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5 Laminar Internal Flow

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5 Laminar Internal Flow 5 Laminar Internal Flow 5.1 Governing Equations These notes will cover the general topic of heat transfer to a laminar flow of fluid that is moving through an enclosed channel. This channel can take the form of a simple circular pipe, or more complicated geometries such as a rectangular duct or an annulus. It will be assumed that the flow has constant thermophysical properties (including density). We will ¯rst examine the case where the flow is fully developed. This condition implies that the flow and temperature ¯elds retain no history of the inlet of the pipe. In regard to momentum conservation, FDF corresponds to a velocity pro¯le that is independent of axial position x in the pipe. In the case of a circular pipe, u = u(r) where u is the axial component of velocity and r is the radial position. The momentum and continuity equations will show that there can only be an axial component to velocity (i.e., zero radial component), and that µ ¶ ³ r ´2 u(r) = 2u 1 ¡ (1) m R with the mean velocity given by Z Z 1 2 R um = u dA = u r dr (2) A A R 0 The mean velocity provides a characteristic velocity based on the mass flow rate; m_ = ½ um A (3) where A is the cross sectional area of the pipe. The thermally FDF condition implies that the dimensionless temperature pro¯le is independent of axial position. The dimensionless temperature T is de¯ned by T ¡ T T = s = func(r); TFDF (4) Tm ¡ Ts in which Ts is the surface temperature (which can be a function of position x) and Tm is the mean temper- ature, de¯ned by Z 1 Tm = u T dA (5) um A A This de¯nition is consistent with a global 1st law statement, in that m_ CP (Tm;2 ¡ Tm;1) = Q_ 1¡2 (6) with CP being the speci¯c heat of the fluid. A di®erential form of this law is dT dA mC_ m = q00 s = q00 P (7) P dx s dx s 00 in which qs is the surface heat flux, which may be a function of x, and As and P are the pipe surface area and perimeter. Note that integration of Eq. (7) over the length of the pipe results in Eq. (6). The thermally FDF condition, in Eq. (4), also states that the heat transfer coe±cient h is constant. The convection law for internal flow is de¯ned by ¯ ¯ ¯ ¯ 00 @T ¯ k(Ts ¡ Tm) dT ¯ qs = h(Ts ¡ Tm) = k ¯ = ¡ ¯ (8) @r R R dr 1 with r = r=R being the dimensionless r coordinate. There is no negative sign in Fourier's law, because the radial coordinate points toward the surface. Rearranging, ¯ ¯ hD dT ¯ NuD = = ¡2 ¯ = constant (9) k dr 1 The governing DE for the temperature distribution in the fluid is µ µ ¶ ¶ @T 1 @ @T @2T ½ C u = kr2T = k r + (10) P @x r @r @r @x2 1 with boundary conditions (for the circular pipe case) ¯ ¯ ¯ ¯ @T ¯ @T ¯ 00 ¯ = 0;T (r = R) = Ts or ¡ k ¯ = qs (11) @r 0 @r R The speci¯c form of the wall BC would obviously depend on the given conditions, i.e., speci¯ed temperature, heat flux, or something more complicated. A convection{type BC is not an option here, as we are trying to determine the heat transfer coe±cient h from a detailed formulation of convective{di®usive energy transport. No BCs are given for the axial direction (we would need two, typically an inlet and an outlet condition, to close the problem) because, as will be developed below, such will not be needed for the FDF case. 5.2 Fully Developed Flow The second term on the right hand side of Eq. (10), which accounts for axial di®usion of heat, can usually be neglected in comparison to the ¯rst for the FDF condition. Nevertheless, Eq. (10) remains a partial DE (2nd order in r, 1st in x), and it would appear that techniques used to solve PDEs must be employed. This, however, will not be the case per the FDF condition: an ODE will emerge once the equation is recast with T as the dependent variable. Doing this, however, requires addressing the convection term on the left hand side. Using Eq. (4), it follows that @T dT dT = s (1 ¡ T ) + m T (12) @x dx dx Rather than substituting this formula into Eq. (10), it is simpler to ¯rst consider the form of the boundary condition at the wall. Two important cases are constant heat flux and constant Ts. 5.2.1 Constant heat flux 