A shell-and-tube heat exchanger with 2-shell passes and 8 -tube passes is used to heat ethyl alcohol \(\left(c_{p}=2670 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in the tubes from \(25^{\circ} \mathrm{C}\) to \(70^{\circ} \mathrm{C}\) at a rate of \(2.1 \mathrm{~kg} / \mathrm{s}\). The heating is to be done by water \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the shell side at \(95^{\circ} \mathrm{C}\) and leaves at \(45^{\circ} \mathrm{C}\). If the overall heat transfer coefficient is \(950 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer surface area of the heat exchanger.

First, we need to determine the heat transfer rate of the ethyl alcohol, which can be found using the following formula: $$q = m_{alcohol} \cdot c_{p,alcohol} \cdot \Delta T_{alcohol}$$ where \(q\) is the heat transfer rate, \(m_{alcohol}\) is the mass flow rate of the ethyl alcohol, \(c_{p,alcohol}\) is the specific heat capacity of the ethyl alcohol, and \(\Delta T_{alcohol}\) is the temperature difference for the ethyl alcohol. Using the given values, we get: $$q = 2.1 \frac{\text{kg}}{\text{s}} \cdot 2670 \frac{\text{J}}{\text{kg} \cdot \text{K}} \cdot (70 ^\circ \text{C}-25 ^\circ \text{C})$$

Next, we need to find the heat transfer rate of the water, which can be calculated using the same formula as before: $$q = m_{water} \cdot c_{p,water} \cdot \Delta T_{water}$$ where \(m_{water}\) is the mass flow rate of the water, \(c_{p,water}\) is the specific heat capacity of the water, and \(\Delta T_{water}\) is the temperature difference for the water. We also know that the heat transfer rate of the water and the ethyl alcohol is supposed to be the same, so: $$m_{water} \cdot c_{p,water} \cdot \Delta T_{water} = m_{alcohol} \cdot c_{p,alcohol} \cdot \Delta T_{alcohol}$$ Rearranging the equation to solve for the mass flow rate of the water, we get: $$m_{water} = \frac{m_{alcohol} \cdot c_{p,alcohol} \cdot \Delta T_{alcohol}}{c_{p,water} \cdot \Delta T_{water}}$$

Now that we have the heat transfer rate, we can determine the required surface area of the heat exchanger. To do this, we will use the overall heat transfer coefficient \(U\) and the expression: $$q = U \cdot A \cdot \Delta T_{lm}$$ where \(A\) is the heat transfer surface area and \(\Delta T_{lm}\) is the logarithmic mean temperature difference between the two fluids in the exchanger. Since we have a 2-shell pass and 8-tube pass heat exchanger, we can assume that the flow is counter-current, meaning that the hot and cold fluids flow in opposite directions. To find the \(\Delta T_{lm}\), we need to calculate the inlet and outlet temperature differences: $$\Delta T_{inlet} = T_{Hin} - T_{Cin}$$ $$\Delta T_{outlet} = T_{Hout} - T_{Cout}$$ where \(T_{Hin}\) and \(T_{Hout}\) are the inlet and outlet temperatures of the hot fluid (water), and \(T_{Cin}\) and \(T_{Cout}\) are the inlet and outlet temperatures of the cold fluid (ethyl alcohol). Now we can determine the \(\Delta T_{lm}\) using: $$\Delta T_{lm} = \frac{\Delta T_{inlet} - \Delta T_{outlet}}{\ln(\frac{\Delta T_{inlet}}{\Delta T_{outlet}})}$$ Finally, we can find the required heat transfer surface area \(A\) using the formula: $$A = \frac{q}{U \cdot \Delta T_{lm}}$$ Once all the calculations are carried out, we will find the total heat transfer surface area of the shell-and-tube heat exchanger.