#### Transcript Chapter 3

Chapter 3 One-Dimensional Steady-State Conduction Chapter 3 1 One-Dimensional Steady-State Conduction • Conduction problems may involve multiple directions and timedependent conditions • Inherently complex – Difficult to determine temperature distributions • One-dimensional steady-state models can represent accurately numerous engineering systems • In this chapter we will Learn how to obtain temperature profiles for common geometries with and without heat generation. Introduce the concept of thermal resistance and thermal circuits Chapter 3 2 The Plane Wall Consider a simple case of onedimensional conduction in a plane wall, separating two fluids of different temperature, without energy generation Cold fluid T ,1 T,2 , h2 Ts ,1 • Temperature is a function of x Ts , 2 • Heat is transferred in the x-direction Must consider – Convection from hot fluid to wall T,1,h1 qx T , 2 – Conduction through wall – Convection from wall to cold fluid Begin by determining temperature distribution within the wall Chapter 3 Hot fluid x=0 x=L x 3 Temperature Distribution • Heat diffusion equation (eq. 2.4) in the x-direction for steady-state conditions, with no energy generation: d dT k 0 dx dx • Boundary Conditions: qx is constant T (0) Ts,1, T ( L) Ts,2 • Temperature profile, assuming constant k: x T ( x) (Ts,2 Ts,1 ) Ts,1 (3.1) L Temperature varies linearly with x Chapter 3 4 Thermal Resistance Based on the previous solution, the conduction hear transfer rate can be calculated: Ts,1 Ts,2 dT kA Ts,1 Ts,2 qx kA dx L L / kA (3.2a) Similarly for heat convection, Newton’s law of cooling applies: (TS T ) qx hA(TS T ) 1 / hA (3.2b) And for radiation heat transfer: qrad (Ts Tsur ) hr A(Ts Tsur ) 1 / hr A (3.2c) Recall electric circuit theory - Ohm’s law for electrical resistance: Electriccurrent Chapter 3 PotentialDifference Resistance 5 Thermal Resistance • We can use this electrical analogy to represent heat transfer problems using the concept of a thermal circuit (equivalent to an electrical circuit). OverallDrivingForce Toverall q Resistance R Compare with equations 3.2a-3.2c The temperature difference is the “potential” or driving force for the heat flow and the combinations of thermal conductivity, convection coefficient, thickness and area of material act as a resistance to this flow: Rt ,cond Chapter 3 L 1 1 , Rt ,conv , Rt ,rad kA hA hr A 6 Thermal Resistance for Plane Wall Cold fluid T ,1 T,2 , h2 Ts ,1 qx Ts , 2 T,1,h1 Hot fluid qx x=0 x T , 2 x=L T,1 Ts ,1 1 / h1 A L / kA Ts ,2 T,2 1 / h2 A In terms of overall temperature difference: qx Rtot Chapter 3 Ts ,1 Ts ,2 T,1 T,2 Rtot 1 L 1 h1 A kA h2 A 7 Composite Walls Express the following geometry in terms of a an equivalent thermal circuit. Chapter 3 8 Composite Walls What is the heat transfer rate for this system? Alternatively qx UAT Rtot T 1 Rt q UA where U is the overall heat transfer coefficient and T the overall temperature difference. 1 1 U Rtot A [(1 / h1 ) ( LA / k A ) ( LB / kB ) ( LC / kC ) (1 / h4 )] Chapter 3 9 Composite Walls (a) Surfaces normal to the xdirection are isothermal For resistances in series: Rtot=R1+R2+…+Rn For resistances in parallel: Rtot=1/R1+1/R2+…+1/Rn Chapter 3 (b) Surfaces parallel to xdirection are adiabatic 10 Example (Problem 3.15 textbook) Consider a composite wall that includes an 8-mm thick hardwood siding (A), 40-mm by 130-mm hardwood studs (B) on 0.65-m centers with glass fiber insulation (D) (paper faced, 28 kg/m3) and a 12-mm layer of gypsum (vermiculite) wall board (C). What is the thermal resistance associated with a wall that is 2.5 m high by 6.5 m wide (having 10 studs, each 2.5 m high?) (Note: Consider the direction of heat transfer to be downwards, along the x-direction) Chapter 3 11 Contact Resistance The temperature drop across the interface between materials may be appreciable, due to surface roughness effects, leading to air pockets. We can define thermal contact resistance: Rt",c T A TB q "x See tables 3.1, 3.2 for typical values of Rt,c Chapter 3 12 Alternative Conduction Analysis When area varies in the x direction and k is a function of temperature, Fourier’s law can be written in its most general form: dT qx k (T ) A( x) dx • For steady-state conditions, no heat generation, one-dimensional heat transfer, qx is constant. qx Chapter 3 x xo dx A( x) T k (T )dT To 13 Example 3.3 Consider a conical section fabricated from pyroceram. It is of circular cross section, with the diameter D=ax, where a=0.25. 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. 