ano T=atb Equation(2. 16)is an expression for the temperature field where a and b are constants of integration For a second order equation, such as(2. 14), we need two boundary conditions to determine a and b One such set of boundary conditions can be the specification of the temperatures at both sides of the slab as shown in Figure 2.3, say T(O)=T1 T(= T2 The condition T(0=T implies that b= T1. The condition T2=T( implies that T2=aL T,or T2-71 L With these expressions for a and b the temperature distribution can be written as Tx)=T,+ (2.17) L This linear variation in temperature is shown in Figure 2. 4 for a situation in which TI> T2 Figure 2.4: Temperature distribution through a slab The heat flux g is also of interest. This is given by dr--k(2-) constant (218) L Muddy points How specific do we need to be about when the one-dimensional assumption is valid? Is it enough to say that dA/dx is small?(MP HT. 2 Why is the th hermal conductivity of light gases such as helium(monoatomic)or hyd (diatomic)much higher than heavier gases such as argon (monoatomic)or nitrogen ogen diatomic)?(MP HT.3)
HT-10 and T = ax + b . (2.16) Equation (2.16) is an expression for the temperature field where a and b are constants of integration. For a second order equation, such as (2.14), we need two boundary conditions to determine a and b. One such set of boundary conditions can be the specification of the temperatures at both sides of the slab as shown in Figure 2.3, say T (0) = T1; T (L) = T2. The condition T (0) = T1 implies that b = T1. The condition T2 = T (L) implies that T2 = aL + T1, or L T T a 2 − 1 = . With these expressions for a and b the temperature distribution can be written as Tx T T T L x 1 2 1 ( ) = + ⎛ − ⎝ ⎜ ⎞ ⎠ ⎟ . (2.17) This linear variation in temperature is shown in Figure 2.4 for a situation in which T1 > T2. T2 T1 T x Figure 2.4: Temperature distribution through a slab The heat flux q& is also of interest. This is given by ( ) constant 2 1 = − = − = − L T T k dx dT q& k . (2.18) Muddy points How specific do we need to be about when the one-dimensional assumption is valid? Is it enough to say that dA/dx is small? (MP HT.2) Why is the thermal conductivity of light gases such as helium (monoatomic) or hydrogen (diatomic) much higher than heavier gases such as argon (monoatomic) or nitrogen (diatomic)? (MP HT.3)
2.2 Thermal resistance circuits There is an electrical analogy with conduction heat transfer that can be exploited in problem oIving. The analog of o is current, and the analog of the temperature difference T- T2, is voltage difference. From this perspective the slab is a pure resistance to heat transfer and we can define where r=L/kA. the thermal resistance. the thermal resistance r increases as l increases as a decreases. and as k decreases The concept of a thermal resistance circuit allows ready analysis of problems such as a composite slab(composite planar heat transfer surface). In the composite slab shown in Figure 2.5, the heat flux is constant with x. The resistances are in series and sum to R=R,+R, If Ti is the temperature at the left, and Tr is the temperature at the right, the heat transfer rate is given by (2.20) RR,+R2 W RI R2 Figure 2.5: Heat transfer across a composite slab(series thermal resistance) Another example is a wall with a dissimilar material such as a bolt in an insulating layer. In this case, the heat transfer resistances are in parallel. Figure 2.6 shows the physical configuration the heat transfer paths and the thermal resistance circuit
HT-11 2.2 Thermal Resistance Circuits There is an electrical analogy with conduction heat transfer that can be exploited in problem solving. The analog of Q& is current, and the analog of the temperature difference, T1 - T2, is voltage difference. From this perspective the slab is a pure resistance to heat transfer and we can define R T T Q 1 − 2 = & (2.19) where R = L/kA, the thermal resistance. The thermal resistance R increases as L increases, as A decreases, and as k decreases. The concept of a thermal resistance circuit allows ready analysis of problems such as a composite slab (composite planar heat transfer surface). In the composite slab shown in Figure 2.5, the heat flux is constant with x. The resistances are in series and sum to R = R1 + R2. If TL is the temperature at the left, and TR is the temperature at the right, the heat transfer rate is given by R1 R2 T T R T T Q L R L R + − = − = & . (2.20) 1 2 TL Q& TR R1 R2 x Figure 2.5: Heat transfer across a composite slab (series thermal resistance) Another example is a wall with a dissimilar material such as a bolt in an insulating layer. In this case, the heat transfer resistances are in parallel. Figure 2.6 shows the physical configuration, the heat transfer paths and the thermal resistance circuit
