 42W / m 2 K 2(Ts − T∞ ) & qA L = Ah(Ts − T∞ ) 2 Mr. Amjed Ahmed 33 Ch 2: Heat Conduction 3rd Year College Of Technical Example 2.8 A plane wall is a composite of two materials, A and B. The wall of material A (k = 75 W/m K) has uniform heat generation 1.5 X 106 W/m3, and thickness 50 mm. The wall material B has no generation with (k = 150 W/m K) and thickness 20 mm. The inner surface of material A is well insulated, while the outer surface of material B is cooled by a water stream with 30°C and heat transfer coefficient 1000 W/m2 K. 1. Sketch the temperature distribution that exists in the composite under steadystate conditions. 2. Determine the maximum temperature To of the insulated surface and the temperature of the cooled surface Ts. Solution Assumptions: 1. Steadystate conditions. 2. Onedimensional conduction in x direction. 3. Negligible contact resistance between walls. 4. Inner surface of A adiabatic. 5. Constant properties for materials A and B. & qL T2 = T∞ + A 2.26 h 1.5 × 10 6 × 0.05 T2 = 30 + = 105o C 1000 q′′ = T1 − T∞ ′′ ′′ Rcond + Rconv ′′ ′′ T1 = T∞ + ( Rcond + Rconv )q′′ q′′ & q= ALA where the resistances for a unit surface area are Hence T1 =115oC From Equation 2.24 the temperature at the insulated surface is
To = T1 + & qL2 A 2k A 1.5 × 10 6 (0.05) 2 = 140 o C 2 × 75 To = 115 + Mr. Amjed Ahmed 34 Ch 2: Heat Conduction 3rd Year College Of Technical 2.6.2 Radial Shapes with Heat Generation
To determine the temperature distribution in the cylinder, we begin with the appropriate form of the heat equation. For constant thermal conductivity is
& 1 ∂ ∂T q (r )+ =0 r ∂r ∂r k & ∂T q2 r =− r + C1 2k ∂r & q2 T (r ) = − r + C1 ln r + C 2 4k A Solid Cylinder
To obtain the constants (C1 & C2), we introduce the following boundary conditions B.C.1 dT/dr=0 at r=0 C1=0 & q2 B.C.2 T=Ts at r=ro C2 = ro + Ts 4k Solving for C1 and C2 and substituting into the general solution, we then obtain &2 qro r2 T (r ) = − (1 − 2 ) + Ts 2.28 4k ro The maximum temperature T=To at r=0 &2 &2 qr qro To = − o + Ts = To − Ts 4k 4k &2 qro in equation 2.28 substitution replace group 4k T (r ) − Ts r2 = 1− 2 To − Ts ro The energy balance given by
E g = Eout & qV = Ah(Ts − T∞ ) &2 qπro L = 2πro hL(Ts − T∞ ) 2.29 2.30 The surface temperature is Ts = T∞ + & qro 2h 2.31 Mr. Amjed Ahmed 35 Ch 2: Heat Conduction 3rd Year College Of Technical B For Hollow Cylinder & q2 r + C1 ln r + C 2 4k To obtain the constants (C1 and C2), we introduce the following boundary conditions & q2 Ti = − B.C.1 T=Ti at r=ri ri + C1 ln ri + C2 4k & q2 B.C.2 T=To at r=ro To = − ro + C1 ln ro + C2 4k Solving for C1 and C2 and substituting into the general solution, we then obtain 2 &2 (T − To ) + q (ri − ro ) / 4k C1 = i ln(ri / ro ) T (r ) = −
2 &2 & q 2 (Ti − To ) + q (ri − ro ) / 4k × ln ro ro − C2 = To + 4k ln(ri / ro ) In which case the Temperature distribution is 2 &2 & q(ri − ro ) ln(r / ro ) ⎡ q ⎤ 2 2 + T (r ) = To + ⎢ 4k (ro − ri ) + (To − Ti )⎥ 4k ln(ro / ri ) ⎣ ⎦
The energy balance given by E g = E out 2.32 & qV = Ah (Ts − T∞ )
2 2 & qπ (ro − ri ) L = 2πro hL (Ts − T∞ ) The surface temperature is 2 &2 q(ro − ri ) Ts = T∞ + 2hro 2.33 :اﺷﺘﻘﺎﻗﺎت ﻣﻬﻤﺔ اﺷﺘﻖ ﻋﻼﻗﺔ ﻟﺘﻮزﻳﻊ درﺟﺔ اﻟﺤﺮارة ﻟﻼﺳﻄﻮاﻧﻪ اﻟﺤﻠﻘﻴﺔ اﻟﻤﺠﻮﻓﺔ؟ اﺷﺘﻖ ﻋﻼﻗﺔ ﻟﺘﻮزﻳﻊ درﺟﺔ اﻟﺤﺮارة ﻟﻼﺳﻄﻮاﻧﺔ اﻟﺤﻠﻘﻴﺔ اﻟﻤﺠﻮﻓﺔ اذا آﺎن اﻟﺴﻄﺢ اﻟﺨﺎرﺟﻲ ﻣﻌﺰول؟ اﺷﺘﻖ ﻋﻼﻗﺔ ﻟﺘﻮزﻳﻊ درﺟﺔ اﻟﺤﺮارة ﻟﻠﻜﺮة؟ Mr. Amjed Ahmed 36 Ch 2: Heat Conduction 3rd Year College Of Technical Example 2.9 A graphitemoderated nuclear reactor. Heat is generated uniformly in uranium rods of 0.05 m diameter at the rate of 7.5 x 107 W/m3. These rods are jacketed by an annulus in which water at an average temperature of 120°C is circulated. The water cools the rods and the average convection heat transfer coefficient is estimated to be 55,000 W/m2 K. If the thermal conductivity of uranium is 29.5 W/m K, determine the center temperature of the uranium fuel rods. Figure 2.9 Nuclear Reactor. Solution The rate of heat flow by conduction at the outer surface equals the rate of heat flow by convection from the surface to the water: & q ro 2h 7 . 5 × 10 7 × 0 . 025 T s = 120 + = 137 o C 2 × 55000 T s = T∞ +
