1.0 Heat Transfer modes Heat transfer processes are classified into three types. The first is conduction, which is defined as transfer of heat occurring through intervening matter without bulk motion of the matter. Figure 1. 1 shows the process pictorially. A solid (a block of metal, say) has one surface at a high temperature and one at a lower temperature. This type of heat conduction can occur, for example, through a turbine blade in a jet engine. The outside surface, which is exposed to gases from the combustor, is at a higher temperature than the inside surface, which has cooling air next to it. The level of the wall temperature is critical for a turbine blade. Heat“ flows” to right(q) Solid Figure 1.1: Conduction heat transfer The second heat transfer process is convection, or heat transfer due o a nowi ing fluid. The fluid can be a gas or a liquid; both have applications in aerospace technology. In convection heat transfer, the heat is moved through bulk transfer of a non-uniform temperature fluid The third process is radiation or transmission of energy through space without the necessary presence of matter. Radiation is the only method for heat transfer in space. Radiation can be important even in situations in which there is an intervening medium; a familiar example is the heat transfer from a glowing piece of metal or from a fire. Muddy points How do we quantify the contribution of each mode of heat transfer in a given situation? (MP HT. 1) 2.0 Conduction heat transfer We will start by examining conduction heat transfer. We must first determine how to relate the heat transfer to other properties(either mechanical, thermal, or geometrical). The answer to this is rooted in experiment, but it can be motivated by considering heat flow along a"bar"between two heat reservoirs at TA, TB as shown in Figure 2. 1. It is plausible that the heat transfer rate @, is
HT-5 1.0 Heat Transfer Modes Heat transfer processes are classified into three types. The first is conduction, which is defined as transfer of heat occurring through intervening matter without bulk motion of the matter. Figure 1.1 shows the process pictorially. A solid (a block of metal, say) has one surface at a high temperature and one at a lower temperature. This type of heat conduction can occur, for example, through a turbine blade in a jet engine. The outside surface, which is exposed to gases from the combustor, is at a higher temperature than the inside surface, which has cooling air next to it. The level of the wall temperature is critical for a turbine blade. Thigh Tlow Solid Heat “flows” to right ( q&) Figure 1.1: Conduction heat transfer The second heat transfer process is convection, or heat transfer due to a flowing fluid. The fluid can be a gas or a liquid; both have applications in aerospace technology. In convection heat transfer, the heat is moved through bulk transfer of a non-uniform temperature fluid. The third process is radiation or transmission of energy through space without the necessary presence of matter. Radiation is the only method for heat transfer in space. Radiation can be important even in situations in which there is an intervening medium; a familiar example is the heat transfer from a glowing piece of metal or from a fire. Muddy points How do we quantify the contribution of each mode of heat transfer in a given situation? (MP HT.1) 2.0 Conduction Heat Transfer We will start by examining conduction heat transfer. We must first determine how to relate the heat transfer to other properties (either mechanical, thermal, or geometrical). The answer to this is rooted in experiment, but it can be motivated by considering heat flow along a "bar" between two heat reservoirs at TA, TB as shown in Figure 2.1. It is plausible that the heat transfer rate Q& , is a
