Question

Hot oil \(\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is to be cooled by water $\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)$ in a 2-shell-passes and 12 -tube-passes heat exchanger. The tubes are thin-walled and are made of copper with a diameter of $1.8 \mathrm{~cm}\(. The length of each tube pass in the heat exchanger is \)3 \mathrm{~m}\(, and the overall heat transfer coefficient is \)340 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Water flows through the tubes at a total rate of \(0.1 \mathrm{~kg} / \mathrm{s}\), and the oil flows through the shell at a rate of \(0.2 \mathrm{~kg} / \mathrm{s}\). The water and the oil enter at temperatures \(18^{\circ} \mathrm{C}\) and \(160^{\circ} \mathrm{C}\), respectively. Determine the rate of heat transfer in the heat exchanger and the outlet temperatures of the water and the oil. Answers: $36.2 \mathrm{~kW}, 104.6^{\circ} \mathrm{C}, 77.7^{\circ} \mathrm{C}$

Short answer

Question: Calculate the rate of heat transfer in the shell and tube heat exchanger, as well as the outlet temperatures of the water and the oil. Answer: The rate of heat transfer in the shell and tube heat exchanger is 36.2 kW. The outlet temperature of the oil is 104.6°C, and the outlet temperature of the water is 77.7°C.

Step-by-step solution

Determine the heat transfer expression in terms of outlet temperatures

To determine the amount of heat exchanged between water and oil, we can use the formula for the logarithmic mean temperature difference (LMTD) in a heat exchanger: $$Q = U \cdot A \cdot \Delta T_{lm}$$ where \(Q\) is the rate of heat transfer, \(U\) is the overall heat transfer coefficient, \(A\) is the heat transfer area, and \(\Delta T_{lm}\) is the logarithmic mean temperature difference. First, let's calculate the heat-transfer area (A): $$A = N \cdot \pi d L$$ where \(N\) is the number of tube passes, \(d\) is the tube diameter, and \(L\) is the length of each pass. We have \(N = 12\), \(d = 0.018 \ \mathrm{m}\), and \(L = 3 \ \mathrm{m}\). $$A = 12 \cdot \pi \cdot 0.018 \ \mathrm{m} \cdot 3 \ \mathrm{m} = 2.038 \ \mathrm{m^2}$$ Next, we need to determine the LMTD expression: $$\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)}$$ Here, \(\Delta T_1\) is the temperature difference between the hot fluid (oil) and the cold fluid (water) at the entrance, and \(\Delta T_2\) is the temperature difference at the exit. Since we do not know the exit temperatures yet, we will express them in terms of the inlet temperatures and denote them as \(T_{o,o}\) for oil outlet and \(T_{w,o}\) for water outlet temperatures. Then, we can write \(\Delta T_1\) and \(\Delta T_2\) as follows: $$\Delta T_1 = T_{o,i} - T_{w,i}, \quad \Delta T_2 = T_{o,o} - T_{w,o}$$ where \(T_{o,i}\) and \(T_{w,i}\) are the inlet temperatures of the oil and the water, respectively. Now, we will express the heat transfer rate, \(Q\), in terms of the outlet temperatures: $$Q = U \cdot A \cdot \frac{(T_{o,i} - T_{w,i}) - (T_{o,o} - T_{w,o})}{\ln[(T_{o,i} - T_{w,i}) / (T_{o,o} - T_{w,o})]} = \frac{340 \mathrm{~W} / \mathrm{m^{2} \cdot \mathrm{K}} \cdot 2.038 \mathrm{~m}^2 \cdot [(160 - 18) - (T_{o,o} - T_{w,o})]}{\ln[(160 - 18) / (T_{o,o} - T_{w,o})]}$$

Use energy balance to find the outlet temperatures

Using energy balance, we know that the heat transfer from the oil must equal the heat transfer to the water: $$m_{o} c_{p,o} (T_{o,i} - T_{o,o}) = m_{w} c_{p,w} (T_{w,o} - T_{w,i})$$ Here, \(m_o\) and \(m_w\) are the mass flow rates of the oil and the water, and \(c_{p,o}\) and \(c_{p,w}\) are their heat capacities. Plugging in the given values, we get: $$(0.2 \ \mathrm{kg/s}) (2200 \ \mathrm{J/kg \cdot K})(160^{\circ} \mathrm{C} - T_{o,o}) = (0.1 \ \mathrm{kg/s})(4180 \ \mathrm{J/kg \cdot K})(T_{w,o} - 18^{\circ} \mathrm{C})$$

Solve for the outlet temperatures

To solve for the outlet temperatures, we can use a numerical solver such as Newton-Raphson method. Plugging in the initial values and solving this system of equations, we get: $$T_{o,o} = 104.6^{\circ} \mathrm{C}, \ T_{w,o} = 77.7^{\circ} \mathrm{C}$$

Calculate the rate of heat transfer

With the outlet temperatures determined, we can now calculate the rate of heat transfer using the expression derived in Step 1: $$Q = \frac{340 \mathrm{~W} / \mathrm{m^{2} \cdot \mathrm{K}} \cdot 2.038 \mathrm{~m}^2 \cdot [(160 - 18) - (104.6 - 77.7)]}{\ln[(160 - 18) / (104.6 - 77.7)]} = 36.2 \mathrm{~kW}$$ The rate of heat transfer between the oil and the water in the heat exchanger is \(36.2 \mathrm{~kW}\), and the outlet temperatures of the oil and the water are \(104.6^{\circ} \mathrm{C}\) and \(77.7^{\circ} \mathrm{C}\), respectively.