There are many surface-convection heat transfer problems in which the surfaceconvection surface area per unit volume Aku/V is very large (e.g., Figure 7.8). Then due to this large specific surface area (or large NTU), a fluid having a temperature 0, and flowing into this porous solid will

Question

There are many surface-convection heat transfer problems in which the surfaceconvection surface area per unit volume $A_{ku}/V$ is very large (e.g., Figure). Then due to this large specific surface area (or large NTU), a fluid having a temperature $\left\langle T_{f}\right\rangle _{0}$ , and flowing into this porous solid will come to thermal equilibrium with the solid over a short distance. Show that by using the fluid flow per unit solid volume $\dot{n}_{f}$ , the energy equation for a porous solid of volume $V_{s}$ , with a uniform temperature $T_{s}$ , is given by

(\dot{n}c_{p})_{f}V_{s}(T_{s}-\left\langle T_{f}\right\rangle _{0})=-(\rho c_{p}V)_{s}\frac{dT_{s}}{dt} +\dot{S}_{s}

integral-volume energy equation for porous solid with very large $A_{ku}/V_{s}$ .

Start from(2.72) and allow for the surface-convection heat transfer mechanism
only.

[latex](\dot{n}c_{p})_{f}V_{s}(T_{s}-\left\langle T_{f}\right\rangle _{0})=-(\rho c_{p}V)_{s}\frac{dT_{s}}{dt} +\dot{S}_{s}[/latex]

integral-volume energy equation for porous solid with very large [latex]A_{ku}/V_{s}[/latex].

Start from(2.72) and allow for the surface-convection heat transfer mechanism
only.

Accepted Answer

Starting from (2.72) and allowing for surface-convection heat transfer mechanism only, we have

[latex] \left\langle Q_{u}\right\rangle _{L-0}=-(\rho c_{p}V)_{s}\frac{dT_{s}}{dt} +\dot{S}_{s} [/latex]

where from [latex]\left\langle Q_{u}\right\rangle _{L-0}\equiv \frac{T_{s}-\left\langle T_{f}\right\rangle _{0}}{\left\langle R_{u}\right\rangle _{L}} [/latex] , [latex]\left\langle R_{u}\right\rangle _{L}=\frac{T_{s}-\left\langle T_{f}\right\rangle _{0}}{(\dot{M}c_{p})_{f}(\left\langle T_{f}\right\rangle _{L}-\left\langle T_{f}\right\rangle _{0})} =\frac{1}{(\dot{M}c_{p})_{f}\epsilon _{he}}=\frac{1}{(\dot{M}c_{p})_{f}(1-e^{-NTU})} [/latex], and [latex]NTU\equiv \frac{R_{u,f}}{\left\langle R_{ku}\right\rangle _{D}} =\frac{1}{\left\langle R_{ku}\right\rangle_{D} (\dot{M}c_{p})_{f}}=\frac{A_{ku}\left\langle Nu\right\rangle _{D}k_{f}/D}{(\dot{M}c_{p})_{f}} [/latex], we have

[latex] \left\langle Q_{u}\right\rangle _{L-0}=\frac{T_{s}-\left\langle T_{f}\right\rangle _{0}}{\left\langle R_{u}\right\rangle _{L}} =(\dot{M}c_{p})_{f}(1-e^{-NTU})(T_{s}-\left\langle T_{f}\right\rangle _{0})[/latex]

[latex] NTU=\frac{A_{ku}\left\langle Nu\right\rangle _{D}k_{f}/D}{(\dot{M}c_{p})_{f}} [/latex]

[latex]=\frac{A_{ku}}{V_{s}} \frac{\left\langle Nu\right\rangle _{D}k_{f}/D}{(\dot{n}c_{p})_{f}} [/latex]

Now for[latex] A_{ku}/V_{s} \gg 1(1/m)[/latex], we have

[latex]NTU → \infty[/latex] .

This limit of very large number of transfer units results in [latex]\left\langle T_{f}\right\rangle _{L}= T_{s}[/latex], where [latex]T_{s}[/latex] is the uniform porous solid temperature. Then

[latex] \left\langle Q_{u}\right\rangle _{L-0}= (\dot{M}c_{ p} )_{f} (1 − e^{−∞})(T_{s} − \left\langle T_{f}\right\rangle _{0}) [/latex]

[latex] =(\dot{M}c_{ p} )_{f} (T_{s} − \left\langle T_{f}\right\rangle _{0}) [/latex]

Now using [latex] (\dot{M}c_{ p} )_{f} = (\dot{n}c_{p} )_{f} V_{s}[/latex] , we have

[latex]\left\langle Q_{u}\right\rangle _{L-0}= (\dot{n}c_{p} )_{f} V_{s} (T_{s} − \left\langle T_{f}\right\rangle _{0}) [/latex]

Using this in the above energy equation, we have

[latex](\dot{n}c_{p} )_{f} V_{s} (T_{s} − \left\langle T_{f}\right\rangle _{0})=-(\rho c_{p}V)_{s}\frac{dT_{s}}{dt} +\dot{s}_{s}V_{s} [/latex]

integral-volume energy equation for porous solid with very large [latex]A_{ku}/V_{s}[/latex]

or

[latex](\dot{n}c_{p} )_{f}(T_{s} − \left\langle T_{f}\right\rangle _{0})= -(\rho c_{p})_{s}\frac{dT_{s}}{dt} +\dot{s}_{s} [/latex]

Question 7.4: There are many surface-convection heat transfer problems in ...

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