Question

Hot water at an average temperature of \(80^{\circ} \mathrm{C}\) and an average velocity of \(1.5 \mathrm{~m} / \mathrm{s}\) is flowing through a \(25-\mathrm{m}\) section of a pipe that has an outer diameter of $5 \mathrm{~cm}\(. The pipe extends \)2 \mathrm{~m}$ in the ambient air above the ground, dips into the ground $(k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( ) vertically for \)3 \mathrm{~m}$, and continues horizontally at this depth for \(20 \mathrm{~m}\) more before it enters the next building.

Short answer

Based on the given information, we are unable to calculate the heat loss in the different sections of the pipe due to missing data on the pipe's thermal conductivity, ambient air temperature, and ground temperature. To proceed, these additional details need to be supplied to calculate the heat loss in each section and to find the total heat loss in the pipe.

Step-by-step solution

Calculate the heat loss in the air section

The formula for heat loss in a cylinder exposed to ambient air is: \(q = 2 \cdot \pi \cdot k \cdot L_{air} \cdot \frac{T_{avg} - T_{ambient}}{r}\) where \(q\) is the heat loss, \(k\) is the thermal conductivity of the pipe, \(L_{air}\) is the length of the pipe exposed to air, \(T_{avg}\) is the average temperature of the hot water, \(T_{ambient}\) is the ambient temperature, and \(r\) is the radius of the pipe. Let's use the given values and calculate the heat loss in the air section of the pipe. We don't know the thermal conductivity of the pipe (\(k\)) and the ambient temperature (\(T_{ambient}\)); therefore, we are not able to calculate the heat loss in the air section of the pipe. Make a note that more information would be needed for this part of the problem.

Calculate the heat loss in the vertical ground section

Again, the formula for heat loss in a cylinder is: \(q = 2 \cdot \pi \cdot k \cdot L_{vertical} \cdot \frac{T_{avg} - T_{ground}}{r}\) where \(L_{vertical}\) is the length of the pipe vertically into the ground, and \(T_{ground}\) is the ground temperature. We don't know the thermal conductivity of the pipe (\(k\)) and the ground temperature (\(T_{ground}\)); therefore, we are not able to calculate the heat loss in the vertical ground section of the pipe. Make a note that more information would be needed for this part of the problem.

Calculate the heat loss in the horizontal ground section

The formula for heat loss in a cylinder horizontally in the ground is: \(q = 2 \cdot \pi \cdot k \cdot L_{horizontal} \cdot \frac{T_{avg} - T_{ground}}{r}\) where \(L_{horizontal}\) is the length of the pipe horizontally in the ground. We don't know the thermal conductivity of the pipe (\(k\)) and the ground temperature (\(T_{ground}\)); therefore, we are not able to calculate the heat loss in the horizontal ground section of the pipe. Make a note that more information would be needed for this part of the problem.

Conclusion

We have attempted to calculate the heat loss in the three sections of the pipe (air, vertical ground, and horizontal ground) using the available information. However, more information is necessary to solve this problem entirely, specifically the thermal conductivity of the pipe and temperatures of the ambient air and ground. If additional information is provided, one can recalculate the heat loss in each section, then sum them up to find the total heat loss in the pipe.