function of the temperature of the two reservoirs, the bar geometry and the bar properties.(Are there other factors that should be considered? If so, what? ) This can be expressed as 0=fiTA, TB, bar geometry, bar properties) It also seems reasonable to postulate that o should depend on the temperature difference TA-TB. If Ta-TB is zero then the heat transfer should also be zero. The temperature dependence can therefore Q=f[(TA-TB), TA, bar geometry, bar propertie (22) Figure 2.1: Heat transfer along a bar An argument for the general form of f2 can be made from physical considerations. On requirement, as said, is f2=0 if TA= TB. Using a MacLaurin series expansion, as follows. e f△D=f0)+ △T+ o(△D If we define AT=TA-TB and f=fi, we find that(for small TA-TB), 21-7B)=Q=1+0 a(Ta-TB)7 TA-TB+ We know that f2(0)=0. The derivative evaluated at TA= TB(thermal equilibrium)is a measurable property of the bar. In addition, we know that 2>0 if T>Taor >0. It also seems reasonable that if we had two bars of the same area, we would have twice the heat transfer so that we can postulate that O is proportional to the area. Finally, although the argument is by no means rigorous, experience leads us to believe that as L increases o should get smaller. All of these lead to the generalization(made by Fourier in 1807)that, for the bar, the derivative in equation(2. 4) has the form T-6
HT-6 function of the temperature of the two reservoirs, the bar geometry and the bar properties. (Are there other factors that should be considered? If so, what?). This can be expressed as Q& = f1 (TA , TB , bar geometry, bar properties) (2.1) It also seems reasonable to postulate that Q& should depend on the temperature difference TA - TB. If TA – TB is zero, then the heat transfer should also be zero. The temperature dependence can therefore be expressed as Q& = f2 [ (TA - TB), TA, bar geometry, bar properties] (2.2) L TA TB Q& Figure 2.1: Heat transfer along a bar An argument for the general form of f2 can be made from physical considerations. One requirement, as said, is f2 = 0 if TA = TB. Using a MacLaurin series expansion, as follows: f( T) f(0) f ( T) T 0 ∆ ∆ = + ∆ ∂ ∂ +L (2.3) If we define ∆T = TA – TB and f = f2, we find that (for small TA – TB), f (T T ) Q f (0) f (T T ) TT . 2A B 2 2 A B TA TB 0 − == + A B ∂ ∂ − ( ) − + ⋅ − = L (2.4) We know that f2(0) = 0 . The derivative evaluated at TA = TB (thermal equilibrium) is a measurable property of the bar. In addition, we know that Q TT f T T A B 2 A B ⋅ > > ∂ ∂ − ( ) 0 0 if or > . It also seems reasonable that if we had two bars of the same area, we would have twice the heat transfer, so that we can postulate that Q& is proportional to the area. Finally, although the argument is by no means rigorous, experience leads us to believe that as L increases Q& should get smaller. All of these lead to the generalization (made by Fourier in 1807) that, for the bar, the derivative in equation (2.4) has the form
pen iIn equation(2.5), k is a proportionality factor that is a function of the material and the temperature, A is the cross-sectional area and L is the length of the bar. In the limit for any temperature difference ATacross a length Ax as both L, Ta-TB-0, we can say dT O=kA (A-n)2k4(B-T)=-A (2.6) L L A more useful quantity to work with is the heat transfer per unit area, defined as 旦 A The quantity q is called the heat flux and its units are Watts/m". The expression in(2.6)can be written in terms of heat flux as Equation 2.8 is the one-dimensional form of Fourier's law of heat conduction. The proportionality constant k is called the thermal conductivity. Its units are W/m-K. Thermal conductivity is a well-tabulated property for a large number of materials. Some values for familiar materials are given in Table 1; others can be found in the references. The thermal conductivity is a function of temperature and the values shown in table l are for room temperature Table 2.1: Thermal conductivity at room temperature for some metals and non-metals Metals Fe Steel kWmK]4203902007050 -metals H,0 Air Engine oil /m-K 0.6 0.026 10is Brick Wood Cork 2004
