Question

Starting with an energy balance on the volume element obtain the three-dimensional transient explicit finite difference equation for a general interior node in rectangular coordinates for T(x, y,z) for the case of constant thermal conductivity and no heat generation. Conductivity and no heat generation.

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Answered 1 year ago
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This is a case of three-dimensional heat transfer for an interior node. Thermal conductivity is constant, there is no heat generation and when the temperature is a function of time, it means that heat transfer is transient.

all sidesQ.+E.gen., element=ΔEelementΔtQ.cond, front+Q.cond, back+Q.cond, left+Q.cond, right+Q.cond, bottom+Q.cond, top+E.gen., element=ΔEelementΔtE.gen., element=0     Q.cond, front+Q.cond, back+Q.cond, left+Q.cond, right+Q.cond, bottom+Q.cond, top=ΔEelementΔt\begin{align*} \sum_{\text{all sides}} \overset{.}{Q} + \overset{.}{E}_{\text{gen., element}} = \frac{\Delta E_{\text{element}}}{\Delta t}\\ \overset{.}{Q}_{\text{cond, front}}+\overset{.}{Q}_{\text{cond, back}}+\overset{.}{Q}_{\text{cond, left}}+\overset{.}{Q}_{\text{cond, right}}+\overset{.}{Q}_{\text{cond, bottom}}+\overset{.}{Q}_{\text{cond, top}}+ \overset{.}{E}_{\text{gen., element}} = \frac{\Delta E_{\text{element}}}{\Delta t}\\ \overset{.}{E}_{\text{gen., element}} = 0\\ \implies\ \overset{.}{Q}_{\text{cond, front}}+\overset{.}{Q}_{\text{cond, back}}+\overset{.}{Q}_{\text{cond, left}}+\overset{.}{Q}_{\text{cond, right}}+\overset{.}{Q}_{\text{cond, bottom}}+\overset{.}{Q}_{\text{cond, top}} = \frac{\Delta E_{\text{element}}}{\Delta t}\\ \end{align*}

Now we will substitute these terms with their expression with respect to coordinates on x, y and z axes (m, n, q).

Q.cond=kAΔTlΔEelementΔt=ρVelementcpTm,n,ri+1Tm,n,riΔtVelement=ΔxΔyΔz     ΔEelementΔt=ρΔxΔyΔzcpTm,n,ri+1Tm,n,riΔtkΔyΔzTm1,n,qTm,n,qΔx+kΔyΔzTm+1,n,qTm,n,qΔx+kΔxΔzTm,n1,qTm,n,qΔy+kΔxΔzTm,n+1,qTm,n,qΔy+kΔxΔyTm,n,q1Tm,n,qΔz+kΔxΔyTm,n,q+1Tm,n,qΔz=ρΔxΔyΔzcpTm,n,ri+1Tm,n,riΔt\begin{align*} \overset{.}{Q}_{\text{cond}} = kA\frac{\Delta T}{l}\\ \frac{\Delta E_{\text{element}}}{\Delta t} = \rho V_{\text{element}}c_p\frac{T_{m,n,r}^{i+1}-T_{m,n,r}^i}{\Delta t}\\ V_{\text{element}} = \Delta x \Delta y \Delta z\\ \implies\ \frac{\Delta E_{\text{element}}}{\Delta t} = \rho \Delta x \Delta y \Delta zc_p\frac{T_{m,n,r}^{i+1}-T_{m,n,r}^i}{\Delta t}\\ \\ k\Delta y\Delta z\frac{T_{m-1,n,q}-T_{m,n,q}}{\Delta x}+k\Delta y\Delta z\frac{T_{m+1,n,q}-T_{m,n,q}}{\Delta x}+k\Delta x\Delta z\frac{T_{m,n-1,q}-T_{m,n,q}}{\Delta y}\\+k\Delta x\Delta z\frac{T_{m,n+1,q}-T_{m,n,q}}{\Delta y} +k\Delta x\Delta y\frac{T_{m,n,q-1}-T_{m,n,q}}{\Delta z}+k\Delta x\Delta y\frac{T_{m,n,q+1}-T_{m,n,q}}{\Delta z} \\ =\rho \Delta x \Delta y \Delta zc_p\frac{T_{m,n,r}^{i+1}-T_{m,n,r}^i}{\Delta t}\\ \end{align*}

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