Repeat Problem 3–76 for a pipe made of copper (k = 386 W/m · °C) instead of cast iron.

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Hot water is flowing through a 15 m section of a copper pipe. The pipe is exposed to cold air and surfaces in the basement. The rate of heat loss from the hot water and the average velocity of the water in the pipe as it passes through the basement are to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal properties are constant.

Properties The thermal conductivity and emissivity of copper are given to be [latex] k = 386 W / m \cdot{ }^{\circ} C ext { and } \varepsilon=0.7[/latex]

Analysis The individual resistances are

[latex] A_{i}=\pi D_{i} L=\pi(0.04 m )(15 m )=1.885 m ^{2}[/latex]

[latex] A_{o}=\pi D_{o} L=\pi(0.046 m )(15 m )=2.168 m ^{2} [/latex]

[latex] R_{i}=\frac{1}{h_{i} A_{i}}=\frac{1}{\left(120 W / m ^{2} \cdot{ }^{\circ} C ight)\left(1.885 m ^{2} ight)}=0.0044^{\circ} C / W [/latex]

[latex] R_{ ext {pipe }}=\frac{\ln \left(r_{2} / r_{1} ight)}{2 \pi k L}=\frac{\ln (2.3 / 2)}{2 \pi\left(386 W / m \cdot{ }^{\circ} C ight)(15 m )} = 0.0000038^{\circ} C / W[/latex]

The outer surface temperature of the pipe will be somewhat below the water temperature. Assuming the outer surface temperature of the pipe to be 80°C (we will check this assumption later), the radiation heat transfer coefficient is determined to be

[latex] h_{r a d}=\varepsilon \sigma\left(T_{2}^{2}+T_{s u r r}^{2} ight)\left(T_{2}+T_{s u r r} ight) [/latex]

[latex] = (0.7)\left(5.67 imes 10^{-8} W / m ^{2} \cdot K ^{4} ight)\left[(353 K )^{2}+(283 K )^{2} ight](353+283) [/latex]

[latex] = 5.167 W / m ^{2} \cdot K [/latex]

Since the surrounding medium and surfaces are at the same temperature, the radiation and convection heat transfer coefficients can be added and the result can be taken as the combined heat transfer coefficient. Then,

[latex] h_{ ext {combined }}=h_{ ext {rad }}+h_{ ext {conv }, 2}=5.167+15=20.167 W / m ^{2} \cdot{ }^{\circ} C[/latex]

[latex] R_{o}=\frac{1}{h_{o} A_{o}}=\frac{1}{\left(20.167 W / m ^{2} \cdot{ }^{\circ} C ight)\left(2.168 m ^{2} ight)}=0.0229^{\circ} C / W[/latex]

[latex] R_{ ext {total }}=R_{i}+R_{ ext {pipe }}+R_{o}=0.004+0.0000038+0.0229 = 0.0273^{\circ} C / W[/latex]

The rate of heat loss from the hot tank water then becomes

[latex] \dot{Q}=\frac{T_{\infty 1} - T_{\infty 2}}{R_{ ext {total }}}=\frac{(90 - 10)^{\circ} C }{0.0273{ }^{\circ} C / W } = 2 9 3 0 W[/latex]

For a temperature drop of 3°C, the mass flow rate of water and the average velocity of water must be

[latex] \dot{Q}=\dot{m} C_{p} \Delta T \longrightarrow \dot{m}=\frac{\dot{Q}}{C_{p} \Delta T}=\frac{2930 J / s }{\left(4180 J / kg \cdot{ }^{\circ} C ight)\left(3{ }^{\circ} C ight)}=0.234 kg / s [/latex]

[latex] \dot{m} = ho V A_{c} \longrightarrow V = \frac{\dot{m}}{ ho A_{c}} = \frac{0.234 kg / s }{\left(1000 kg / m ^{3} ight)\left[\frac{\pi(0.04 m )^{2}}{4} ight]}=0.186 m / s[/latex]

Discussion The outer surface temperature of the pipe is

[latex] \dot{Q}=\frac{T_{\infty 1}-T_{s}}{R_{i}+R_{ ext {pipe }}} ightarrow 2930 W =\frac{\left(90 - T_{s} ight)^{\circ} C }{(0.0044+0.0000)^{\circ} C / W } ightarrow T_{s} = 77^{\circ} C[/latex]

which is very close to the value assumed for the surface temperature in the evaluation of the radiation resistance. Therefore, there is no need to repeat the calculations.

Question 3.77: Repeat Problem 3–76 for a pipe made of copper (k = 386 W/m ·...

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