HT-7 ∂ ∂ − ( ) = − = f T T kA L 2 A B TA TB 0 . (2.5) In equation (2.5), k is a proportionality factor that is a function of the material and the temperature, A is the cross-sectional area and L is the length of the bar. In the limit for any temperature difference ∆T across a length ∆x as both L, TA - TB → 0, we can say ( ) ( ) dx dT kA L T T kA L T T Q kA A B B A = − − = − − = & . (2.6) A more useful quantity to work with is the heat transfer per unit area, defined as q A Q & & = . (2.7) The quantity q& is called the heat flux and its units are Watts/m2 . The expression in (2.6) can be written in terms of heat flux as dx dT q& = −k . (2.8) Equation 2.8 is the one-dimensional form of Fourier's law of heat conduction. The proportionality constant k is called the thermal conductivity. Its units are W / m-K. Thermal conductivity is a well-tabulated property for a large number of materials. Some values for familiar materials are given in Table 1; others can be found in the references. The thermal conductivity is a function of temperature and the values shown in Table 1 are for room temperature. Table 2.1: Thermal conductivity at room temperature for some metals and non-metals Metals Ag Cu Al Fe Steel k [W/m-K] 420 390 200 70 50 Non-metals H20 Air Engine oil H2 Brick Wood Cork k [W/m-K] 0.6 0.026 0.15 0.18 0.4 -0 .5 0.2 0.04
2. 1 Steady-State One-Dimensional Conduction Insulated (no heat transfer (x) t o(x+dx) Figure 2.2: One-dimensional heat conductio For one-dimensional heat conduction( temperature depending on one variable only), we can devise a basic description of the process. The first law in control volume form(steady flow energy equation) with no shaft work and no mass flow reduces to the statement that 2o for all surfaces=0 (no heat transfer on top or bottom of figure 2.2). From equation(2. 8), the heat transfer rate in at the left(at x)is A (2.9) The heat transfer rate on the right is our+ dr)=a(x)+dg dr+ (2.10) Using the conditions on the overall heat flow and the expressions in(2.9)and (2.10) x)-cx)+2(x)dx+…|=0 (211) Taking the limit as dx approaches zero we obtain de(x) 0
HT-8 2.1 Steady-State One-Dimensional Conduction Q( )x & Q( ) x + dx & dx x Insulated (no heat transfer) Figure 2.2: One-dimensional heat conduction For one-dimensional heat conduction (temperature depending on one variable only), we can devise a basic description of the process. The first law in control volume form (steady flow energy equation) with no shaft work and no mass flow reduces to the statement that ΣQ& for all surfaces = 0 (no heat transfer on top or bottom of figure 2.2). From equation (2.8), the heat transfer rate in at the left (at x) is Qx kA ˙ dT dx x ( ) = − ⎛ ⎝ ⎞ ⎠ . (2.9) The heat transfer rate on the right is ˙ ˙ ˙ Q x dx Q x dQ dx dx x ( ) + = ( ) + +L. (2.10) Using the conditions on the overall heat flow and the expressions in (2.9) and (2.10) ˙ ˙ ˙ Q Q Q x x d dx ( ) − ( ) + ( ) x dx 0 + ⎛ ⎝ ⎜ ⎞ ⎠ L⎟ = . (2.11) Taking the limit as dx approaches zero we obtain dQ x dx ˙( ) = 0 , (2.12a) or
d/ dT d (2.12b) If k is constant (i.e. if the properties of the bar are independent of temperature), this reduces to d dT 0 (213a) or(using In ru dt+(lda dr dx dx (2.13b) Equations(2. 13a)or(2. 13b)describe the temperature field for quasi-one-dimensional steady state (no time dependence) heat transfer. We now apply this to some examples Example 2. 1 Heat transfer through a plane slab T=T x=0 L Figure 2.3: Temperature boundary conditions for a slab For this configuration, the area is not a function of x, 1. e. A= constant. Equation(2. 13)thus became =0 (214) Equation(2. 14) can be integrated immediately to yield (215) HT-9
HT-9 d dx kA dT dx 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = . (2.12b) If k is constant (i.e. if the properties of the bar are independent of temperature), this reduces to d dx A dT dx 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = (2.13a) or (using the chain rule) d T dx 1 A dA dx dT dx 0 2 2 + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = . (2.13b) Equations (2.13a) or (2.13b) describe the temperature field for quasi-one-dimensional steady state (no time dependence) heat transfer. We now apply this to some examples. Example 2.1: Heat transfer through a plane slab x T = T1 T = T2 Slab x = 0 x = L Figure 2.3: Temperature boundary conditions for a slab For this configuration, the area is not a function of x, i.e. A = constant. Equation (2.13) thus became 0 2 2 = dx d T . (2.14) Equation (2.14) can be integrated immediately to yield a dx dT = (2.15)