Question 6.3: In rapid cooling of plastic sheets formed by molding, water ...

(a) The thermal circuit diagram is shown in Figure (b). The energy equation is the surface energy equation , which reduces to

(b) The surface-convection heat transfer rate is given by A_{ku}=Lw,       \left\langle Q_{ku}\right\rangle _{L}=\frac{T_{s}-T_{f,\infty }}{\left\langle R_{ku}\right\rangle_{L} } =A_{ku}\left\langle Nu\right\rangle _{L}\frac{k_{f}}{L} (T_{s}-T_{f,\infty }), i.e.,

where A_{ku} = L^{2}.
The Nusselt number is related to the Reynolds number Re_{L}. The Reynolds number is given by Re_{L}\equiv \frac{F_{u}}{F_{\mu }} \equiv \frac{u_{f,\infty }L}{v_{f}}       ,v_{f}=(\mu /\rho )_{f} as

For water we use Table, and find that for T = 330 K

ν_{f} = 505 × 10^{−9}m^{2}/s          Table
k_{f} = 0.648 × W/m-K           Table
Pr = 3.22          Table

Then

(i) Re_{L}=\frac{0.5(m/s) × 0.2(m)}{505 × 10^{−9} (m^{2}/s)} = 1.980 × 10^{5} < Re_{L,t} = 5 × 10^{5}       laminar-flow regime

(ii) Re_{L}=\frac{5(m/s) × 0.2(m)}{505 × 10^{−9} (m^{2}/s)} = 1.980 × 10^{6} > Re_{L,t} = 5 × 10^{5}   turbulent-flow regime

Based on this magnitude for Re_{L}, we use \left\langle Nu\right\rangle _{L}=0.332Pr^{1/3}\left(\frac{u_{f,\infty }}{v_{f}} \right) ^{1/2}\int_{0}^{L}{x^{-1/2}dx}=2Nu_{L} =0.664Re_{L}^{1/2}Pr^{1/3}=\frac{L}{A_{ku}\left\langle R_{ku}\right\rangle _{L}k_{f}} ,Re_{L}\leq Re_{L,t}=5\times 10^{5} and\left\langle\overline{Nu} \right\rangle _{L}\equiv \left\langle Nu\right\rangle _{L}=(0.037Re^{4/5}_{L}-871)Pr^{1/3}    for       5\times 10^{5}\lt Re_{L}\lt 10^{8},    and     0.6\lt Pr\lt 60 for \left\langle Nu\right\rangle _{L} i.e.,

(i) \left\langle Nu\right\rangle _{L}= 0.664Re^{1/2} _{L} Pr^{1/3} = 0.664(1.980 × 10^{5} )^{ 1/2} (3.22)^{1/3}

= 436.1

(ii) \left\langle Nu\right\rangle _{L}= (0.037Re^{4/5}_{L} − 871)Pr^{1/3}

= 4.666 × 10^{3}                  average laminar-turbulent Nusselt number.

Now returning to   \left\langle Q_{ku}\right\rangle _{L} , we have

(i)  \left\langle Q_{ku}\right\rangle _{L} = (0.2)^{2} (m^{2}) × 436.1 × 10^{3} ×\frac{0.648(W/m-K)}{0.2(m)} × (95 − 20)(K)

(ii) \left\langle Q_{ku}\right\rangle _{L} =(0.2)^{2} (m^{2}) × 4.666 × 10^{3} ×\frac{0.648(W/m-K)}{0.2(m)}× (95 − 20)(K)

(c) The thermal boundary-layer thickness is given by \frac{\delta _{v}}{\delta _{\alpha }} =Pr^{1/3}    ,\delta _{v}=5.0\left(\frac{v_{f}L}{u_{f,\infty }} \right) ^{1/2}=5.0LRe_{L}^{-1/2},    \delta _{\alpha }=5.0\left(\frac{v_{f}L}{u_{f,\infty }} \right) ^{1/2}\left(\frac{\alpha _{f}}{v_{f}} \right) ^{1/3} and \overline{\delta } _{v}\equiv \delta _{v}0.37LRe_{L}^{-1/5}=0.37\left(\frac{v_{f}}{u_{f,\infty }} \right) ^{1/5}L^{4/5},      \frac{\overline{\delta }_{v} }{\overline{\delta }_{\alpha } } \simeq 1 as

(i)  \delta _{\alpha }=5\left(\frac{v_{f}L}{u_{f,\infty }} \right) ^{1/2}Pr^{-1/3}

(ii)  \delta _{\alpha }=\delta _{v }=0.37\left(\frac{v_{f}}{u_{f,\infty }} \right) ^{1/5}L^{4/5}

= 2.758 mm.

(d) The average surface-convection resistance \left\langle R_{ku}\right\rangle _{L} is found from A_{ku}=Lw,       \left\langle Q_{ku}\right\rangle _{L}=\frac{T_{s}-T_{f,\infty }}{\left\langle R_{ku}\right\rangle_{L} } =A_{ku}\left\langle Nu\right\rangle _{L}\frac{k_{f}}{L} (T_{s}-T_{f,\infty }), i.e.,

(i) \left\langle R_{ku}\right\rangle _{L}=\frac{T_{s}-T_{f,\infty }}{\left\langle Q_{ku}\right\rangle_{L} } =\frac{(95 − 20)(^{\circ }C)}{4.239 × 10^{3} (W)} = 0.01769^{\circ }C/W 

(ii)  \left\langle R_{ku}\right\rangle _{L}=\frac{T_{s}-T_{f,\infty }}{\left\langle Q_{ku}\right\rangle_{L} } =\frac{(95 − 20)(^{\circ }C)}{4.535 × 10^{4} (W)} = 0.001654^{\circ }C/W

Figure (b) shows the average heat flux vector \left\langle q_{ku}\right\rangle _{L}, which is perpendicular to the surface (between Ts is uniform over the surface). The thermal circuit diagram is also shown