00 From Eqs. (8) and (12), the qs = constant case results in @T dT dT = s = m (13) @x dx dx 00 The di®erential energy balance in Eq. (7) can be used to relate the axial temperature derivative to qs , and 00 Eq. (8) can then be used to replace qs with the convection law. Likewise, the radial conduction (or di®usion) term on the right can be made dimensionless via introduction of T and r. Even though both Ts and Tm will be functions of x for this case, they can be brought inside the radial derivatives (either multiplicatively or additively) because they are not functions of r. After rearranging terms and using the circular pipe con¯guration, the result is µ ¶ 1 d dT ¡u Nu = r (14) D r dr dr in which u = u=um is the dimensionless velocity pro¯le. The DE is ordinary, as opposed to partial, because of T being a function solely of r. The boundary conditions are ¯ ¯ @T ¯ ¯ = 0; T (1) = 0 (15) @r 0 Equation (14) is an inhomogeneous ODE, with two homogeneous BCs in Eq. (15). The solution, which can be obtained by direct integration, is Nu ¡ ¢ T = D 3 ¡ 4r2 + r4 (16) 8 The Nusselt number { which is currently unknown and is sought from the analysis { is obtained from the mean temperature condition. On a dimensionless basis, the mean temperature is Z 1 T m = 2 u T r dr = 1 (17) 0 and this results in 48 Nu = ¼ 4:364; constant q00 (18) D 11 s 2 5.2.2 Constant surface temperature From Eq. (12), the Ts = constant case results in @T dT = m T (19) @x dx 00 By comparison with Eq. (13), the only change from the qs = constant formulation is that the convection term will now contain T , i.e., µ ¶ 1 d dT ¡u Nu T = r (20) D r dr dr The boundary conditions are the same as before, and the mean temperature condition of Eq. (17) holds as well. Equation (20) is an eigenvalue problem, the eigenvalue being the Nusselt number NuD. Observe that both the DE and the BCs are homogeneous1. Therefore, there can only exist non{trivial (i.e., non{zero) solutions to this problem for a unique value (or values) of the eigenvalue NuD. A simple example of an eigenvalue problem is the homogeneous boundary value problem f 00(x) + c2f(x) = 0; f 0(x = 0) = 0; f(x = 1) = 0 The solution to this problem is f = cos(c x) and the solution also has an in¯nite number of eigenvalues c: c = ¼=2; 3¼=2; 5¼=2;::: The problem we face in Eq. (20) is much the same. Unlike the simple example given above, however, there is no general solution to the ODE in Eq. (20) { at least not in terms of commonly{known mathematical functions like sin and cos. It turns out that a general solution to Eq. (20), in the form of ordinary Bessel functions, can be found for the special case of u = 1 (a case known as slug flow). Our interest, however, is for the laminar, parabolic flow case. An iterative strategy can be used to determine NuD for the constant temperature case. The approach is to write Eq. (20) as µ ¶ 1 d dT ¡u Nu T = r n (21) Dn n¡1 r dr dr th where T n refers to the n iterated solution to Eq. (21). The BCs on T n are the same as before. Note that T n(r) will be a function of T n¡1 and NuDn. In addition, each iterate T n must satisfy the mean condition Z 1 2 u T n r dr = 1 (22) 0 and this provides a relation to determine NuDn. For a su±ciently large number of iterations n, the procedure will give NuDn¡1 ¼ NuDn = NuD, i.e., the iterated Nusselt numbers will converge to a constant. A starting point in the process is to let T 0 = 1. You should be able to show, using the previous section 00 results, that the ¯rst iteration gives NuD1 = 48=11, i.e., the constant qs result. Each iterated solution T n can be obtained by direct integration of Eq. (21). For the given T 0, each solution T n will be a polynomial in r of degree 4n { and you can see that the algebra can quickly get complicated. It helps to use a symbolic package such as Mathematica or Maple to solve this problem. The following illustrates how the problem would be solved in mathematica. The ¯rst two lines set the velocity pro¯le and the T 0 = 1 condition; u[r_] := 2(1 - r^2); tnm1[r_] := 1 and the next set of instructions ¯nd the T n = tn[r] solution and NuDn value using the current value of T n¡1 = tnm1[r].
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