The maximum temperature from equation 2.29 &2 qr 7.5 × 10 7 × (0.025) 2 To = o + Ts = + 137 = 534 o C 4k 4 × 29.5 Mr. Amjed Ahmed 37 Ch 2: Extended Surfaces 3rd Year College of Technical 2.7 Heat Transfer In Extended Surfaces
Extended surfaces have wide industrial application as fins attached to the walls of heat transfer equipment in order to increase the rate of heating or cooling q = h As (Ts T∞). Fins come in many shapes and forms, some of which are shown in Fig 2.11. (a) Bare surface (b) Finned surface Figure 2.10 Use of fins to enhance heat transfer from a plane wall. Figure 2.11 uniform Fin configurations (a) Rectangular Fin, (b)& (c)Pin Fin The selection of fins is made on the basis of thermal performance and cost. the fins is stronger when the fluid is a gas rather than a liquid. The selection of suitable fin geometry requires a compromise among: • A cost and weight are available space • Pressure drop of the heat transfer fluid • Heat transfer characteristics of the extended surface. Figure 2.12 nonuniform Fin configurations (a) Parabolic (b) Triangular (c) Annular fin (d) Pin fin. Mr. Amjed Ahmed 38 Ch 2: Extended Surfaces 3rd Year College of Technical Consider a pin fin having the shape of a rod whose base is attached to a wall at surface temperature Ts. The fin is cooled along its surface by a fluid at temperature T∞ To derive an equation for temperature distribution, we make a heat balance for a small element of the fin. Heat flows by conduction into the left face of the element, while heat flows out of the element by conduction through the right face and by convection from the surface. Assumptions 1. The fin has a uniform crosssectional area 2. The fin is made of a material having uniform conductivity (k = constant) 3. The heat transfer coefficient between the fin and the fluid is constant (h=constant). 4. One dimensional steady state condition only. 5. Non heat generation(q=0). 6. Radiation is negligible. D
Figure 2.12 Schematic Diagram of a Pin Fin Protruding from a Wall Ein = Eout qx = qx+dx +qconv dq q x +dx = q x + x dx dx In symbolic form, this equation becomes dT ( x ) dT (x ) − kA = − kA + hc dAs (T ( x) − T∞ ) dx x dx x + dx dAs= Pdx Where P is the perimeter of the fin Pdx is the fin surface area between x and x+dx. A Cross section area of fin If k and h are uniform, Eq. 2.34 simplifies to the form d 2T ( x ) hP [T ( x) − T∞ ] = 0 − dx 2 kA 2.34 2.35 It will be convenient to define an excess temperature of the fin above the environment, θ(x) = [T(x)  T∞], and transform Eq. 2.35 into the form d 2θ (x ) − m 2θ = 0 dx 2 2.36 Where m2= hP/kA. Mr. Amjed Ahmed 39 Ch 2: Extended Surfaces 3rd Year College of Technical Last equation is a linear, homogeneous, secondorder differential equation whose general solution is of the form θ(x) = C1 e mx + C2 e –mx 2.37 To evaluate the constants C1 and C2 it is necessary to specify appropriate boundary conditions. B.C.1 θ(0) = (Ts – T∞) at x=0 θs = C1 + C2 2.38 A second boundary condition depends on the physical condition at the end of the fin. we will treat the following Four Cases: Case1: The fin is very long and the temperature at the end approaches the fluid temperature: θ(∞) = (T∞ – T∞) = 0 at x=∞ Case2: The end of the fin is insulated: dθ ( x ) at =0 dx Case3: The temperature at the end of the fin is fixed: θ(L) = (TL – T∞) at Case4: The tip loses heat by convection dθ ( x ) −k = hθ ( L ) dx x= L x=L x=L at x=L
Case 2 X L X L Figure 2.13 Representations of Four Boundary Conditions at the Tip of a Fin Mr. Amjed Ahmed 40 Ch 2: Extended Surfaces 3rd Year College of Technical Case 1
The second boundary condition is: at x=∞ B.C.2 θ(∞) = (T∞ – T∞) = 0 m∞ –m ∞ 0= C1 e + C2 e B.C.1 θ(0) = (Ts – T∞) at x=0 θs = C1 + C2 Differentiating C1=0 C2= θs
2.39 θ(x)= θs emx ∂θ = −mθ s e −mx 2.40 ∂x Since the heat conducted across the root of the fin must equal the heat transferred by convection from the surface of the rod to the fluid, dT q fin = − kA = h P(T(x) − T∞ )dx 2.41 dx x =0
The rate of heat flow can be obtained by Two different methods. Method 1. By left term in equation 2.41 substituting Eq. 2.40 for x= 0 yields dθ 2.42 q fin = − kA = − kA[− mθ (0 )e (−m )0 ] dx x =0 ⎡ hP ⎤ q fin = kA[mθs ] = kA⎢ θs ⎥ ⎢ kA ⎥ ⎦ ⎣
