(a) The thermal circuit diagram is shown in Figure (b).
(b) The energy equation for the uniform temperature pipe-water segment is Q\mid_{A,1}=\int_{A_{1}}{(q_{ku}.s_{n})dA} =-(\rho c_{p}V)_{1}\frac{dT_{1}}{dt} +\dot{S}_{1} for Bi_{L}\lt 0.1, with Q_{1} =\dot{ S}_{ 1} = 0. Then we have
Q\mid_{A,1}=Q_{ku,1-\infty }+Q_{r,1-\infty }=-(\rho c_{p}V)_{1}\frac{dT_{1}}{dt}
The volume V is occupied by water (w) and brass (b), i.e.,
(\rho c_{p}V)_{1}=(\rho c_{p})_{w}\pi \frac{(D-2l)^{2}}{4}L +(\rho c_{p})_{b}\pi \frac{[D^{2}-(D-2l)^{2}]}{4}L
For surface radiation, for a two-surface enclosure, with A_{r,∞} \ll A_{r,1} and F_{1-∞} = 1, we have Q_{r,1-2}=A_{r,1}\epsilon _{r,1}[E_{b,1}(T_{1})-E_{b,2}(T_{2})] for F_{1-2}=1 with \frac{A_{r,1}}{A_{r,2}}\simeq 0 or \epsilon _{r,2} =1, i.e.,
Q_{r,1-∞} = A_{r,1}\epsilon_{r,1}σ_{SB}(T^{ 4}_{ 1} − T^{ 4}_{ s,∞}), A_{r,1} = πDL = A_{ku}.
The surface convection is given by \left\langle Q_{ku}\right\rangle _{L( or D)}=\frac{T_{s}-T_{f,\infty }}{\left\langle R_{ku}\right\rangle_{L(or D)} } =A_{ku}\frac{k_{f}}{L(or D)}\left\langle Nu\right\rangle _{L(or D)}(T_{s}-T_{f,\infty }), Re_{L(or D)}=\frac{u_{f,\infty }L(or D)}{v_{f}} , i.e.,
Q_{ku,1-\infty }=A_{ku}\left\langle Nu\right\rangle _{L}\frac{k_{f}}{L}(T_{1}-T_{f,\infty })
The Nusselt number is found from Table, where it is suggested that we use the correlation for the vertical plates, i.e.,
\left\langle Nu\right\rangle _{L}=[(\left\langle Nu_{L,l}\right\rangle) ^{6}+(\left\langle Nu_{L,t}\right\rangle )^{6}]^{1/6}
\left\langle Nu_{L,l}\right\rangle =\frac{2.8}{ln[(1 + 2.8/(a_{1}Ra^{1/4} )]}
\left\langle Nu_{L,t}\right\rangle =\frac{0.13Pr^{0.22}}{(1 + 0.61Pr^{0.81})^{ 0.42}} Ra_{L}^{1/3}
a_{1}=\frac{4}{3}\frac{0.503}{[1 + (0.492/Pr)^{9/16}] ^{4/9}}
Ra_{L}=\frac{g\beta _{f}(T_{f}-T_{f,\infty })L^{3}}{v_{f}\alpha _{f}}
The thermophysical properties are, at T = 300 K,
water : \rho _{w} = 997 kg/m^{3} Table
c_{p,w} = 4,179 J/kg-K Table
brass : \rho _{b} = 8,933 kg/m^{3} Table
c_{p,b} = 355 J/kg-K Table
\epsilon_{r,1} = 0.60 Table
air : k_{f} = 0.0267 W/m-K Table
β_{f} =\frac{1}{300} 1/K ideal gas \beta _{f}=\frac{1}{T_{f}}, p=\rho _{f}\frac{R_{g}}{M}T_{f}
ν_{f} = 1.566 × 10^{−5} m^{2}/s Table
α_{f} = 2.257 × 10^{−5} m^{2}/s Table
Pr = 0.69 Table
other parameters : T_{f,∞} = T_{s,∞} = 293.15 K
D = 0.06 m , l = 10^{−3} m , L = 0.3 m.
The results for T_{1}(t) are plotted in Figure (c). The results with and without the surface radiation are shown. The surface radiation accelerates the cooldown. Note that within about one hour, the water temperature drops to near 30^{\circ }C. This is the average surface temperature for the human skin. Any further drop in the temperature results in a cold sensation when this water flows over the human hands.
(c) From the case of no surface radiation, Q_{r,1-∞} = 0, and 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}}), for \dot{S}_{ 1} − Q_{1} = 0 (a_{1} = 0), we have
T_{1}(t) = T_{f,∞} + [T_{1}(t = 0) − T_{f,∞}]e^{−t/\tau_{1}}
\tau_{1}= (ρc_{p}V)_{1}\left\langle R_{ku}\right\rangle _{L}
The solution is valid for a constant \left\langle R_{ku}\right\rangle _{L}. Here we use \left\langle R_{ku}\right\rangle _{L}(t = 0) = 4.353^{\circ }C/W.
\tau_{1} = 2307.2 × 3.997 = 9, 222 s
Now, after t = 2 hr = 7,200 s, we have
T^{1}(t = 2 hr) = 293.15(K) + (318.15 − 293.15)(K) e^{−[7,200 (s)]/[9,222 (s)]}
= 304.6 K = 31.45^{\circ}C.
This is what is found from the use of a solver (e.g., MATLAB) by setting \epsilon _{r,1} = 0, and using a constant \left\langle R_{ku}\right\rangle _{L}(t = 0), as shown in Figure (c). The effect of variable \left\langle R_{ku}\right\rangle _{L} is less profound compared to that of surface emissivity.
