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)\( in the tubes from \)20^{\circ} \mathrm{C}\( to \)70^{\circ} \mathrm{C}\( at a rate of \)4.5 \mathrm{~kg} / \mathrm{s}$. Heat is supplied by hot oil \(\left(c_{p}=2300 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the shell side at \(170^{\circ} \mathrm{C}\) at a rate of $10 \mathrm{~kg} / \mathrm{s}$. For a tube-side overall heat transfer coefficient of \(350 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer surface area on the tube side. Answer: \(25.7 \mathrm{~m}^{2}\)

Short answer

Question: Determine the heat transfer surface area on the tube side of a shell-and-tube heat exchanger given the flow rates and specific heat capacities of the water and oil, the initial and final temperatures of the water, and the overall heat transfer coefficient on the tube side. Answer: The heat transfer surface area on the tube side of the shell-and-tube heat exchanger is approximately 25.7 m².

Step-by-step solution

Calculate the heat transfer rate required for the water to reach the desired temperature

Since we know the mass flow rate, specific heat capacity, and temperature difference of the water, we can find the heat transfer rate using the following equation: Q = m_water * c_p_water * ΔT_water Where Q represents the heat transfer rate (in watts), m_water is the mass flow rate of water (in kg/s), c_p_water is the specific heat capacity of water (in J/kg·K), and ΔT_water is the temperature difference of the water (in Kelvin). Q = (4.5 kg/s) * (4180 J/kg·K) * (70 - 20) K Q = 945900 W

Calculate the logarithmic mean temperature difference (LMTD)

Next, we need to calculate the logarithmic mean temperature difference. Since there are two shell passes and 12 tube passes, the fluid temperatures are as follows: T1_hot = 170°C, T2_hot = unknown, T1_cold = 20°C, T2_cold = 70°C We use the following equation to find the LMTD: LMTD = (ΔT1 - ΔT2) / ln(ΔT1/ΔT2) Where ΔT1 = T1_hot - T1_cold and ΔT2 = T2_hot - T2_cold To find T2_hot, we use the heat transfer rate equation for the hot oil: Q = m_oil * c_p_oil * (T1_hot - T2_hot) Where m_oil is the mass flow rate of oil (in kg/s) and c_p_oil is the specific heat capacity of oil (in J/kg·K) T2_hot = T1_hot - Q / (m_oil * c_p_oil) = 170 - (945900 W) / (10 kg/s * 2300 J/kg·K) T2_hot ≈ 131.14°C Now, we can find the LMTD: LMTD = ((170 - 20) - (131.14 - 70)) / ln((170 - 20) / (131.14 - 70)) LMTD ≈ 78.21 K

Calculate the heat transfer surface area

Now, we can find the heat transfer surface area on the tube side using the following equation: A = Q / (U * LMTD) Where A is the heat transfer surface area (in m²), U is the overall heat transfer coefficient (in W/m²·K), and LMTD is the logarithmic mean temperature difference (in Kelvin). A = 945900 W / (350 W/m²·K * 78.21 K) ≈ 25.7 m² So, the heat transfer surface area on the tube side is approximately 25.7 m².