Equation(2. 24)is a second order differential equation for T. Integrating this equation once gives dT (2.26) where a is a constant of integration. Equation(2. 26) can be written as du (2.27) where both sides of equation(2.27)are exact differentials. It is useful to cast this equation in terms of a dimensionless normalized spatial variable so we can deal with quantities of order unity. To do this, divide through by the inner radius, r'l dt=a (/n) (2.28) Integrating(2.28) yields T=aln(-+b (229) To find the constants of integration a and b, boundary conditions are needed. These will be taken to be known temperatures T, and T2 at r and r? respectively. Applying T= T at r=r gives T= b Applying T=T2 atr = d T F In(2/ri) The temperature distribution is thus In(r/r) h(2/r) (230) As said, it is generally useful to put expressions such as(2.30) into non-dimensional and normalized form so that we can deal with numbers of order unity(this also helps in checking whether results are consistent). If convenient, having an answer that goes to zero at one limit is also useful from the perspective of ensuring the answer makes sense. Equation(2.30)can be put in non- dimensional fo
HT-15 Equation (2.24) is a second order differential equation for T. Integrating this equation once gives a dr dT r = . (2.26) where a is a constant of integration. Equation (2.26) can be written as r dr dT = a (2.27) where both sides of equation (2.27) are exact differentials. It is useful to cast this equation in terms of a dimensionless normalized spatial variable so we can deal with quantities of order unity. To do this, divide through by the inner radius, r1 ( ) ( )1 1 / / r r d r r dT = a (2.28) Integrating (2.28) yields T a r r b 1 = ⎛ ⎝ ⎜ ⎞ ⎠ ln ⎟ + . (2.29) To find the constants of integration a and b, boundary conditions are needed. These will be taken to be known temperatures T1 and T2 at r1 and r2 respectively. Applying T = T1 at r = r1 gives T1 = b. Applying T = T2 at r = r2 yields 1 1 2 2 ln T r r T = a + , or ( ) 2 1 2 1 ln r /r T T a − = . The temperature distribution is thus ( ) ( ) ( ) 1 2 1 1 2 1 ln / ln / T r r r r T = T − T + . (2.30) As said, it is generally useful to put expressions such as (2.30) into non-dimensional and normalized form so that we can deal with numbers of order unity (this also helps in checking whether results are consistent). If convenient, having an answer that goes to zero at one limit is also useful from the perspective of ensuring the answer makes sense. Equation (2.30) can be put in nondimensional form as
72-7hn(2/) (231) The heat transfer rate, 0, is given by -kA (2-)1,-k(G-7) n(/r)In per unit length. The thermal resistance R is given by 2/) R (232) R The cylindrical geometry can be viewed as a limiting case of the planar slab problem. to make the connection, consider the case when <<1. From the series expansion for In(1+x) we recall that (233) (Look it up, try it numerically, or use the binomial theorem on the series below and integrate term by term =1-x+x2+…) The logarithms in Equation(2.31)can thus be written as Inl|+r-sr- and In 2=2- (234) in the limit of (r2-rn)<<rl. USing these expressions in equation(2.30)gives T1 (235) With the substitution of r-r=x, and r-ri=l we obtain T+(2-7) (2.36) HT-16
HT-16 ( ) ( ) 2 1 1 2 1 1 ln / ln / r r r r T T T T = − − . (2.31) The heat transfer rate, Q& , is given by ( ) ( ) ( ) ( ) 2 1 1 2 2 1 1 2 1 1 ln / 2 1 ln / 2 r r k T T r r r T T r k dr dT Q kA − = − = − =− π π & per unit length. The thermal resistance R is given by ( ) k r r R 2π ln / 2 1 = (2.32) R T T Q 1 − 2 = & . The cylindrical geometry can be viewed as a limiting case of the planar slab problem. To make the connection, consider the case when 1 1 2 1 << − r r r . From the series expansion for ln (1 + x) we recall that ln 1+ x x - x 2 + x 3 + 2 3 ( ) ≈ K (2.33) (Look it up, try it numerically, or use the binomial theorem on the series below and integrate term by term. = − + +K + 2 1 1 1 x x x ) The logarithms in Equation (2.31) can thus be written as ln ln 1 r r r r r r r r r r r 1 1 1 1 2 1 2 1 1 + ⎛ − ⎝ ⎜ ⎞ ⎠ ⎟ ≅ − ≅ − and (2.34) in the limit of (r2 – r1) << r1. Using these expressions in equation (2.30) gives ( )( ) ( ) 1 2 1 1 2 1 T r r r r T T T + − − = − . (2.35) With the substitution of r – r1 = x, and r2 – r1 = L we obtain ( )L x T = T1 + T2 −T1 (2.36)