q fin = h PAk ⋅ θ s
Method 2 . By right term in equation 2.41
∞ q fin = hP (T(x) − T∞ )dx = hP ∫ θ s e −mx dx
0 q fin = h Pθ s e − mx ∞ m = h PAk ⋅ θ s 2.43 0 Case 2
The second boundary condition is : θ s = C1 + C2 B.C. 1 B.C.2 dT =0 dx at x=L dθ ( x ) = mC1e mL − mC2 e −mL = 0 dx x= L θ(x) = C1 e mx + C2 e –mx mC1e mL = mC 2 e − mL
Substituting in B.C.1 C1 = C2e −2 mL
C2 = θ s = C2 e −2 mL + C2 θs
1 + e −2 mL Mr. Amjed Ahmed 41 Ch 2: Extended Surfaces 3rd Year College of Technical C1 = θs
1+ e
θs − 2 mL e −2 mL
mx C1 =
θs
e mx θs
1 + e 2 mL Substituting the above relations for C1 and C2 into Eq.(2.37) θ ( x) =
θ ( x) = 1+ e
1+ e 2 mL e
e +
⎛ ⎜ ⎜ ⎝ θs
2 mL mx θ ( x) = θs
e
mL 1+ e ⎛e θs e mL ⎞ ⎟+ e mx ⎜ mL ⎟ ⎜e − 2 mL e ⎠ 1+ e ⎝ − 2 mL mL mL +e ⎛e ⎞ e θ ( x) = θ s ⎜ mL mL + mL − mL ⎟ ⎜e +e ⎟ e +e ⎝ ⎠ m(L x) m(L x) ⎞ ⎛e ⎛ (e m(Lx) + e m(Lx) ) / 2 ⎞ +e ⎟ = θs ⎜ θ ( x) = θ s ⎜ ⎟ ⎜ e mL + e mL ⎟ ⎜ (e mL + e mL ) / 2 ⎟ ⎠ ⎠ ⎝ ⎝ e
mL
m(L x) m(L x) +e mL e mx e mL + θs ⎞ ⎟ ⎟ ⎠ − mL e mx e mL Noting that Sinh(mL) = e mL − e − mL 2 Cosh(mL) = e mL + e − mL 2 The temperature distribution is: ⎛ cosh m( L − x) ⎞ θ ( x) = θ s ⎜ ⎜ cosh(mL) ⎟ ⎟ ⎝ ⎠ 2.44 The heat loss from the fin can be found by substituting the temperature gradient at the root into Eq.(2.37), we get q fin = − kA − m sinh m( L − x) dθ ( x ) = θs dx cosh( mL) − m sinh mL dθ ( x ) = θs = −θ s m tan mL dx x=0 cosh mL dθ
dx
x =0 q fin = hPAk ⋅θ s tanh (mL) 2.45 Mr. Amjed Ahmed 42 Ch 2: Extended Surfaces 3rd Year College of Technical Case 3
The second boundary condition is : θ s = C1 + C 2 B.C. 1 C 2 = θ s − C1
2.37 B.C.2 θ(x)= θL at x=L Substituting in B.C.2 θ(x) = C1 e mx + C2 e –mx θ(L) = C1 e mL + C2 e –mL θ(L) =C1 e mL +( θs –C1) e –mL e mL − e −mL θ ( L ) − θ s ⋅ e − mL θ s (e mL − e − mL ) − θ ( L ) + θ s ⋅ e − mL C2 = θ s − = e mL − e −mL e mL − e −mL θ s ⋅ e mL − θ ( L ) C 2 = mL e − e −mL
Substituting the above relations for C1 and C2 into Eq.(2.37) θ ( L ) − θ s ⋅ e − mL mx θ s ⋅ e mL − θ ( L ) –mx θ(x) = e + mL − mL e e mL − e −mL e −e mx − mx ⎡ (θ ( L ) / θ s )(e − e ) + e m ( L − x ) − e − m ( L− x ) ⎤ θ(x) = θ s ⎢ ⎥ e mL − e −mL ⎢ ⎥ ⎣ ⎦ C1 = θ ( L ) − θ s ⋅ e − mL ⎡ θ ( L ) e mx − e − mx e m ( L− x ) − e − m ( L− x ) ⎤ )+( )⎥ ⎢ ( θ )( 2 2 ⎢s ⎥ θ(x) = θ s e mL − e −mL ⎥ ⎢ ⎢ ⎥ 2 ⎣ ⎦ The temperature distribution is: ⎡ (θ ( L ) / θ s ) sinh mx + sinh m( L − x) ⎤ θ(x) = θ s ⎢ 2.46 ⎥ sinh mL ⎣ ⎦ The heat loss from the fin can be found by substituting the temperature gradient at the root into Eq.(2.37), we get ⎡ {(θ / θ )m ⋅ − m cosh mL}(sinh mL )⎤ dθ ( x ) = θ s ⎢ ( L) s ⎥ dx x=0 (sinh mL )2 ⎣ ⎦ ⎡ (θ / θ ) − cosh mL ⎤ dθ ( x ) = mθ s ⎢ ( L ) s ⎥ dx x=0 sinh mL ⎣ ⎦
q fin = − kA
q fin = ⎡ {(θ / θ )m ⋅ cosh mx + ( − m) cosh m( L − x)}(sinh mL ) − 0 ⎤ dθ ( x ) = θ s ⎢ ( L) s ⎥ dx x=0 (sinh mL )2 ⎣ ⎦ ⎡ − (θ ( L ) / θ s ) + cosh mL ⎤ = − mθ s kA⎢ ⎥ sinh mL x =0 ⎣ ⎦ ⎡ cosh mL − (θ ( L ) / θ s ) ⎤ hP θ s kA⎢ ⎥ kA sinh mL ⎣ ⎦ dθ dx Mr. Amjed Ahmed 43 Ch 2: Extended Surfaces
⎡ cosh mL − (θ ( L ) / θ s ) ⎤ q fin = M ⎢ ⎥ sinh mL ⎣ ⎦ M = hPAk ⋅ θ s 3rd Year College of Technical 2.47 Noting that Case 4
The second boundary condition is: θ s = C1 + C 2 B.C. 1
B.C.2
−k dθ ( x ) = hθ ( L ) dx x = L C 2 = θ s − C1 dθ ( x ) = mC1e mL − mC 2 e −mL dx x= L
Substituting above equations in B.C.2 − k ( mC1e mL − mC 2 e − mL ) = h(C1 e mL + C 2 e mL ) Substituting B.C.2 − k ( mC1e mL − m(θ s − C1 )e − mL ) = h (C1 e mL + (θ s − C1 ) e
C1 = C1 = C1 = − e 2mL ( h / km)) e 2mL + ( h / km)e 2mL − (h / km ) + 1
2mL θ(x) = C1 e mx + C2 e –mx θ(L) = C1 e mL + C2 e –mL (2.37) mL ) θ s (e − ( h / km )e 2mL ) e 2mL + ( h / km)e 2mL + 1 − ( h / km)
2mL θ s (e θ s (e − ( h / km )e 2mL ) e 2mL + ( h / km)e 2mL + 1 − ( h / km)
2mL C2 = θ s − C2 = θ s e 2mL (1 − (h / km ))
e
2mL + ( h / km)e 2mL + 1 − ( h / km) θ s (e
e e 2mL + ( h / km )e + 1 − ( h / km )) − θ s e 2mL (1 − ( h / km)) e 2mL + ( h / km)e 2mL + 1 − (h / km )
2mL C2 = θ s C2 = θ s 2mL + ( h / km)e 2mL + 1 − ( h / km ) − e 2mL + e 2mL ( h / km)) e 2mL + ( h / km)e 2mL + 1 − ( h / km) −e
