A shell-and-tube heat exchanger with 2-shell passes and 12 -tube passes is used to heat water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) with ethylene glycol \(\left(c_{p}=2680 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\). Water enters the tubes at \(22^{\circ} \mathrm{C}\) at a rate of \(0.8 \mathrm{~kg} / \mathrm{s}\) and leaves at \(70^{\circ} \mathrm{C}\). Ethylene \(\mathrm{glycol}\) enters the shell at \(110^{\circ} \mathrm{C}\) and leaves at \(60^{\circ} \mathrm{C}\). If the overall heat transfer coefficient based on the tube side is \(280 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer and the heat transfer surface area on the tube side.

We can find the rate of heat transfer using the equation: \(\dot{Q} = \dot{m}_{water} \cdot c_{p_{water}} \cdot ( T_{out_{water}} - T_{in_{water}} )\) where \(\dot{Q}\) = rate of heat transfer (W) \(\dot{m}_{water}\) = mass flow rate of water (0.8 kg/s) \(c_{p_{water}}\) = specific heat capacity of water (4180 J/kg·K) \(T_{out_{water}}\) = outlet temperature of water (70°C) \(T_{in_{water}}\) = inlet temperature of water (22°C) Now, let's plug in the values: \(\dot{Q} = 0.8 \cdot 4180 \cdot (70-22)\) \(\dot{Q} = 161216 \mathrm{~W}\) The heat transfer rate is 161216 W.

Now we can calculate the heat transfer surface area using the LMTD method. First, we need to find the temperature differences at the inlet and outlet of the heat exchanger: $\Delta T_{in} = T_{in_{glycol}} - T_{in_{water}} = 110 - 22 = 88 \mathrm{~K}$ $\Delta T_{out} = T_{out_{glycol}} - T_{out_{water}} = 60 - 70 = -10 \mathrm{~K}$ Next, we can find the log mean temperature difference (LMTD) using the equation: \(LMTD = \frac{\Delta T_{in} - \Delta T_{out}}{\ln \left( \frac{\Delta T_{in}}{\Delta T_{out}} \right)}\) Plug in the values for \(\Delta T_{in}\) and \(\Delta T_{out}\): \(LMTD = \frac{88 - (-10)}{\ln \left( \frac{88}{-10} \right)} \approx 29.07 \mathrm{~K}\) Now, we can calculate the heat transfer surface area on the tube side using the overall heat transfer coefficient: \(A = \frac{\dot{Q}}{U \cdot LMTD}\) where \(A\) = heat transfer surface area (m²) \(U\) = overall heat transfer coefficient (280 W/m²·K) Plug in the values for \(\dot{Q}\), \(U\), and \(LMTD\): \(A = \frac{161216}{280 \cdot 29.07} \approx 19.69 \mathrm{~m}^2\) The heat transfer surface area on the tube side is approximately 19.69 m².