Hot water at an average temperature of 90°C is flowing through a 15-m section of a cast iron pipe (k=52 W/m · °C) whose inner and outer diameters are 4 cm and 4.6 cm, respectively. The outer surface of the pipe, whose emissivity is 0.7, is exposed to the cold air at 10°C in the basement, with a heat

Question

Hot water at an average temperature of 90°C is flowing through a 15-m section of a cast iron pipe ([latex]k=52 W / m \cdot{ }^{\circ} C[/latex]) whose inner and outer diameters are 4 cm and 4.6 cm, respectively. The outer surface of the pipe, whose emissivity is 0.7, is exposed to the cold air at 10°C in the basement, with a heat transfer coefficient of [latex]15 W / m ^{2} \cdot{ }^{\circ} C[/latex]. The heat transfer coefficient at the inner surface of the pipe is [latex]120 W / m ^{2} \cdot{ }^{\circ} C[/latex]. Taking the walls of the basement to be at 10°C also, determine the rate of heat loss from the hot water. Also, determine the average velocity of the water in the pipe if the temperature of the water drops by 3°C as it passes through the basement.

Accepted Answer

Hot water is flowing through a 3-m section of a cast iron 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 center line and no variation in the axial direction. 3 Thermal properties are constant.

Properties The thermal conductivity and emissivity of cast iron are given to be [latex] k = 52 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_{1} L}=\frac{\ln (2.3 / 2)}{2 \pi\left(52 W / m \cdot{ }^{\circ} C ight)(15 m )}=0.00003{ }^{\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_{ ext {combined }} 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}=0.0044+0.00003+0.0229=0.0273^{\circ} C / W[/latex]

The rate of heat loss from the hot water pipe 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 2 7 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{2927 J / s }{\left(4180 J / kg \cdot{ }^{\circ} C ight)\left(3{ }^{\circ} C ight)}=0.233 kg / s[/latex]

[latex] \dot{m}= ho V A_{c} \longrightarrow V=\frac{\dot{m}}{ ho A_{c}}=\frac{0.233 kg / s }{\left(1000 kg / m ^{3} ight) \frac{\pi(0.04 m )^{2}}{4}}=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 2927 W =\frac{\left(90 - T_{s} ight)^{\circ} C }{(0.0044+0.00003)^{\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 3.76: Hot water at an average temperature of 90°C is flowing throu...

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