2mL + e 2mL (h / km ) + e 2mL ( h / km) + 1 − ( h / km)) e 2mL + ( h / km )e 2mL + 1 − ( h / km) Substituting the above relations for C1 and C2 into Eq.(2.37) θ (e 2mL − (h / km)e 2mL) θ(x) = 2mL s e mx + 2mL e + (h / km)e + 1 − (h / km)
2mL θs e 2mL −e 2mL + e 2mL ( h / km) + e 2mL ( h / km) + 1 − ( h / km)) –mx e e 2mL + ( h / km )e 2mL + 1 − ( h / km) (h / km)e m ( L− x )  (h / km)e − m ( L− x ) + e m ( L− x ) + e − m ( L − x ) e mL + e −mL + (h / km)e mL − (h / km)e −mL The temperature distribution is: θ ( x) = θ s Mr. Amjed Ahmed 44 Ch 2: Extended Surfaces 3rd Year College of Technical θ ( x) = θ s (h / km) sinh m( L − x) + coshm( L − x) (h / km) sinh mL + cosh mL 2.48 The heat loss from the fin can be found by substituting the temperature gradient at the root into Eq.(2.37), we get dθ ( x ) (− m(h / km) cosh mL − ( − m sinh mL))(cosh mL + ( h / km) sinh mL) − 0 = θs dx x=0 ((h / km) sinh mL + cosh mL) 2 dθ ( x ) (h / km) cosh mL − sinh mL) = − mθ s dx x=0 ( h / km) sinh mL + cosh mL
q fin = − kA dθ dx = mθ s kAs
x =0 ( h / km) cosh mL − sinh mL ) ( h / km) sinh mL + cosh mL q fin = hP ( h / km) cosh mL − sinh mL ) θ s kA kA ( h / km) sinh mL + cosh mL ( h / km) cosh mL − sinh mL ) q fin = M ( h / km) sinh mL + cosh mL
M = hPAk ⋅ θs 2.49 Noting that Table 1 Temperature distribution and rate of heat transfer for fins θ = T − T∞
M = hPAk ⋅ θs θ s = θ (0) = Ts − T∞
m2 = hP kA m= hP kA P A : Perimeter of the fin : Cross section area of fin Mr. Amjed Ahmed 45 Ch 2: Extended Surfaces 3rd Year College of Technical 2.7.1 Fin Performance
The heat transfer effectiveness of a fin is measured by a parameter called fin effectiveness and the fin efficiency, which is defined as i Fin Effectiveness ε. A ratio of the fin heat transfer rate to the heat transfer rate
2.52 that would exist without the fin. q fin q fin = εf = qwithou ⋅ fin hAc (Ts − T∞ ) where Ac is the fin crosssectional area at the base. the use of fins may rarely be justified unless ε >= 2. ii Fin Efficiency η
ηf =
q fin qmax
2.53 qmax = hA f (Tb − T∞ ) = hPLθ b 2.54
Rectangular Where Af is the surface area of the fin is A f = 2 wLc
A f = 2 w L2 + (t / 2) A f = 2π (r − r
2 2c [ 2 1/ 2 2 1/ 2 A f = 2.o5w L + (t / 2)
2
2 1 [ ) Triangular Parabolic Annular Where as for a fin of rectangular cross section (length L & thickness t) and an adiabatic end (Case 2) is M tanh mL tanh mL ηf = = hPLθb mL 2.55 a corrected fin length of the form Lc = L + (t/2). tanh mLc tanh h PL2 / kA ηf = or ηf = mLc h PL2 / kA A fin efficiency for a circular pin fin (Diameter D & Length L) and an adiabatic end (Case 2) is tanh 4 L2 h / kD 2.56 ηf = 4 L2 h / kD In Figures 2.14 and 2.15 fin efficiencies are plotted as a function of the parameter 3/ 2 Lc (h / kAp )1 / 2 inferred for the straight and the annular fins. Fin efficiencies obtained from the figures may be used to calculate the actual fin heat transfer rate from the expression
q f = η f qmax = η f hA f θ b
2.57 Mr. Amjed Ahmed 46 Ch 2: Extended Surfaces 3rd Year College of Technical Figure 2.14 Efficiency of straight fins (rectangular, triangular, and parabolic profiles). Figure 2.15 Efficiency of annular fins of rectangular profile. Mr. Amjed Ahmed 47 Ch 2: Extended Surfaces 3rd Year College of Technical Example 2.10 Consider a copper pin fin 0.25 cm in diameter k = 396 W/m K that protrudes from a wall at 95°C into ambient air at 25°C. The heat transfer is mainly by natural convection with a coefficient equal to 10 W/m2 K. Calculate the heat loss, assuming that : (a) the fin is "infinitely long" (b) the fin is 2.5 cm long and the coefficient at the end is the same as around the circumference. (c) how long would the fin have to be for the infinitely long solution to be correct within 5 %? Solution (a) A heat loss for the "Infinitely long" fin is q fin = − kA − mθ (0 )e (− m )0 = hPAk θ s T= 25 C q= [(10 W/m K) л(0.0025 m)(396 W/m K) (л /4(0.0025 m)2 ]0.5 (9525)°C q = 0.865 W (b) The equation for the heat loss from the finite fin is case 4: sinh mL + (h / mk ) cosh mL q fin = hPAkθ s = 0.140 W cosh mL + ( h / mk ) sinh mL
2 ( ) (c) For the two solutions to be within 5%, it is necessary that
sinh mL + ( h / mk ) cosh mL >= 0.95 cosh mL + (h / mk ) sinh mL This condition is satisfied when mL > 1.8 or L > 28.3 cm.
