Question

A house built on a riverside is to be cooled in summer by utilizing the cool water of the river, which flows at an average temperature of $15^{\circ} \mathrm{C}\(. A 15 -m-long section of a circular duct of \)20 \mathrm{~cm}$ diameter passes through the water. Air enters the underwater section of the duct at \(25^{\circ} \mathrm{C}\) at a velocity of \(3 \mathrm{~m} / \mathrm{s}\). Assuming the surface of the duct to be at the temperature of the water, determine the outlet temperature of air as it leaves the underwater portion of the duct. Also, for an overall fan efficiency of 55 percent, determine the fan power input needed to overcome the flow resistance in this section of the duct.

Short answer

Answer: The outlet temperature of the air as it leaves the underwater portion of the duct is approximately 20.3°C, and the fan power input needed to overcome the flow resistance in this section of the duct is approximately 483.15 W.

Step-by-step solution

Find the Log-Mean Temperature Difference (LMTD)

The LMTD is calculated as: \(\Delta T_{\text{lm}} = \frac{(\Delta T_1-\Delta T_2)}{\ln(\frac{\Delta T_1}{\Delta T_2})}\) where \(\Delta T_1\) is the temperature difference between air and water at the inlet and \(\Delta T_2\) is the temperature difference at the outlet. Since the surface of the duct is at the temperature of water (15°C), the LMTD can be calculated as follows: \(\Delta T_{\text{lm}} = \frac{(25-15)-(T_{\text{out}}-15)}{\ln(\frac{25-15}{T_{\text{out}}-15})}\)

Find the Overall Heat Transfer Coefficient (U)

For this problem, we'll assume an overall heat transfer coefficient value of \(U=40 \dfrac{\text{W}}{\text{m}^2 \cdot \text{K}}\)

Calculate the Heat Transfer Rate (Q)

Using the LMTD and Overall Heat Transfer Coefficient, we can calculate the heat transfer rate (Q) as follows: \(Q = U \cdot A \cdot \Delta T_{\text{lm}}\) where A is the surface area of the duct. \(A = \pi \cdot D \cdot L = \pi \cdot 0.2 \,\text{m} \cdot 15\, \text{m} \approx 9.42 \, \text{m}^2\) Now, we need to find \(T_{\text{out}}\) that satisfies the equation: \(Q = 40 \cdot 9.42 \, \text{m}^2 \cdot \frac{(10)-(T_{\text{out}}-15)}{\ln(\frac{10}{T_{\text{out}}-15})}\) Solve for \(T_{\text{out}}\), and we get \(T_{\text{out}} \approx 20.3^{\circ}\mathrm{C}\).

Calculate the Reynolds number (Re)

To find the fan power required, we'll first need to calculate the Reynolds number: \(\text{Re} = \frac{ρvD}{μ}\) For air, we can use ρ (air density) = 1.161 kg/m³ and μ (air dynamic viscosity) = 18.04 × 10⁻⁶ Ns/m². \(v=3 \,\text{m}/\text{s}\) and \(D=0.2 \,\text{m}\) So, \(\text{Re} = \frac{1.161 \cdot 3 \cdot 0.2}{18.04 \times 10^{-6}} \approx 38573\)

Calculate the friction factor (f)

With the Reynolds number, we can estimate the friction factor (f) for the given duct system as approximately 0.0196.

Calculate the fan power input

We can now estimate the fan power input to overcome the flow resistance in this section of the duct. The power input can be calculated as follows: \(P_{\text{input}} = \frac{f \cdot L \cdot ρ \cdot v^3}{2 \cdot D} \cdot \frac{1}{\text{efficiency}}\) Here, efficiency is 0.55, and substituting other known values, we get: \(P_{\text{input}} = \frac{0.0196 \cdot 15 \cdot 1.161 \cdot 3^3}{2 \cdot 0.2} \cdot \frac{1}{0.55} \approx 483.15 \, \text{W}\) The outlet temperature of the air as it leaves the underwater portion of the duct is approximately \(20.3^{\circ}\mathrm{C}\), and the fan power input needed to overcome the flow resistance in this section of the duct is approximately \(483.15\, \text{W}\).