Question

Hot water is flowing at an average velocity of \(1.5 \mathrm{~m} / \mathrm{s}\) through a cast iron pipe \((k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are \(3 \mathrm{~cm}\) and \(3.5 \mathrm{~cm}\), respectively. The pipe passes through a \(15-\mathrm{m}\)-long section of a basement whose temperature is \(15^{\circ} \mathrm{C}\). If the temperature of the water drops from \(70^{\circ} \mathrm{C}\) to \(67^{\circ} \mathrm{C}\) as it passes through the basement and the heat transfer coefficient on the inner surface of the pipe is \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the combined convection and radiation heat transfer coefficient at the outer surface of the pipe.

Short answer

The combined convection and radiation heat transfer coefficient at the outer surface of the pipe, \(h_2\), can be found by rearranging the equation from Step 5: \(h_{2} = \frac{q \cdot \Delta x}{k \cdot A_{circum} \cdot L \cdot \Delta T}\). By plugging in the given values and solving for \(h_2\), you can find the combined convection and radiation heat transfer coefficient at the outer surface of the pipe.

Step-by-step solution

Calculate the temperature difference and pipe section length

We are given the inlet temperature \(T_{1} = 70^{\circ} \mathrm{C}\), the outlet temperature \(T_{2} = 67^{\circ} \mathrm{C}\), and the pipe length L = \(15 \mathrm{m}\).

Calculate the heat conduction through the pipe wall

First, we will calculate the heat conduction through the pipe wall using Fourier's Law of heat conduction: \(q = k\frac{A\Delta T}{\Delta x}\), where q is the heat transfer rate, k is the thermal conductivity, A is the pipe cross-sectional area, \(\Delta T\) is the temperature difference, and \(\Delta x\) is the thickness of the pipe wall. The pipe wall thickness can be calculated as \(\Delta x = \frac{d_{2} - d_{1}}{2}\), where \(d_{1}\) is the inner diameter and \(d_{2}\) is the outer diameter. Here, \(d_{1} = 3 \mathrm{cm}\) and \(d_{2} = 3.5 \mathrm{cm}\), so \(\Delta x = 0.25 \mathrm{cm}\). Next, we will find the temperature drop across the pipe wall, \(\Delta T = T_{water} - T_{wall}\). The average water temperature, \(T_{water}\), can be calculated as: \(\frac{T_{1} + T_{2}}{2}\). Thus, \(T_{water} = 68.5^{\circ} \mathrm{C}\). Using Fourier's Law, we can calculate q: \(q = \frac{k \cdot A_{circum} \cdot L \cdot (T_{water} - T_{basement})}{\Delta x}\)

Calculate the heat transfer rate from the inner surface of the pipe

We are given the heat transfer coefficient on the inner surface of the pipe, \(h_{1} = 400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). We can now calculate the heat transfer rate from the inner surface of the pipe, \(q_{1}\), using the formula: \(q_{1} = h_{1} \cdot A_{1} \cdot (T_{water} - T_{basement})\)

Calculate the heat transfer rate from the outer surface of the pipe

We want to find heat transfer rate at the outer surface of the pipe, which is equal to the heat conducted through the pipe wall: \(q = q_{1} = q_{2}\), where \(q_{2}\) is the heat transfer rate at the outer surface of the pipe.

Determine the combined convection and radiation heat transfer coefficient

Now, we can calculate the combined convection and radiation heat transfer coefficient at the outer surface of the pipe, \(h_{2}\): \(q_{2} = h_{2} \cdot A_{2} \cdot (T_{wall} - T_{basement})\) From Step 4, we know that \(q_{2} = q_{1} = q\). Therefore, we can rearrange the equation to solve for \(h_{2}\): \(h_{2} = \frac{q \cdot \Delta x}{k \cdot A_{circum} \cdot L \cdot \Delta T}\). Plug in the given values, and the combined convection and radiation heat transfer coefficient at the outer surface of the pipe, \(h_{2}\), can be found.