(a) The thermal circuit diagram is shown in Figure (c).
(b) The rate of surface-convection heat transfer is that 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.,
\left\langle Q_{ku}\right\rangle _{L} =A_{ku}\left\langle Nu\right\rangle _{L}\frac{k_{f}}{L} (T_{s}-T_{f,\infty })
Where
A_{ku}=4L^{2}
For a single, normal jet, the Nusselt number is found from \left\langle Nu\right\rangle _{L}=2Re_{D}^{1/2}Pr^{0.42}(1+0.005Re_{D}^{0.55})^{1/2}\frac{1-1.1D/L}{1+0.1(L_{n}/D-6)D/L} for 7.5\gt \frac{L}{D}\gt 2.5 , i.e.,
\left\langle Nu\right\rangle _{L}=2Re_{D}^{1/2}Pr^{0.42}(1+0.005Re_{D}^{0.55})^{1/2}\frac{1-1.1D/L}{1+0.1(L_{n}/D-6)D/L}
where from Re_{D}=\frac{\left\langle u_{f}\right\rangle D}{v_{f}}
Note that here L is the radius of the circle on the surface. As in Example 6.3, the thermophysical properties are obtained from Table, and we have
ν_{f} = 505 × 10^{−9} m^{2}/s Table
k_{f} = 0.648 W/m-K Table
Pr = 3.22 Table
Then
Re_{D}=\frac{25(m/s) × 0.025(m)}{505 × 10^{−9} (m/s^{2})} = 1.238 × 10^{6}
Note that this Re_{D} is outside the range of the data used in the correlation \left\langle Nu\right\rangle _{L}=2Re_{D}^{1/2}Pr^{0.42}(1+0.005Re_{D}^{0.55})^{1/2}\frac{1-1.1D/L}{1+0.1(L_{n}/D-6)D/L} for 7.5\gt \frac{L}{D}\gt 2.5 . Due to lack of an alternative correlation, we will use \left\langle Nu\right\rangle _{L}=2Re_{D}^{1/2}Pr^{0.42}(1+0.005Re_{D}^{0.55})^{1/2}\frac{1-1.1D/L}{1+0.1(L_{n}/D-6)D/L} for 7.5\gt \frac{L}{D}\gt 2.5 . Now we evaluate \left\langle Nu\right\rangle _{L} and we note that 2L = 20 cm.
\left\langle Nu\right\rangle _{L}= 2(1.238 × 10^{6} )^{1/2}(3.22)^{0.42}[1 + 0.005(1.238 × 10^{6})^{0.55}]^{1/2}×
\frac{1 − 1.1[0.025(m)/0.10(m)]}{1 + 0.1[0.10(m)/0.025(m) − 6][0.025(m)/0.10(m)] }
= 3.637 × 10^{9} × 3.496 × 0.7632
= 9.702 × 10^{4}
The surface-convection heat transfer rate is
\left\langle Q_{ku}\right\rangle _{L}= (0.2)^{2}(m^{2} ) × 9.702 × 10^{4} ×\frac{0.648(W/m-K)}{0.1(m)} (95 − 20)(K)
= 1.886 × 10^{6} W.
(c) The average surface-convection resistance \left\langle R_{ku}\right\rangle _{L} 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.,
\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)}{1.886 × 10^{6} (W)} = 3.976 × 10^{−5\circ }C/W
A_{ku} \left\langle R_{ku}\right\rangle _{L}= (0.2)^{2}(m^{2}) × 3.976 × 10^{5} (^{\circ }C/W) = 1.590 × 10^{−6\circ }C/(W/m^{2}).
(d) Although the amount of water used in the impinging-jet cooling is much less than that for parallel flow, for the conditions used, the heat transfer rate is larger for the single jet, i.e.,
\left\langle Q_{ku}\right\rangle _{L}= 4.535 × 10^{4} W forced, turbulent parallel flow
\left\langle Q_{ku}\right\rangle _{L}= 1.886 × 10^{6} W forced, turbulent single jet.
