(a) The thermal circuit diagram is shown in Figure (b) . Note that Q_{k,1-1^{\prime }} flows through all the resistances and at surface 2 is labeled as Q_{2} . Also note that T_{ 1}^{\prime} and T _{2}^{\prime} are not initially given. In other words, by using the overall resistance R_{k,\sum} , we have from Q_{1}+Q_{k,1-1^{\prime }}=0 or Q_{1}=-Q_{k,1-1^{\prime }}
Q_{1}+Q_{k,1-1^{\prime }}=0 energy equation for node T_{1}
Q_{1}=-Q_{k,1-1^{\prime }}=-\frac{T_{1}-T_{1^{\prime }}}{R_{k,1-1^{\prime }}}= -\frac{T_{1}-T_{2}}{R_{k,\sum }}
R_{k,\sum }=R_{k,1-1^{\prime } }+R_{k,1^{\prime }-2^{\prime } }+R_{k,2^{\prime }-2 } overall resistance between nodes T_{1} and T_{2}.
So, we can solve for Q_{1} without evaluation T_{1^{\prime}} .
(b) We now evaluate Q_{1} using the above energy equation. The resistances are all for radial geometries. From Table, for cylindrical shells we have
R_{k,1-1^{\prime } }=\frac{\ln \frac{R_{1}+l_{c}}{R_{1}} }{2\pi Lk_{c}}
R_{k,1^{\prime }-2^{\prime } }=\frac{\ln \frac{R_{2}}{R_{1}+l_{c}} }{2\pi L\left\langle k\right\rangle }
R_{k,2^{\prime }-2 }=\frac{\ln \frac{R_{2}+l_{t}}{R_{2}} }{2\pi L k_{t} }
The thermal conductivities for copper (Table), polyurethane foam formed in place (Table), and Teflon (Table), are
copper: k_{c} = 385 W/m-K at T = 300 K Table
foam: \left\langle k\right\rangle = 0.026 W/m-K at T = 298 K Table ]
Teflon: k_{t} = 0.26 W/m-K at T = 293 K Table .
Using the numerical values, we have
R_{k,1-1^{\prime } }=\frac{\ln \frac{0.039(m)}{0.037(m)} }{2\pi \times 1(m)\times 385(W/m-K) }=2.177\times 10^{-5\circ }C/W
R_{k,1^{\prime }-2^{\prime } }=\frac{\ln \frac{0.074(m)}{0.039(m)} }{2\pi \times 1(m)\times 0.026(W/m-K) }=3.755^{\circ }C/W
R_{k,2^{\prime }-2 }=\frac{\ln \frac{0.074(m)}{0.072(m)} }{2\pi \times 1(m)\times 0.026(W/m-K) }=1.678\times 10^{-2\circ }C/W
Then Q_{1}=-\frac{(45-25)(^{\circ }C)}{(2.177\times 10^{-5}+3.755+1.678\times 10^{-2})(^{\circ }C/W)}=-5.303W
This is relatively low heat loss from the water and this insulation is considered a good insulation. Note that the foam resistance is dominant (lowest k and largest thickness among the three layers).
(c) The temperatures T_{1^{\prime}} and T_{2^{\prime}} are found by either writing the nodal energy equation or using Figure Ex. 3.8(b) and noting that
Q_{1}=-\frac{T_{1}-T_{2}}{R_{k,1-1^{\prime }}} or T_{1^{\prime }} =T_{1}+(Q_{1}R_{k,1-1^{\prime }})
Q_{1}=-\frac{T_{1^{\prime }}-T_{2^{\prime }}}{R_{k,1^{\prime }-2^{\prime }}} or T_{2^{\prime }} =T_{1^{\prime }}+(Q_{1}R_{k,1^{\prime }-2^{\prime }})
Using the numerical results, we have
T_{1^{\prime }}=45(^{\circ }C)-[-5.303(W)\times 2.177\times 10^{-5}(^{\circ }C/W)]=45(^{\circ }C) (no noticeable change in temperature across the copper tube)
T_{2^{\prime }}=45(^{\circ }C)-[-5.303(W)\times 3.755(^{\circ }C/W)]=25.08(^{\circ }C) (close to T_{2} ).
