Hot water at \(90^{\circ} \mathrm{C}\) enters a \(15-\mathrm{m}\) section of a cast iron pipe \((k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are 4 and \(4.6 \mathrm{~cm}\), respectively, at an average velocity of \(1.2 \mathrm{~m} / \mathrm{s}\). The outer surface of the pipe, whose emissivity is \(0.7\), is exposed to the cold air at \(10^{\circ} \mathrm{C}\) in a basement, with a convection heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the walls of the basement to be at \(10^{\circ} \mathrm{C}\) also, determine \((a)\) the rate of heat loss from the water and \((b)\) the temperature at which the water leaves the basement.

In order to find the heat loss, we will first find the inner and outer surface areas of the pipe. These can be calculated using the following formulas: \(A_{in}=2\pi r_{in} L\), and \(A_{out}=2\pi r_{out} L\) where \(r_{in}\) and \(r_{out}\) are the radii of the inner and outer diameters, and \(L\) is the length of the pipe. From the given data, we know that the inner diameter is \(4 \mathrm{~cm}\) and the outer diameter is \(4.6 \mathrm{~cm}\). Therefore, the radii are \(r_{in}=2 \mathrm{~cm}\) and \(r_{out}=2.3 \mathrm{~cm}\). The length of the pipe is \(15 \mathrm{~m}\). Putting these values in the formulas, we can find the areas.

To find the rate of heat loss through conduction, we'll use the formula for conduction through a cylinder: \(q_{cond}=\frac{2\pi Lk(T_{in}-T_{out})}{ln(r_{out}/r_{in})}\) where \(q_{cond}\) is the conduction heat transfer rate, \(k\) is the pipe's thermal conductivity, \(T_{in}\) and \(T_{out}\) are the temperatures at the inner and outer surfaces of the pipe, and \(ln\) is the natural logarithm. We are given that the hot water enters the pipe at \(90^{\circ} \mathrm{C}\), so \(T_{in} = 90^{\circ} \mathrm{C}\). The surrounding air temperature is \(10^{\circ} \mathrm{C}\), and as the pipe is exposed to this air, \(T_{out} = 10^{\circ} \mathrm{C}\). The thermal conductivity of the cast-iron pipe is \(k = 52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Now, we can plug these values into the formula and calculate the conduction heat transfer rate, \(q_{cond}\).

Now, we'll find the rate of heat loss through convection from the outer surface of the pipe to the cold air. For that, we'll use the formula for convection heat transfer: \(q_{conv} = hA_{out}(T_{out}-T_{\infty})\) where \(q_{conv}\) is the convection heat transfer rate, \(h\) is the convection heat transfer coefficient, \(T_{\infty}\) is the air temperature, and \(A_{out}\) is the outer surface area of the pipe. From the given data, we know that the air temperature is \(10^{\circ} \mathrm{C}\), and the convection heat transfer coefficient is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). We have already calculated \(A_{out}\) in Step 1. Plugging these values into the formula, we can find the convection heat transfer rate, \(q_{conv}\).

The total heat loss rate from the water can be found by equating the conduction and convection heat transfer rates: \(q_{loss} = q_{cond} = q_{conv}\) We already calculated \(q_{cond}\) and \(q_{conv}\) in the previous steps, so we can now find the total heat loss rate from the water, \(q_{loss}\). To find the temperature of the water leaving the basement, we'll apply the concept of conservation of energy: the heat loss from the water should equal the heat gained by the surrounding air through convection. Therefore, using the formula for the convection heat transfer rate, we can find the final temperature of the water: \(q_{conv} = m \cdot c \cdot (T_{out}-T_{in})\) where \(m\) is the mass flow rate of the water, and \(c\) is the specific heat capacity of water. We can rearrange this formula to find the temperature of the water leaving the basement, \(T_{out}\), given all other parameters and the previously calculated \(q_{conv}\).