Question

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.

Short answer

Based on the given information and calculations, the rate of heat transfer from the pipe to the air in the rooms is (a) approximately 198.9 W, and the temperature drop of the hot water as it flows through the 5-m-long section of the wall is (b) approximately 0.24°C.

Step-by-step solution

Calculate the thermal resistance of the insulation

First, we need to calculate the thermal resistance of the insulation around the pipe, since the heat transfer rate depends on this resistance. The formula for thermal resistance of a cylindrical system is: $$ R = \frac{\ln(\frac{r_{2}}{r_{1}})}{2 \pi k L} $$ where \(r_{1}\) is the outer radius of the pipe, \(r_{2}\) is the outer radius of the insulation, \(k\) is the thermal conductivity of the insulation, and \(L\) is the length of the pipe.

Calculate the outer radius of the insulation

In our case, we have an outer diameter of the pipe of \(2.5\,\text{cm}\), and a wall thickness of \(14\,\text{cm}\). So, the outer radius of the pipe is: $$ r_{1} = \frac{2.5}{2} = 1.25\,\text{cm} = 0.0125\,\text{m} $$ Then, the outer radius of the insulation is: $$ r_{2} = r_{1} + 0.14\,\text{m} = 0.1525\,\text{m} $$

Calculate the thermal resistance of the insulation

Now, we can plug the values into the resistance equation above and get: $$ R = \frac{\ln(\frac{0.1525}{0.0125})}{2 \pi (0.035 \frac{\text{W}}{\text{m} \cdot \text{K}}) \cdot 5\text{m}} = 0.176\,\text{K/W} $$

Calculate the heat transfer rate

Now, we can calculate the heat transfer rate using the thermal resistance, with the formula: $$ q = \frac{T_{1} - T_{2}}{R} $$ where \(T_{1}\) is the temperature of the hot water in the pipe, and \(T_{2}\) is the temperature of the surfaces of the wall. Plugging in the values, we get: $$ q = \frac{53^{\circ} \mathrm{C} - 18^{\circ} \mathrm{C}}{0.176\,\text{K/W}} = 198.9\,\text{W} $$ So, the rate of heat transfer from the pipe to the air in the rooms is \((a) \approx 198.9\,\text{W}\).

Calculate the temperature drop of the hot water

We can now use the energy balance to find the temperature drop of the hot water as it flows through the 5-m-long section of the wall. We know that: $$ q = \dot{m} c_{p} \Delta T $$ where \(\dot{m}\) is the mass flow rate of the hot water, \(c_{p}\) is the specific heat capacity of water, and \(\Delta T\) is the temperature drop. The mass flow rate can be found using \(\dot{m} = \rho v A\), where \(\rho\) is the density of water, \(v\) is the average velocity of the hot water, and \(A\) is the cross-sectional area of the pipe. The area is given by: \(A = \pi r_{1}^{2}\). We have: $$ \dot{m} = (1000 \frac{\text{kg}}{\text{m}^3})(0.4 \frac{\text{m}}{\text{s}})(\pi (0.0125\,\text{m})^2) \approx 0.0196\,\text{kg/s} $$ Now, we plug this value into the energy balance equation and solve for the temperature drop: $$ \Delta T = \frac{q}{\dot{m} c_{p}} = \frac{198.9\,\text{W}}{(0.0196\,\text{kg/s})(4178\,\frac{\text{J}}{\text{kg} \cdot \text{K}})} \approx 0.24\,\text{K} $$ So, the temperature drop of the hot water as it flows through this 5-m-long section of the wall is \((b) \approx 0.24\,^{\circ}\mathrm{C}\).