which is the same as equation(2. 17). The plane slab is thus the limiting case of the cylinder if (r r1/r<<l, where the heat transfer can be regarded as taking place in(approximately )a planar slab To see when this is appropriate, consider the expansion In(1 which is the ratio of heat flux for a cylinder and a plane slab Table 2.2: Utility of plane slab approximation 2.3 In(1 95.91.87.84.81 For 10% error, the ratio of thickness to inner radius should be less than 0.2, and for 20% error the thickness to inner radius should be less than 0.5 A second example is the spherical shell with specified temperatures T(ri= T and T(r2) T2, as sketched in Figure 2.10 T1、 Figure 2.10: Spherical shell The area is now A(r)=4tr, so the equation for the temperature field is (237) ntegrating equation(2.37)once yield a (238) Integrating again gives HT-17
HT-17 which is the same as equation (2.17). The plane slab is thus the limiting case of the cylinder if (r - r1) / r << 1, where the heat transfer can be regarded as taking place in (approximately) a planar slab. To see when this is appropriate, consider the expansion ( ) x ln 1+ x , which is the ratio of heat flux for a cylinder and a plane slab. Table 2.2: Utility of plane slab approximation x .1 .2 .3 .4 .5 ( ) x ln 1+ x .95 .91 .87 .84 .81 For < 10% error, the ratio of thickness to inner radius should be less than 0.2, and for 20% error, the thickness to inner radius should be less than 0.5. A second example is the spherical shell with specified temperatures T (r1) = T1 and T (r2) = T2, as sketched in Figure 2.10. r1 T T2 1 r2 Figure 2.10: Spherical shell The area is now 2 A(r) = 4πr , so the equation for the temperature field is d dr r dT dr 0 ⎛ 2 ⎝ ⎜ ⎞ ⎠ ⎟ = . (2.37) Integrating equation (2.37) once yields 2 a /r dr dT = . (2.38) Integrating again gives
or, normalizing the spatial variable /n) (239) where a and b are constants of integration. As before, we specify the temperatures at r= rI and r r2. Use of the first boundary condition gives T(=T=a'+b. applying the second boundary condition gives T(2)=7 b 2/r) Solving for aand b T T1-72 i1/r2 In non-dimensional form the temperature distribution is thus (2.41) 721-G/n2
HT-18 b r a T = − + or, normalizing the spatial variable ( ) b r r a T + ′ = 1 / (2.39) where a′ and b are constants of integration. As before, we specify the temperatures at r = r1 and r = r2. Use of the first boundary condition gives T( ) r1 = T1 = a′ + b . Applying the second boundary condition gives ( ) ( ) b r r a T r T + ′ = = 2 1 2 2 / Solving for a′ and b, . 1 / 1 / 1 2 1 2 1 1 2 1 2 r r T T b T r r T T a − − = − − − ′ = (2.40) In non-dimensional form the temperature distribution is thus: ( ) ( ) 1 2 1 1 2 1 1 / 1 / r r r r T T T T − − = − − (2.41)
3.0 Convective Heat Transfer The second type of heat transfer to be examined is convection, where a key problem is determining the boundary conditions at a surface exposed to a flowing fluid. An example is the wall temperature in a turbine blade because turbine temperatures are critical as far as creep(and thus blade) life. A view of the problem is given in Figure 3. 1, which shows a cross-sectional view of a turbine blade. There are three different types of cooling indicated, all meant to ensure that the metal is kept at a temperature much lower than that of the combustor exit flow in which the turbine blade operates. In this case, the turbine wall temperature is not known and must be found as part of the solution to the problem Figure 3. 1: Turbine blade heat transfer configuration To find the turbine wall temperature, we need to analyze convective heat transfer, which means we need to examine some features of the fluid motion near a surface. The conditions near a surface are illustrated schematically in Figure 3.2 C Velocity distribution c=0 at surface (velocity) T TM Figure 3. 2: Temperature and velocity distributions near a surface. HT-19
HT-19 3.0 Convective Heat Transfer The second type of heat transfer to be examined is convection, where a key problem is determining the boundary conditions at a surface exposed to a flowing fluid. An example is the wall temperature in a turbine blade because turbine temperatures are critical as far as creep (and thus blade) life. A view of the problem is given in Figure 3.1, which shows a cross-sectional view of a turbine blade. There are three different types of cooling indicated, all meant to ensure that the metal is kept at a temperature much lower than that of the combustor exit flow in which the turbine blade operates. In this case, the turbine wall temperature is not known and must be found as part of the solution to the problem. Figure 3.1: Turbine blade heat transfer configuration To find the turbine wall temperature, we need to analyze convective heat transfer, which means we need to examine some features of the fluid motion near a surface. The conditions near a surface are illustrated schematically in Figure 3.2. T c∞ Ve locity distribution; c = 0 at surface c (velocity) δ′ T∞ Tw y y Figure 3.2: Temperature and velocity distributions near a surface