Question

A heat exchanger contains 400 tubes with inner diameter of \(23 \mathrm{~mm}\) and outer diameter of \(25 \mathrm{~mm}\). The length of each tube is $3.7 \mathrm{~m}\(. The corrected log mean temperature difference is \)23^{\circ} \mathrm{C}$, while the inner surface convection heat transfer coefficient is \(3410 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the outer surface convection heat transfer coefficient is \(6820 \mathrm{~W} / \mathrm{m}^{2}\), \(\mathrm{K}\). If the thermal resistance of the tubes is negligible, determine the heat transfer rate.

Short answer

Answer: Using the given information and the calculations in the solution, the heat transfer rate in the heat exchanger can be found by the formula: \(q = U \cdot 400 A_{1} \cdot 23 \mathrm{~K}\) First, calculate the surface area of one tube (\(A_{1}\)): \(A_{1} = 2 \pi (0.0115 \mathrm{~m})(3.7 \mathrm{~m})\) Next, find the overall heat transfer coefficient (U): \(U = \left(\frac{1}{3410} + \frac{1}{6820}\right)^{-1}\) Finally, determine the heat transfer rate (q) using these values: \(q = U \cdot 400 A_{1} \cdot 23 \mathrm{~K}\) Perform these calculations to find the heat transfer rate in Watts (W).

Step-by-step solution

Determine the total surface area of the tubes

To find the total surface area of the tubes, we first need to calculate the surface area of one tube and then multiply it by the total number of tubes. The surface area of one tube can be found using the formula: \(A_{1} = 2 \pi r L\) Where \(r\) is the inner radius (half of the inner diameter) and \(L\) is the length of the tube. Let's calculate \(A_{1}\): \(A_{1} = 2 \pi (0.0115 \mathrm{~m})(3.7 \mathrm{~m})\) Now we can calculate the total surface area \(A_{total}\) by multiplying by the number of tubes: \(A_{total} = 400 A_{1}\)

Calculate the overall heat transfer coefficient

Since the thermal resistance of the tubes is negligible, we can calculate the overall heat transfer coefficient \(U\) using the convection heat transfer coefficients of the inner (\(h_{in}\)) and outer (\(h_{out}\)) surfaces: \(\frac{1}{U} = \frac{1}{h_{in}} + \frac{1}{h_{out}}\) Let's calculate \(U\): \(U = \left(\frac{1}{3410} + \frac{1}{6820}\right)^{-1}\)

Determine the heat transfer rate using the overall heat transfer coefficient

Now we can determine the heat transfer rate \(q\) using the overall heat transfer coefficient (\(U\)), the total surface area (\(A_{total}\)), and the corrected log mean temperature difference (\(\Delta T_{m}\)): \(q = U A_{total} \Delta T_{m}\) Let's plug in the values to find \(q\): \(q = U \cdot 400 A_{1} \cdot 23 \mathrm{~K}\) The heat transfer rate \(q\) will be in Watts (W).