(a) The thermal circuit diagram is shown in Figure (b).
(b) We will use the correlation in Table for discontinuous solid surfaces. First we need to evaluate the particle diameter from D_{p}\equiv \frac{6V_{s}}{A_{ku}} =\frac{6(1-\epsilon )V}{A_{ku}} , i.e.,
D_{p}=\frac{6V_{s}}{A_{ku}}
Here the tubes should be treated as solid (because the surface convection under study occurs on the outside surface), and we have for each tube
V_{s} = πD^{2} w/4
A_{ku} = πDw.
Then
D_{p}=\frac{6πD^{2} w}{4πDw} =\frac{6}{4}D=\frac{3}{2}D= 1.5 × (0.01)(m)
= 1.5 × 10^{−2} m.
The porosity \epsilon is defined by \epsilon \equiv \frac{V_{f}}{V} \equiv \frac{V_{f}}{V_{s}+V_{f}} , or \frac{V_{s}}{V_{s}+V_{f}}=1-\epsilon as
\epsilon = \frac{V_{f}}{V_{s}+V_{f}}=\frac{Llw − N_{t}πD^{2} w/4}{Llw}
=1-\frac{N_{t}πD^{2}}{4Ll}
where N_{t} is the number of tubes, which here is N_{t} = 11 × 11 = 121. Then
\epsilon =1-\frac{121 × π × (10^{−2})^{2}(m)^{2}}{4 × (0.015)^{2} (m)^{2}} = 0.5776.
The particle Reynolds number is defined by Re_{D,p}\equiv \frac{\rho _{f}\left\langle u_{f}\right\rangle D_{p} }{\mu _{f}(1-\epsilon )} as
Re_{D,p}=\frac{\rho _{f}\left\langle u_{f}\right\rangle D_{p} }{\mu _{f}(1-\epsilon )}
where \left\langle u_{f}\right\rangle is found from \dot{M}_{ f} in \dot{M}_{f}=A_{u}\dot{m}_{f}=A_{u}\rho _{f}\left\langle u_{f}\right\rangle , i.e.,
\dot{M}_{f}=A_{u}\rho _{f}\left\langle u_{f}\right\rangle
or
\left\langle u_{f}\right\rangle=\frac{\dot{M}_{f}}{A_{u}\rho _{f}} ,A_{u}= l\times w
Then, using this in the expression for Re_{D,p} , we have
Re_{D,p}=\frac{\dot{M}_{f}D_{p}}{\mu _{f}(1-\epsilon )lw}
From Table and at T = 350 K, for ethylene glycol we have
\rho _{f} = 1,079 kg/m^{3} Table
c_{p,f} = 2,640 J/kg-K Table
ν_{f} = 3.25 × 10^{−6} m^{2}/s Table
k_{f} = 0.261 W/m-K Table
Pr = 35.2 Table
Then
Re_{D,p}=\frac{5(kg/s) × 1.5 × 10^{−2} (m)}{3.25 × 10^{−6}(m^{2}/s) × 1,079(kg/m^{3} )(1 − 0.5776)(0.15)^{2}(m)^{2}}
= 2.251 × 10^{3}
(c) The correlation for \left\langle Nu\right\rangle _{D,p},p 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} .
Then
\left\langle Nu\right\rangle _{D,p}= 2 + [0.4 × (2.251 × 10^{3} )^{ 1/2} + 0.2 × (2.251 × 10^{3} )^{ 2/3} ](35.2)^{0.4}
= 223.6.
Note the rather large \left\langle Nu\right\rangle _{D,p} .
(d) The NTU is defined by 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}} and, using the definition of \left\langle Nu\right\rangle _{D,p} (i.e., therelation between \left\langle Nu\right\rangle _{D,p} and \left\langle R_{ku}\right\rangle _{L}), as given by \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
NTU =\frac{1}{(\dot{M}c_{p})_{f}\left\langle R_{ku}\right\rangle_{D} }=\frac{A_{ku}}{(\dot{M}c_{p})_{f}} \frac{\left\langle Nu\right\rangle _{D,p}k_{f}}{D_{p}} \frac{1-\epsilon }{\epsilon }
where
A_{ku} = N_{t}πDw
Then
NTU =\frac{11 × 11 × π × 0.01(m) × 0.15(m) × 223.6 × 0.261(W/m-K) × (1 − 0.5776)}{5(kg/s) × 2,640(J/kg-^{\circ }C) × 0.015(m) × 0.5776}
= 0.1231.
(e) The heat transferred rate is given by \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}} and \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})} , i.e.,
\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}}
= − (\left\langle T_{f}\right\rangle _{0}− T_{s})\dot{M}c_{ p} \epsilon _{he}
= − (\left\langle T_{f}\right\rangle _{0} − T_{s})\dot{M}c_{ p} (1 − e^{−NTU})
= − [90(^{\circ }C) − 45(^{\circ }C)] × 5(kg/s) × 2,640(J/kg-^{\circ } C) × (1 − e^{−0.1231})
= − 6.878 × 10^{4} W
(f) The exit temperature is determined from \epsilon _{h,e}\equiv \frac{\left\langle T_{f}\right\rangle _{L}-\left\langle T_{f}\right\rangle _{0}}{T_{s}-\left\langle T_{f}\right\rangle _{0}} =\frac{\left\langle T_{f}\right\rangle _{0}-\left\langle T_{f}\right\rangle _{L}}{\left\langle T_{f}\right\rangle _{0}-T_{s}} =1-e^{-NTU} , i.e.,
\frac{\left\langle T_{f}\right\rangle _{0}-\left\langle T_{f}\right\rangle _{L}}{\left\langle T_{f}\right\rangle _{0}-T_{s}} =1-e^{-NTU}
or
\left\langle T_{f}\right\rangle _{L} = \left\langle T_{f}\right\rangle _{0} − (1 − e^{−NTU})(\left\langle T_{f}\right\rangle _{0} − T_{s})
= 90(^{\circ }C) − (1 − e^{−0.1231})(90 − 45)(^{\circ} C)
= 90(^{\circ}C) − 5.212(^{\circ} C) = 84.79^{\circ} C
