(a) Figure (b) shows the thermal circuit diagram.
(b) The energy equation for the solid, from Figure(b), is
\left\langle Q_{u}\right\rangle _{L-0}=-(\rho c_{p}V)_{s}\frac{dT_{s}(t)}{dt}
\frac{T_{s}-\left\langle T_{f}\right\rangle_{0}} {\left\langle R_{u}\right\rangle_{L} } =-(\rho c_{p}V)_{s}\frac{dT_{s}(t)}{dt}
The average convection resistance \left\langle R_{u}\right\rangle_{L} is defined in Table, i.e,
\left\langle R_{u}\right\rangle_{L}=\frac{1}{(\dot{M}c_{p})_{f}(1-e^{-NTU})}
NTU=\frac{1}{\left\langle R_{ku}\right\rangle _{D}(\dot{M}c_{p})_{f}}
The energy equation is similar to Q\mid_{A,1}=Q_{1}+Q_{ku,1-\infty }=Q_{1}+\frac{T_{1}(t)-T_{f,\infty }}{\left\langle R_{ku}\right\rangle_{L or D} } =-(\rho c_{p}V)_{1}\frac{dT_{1}(t)}{dt}+\dot{S}_{1} , i.e., with Q_{s} and \dot{S}_{s} both equal to zero, then T_{1}(t)=T_{f,\infty }+[T_{1}(t=0)-T_{f,\infty }]e^{-t/\tau _{1}}+a_{1}\tau _{1}(1-e^{-t/\tau _{1}}) is the solution,
T_{s}(t)=\left\langle T_{f}\right\rangle_{0}+(T_{s}(t=0)-\left\langle T_{f}\right\rangle_{0} )e^{-t/\tau _{s}}
\tau _{s}=(\rho c_{p}V)_{s}\left\langle R_{u}\right\rangle _{L}
(c) The Nusselt number for porous media is given in Table as
\left\langle Nu\right\rangle _{D,p}= 2 + (0.4Re^{1/2}_{ D,p} + 0.2Re^{2/3}_{ D,p} )Pr^{0.4}
Re_{D,p}=\frac{\rho _{f}\left\langle u_{f}\right\rangle D_{p} }{\mu _{f}(1-\epsilon )} =\frac{\dot{M}_{f}D_{p}}{A_{u}\mu _{f}(1-\epsilon )}
From Table, at T = 1,500 K, we have
\mu_{f} = ρ_{f} ν_{f} = 0.235(kg/m^{3}) × 2.29 × 10^{−4}(m^{2}/s) = 5.382 × 10^{−5} Pa-s Table
k_{f} = 0.0870 W/m-K Table
c_{p} = 1,202 J/kg-K Table
Pr = 0.70 Table
Then
Re_{D,p}=\frac{10^{−3} (kg/s) × 2 × 10^{−4} (m) }{\frac{π × (3 × 10^{−2} )^{2} (m^{2})}{4}× 5.382 × 10^{−5} (Pa-s)(1 − 0.25) }
= 7.014
\left\langle Nu\right\rangle _{D,p}=2 + [0.4 × (7.014)^{1/2} + 0.2 × (7.014)^{2/3} ] × (0.7)^{0.4}
= 3.554.
(d) From Table, or from \left\langle Nu\right\rangle _{D,p}\equiv \frac{D_{p}}{A_{ku}\left\langle R_{ku}\right\rangle_{D,p}k_{f} } \left(\frac{\epsilon }{1-\epsilon } \right) , we have
\left\langle R_{ku}\right\rangle_{D}= \frac{D_{p}}{A_{ku}\left\langle Nu\right\rangle _{D,p}k_{f} } \left(\frac{\epsilon }{1-\epsilon } \right)
Then
NTU=\frac{A_{ku}\left\langle Nu\right\rangle _{D,p}k_{f}(1-\epsilon )}{D_{p}\epsilon (\dot{M}c_{p})_{f}}
=\frac{0.46(m^{2}) × 3.554 × 0.0870(W/m-K) × (1 − 0.25)}{2 × 10^{−4} (m) × 0.25 × 10^{−3} (kg/s) × 1,202(J/kg-K)} = 1,775.
(e) The average convection resistance \left\langle R_{u}\right\rangle _{L} is
\left\langle R_{u}\right\rangle _{L}=\frac{1}{(\dot{M}c_{p})_{f}(1-e^{-NTU})}
=\frac{1}{10^{−3}(kg/s) × 1,202(J/kg-K)(1 − e^{−1,775})} = 0.8319^{\circ }C/W.
The time constant \tau _{s} is
\tau _{s}=(\rho c_{p}V)_{s}\left\langle R_{u}\right\rangle_{L} = (\rho c_{p} )_{s}\left(\frac{\pi D^{2}}{4}L \right) (1-\epsilon )\left\langle R_{u}\right\rangle _{L}
where we have used the volume of the solid. From Table, for stainless steel 316, we have
stainless steel 316: \rho _{s}= 8,238 kg/s Table
c_{p,s} = 486 J/kg-K Table
Then
\tau _{s}= 8,238(kg/m^{3}) × 486(J/kg-K) × \frac{π × (3 × 10^{−2} )^{2}(m^{2} )}{4}× 2 × 10^{−2} (m)
× (1 − 0.25) × 0.8319(^{\circ }C/W) = 35.30 s.
Then
T_{s}(t = t_{o} = 5 s) = 1,600^{\circ } C + [40^{\circ }C − 1,600^{\circ }C]e^{−5/35.50} = 246.0^{\circ }C
The instantaneous fluid stream outlet temperature is determined from \frac{T_{s}-\left\langle T_{f}\right\rangle _{L}}{T_{s}-\left\langle T_{f}\right\rangle _{0}} =\frac{\left\langle T_{f}\right\rangle _{L}-T_{s}}{\left\langle T_{f}\right\rangle _{0}-T_{s}} =e^{-NTU}, i.e.,
\left\langle T_{f}\right\rangle _{L}(t = t_{o} = 5 s) = T_{s}(t) + [\left\langle T_{f}\right\rangle _{0} − T_{s}(t_{o})]e^{−NTU}
= 246.0(^{\circ }C) + [1,600(^{\circ }C) − 246.0(^{\circ }C)]e^{−1,775} = 246.0^{\circ }C.