Example 2.11 To increase the heat dissipation from a 2.5 cm OD tube, circumferential fins made of aluminum (k = 200 W/m K) are soldered to the outer surface. The fins are 0.1 cm thick and have an outer diameter of 5.5 cm. If the tube temperature is 100°C, the environmental temperature is 25°C, and the heat transfer coefficient between the fin and the environment is 65 W/m2 K, calculate the rate of heat loss from two fins. Solution a parameters required to obtain the fin efficiency curve in Fig. 2.15 are Mr. Amjed Ahmed 48 Ch 2: Extended Surfaces 3rd Year College of Technical Example 2.12 The cylinder barrel of a motorcycle is constructed of 2024T6 aluminum alloy (k = 186 W/m K) and is of height H = 0.15 m and OD = 50 mm. Under typical operating conditions the outer surface of the cylinder is at a temperature of 500 K and is exposed to ambient air at 300 K, with a convection coefficient of 50 W/m2 K. Annular fins of rectangular profile are typically added to increase heat transfer to the surroundings. Assume that five (N=5) such fins, which are of thickness t = 6 mm, length L = 20 mm and equally spaced, are added. What is the increase in heat transfer due to addition of the fins? Solution Assumptions: 1. Steadystate conditions. 2. Onedimensional radial conduction in fins. 3. Constant properties. 4. No internal heat generation. 5. Negligible radiation exchange with surroundings. 6. Uniform convection coefficient over outer surface (with or without fins). With the fins in place, the heat transfer rate is q f = N η f q max = N η f hA f θ b q=qf+qb q f = Nη f h 2π r22c − r12 (Tb − T∞ ) ( ) Heat. transfer from the exposed cylinder surface is q = hAb (Tb − T∞ )
Hence Ab = ( H − Nt ) 2π r1 q = Nη f h 2π r22c − r12 (Tb − T∞ ) + h( H − Nt )2πr (Tb − T∞ ) ( ) The fin efficiency may be obtained from Figure 2.19 with Hence q = 5 (100.22) + 188.5 = 690 W Without the fins, the heat transfer rate is
q f = hAwo (Tb − T∞ ) Awo = H ( 2πr1 ) Hence qwo = 50 W/m2 K (0.15 x л x 0.025) m2 (200 K) = 236 W Mr. Amjed Ahmed 49 Ch 3: Unsteady State Conduction 3rd Year College of Technical Chapter Three Unsteady State Conduction
3.1 Introduction
To determine the time dependence of the temperature distribution within a solid during a transient process,. One such approach may be used under conditions for which temperature gradients within the solid are small. It is termed the lumped capacitance method. 3.2 The Lumped Capacitance Method
The lumped capacitance method is the assumption that the temperature of the solid is spatially uniform at any instant during the transient process.(The temperature gradients within the solid are negligible). From Fourier's law, heat conduction in the absence of a temperature gradient implies the existence of infinite thermal conductivity Figure 3.1 Cooling of a hot metal forging.( Rcond << Rconv ) Applying energy conservation to the control volume. the energy terms
Ein + Eg − Eout = Est − Eout = E st − hAs (T − T∞ ) = ρVC p Assume dθ dt Separating variables and integrating equation, we then obtain t ρVC p θ dθ − ∫ dt = ∫ hAs θi θ 0 ρVC p θ i ρLc C p Ti − T∞ or 3.1 t= ln t= ln hAs θ h T − T∞ This equation used to determine the time required for the solid to reach some temperature − hAsθ = ρVC p
hAs θ = exp ( − )t θi ρVC p θ = (T − T∞ ) dT dt dθ / dt = dT / dt or T − T∞ h = exp ( − )t Ti − T∞ ρLc C p 3.2 This Equation used to compute the temperature reached by the solid at some time Where θ i = (Ti − T∞ ) and exponent group is
Mr. Amjed Ahmed 50 Ch 3: Unsteady State Conduction 3rd Year College of Technical hAs hL h h k Lc k t k t= t= t = ( c )( )( 2 ) = Biα = BiFo ρVC p ρC p Lc ρC p Lc k Lc k ρC p Lc ρC p 3.3 Where Lc is the characteristic length as the ratio of the solid's volume to surface area Lc=V/As . Lc = L/2 for a plane wall of thickness 2L. Lc = r/2 for a long cylinder (end edge are negligible) Lc = r/3 for a sphere Lc = rori for a long annular cylinder(end edge are negligible).
