A test is conducted to determine the overall heat transfer coefficient in a shell-and-tube oil-to-water heat exchanger that has 24 tubes of internal diameter \(1.2 \mathrm{~cm}\) and length \(2 \mathrm{~m}\) in a single shell. Cold water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes at \(20^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\) and leaves at \(55^{\circ} \mathrm{C}\). Oil \(\left(c_{p}=2150 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flows through the shell and is cooled from \(120^{\circ} \mathrm{C}\) to \(45^{\circ} \mathrm{C}\). Determine the overall heat transfer coefficient \(U_{i}\) of this heat exchanger based on the inner surface area of the tubes.

First, we will find the heat transfer rate \((Q)\) for the water flowing through the tubes. The formula to compute the heat transfer rate is: $$Q = m_w c_{p_w}(T_{out,w} - T_{in,w})$$ Where: - \(Q\) is the heat transfer rate, in Watts - \(m_w\) is the mass flow rate of water, in kg/s - \(c_{p_w}\) is the specific heat capacity of water, in J/kg·K - \(T_{out,w}\) is the outlet temperature of the water, in °C - \(T_{in,w}\) is the inlet temperature of the water, in °C Given values: - \(m_w = 3 \hspace{1mm} \text{kg/s}\) - \(c_{p_w} = 4180 \hspace{1mm} \text{J/kg·K}\) - \(T_{in,w} = 20^{\circ}\mathrm{C}\) - \(T_{out,w} = 55^{\circ}\mathrm{C}\) Now, we will compute the heat transfer rate. $$Q_w = (3 \hspace{1mm}\text{kg/s})(4180\hspace{1mm}\text{J/kg·K})(55^{\circ}\mathrm{C}-20^{\circ}\mathrm{C})$$

Next, we will find the heat transfer rate \((Q)\) for the oil flowing through the shell. The formula to compute the heat transfer rate is: $$Q = m_o c_{p_o}(T_{in,o} - T_{out,o})$$ Where: - \(Q\) is the heat transfer rate, in Watts - \(m_o\) is the mass flow rate of oil, in kg/s - \(c_{p_o}\) is the specific heat capacity of oil, in J/kg·K - \(T_{out,o}\) is the outlet temperature of the oil, in °C - \(T_{in,o}\) is the inlet temperature of the oil, in °C Given values: - \(c_{p_o} = 2150 \hspace{1mm} \text{J/kg·K}\) - \(T_{in,o} = 120^{\circ}\mathrm{C}\) - \(T_{out,o} = 45^{\circ}\mathrm{C}\) As the heat transfer rate should be equal for both oil and water, which we found in step 1. To find the mass flow rate of the oil, we can rewrite the formula above to solve for \(m_o\): $$m_o = \frac{Q}{c_{p_o}(T_{in,o} - T_{out,o})}$$ Now, we can compute the mass flow rate of oil.

In this step, we will compute the logarithmic mean temperature difference \((\Delta T_{lm})\), which represents the average temperature difference between the two fluids. The formula for \(\Delta T_{lm}\) is: $$\Delta T_{lm} = \frac{(T_{in,o} - T_{out,w}) - (T_{out,o} - T_{in,w})}{\ln \left( \frac{T_{in,o} - T_{out,w}}{T_{out,o} - T_{in,w}} \right)}$$ Using the given values and computed temperatures, calculate the logarithmic mean temperature difference.

Now, we will compute the inner surface area of the tubes. The formula for the surface area of a single tube is: $$A_{tube} = \pi d_i L$$ Where: - \(A_{tube}\) is the surface area of a single tube, in m² - \(d_i\) is the inner diameter of the tube, in m - \(L\) is the length of the tube, in m Given values: - Tube diameter, \(d_i = 1.2 \hspace{1mm} \text{cm} = 0.012 \hspace{1mm} \text{m}\) - Tube length, \(L = 2 \hspace{1mm} \text{m}\) - Number of tubes, \(n = 24\) Compute the inner surface area of a single tube and then multiply by the number of tubes to obtain the total inner surface area \((A_i)\).

Finally, we will compute the overall heat transfer coefficient \((U_i)\) based on the inner surface area of the tubes. The formula for \((U_i)\) is: $$U_i = \frac{Q}{A_i \Delta T_{lm}}$$ Use the heat transfer rate, inner surface area, and logarithmic mean temperature difference found earlier to compute \(U_i\).