Question

A shell-and-tube heat exchanger with two 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 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.

Short answer

Answer: The rate of heat transfer is 161,328 W, and the heat transfer surface area on the tube side is 16.08 m².

Step-by-step solution

Calculate the heat gained by water

We know that the heat gained by the water can be calculated using the following equation: \(Q = m_{w} * c_{p_{w}} * (T_{out_{w}} - T_{in_{w}})\) where \(m_{w}\) is the mass flow rate of water, \(c_{p_{w}}\) is the specific heat capacity of water, and \(T_{in_{w}}\) and \(T_{out_{w}}\) are the inlet and outlet temperatures of water, respectively. Plugging in the given values: \(Q = 0.8 \mathrm{\ kg/s} * 4180 \mathrm{\ J/kg\cdot K} * (70^{\circ}\mathrm{C} - 22^{\circ}\mathrm{C})\) \(Q = 161328 \mathrm{\ W}\)

Calculate the heat lost by ethylene glycol

The heat lost by ethylene glycol can be calculated using the same equation: \(Q = m_{g} * c_{p_{g}} * (T_{in_{g}} - T_{out_{g}})\) where \(m_{g}\) is the mass flow rate of ethylene glycol, \(c_{p_{g}}\) is the specific heat capacity of ethylene glycol, and \(T_{in_{g}}\) and \(T_{out_{g}}\) are the inlet and outlet temperatures of ethylene glycol, respectively. Since we do not know the mass flow rate of ethylene glycol, we can equate the heat gained by water to the heat lost by ethylene glycol: \(161328 \mathrm{\ W} = m_{g} * 2680 \mathrm{\ J/kg\cdot K} * (110^{\circ}\mathrm{C} - 60^{\circ}\mathrm{C})\) Solving for \(m_{g}\): \(m_{g} = \dfrac{161328 \mathrm{\ W}}{ 2680 \mathrm{\ J/kg\cdot K} * 50}\), \(m_{g} = 1.202 \mathrm{\ kg/s}\)

Calculate the LMTD

The logarithmic mean temperature difference (LMTD) is given by the following equation: \(LMTD = \dfrac{(T_{1h} - T_{2c}) - (T_{2h} - T_{1c})}{\ln \dfrac{T_{1h} - T_{2c}}{T_{2h} - T_{1c}}}\) where \(T_{1h}\) and \(T_{2h}\) are the inlet and outlet temperatures of the hot fluid (ethylene glycol), and \(T_{1c}\) and \(T_{2c}\) are the inlet and outlet temperatures of the cold fluid (water). Plugging in the given values: \(LMTD = \dfrac{(110-22) - (60-70)}{\ln \dfrac{110-22}{60-70}}\) \(LMTD = \dfrac{88 + 10}{\ln \dfrac{88}{-10}}\) \(LMTD = 36.39 ^{\circ}\mathrm{C}\)

Calculate the heat transfer surface area

Now we can use the heat transfer equation to determine the surface area: \(Q = U * A * LMTD\) We know the values of \(Q\), \(U\), and \(LMTD\), so we can solve for \(A\): \(A = \dfrac{Q}{U * LMTD}\) \(A = \dfrac{161328 \mathrm{\ W}}{280 \mathrm{\ W/m^{2}\cdot K} * 36.39 ^{\circ}\mathrm{C}}\) \(A = 16.08 \mathrm{\ m^{2}}\) The rate of heat transfer is \(161328 \mathrm{\ W}\), and the heat transfer surface area on the tube side is \(16.08 \mathrm{\ m^{2}}\).