(a) The thermal circuit diagram is shown in Figure (b).
(b) The integral-volume energy equation for the wire is Q\mid_{A}=\int_{A}{(q.s_{n})dA} =Q_{H-1}+Q_{H-2}= \dot{S}_{e,J} , i.e.,
Q\mid_{A,w}= \dot{S}_{e,J} energy equation for wire,
where, using Q_{w} = 0 , we have
Q\mid_{A,w}= \frac{T_{w}-T_{f,\infty }}{R_{k,1-2}+\left\langle R_{ku}\right\rangle _{D,2}} =\frac{T_{w}-T_{f,\infty }}{R_{\sum}} surface heat transfer rate
where from Table
R_{k,1-2}=\frac{ln(R_{2}/R_{1})}{2\pi Lk_{s}} Table
and from \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}} and noting that A_{ku} = πD_{2}L = 2πR_{2}L, we have
\left\langle R_{ku}\right\rangle _{D,2}=\frac{D_{2}}{A_{ku}\left\langle Nu\right\rangle _{D,2}k}
=\frac{2R_{2}}{2\pi R_{2}L\left\langle Nu\right\rangle _{D,2}k_{f}} =\frac{1}{\pi L\left\langle Nu\right\rangle _{D,2}k_{f}}
with D_{2} = 2R_{2}.
(c) To minimize R_{\sum} , we differentiate R_{\sum} and set the resultant equal to zero, i.e.,
\frac{d}{dR_{2}} R_{\sum}=\frac{d}{dR_{2}} \left[\frac{ln(R_{2}/R_{1})}{2\pi Lk_{s}}+\frac{1}{\pi Lk_{f}\left\langle Nu\right\rangle _{D,2}} \right] =0
We then substitute the h, which is independent of R_{2}, and write Nu_{L}=\frac{q_{ku}L}{(T_{s}-T_{f,\infty })k_{f}}\equiv \frac{hL}{k_{f}} or h\equiv Nu_{L}\frac{k_{f}}{L} as the relation between \left\langle Nu\right\rangle _{D,2} and D_{2}, and finally perform differentiation.
(d) If we assume that
k_{f}\left\langle Nu\right\rangle _{D,2}=hD_{2} assuming that h is independent of D_{2},
where h is the heat transfer coefficient, substituting for \left\langle Nu\right\rangle _{D,2} and differentiating we have,
\frac{d}{dR_{2}} \left[\frac{ln(R_{2}/R_{1})}{2\pi Lk_{s}} +\frac{1}{2\pi LhR_{2}} \right] =0
The result is
\frac{1}{2\pi Lk_{s}R_{2}}-\frac{1}{2\pi LhR_{2}^{2}} =0
Solving for R_{2}, we have
R_{2}=\frac{k_{s}}{h} =R_{c} for h independent of D_{2} ,
where R_{c} is called the critical radius (i.e., corresponding to minimum R_{∑} ).
(e) We note that, in general, h depends on D_{2}. For example, in forced flow, from Table, we have for cylinders
\frac{hD_{2}}{k_{f}} =\left\langle Nu\right\rangle _{D,2}=a_{1}Re_{D,2}^{a_{2}}Pr^{1/3}, a_{2}\lt 1
Note that from Table, for example for Re_{D,2} < 4 × 10^{3}, a_{2} is 0.466. Then h is proportional to D^{a_{2}−1}_{2} .
Now if we use the relation
h=\frac{k_{f}}{D_{2}}a_{1}Re_{D,2}^{a_{2}} Pr^{1/3}\equiv a_{R}D_{2}^{a_{2}-1}, a_{R}=k_{f}a_{1}\frac{u_{f,\infty }^{a_{2}}}{v_{f}^{a_{2}}} Pr^{1/3}
where a_{R} contains all the other constants (i.e., a_{1}, a_{2}, Pr, etc.), then we can write
k_{f}\left\langle Nu\right\rangle _{D,2}=a_{R}D_{2}^{1+(a_{2}-1)}
=a_{R}D_{2}^{a_{2}}
By substituting for k_{f}\left\langle Nu\right\rangle _{D,2}, we have
\frac{d}{dR_{2}} \left[\frac{lnR_{2}/R_{1}}{2\pi Lk_{s}}+\frac{1}{2^{a_{2}}\pi La_{R}R_{2}^{a_{2}}} \right]
After differentiation, we have
\frac{1}{2\pi Lk_{f}R_{2}} +\frac{-a_{2}}{2^{a_{2}}\pi La_{R}R_{2}^{a_{2+1}}} =0
or
\frac{1}{2k_{s}}-\frac{a_{2}}{2^{a_{2}}a_{R}R_{2}^{a_{2}-1}} =0
Solving for R_{2}, we have
R_{2}=R_{c}=\left(\frac{a_{2}k_{s}}{2^{a_{2}-1}a_{R}} \right) ^{1/a_{2}}=\left(\frac{a_{2}k_{s}v_{f}^{a_{2}}}{2^{a_{2}-1}k_{f}a_{1}u_{f,\infty }^{a_{2}}Pr^{1/3}} \right) ^{1/a_{2}}
For the case of a_{2} = 1 , which gives a_{R} = h , we obtain R_{2} = R_{c} = k_{s}/h , as before. For 0 < a_{2} < 1 , a different critical radius is found. Note that the unit of a_{R} depends on a_{2} .