FO = kα is termed the Fourier number It is a dimensionless time and substituting ρC p
θi T − T∞ = = exp ( − BiFo) θ Ti − T∞ equation 3.3 into 3.2, we obtain 3.4 The difference between the solid and fluid temperatures must decay exponentially to zero as approaches infinity time. The quantity ρVCp/hAs may be interpreted as a thermal time constant. as 1 τt = ( )( ρVC p ) = Rt Ct 3.5 hAs where Rt is the resistance to convection heat transfer Ct is the lumped thermal capacitance of the solid. Any increase in Rt or Ct, will cause a solid to respond more slowly to changes in its thermal environment and will increase the time required to reach thermal equilibrium (θ = 0). Figure 3.2 Transient temperature response of lumped capacitance solids 3.2.1 Energy Transfer between a Solid and Surrounding
To determine the total energy transfer Q occurring up to some time t
Q = ∫ qdt = hAs ∫ (T − Ts )dt = hAs ∫ θdt
0 0 0 t t t Substitution equation 3.2
Q = hAs ∫ θi exp ( −
0 t hAs )t ρVC p dt Q = ρVC pθ i (1 − exp ( − hAs )t ) ρVC p 1 Q = ρVC pθ i (1 − exp(− t )) τ 3.6 Mr. Amjed Ahmed 51 Ch 3: Unsteady State Conduction 3rd Year College of Technical 3.2.2 A dimensionless group Biot number :
Applying energy balance to the surface under steady state Ein=Eout kA hA(Ts 2 − T∞ ) = (Ts1 − Ts 2 ) L Ts1 − Ts 2 L / kA hL = = Ts 2 − T∞ 1 / hA k hL is dimensionless group Biot number (Bi). Where k hL R Bi = c = cond k Rconv Figure 3.3 Transient temperature distribution for different Biot No. in a plane wall cooled by convection. Applicability of Lumped Capacity Analysis
When confronted with transient conduction problems, the very first thing that one should do is calculate the Biot number. If the following condition is satisfied hL Bi = c ≤ 0.1 k the error associated with using the lumped capacitance method is small. Mr. Amjed Ahmed 52 Ch 3: Unsteady State Conduction 3rd Year College of Technical Example 3.1 A thermocouple junction, which may be approximated as a sphere, is to be used for temperature measurement in a gas stream. The convection coefficient between the junction surface and the gas is known to be h = 400 W/m2 K, and the junction properties are k = 20 W/m K, Cp = 400 J/kg K, and ρ = 8500 kg/m3. Determine the junction diameter needed for the thermocouple to have a time constant of 1 s. If the junction is at 25°C and is placed in a gas stream that is at 200°C, how long will it take for the junction to reach 199°C? Solution Assumptions: 1. Temperature of junction is uniform at any instant. 2. Radiation exchange with the surroundings is negligible. 3. Losses by conduction through the leads are negligible. 4. Constant properties. 5. Using the lumped capacitance method. As = лD2 and V = лD3/6 for a sphere 1 1 ρπD 3 τt = ( Cp )( ρVC p ) = hAs hDπ 2 6 D= 6hτ = 7.07 × 10 −4 m = 0.71mm ρC p Bi = hLc 400 × 0.000353 = = 0.000235 k 3 × 20 the lumped capacitance method may be used to an excellent approximation. Lc=r/3 2. The time required for the junction to reach T = 199°C is
t= ρLc C p
h ln Ti − T∞ T − T∞ t= 8500 × 7.06 × 10 −4 × 400 25 − 200 ln = 5.2 s 6 × 400 199 − 200 Mr. Amjed Ahmed 53 Ch 3: Unsteady State Conduction 3rd Year College of Technical 3.3 Transient Heat flow in a SemiInfinite Solid
If a thermal change is suddenly imposed at this surface, a onedimensional temperature wave will be propagated by conduction within the solid. The appropriate equation is ∂ 2T 1 ∂T = 0≤ x≤L ∂x 2 α ∂t To solve this equation we must specify two boundary conditions and the initial temperature distribution. For the initial condition we shall specify that the temperature inside the solid is uniform at Ti, that is, B.C.1 T(x, 0) = Ti.