1. Derive an expression for the temperature distribution T(x) in symbolic form, assuming one-dimensional conditions. Sketch the temperature distribution 2. Calculate the heat rate, qx, through the cone. T2 T1 x2 Chapter 3 x1 x 14 Radial Systems-Cylindrical Coordinates Consider a hollow cylinder, whose inner and outer surfaces are exposed to fluids at different temperatures Temperature distribution Chapter 3 15 Temperature Distribution • Heat diffusion equation (eq. 2.5) in the r-direction for steady-state conditions, with no energy generation: 1 d dT kr 0 r dr dr • Fourier’s law: qr kA • Boundary Conditions: dT dT k (2rL) const dr dr T ( r1 ) Ts,1, T ( r2 ) Ts,2 • Temperature profile, assuming constant k: (Ts ,1 Ts ,2 ) r T (r) ln Ts ,2 Logarithmic temperature distribution (see previous slide) ln(r1 / r2 ) r 2 Chapter 3 16 Thermal Resistance Based on the previous solution, the conduction hear transfer rate can be calculated: • Fourier’s law: qx dT dT qr kA k (2rL) const dr dr 2Lk Ts ,1 Ts ,2 ln(r2 / r1 ) Ts,1 Ts,2 ln(r2 / r1 ) /(2Lk ) Ts ,1 Ts ,2 Rt ,cond In terms of equivalent thermal circuit: qx Rtot Chapter 3 T,1 T,2 Rtot 1 ln(r2 / r1 ) 1 h1 (2r1L) 2kL h2 (2r2 L) 17 Composite Walls Express the following geometry in terms of a an equivalent thermal circuit. Chapter 3 18 Composite Walls What is the heat transfer rate? where U is the overall heat transfer coefficient. If A=A1=2r1L: 1 U 1 r1 r r r r r r 1 ln 2 1 ln 3 1 ln 4 1 h1 k A r1 k B r2 kC r3 r4 h4 alternatively we can use A2=2r2L, A3=2r3L etc. In all cases: 1 U1 A1 U 2 A2 U 3 A3 U 4 A4 Rt Chapter 3 19 Example (Problem 3.37 textbook) A thin electrical heater is wrapped around the outer surface of a long cylindrical tube whose inner surface is maintained at a temperature of 5°C. The tube wall has inner and outer radii of 25 and 75 mm respectively, and a thermal conductivity of 10 W/m.K. The thermal contact resistance between the heater and the outer surface of the tube (per unit length of the tube) is R’t,c=0.01 m.K/W. The outer surface of the heater is exposed to a fluid of temperature –10°C and a convection coefficient of h=100 W/m2 .K. Determine the heater power per unit length of tube required to maintain the heater at To=25°C. Chapter 3 20 Spherical Coordinates • Fourier’s law: dT qr kA dr 2 dT k (4r ) dr • Starting from Fourier’s law, acknowledging that qr is constant, independent of r, and assuming that k is constant, derive the equation describing the conduction heat transfer rate. What is the thermal resistance? Chapter 3 21 For steady-state, one dimensional conditions with no heat generation; The appropriate form of Fourier’s equation is Q = -k A dT/dr = -k(4πr2) dT/dr Note that the cross sectional area normal to the heat flow is A= 4πr2 (instead of dx) where r is the radius of the sphere Chapter 3 22 Equation 2.3-1 may be expressed in the integral form T2 Q r 2 dr = - T1 k (T )dT r1 4 r2 For constant thermal conductivity, k 4k (T 1 T 2) Q= 1 / r1 1 / r 2 = r1r 2 (T 1 T 2 r 2 r1 Generally, this equation can be written in terms of Q = T 2 T1 Rsphere, cond where R = Chapter 3 1 1 1 4k r1 r 2 23 Example: Consider a hollow steel sphere of inside radius r1 = 10 cm and outside radius, r2 = 20 cm. The thermal conductivity of the steel is k = 10 W/moC. The inside surface is maintained at a uniform temperature of T1 = 230 oC and the outside surface dissipates heat by convection with a heat transfer coefficient h = 20 W/m2oC into an ambient at T = 30oC. Determine the thickness of asbestos insulation (k=0.5 W/mK) required to reduce the heat loss by 50%. Chapter 3 24 Example (Problem 3.69 textbook) One modality for destroying malignant tissue involves imbedding a small spherical heat source of radius ro within the tissue and maintaining local temperatures above a critical value Tc for an extended period. Tissue that is well removed from the source may be assumed to remain at normal body temperature (Tb=37°C). Obtain a general expression for the radial temperature distribution in the tissue under steady-state conditions as a function of the heat rate q. If ro=0.5 mm, what heat rate must be supplied to maintain a tissue temperature of T>Tc=42°C in the domain 0.5<r<5 mm? The tissue thermal conductivity is approximately 0.5 W/m.K. Chapter 3 25 Summary • We obtained temperature distributions and thermal resistances for problems involving steady-state, onedimensional conduction in orthogonal, cylindrical and spherical coordinates, without energy generation • Useful summary in Table 3.3 Chapter 3 26