Hot water at an average temperature of \(53^{\circ} \mathrm{C}\) and an average velocity of \(0.4 \mathrm{~m} / \mathrm{s}\) is flowing through a \(5-\mathrm{m}\) section of a thin-walled hot-water pipe that has an outer diameter of \(2.5 \mathrm{~cm}\). The pipe passes through the center of a \(14-\mathrm{cm}\)-thick wall filled with fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). If the surfaces of the wall are at \(18^{\circ} \mathrm{C}\), determine \((a)\) the rate of heat transfer from the pipe to the air in the rooms and \((b)\) the temperature drop of the hot water as it flows through this 5 -m-long section of the wall. Answers: \(19.6 \mathrm{~W}, 0.024^{\circ} \mathrm{C}\)

To find the surface area of the pipe, use the formula for the surface area of a cylinder: \(A=2\pi r L\), where \(r\) is the outer radius of the pipe, and \(L\) is the length of the pipe. Here, \(r = \frac{2.5}{2} \times 10^{-2} \mathrm{~m}\) and \(L = 5 \mathrm{~m}\). \(A = 2 \pi \left(\frac{2.5}{2} \times 10^{-2}\right) (5)\) Calculate the surface area of the pipe.

Using the formula for heat transfer through a cylindrical wall: \(q=\frac{2\pi k L(T_{1}-T_{2})}{\ln{\frac{r_{2}}{r_{1}}}}\), where \(k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is the thermal conductivity of the insulation, \(L=5\mathrm{~m}\) is the length of the section, \((T_{1}-T_{2})=(53-18)=35 ^{\circ}\mathrm{C}\) is the temperature difference between the pipe's surface and the air, \(r_{1}=\frac{2.5}{2}\times10^{-2}\) m is the outer radius and \(r_{2}=(\frac{2.5}{2} + 14) \times 10^{-2}\) m is the outer radius plus wall thickness. Calculate the heat transfer rate using the formula.

As we already found the heat transfer rate \(q\), the next step is to find the temperature drop of the hot water using the energy balance equation: \(\Delta T = \frac{q}{\dot{m}C_p}\), where \(\Delta T\) is the temperature drop, \(\dot{m}\) is the mass flow rate of the hot water, and \(C_p\) is the specific heat capacity of water. First, we need to find the volumetric flow rate: \(V = A_v \times v \), where \(A_v = \pi(r_1)^2\) is the cross-sectional area of the pipe, and \(v=0.4\mathrm{~m}/\mathrm{s}\) is the flow velocity. \(\dot{m}= \rho V\), where \(\rho\) is the density of water, approximately \(1000 \mathrm{~kg}/\mathrm{m}^{3}\). Finally, for water, we have \(C_p \approx 4180 \mathrm{~J}/\mathrm{kg} \cdot \mathrm{K}\). Calculate \(\Delta T\) by plugging in the values. Once you've followed these steps, you'll find that the rate of heat transfer from the pipe to the air in the room is approximately \(19.6\mathrm{~W}\), and the temperature drop of the hot water as it flows through the 5m-long section of the wall is approximately \(0.024^{\circ}\mathrm{C}\).