R k R Figure 2.6: Heat transfer for a wall with dissimilar materials(Parallel thermal resistance) For this situation, the total heat flux o is made up of the heat flux in the two parallel paths 0=0+0, with the total resistance given by 111 R RI R2 (2.21) bog More complex configurations can also be examined; for example, a brick wall with insulation th sides 0.1m T1=150°C T4=10° Insulation 0.03 Figure 2.7: Heat transfer through an insulated wall The overall thermal resistance is given by R=R,+R+r- 4 L2,L3 (2.22) K,, k2A, k3A Some representative values for the brick and insulation thermal conductivity are
HT-12 R2 Q& k2 k1 k1 R1 model Figure 2.6: Heat transfer for a wall with dissimilar materials (Parallel thermal resistance) For this situation, the total heat flux Q& is made up of the heat flux in the two parallel paths: Q Q1 Q2 & & & = + with the total resistance given by: 1 2 1 1 1 R R R = + . (2.21) More complex configurations can also be examined; for example, a brick wall with insulation on both sides. T2 Brick 0.1 m T1 = 150 °C T4 = 10 °C T2 T3 Insulation 0.03 m T1 T3 T4 R1 R2 R3 Figure 2.7: Heat transfer through an insulated wall The overall thermal resistance is given by 3 3 3 2 2 2 1 1 1 1 2 3 k A L k A L k A L R = R + R + R = + + . (2.22) Some representative values for the brick and insulation thermal conductivity are:
krick=k=0.7 W/m-K kinsulation =k,=k3=0.07 W/m-K Using these values, and noting that A1=A2=A3= A, we obtain: L 0.03m AR,=AR, 0.07 W/mK =0.42m2KW AR、=2=01m=0.14m2K k, 0.7 W/mK This is a series circuit so T-T 140K A onstant throughout =142Wm RA 0.98m- K/W T-T Figure 2.8: Temperature distribution through an insulated wall The temperature is continuous in the wall and the intermediate temperatures can be found from applying the resistance equation across each slab, since o is constant across the slab. For example, to find T2 T-T R 4142 W/m2 This yields T1-T2=60K or T2=90C The same procedure gives T3=70C. As sketched in Figure 2.8, the larger drop is across the insulating layer even though the brick layer is much thicker. Muddy points What do you mean by continuous? (MP HT. 4) Why is temperature continuous in the composite wall problem? Why is it continuous at the interface between two materials? (MP HT.5 HT-13
HT-13 kbrick = k2 = 0.7 W/m-K kinsulation = k1 = k3 = 0.07 W/m-K Using these values, and noting that A1 = A2 = A3 = A, we obtain: 0.42 m K/W 0.07 W/m K 0.03 m 2 1 1 1 = 3 = = = k L AR AR 0.14 m K/W 0.7 W/m K 0.1m 2 2 2 2 = = = k L AR . This is a series circuit so 2 2 1 4 142 W/m 0.98 m K/W 140 K constant throughout = = − = = = RA T T A Q q & & x 1 4 4 T T T T − − 0 1.0 1 2 3 4 Figure 2.8: Temperature distribution through an insulated wall The temperature is continuous in the wall and the intermediate temperatures can be found from applying the resistance equation across each slab, since Q& is constant across the slab. For example, to find T2: 2 1 1 2 = 142 W/m − = R A T T q& This yields T1 – T2 = 60 K or T2 = 90 °C. The same procedure gives T3 = 70 °C. As sketched in Figure 2.8, the larger drop is across the insulating layer even though the brick layer is much thicker. Muddy points What do you mean by continuous? (MP HT.4) Why is temperature continuous in the composite wall problem? Why is it continuous at the interface between two materials? (MP HT.5)
Why is the temperature gradient dT/dx not continuous?(MP HT.6) Why is AT the same for the two elements in a parallel thermal circuit? Doesn't the relative area of the bolt to the wood matter?(MP HT7) 2. 3 Steady Quasi-One-Dimensional Heat Flow in Non-Planar Geometry The quasi one-dimensional equation that has been developed can also be applied to non-planar geometries. An important case is a cylindrical shell, a geometry often encountered in situations where fluids are pumped and heat is transferred. The configuration is shown in Figure 2.9 control volume Figure 2.9: Cylindrical shell geometry notation For a steady axisymmetric configuration, the temperature depends only on a single coordinate(r) and Equation(2. 12b)can be written as (2.23) or, since A= 2Tr 0. dr dr (2.24) The steady-flow energy equation(no flow, no work) tells us that o=0.or dh (2.25) The heat transfer rate per unit length is given by Q=-k·2πr
HT-14 Why is the temperature gradient dT/dx not continuous? (MP HT.6) Why is ∆T the same for the two elements in a parallel thermal circuit? Doesn't the relative area of the bolt to the wood matter? (MP HT.7) 2.3 Steady Quasi-One-Dimensional Heat Flow in Non-Planar Geometry The quasi one-dimensional equation that has been developed can also be applied to non-planar geometries. An important case is a cylindrical shell, a geometry often encountered in situations where fluids are pumped and heat is transferred. The configuration is shown in Figure 2.9. r1 control volume r1 r2 r2 Figure 2.9: Cylindrical shell geometry notation For a steady axisymmetric configuration, the temperature depends only on a single coordinate (r) and Equation (2.12b) can be written as k d dr A r dT dr ( ) 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = (2.23) or, since A = 2π r, d dr r dT dr 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = . (2.24) The steady-flow energy equation (no flow, no work) tells us that Qin Qout & & = or = 0 dr dQ& (2.25) The heat transfer rate per unit length is given by Q k2 r dT dr ⋅ =− ⋅ π