Question 7.12: A transparent gas with mean velocity um and constant physica...
A transparent gas with mean velocity u_m and constant physical properties flows through a circular tube of inner diameter D_i and length I (Figure 7.21). A specified heat flux q_e(x) = q_{max} sin(πx/I) is applied along the tube length. The heat transfer coefficient h between the gas and the tube interior surface is assumed independent of x. The tube wall is thin and has thermal conductivity k_n. The gas enters the tube from a large plenum at T_{g,1}. The gas leaves the tube at T_{g,2} and enters a mixing plenum that is also at T_{g,2} . The tube interior surface is diffuse-gray with emissivity ϵ. Set up the energy equations and boundary conditions to determine the wall temperature T(x). Put the equations into a finite-difference form for numerical solution.
Following Examples 7.8 and 7.9, the governing energy equations are
\epsilon \left\{Θ^4(X)+\int_{Z=0}^{X}{\left[\frac{1-\epsilon }{\epsilon }\Phi (Z)-Θ^4(Z) \right]dF_{dX-dZ}(X-Z)+ \int_{Z=X}^{L}{\left[\frac{1-\epsilon }{\epsilon }\Phi (Z)-Θ^4(Z) \right]dF_{dX-dZ}(Z-X)}-Θ_{r,1}^4F_{dX-1}(X)-Θ_{r,2}^4F_{dX-2}(L-X)} \right\}=\Phi (X) (7.42)
\Phi (X)=\sin \left\lgroup\frac{\pi X}{L} \right\rgroup+H[Θ_g(X)-Θ(X)]+N\frac{d^2Θ}{dX^2} (7.43)
Θ_g(X)=Ste^{-StX} \int_{0}^{X}{e^{StZ}Θ(Z)dZ+Θ_{g,1}e^{-StX}} (7.44)
In this form Θ=T(σ/q_{max})^{1/4} and the other parameters are as defined in Examples 7.7 and 7.9 (the reference value of q_e is q_{max} and the reference length is D_i). Substituting the last two equations to eliminate Φ(X) and Θ_g(X) in Equation 7.42 results in
\epsilon Θ^4(X)+\int_{0}^{L}{\left[(1-\epsilon )\left\{\sin \left\lgroup\frac{\pi Z}{L} \right\rgroup+H[Ste^{-StZ}\int_{0}^{Z}{e^{St\xi}} Θ(\xi ) d\xi +Θ_{g,1}e^{-StZ}-Θ(Z)]+N\frac{d^2Θ}{dZ^2} \right\}- \epsilon Θ^4(Z)\right] dF_{dX-dZ}(\left|X-Z\right| )} (7.45)
=N\frac{d^2Θ}{dX^2}+\sin \left\lgroup\frac{\pi X}{L} \right\rgroup+H[Ste^{StX}\int_{0}^{X}{e^{St}}Θ(Z) dZ +Θ_{g,1}e^{-StX}-Θ(X)]+Θ_{r,1}^4dF_{dX-1}(X)+Θ_{r,2}^4dF_{dX-2}(L-X)
For the boundary conditions required by the d^2Θ/dX^2 term, both end edges of the tube are assumed to have negligible heat losses, so (dΘ/dX)_{X=0}=(dΘ/dX)_{X=L}=0. The condition Θ_g(X=0)=Θ_{g,1} was used in deriving Equation 7.44. To proceed with the numerical solution, define
f(X,Z)=\left\lgroup\frac{1-\epsilon }{\epsilon } \left\{\sin \left\lgroup\frac{\pi Z}{L} \right\rgroup+H\left[Ste^{-StZ}\int_{0}^{Z}{}g(\xi ) d\xi +Θ_{g,1}e^{-StZ}-Θ(Z)\right]+N\frac{d^2Θ}{dZ^2} \right\}-Θ^4(Z) \right\rgroup\frac{dF_{dX-dZ}(\left|X-Z\right| )}{dZ} (7.46)
where g(\xi)=e^{St\xi}Θ(\xi ) . Equation 7.45 becomes
\epsilon \left[Θ^4(X)+\int_{0}^{L}{f(X,Z)dZ} \right]=N\frac{d^2Θ}{dZ^2}+\sin \left\lgroup\frac{\pi X}{L} \right\rgroup+ H\left[Ste^{-StX}\int_{0}^{X}{}g(Z ) dZ +Θ_{g,1}e^{-StX}-Θ(X)\right]+Θ_{g,1}^4F_{dX-1}(X)+Θ_{g,2}^4F_{dX-2}(L-X) (7.47)
Numerical integration is applied to each integral, and a set of nonlinear algebraic equations is obtained as in Example 7.11. At X_i = iΔX, Equation 7.47 becomes, by use of the trapezoidal rule and having ΔX = ΔZ, where ΔX = L/I (I is the number of ΔX increments),
\epsilon \left\{Θ_i^4(X)+\Delta X\left[\frac{1}{2}f_0(X_i)+\sum\limits_{j=1}^{I-1}{f_i(X_i)+\frac{1}{2}f_I(X_i) } \right] \right\}=N\frac{Θ_{i+1}-2Θ_i+Θ_{i-1}}{(\Delta X)^2}+\sin \left\lgroup\frac{X_i}{L} \right\rgroup +H\left\{Ste^{-StX_i}\Delta X\left[\frac{1}{2}g_0+\sum\limits_{j=1}^{i-1}{g_j}+\frac{1}{2}g_i \right]+Θ_{g,1}e^{-StX_i}-Θ_i \right\} +Θ_{r,1}^4F_{dX_i-1}(X_i)+Θ_{r,2}^4F_{dX_i-2}(L-X_i) (7.48)
Note that, for the reservoir temperatures for the specified conditions of this example,Θ_{r,1}= Θ_{g,1} and Θ_{r,2}= Θ_{g,2}.By gathering terms after expansion as in Example 7.11, the full set of equations can be written as
(A_{00}Θ_0+B_{00}Θ_0^4)+(A_{01}Θ_1+B_{01}Θ_1^4)+ —- +—-+(A_{0I}Θ_I+B_{0I}Θ_0I^4)=C_0
(A_{10}Θ_0+B_{10}Θ_0^4)+(A_{11}Θ_1+B_{11}Θ_1^4)+ —- +—-+(A_{1I}Θ_I+B_{1I}Θ_1I^4)=C_1
—– + —– + —– +—-+ —– = — (7.49a)
(A_{i0}Θ_0+B_{i}Θ_0^4)+—-+(A_{ij}Θ_j+B_{ij}Θ_j^4) +—-+(A_{iI}Θ_I+B_{iI}Θ_iI^4)=C_i
(A_{I0}Θ_0+B_{I0}Θ_0^4)+—-+(A_{ij}Θ_j+B_{ij}Θ_j^4) +—-+(A_{II}Θ_I+B_{II}Θ_iI^4)=C_I
This has the matrix form
[A_{ij} ][Θ_j ] [B_{ij} ][ Θ_j^4] =[C_i ] (7.49b)
Once Θ_j is determined from the solution of Equation 7.49, the heat flux Φ_j can be found from Equation 7.43 if required.