Assumptions: 1. One Dimensional 2. Extended body to infantry Figure 3.4 Schematic Diagram and Nomenclature for Transient Conduction in a SemiInfinite Solid. Closedform solutions have been obtained for Three Cases of changes in surface conditions, instantaneously applied at t = 0: These three cases are
Case 1 Change in surface temperature: a sudden change in surface temperature T (0, t ) = Ts
T ( x, t ) − Ts ⎛x ⎞ = erf ⎜ m ⎟ Ti − Ts ⎝ 2 αt ⎠ 3.8 3.9 Case 2 Constant surface heat flux: a sudden application of a specified heat flux q''s =q''o as, for example, exposing the surface to radiation 3.10 Case 3. Surface convection a sudden exposure of the surface to a fluid at a different temperature through a uniform and constant heat transfer coefficient h 3.11 3.12 the specific temperature histories computed from Eq. (3.12) are plotted in next Fig. Mr. Amjed Ahmed 54 Ch 3: Unsteady State Conduction 3rd Year College of Technical Figure 3.5 Dimensionless Transient Temperatures B.C. 1 B.C. 2 Figure 2 6 Transient Temperature Distributions in a SemiInfinite Solid where erf is the Gaussian error function, which is encountered frequently in engineering and is defined as
3.13 Values of this function are tabulated in the appendix. The complementary error function, erfc(w), is defined as erfc(w)=1erf(w) Mr. Amjed Ahmed 55 Ch 3: Unsteady State Conduction 3rd Year College of Technical Table 3.1 The Error Function
x 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36 0.38 0.40 0.42 0.44 0.46 0.48 0.50 0.52 0.54 0.56 0.58 0.60 0.62 0.64 0.66 0.68 0.70 0.72 0.74 erf(x) 0.00000 0.02256 0.04511 0.06762 0.09008 0.11246 0.13476 0.15695 0.17901 0.20094 0.22270 0.24430 0.26570 0.28690 0.30788 0.32863 0.34913 0.36936 0.38933 0.40901 0.42839 0.44749 0.46622 0.48466 0.50275 0.52050 0.53790 0.55494 0.57162 0.58792 0.60386 0.61941 0.63459 0.64938 0.66378 0.67780 0.69143 0.70468 x 0.76 0.78 0.80 0.82 0.84 0.86 0.88 0.90 0.92 0.94 0.96 0.98 1.00 1.02 1.04 1.06 1.08 1.10 1.12 1.14 1.16 1.18 1.20 1.22 1.24 1.26 1.28 1.30 1.32 1.34 1.36 1.38 1.40 1.42 1.44 1.46 1.48 1.50 erf(x) 0.71754 0.73001 0.74210 0.75381 0.76514 0.77610 0.78669 0.79691 0.80677 0.81627 0.82542 0.83423 0.84270 0.85084 0.85865 0.86614 0.87333 0.88020 0.88679 0.89308 0.89910 0.90484 0.91031 0.91553 0.92050 0.92524 0.92973 0.93401 0.93806 0.94191 0.94556 0.94902 0.95228 0.95538 0.95830 0.96105 0.96365 0.96610 x 1.52 1.54 1.56 1.58 1.60 1.62 1.64 1.66 1.68 1.70 1.72 1.74 1.76 1.78 1.80 1.82 1.84 1.86 1.88 1.90 1.92 1.94 1.96 1.98 2.00 2.10 2.20 2.30 2.40 2.50 2.60 2.70 2.80 2.90 3.00 3.20 3.40 3.60 erf(x) 0.96841 0.97059 0.97263 0.97455 0.97635 0.97804 0.97962 0.98110 0.98249 0.98379 0.98500 0.98613 0.98719 0.98817 0.98909 0.98994 0.99074 0.99147 0.99216 0.99279 0.99338 0.99392 0.99443 0.99489 0.99532 0.997020 0.998137 0.998857 0.999311 0.999593 0.999764 0.999866 0.999925 0.999959 0.999978 0.999994 0.999998 1.000000 Mr. Amjed Ahmed 56 Ch 3: Unsteady State Conduction 3rd Year College of Technical Example 3.2 Estimate the minimum depth xm at which one must place a water main below the surface to avoid freezing. The soil is initially at a uniform temperature of 20°C. Assume that under the worst conditions anticipated it is subjected to a surface temperature of 15°C for a period of 60 days. Use the following properties for soil (300 K): ρ = 2050 kg/m3 k = 0.52 W/m K Cp= 1840 J/kg K α =0.138 x 106 m2/s Solution To simplify the problem assume that 1. Conduction is onedimensional 2. The soil is a semiinfinite medium 3. The soil has uniform and constant properties. The prescribed conditions correspond to those of Case 1, the temperature distribution in the soil is
T ( x, t ) − Ts ⎛x ⎞ = erf ⎜ m ⎟ Ti − Ts ⎝ 2 αt ⎠ ⎛x ⎞ 0 − (−15) = 0.43 = erf ⎜ m ⎟ ⎜ ⎟ 20 − (−15) ⎝ 2 αt ⎠ From Table 43 we find by interpolation that when xm / 2 αt = 0.4 to satisfy the above relation. Thus xm = 0.4 × 2 αt = 0.68m
Another Solution: To use Fig. 2.35, first calculate T ( x, t ) − Ts 0 − 20 = = 0.57 − 15 − 20 T∞ − Ts and h αt / k = ∞ Then enter the curve Fig.(3.5) obtain xm / 2 αt = 0.4, the same result as above. Mr. Amjed Ahmed 57 Ch 3: Unsteady State Condiction 3rd Year College of Technical 3.4 Heisler Charts For Transient Heat Conduction
The temperature distribution and the heat flow have been calculated and the results are available in the form of charts. we shall illustrate the application of some of these charts to typical problems of transient heat conduction in solids (OneDimensional) having a Bi > 0.1. Three simple geometries for which results have been prepared in graphic form are: 1. An infinite plate of width 2L (see Fig. 3.7) a) Calculate T(0, t) from Figure 3.7a :(Midplate temperature vs time for an infinite plate) b) After that, calculate surface temperature T(x, t) from Figure 3.7b c) Calculate total heat transfer Q at any time from Figure 3.7c , note that: Qo≡ρC V(TiT∞)= ρC Vθi 2. An infinitely long cylinder of radius ro (see Fig. 3.8) 3. A sphere of radius ro (see Fig. 3.9) One boundary condition for all three geometries are similar requires that the temperature gradient at the midplane of the plate, the axis of the cylinder, and the center of the sphere be equal to zero. Physically, this corresponds to no heat flow at these locations. The other boundary condition requires that the heat conducted to or from the surface be transferred by convection to or from a fluid at temperature through a uniform and constant heat transfer coefficient dT hA(Ts − T∞ ) = kA dx Applicability of the Heisler Charts
The calculations for the Heisler charts were performed by truncating the infinite series solutions for the problems into a few terms. This restricts the applicable the charts to values of the Fourier number greater than 0.2. FO = kα > 0.2 ρC p Mr. Amjed Ahmed 58 Ch 3: Unsteady State Condiction 3rd Year College of Technical (a) (b) (c)
Figure 3.7 Dimensionless Transient Temperatures and Heat Flow in an Infinite Plate of Width 2L Mr. Amjed Ahmed 59 Ch 3: Unsteady State Condiction 3rd Year College of Technical (a) (b) (c) Figure 3.8 Dimensionless Transient Temperatures and Heat Flow for a Long Cylinder. Mr. Amjed Ahmed 60 Ch 3: Unsteady State Condiction 3rd Year College of Technical (a) (b) (c) Figure 3.9 Dimensionless transient temperatures and heat flow for a sphere. Mr. Amjed Ahmed 61 Ch 3: Unsteady State Condiction 3rd Year College of Technical Example 3.3 In a fabrication process, steel components are formed hot and then quenched in water. Consider a 2.0 m long, 0.2 m diameter steel cylinder (k = 40 W/m K, α = 1.0 x105 m2/s), initially at 400°C, that is suddenly quenched in water at 50°C. If the heat transfer coefficient is 200 W/m2 K, calculate the following 20 min after immersion: 1. the center temperature 2. the surface temperature 3. the heat transferred to the water during the initial 20 min Solution Since the cylinder has a length 10 times the diameter, we can neglect end effects. we calculate first the Biot number hr 200 × 0.1 Bi = o = = 0.5 > 0.1 k 40 1. we cannot use the lumpedcapacitance method. To use the chart solution we calculate the appropriate dimensionless parameters: αt and Bi2 FO = (0.52)(1.2) = 0.3 Fo = 2 = 1.2 ro The dimensionless centerline temperature for 1/Bi = 2.0 and Fo = 1.2 from Fig. 2.38(a) is T ( 0 , t ) − T∞ T (0, t ) − 50 = 0 . 35 = 0.35 Ti − T∞ 400 − 50 T(0,t) = 172.5 C 2. The surface temperature at r/ro =1.0 and t = 1200 s is obtained from Fig. 3.8(b) in terms of the centerline temperature: T (ro , t ) − T∞ = 0.8 T (0, t ) − T∞ T (ro , t ) − 50 = 0.8 172.5 − 50 and the surface temperature after 20 min is: T(ro, t) = 148°C
3. The initial amount of internal energy stored in the cylinder per unit length is 2 2 Qi = C pπro (Ti − T∞ ) = (kα / Fo )πro (Ti − T∞ ) = 4.4 × 10 7 W / m Then the amount of heat transferred from the steel rod to the water can be obtained from Fig. 3.8(c). Since Q(t)/Qi = 0.61 2m × 4.4 × 10 7 W ⋅ s / m Q(t ) = 0.61 × = 14.9kW ⋅ hr 3600hr Mr. Amjed Ahmed 62 Ch 3: Unsteady State Condiction 3rd Year College of Technical Example 3.4 A large concrete wall 50 cm thick is initially at 60°C. One side of the wall is insulated. The other side is suddenly exposed to hot combustion gases at 900°C through a heat transfer coefficient of 25 W/m2 K. Determine (a) the time required for the insulated surface to reach 600°C. (b) the temperature distribution in the wall at that instant (c) the heat transferred during the process. The following average physical properties are given: k = 1.25 W/m K , Cp=837 J/kg K , ρ = 500 kg/m3 , α =0.30 x 105 m2/s Solution (a). that the wall thickness is equal to L since the insulated surface corresponds to the center plane of a slab of thickness 2L when both surfaces experience a thermal change. The temperature ratio for the insulated face at the time sought is Ts (t ) − T∞ 600 − 900 = = 0.357 60 − 900 Ts (0) − T∞ x =0 1 αt and = 0.1 Fo = 2 = 0.7 Bi L From Fig. 2.37(a) we find that 0.7 × 0.52 t= = 58333s = 16.2hr 0.3 × 10 −5 (b). The temperature distribution in the wall 16 hr after the transient was initiated can be obtained from Fig. 2.37(b) for various values of x/L, as shown below: Bi=10 , Assume of positions From the above dimensionless data we can obtain the temperature distribution as a function of distance from the insulated surface: Mr. Amjed Ahmed 63 Ch 3: Unsteady State Condiction 3rd Year College of Technical Temeratue Distrbution
861 777 708 651 612 900 850 800 750 700 600 650 600 550 500 0 0.5 0.4 0.3 0.2 0.1 x
(c). The heat transferred to the wall per square meter of surface area during the transient can be obtained from Fig. 3.7(c). for Bi = 10 and Bi2 Fo = 70 is Q(t)/Qi=0.70. Q(t) = C p ρL(Ti  T∞ )837 × 500 × 0.5 × (840 ) = 1.758 × 108 J/m 2 The minus sign indicates that the heat was transferred into the wall and the internal energy increased during the process. Mr. Amjed Ahmed T(x) 64 Ch 3: Unsteady State Condiction 3rd Year College of Technical Start Input h, k, ρ, T∞, Ts Cp, V, As
Calculate Bi = hLc and α = k k ρC p
Yes No Bi<0.1
kα ρC p
Yes Calculate FO = Output Figure 3.10 Flow Chart for the solution of Unsteady state conduction problem. Mr. Amjed Ahmed Processing No Fo>0.2 The Lumped Capacitance Method SemInfinite Body The Infinite Body Heisler Charts
t= ρLc C p
h ln Ti − T∞ T − T∞ h=Const Eq. 3.11 Eq. 3.12 T(0,t)=Ts Eq. 3.8 Eq. 3.9 T − T∞ h = exp ( − )t Ti − T∞ ρLc C p Q = ρVC pθ i (1 − exp(−t / τ )) A plate wall Fig. 2.37 A long Cylinder Fig. 2.38 qo=Const Eq. 3.10 A sphere Fig. 2.39 T, t, Q End 65 ...
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