Question

In a parallel-flow heat exchanger, hot fluid enters the heat exchanger at a temperature of \(150^{\circ} \mathrm{C}\) and a mass flow rate of $3 \mathrm{~kg} / \mathrm{s}$. The cooling medium enters the heat exchanger at a temperature of \(30^{\circ} \mathrm{C}\) with a mass flow rate of $0.5 \mathrm{~kg} / \mathrm{s}\( and leaves at a temperature of \)70^{\circ} \mathrm{C}\(. The specific heat capacities of the hot and cold fluids are \)1150 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\( and \)4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}$, respectively. The convection heat transfer coefficient on the inner and outer sides of the tube are $300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( and \)800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(, respectively. For a fouling factor of \)0.0003 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\( on the tube side and \)0.0001 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}$ on the shell side, determine (a) the overall heat transfer coefficient, \((b)\) the exit temperature of the hot fluid, and (c) the surface area of the heat exchanger.

Short answer

In a parallel-flow heat exchanger, the overall heat transfer coefficient (Uo) can be calculated using the formula: \(U_o = \frac{1}{\frac{1}{h_i} + RF_i + \frac{1}{h_o} + RF_o}\), where: - \(U_o\) is the overall heat transfer coefficient (\(W/m^2\cdot K\)), - \(h_i\) is the convection heat transfer coefficient on the inner side of the tube (\(W/m^2\cdot K\)), - \(RF_i\) is the fouling factor on the inner side of the tube (\(m^2\cdot K/W\)), - \(h_o\) is the convection heat transfer coefficient on the outer side of the tube (\(W/m^2\cdot K\)), and - \(RF_o\) is the fouling factor on the outer side of the tube (\(m^2\cdot K/W\)). This relationship shows that the overall heat transfer coefficient is inversely proportional to the sum of the inverse convection heat transfer coefficients and their respective fouling factors. In other words, lower convection coefficients or higher fouling factors will result in a lower overall heat transfer coefficient, leading to lower heat transfer in the heat exchanger.

Step-by-step solution

Calculate the heat transfer rate for the cold fluid.

To calculate the heat transfer rate for the cold fluid, we can use the formula: \(Q = m_cC_{p,c}(T_{c, out} - T_{c, in})\) where: \(Q\) is the heat transfer rate (\(W\)), \(m_c\) is the mass flow rate of the cold fluid (\(kg/s\)), \(C_{p,c}\) is the specific heat capacity of the cold fluid (\(J/kg\cdot K\)), \(T_{c, out}\) is the outlet temperature of the cold fluid (\(^{\circ}C\)), and \(T_{c, in}\) is the inlet temperature of the cold fluid (\(^{\circ}C\)).

Calculate the overall heat transfer coefficient.

To calculate the overall heat transfer coefficient, we can use the formula: \(U_o = \frac{1}{\frac{1}{h_i} + RF_i + \frac{1}{h_o} + RF_o}\) where: \(U_o\) is the overall heat transfer coefficient (\(W/m^2\cdot K\)), \(h_i\) is the convection heat transfer coefficient on the inner side of the tube (\(W/m^2\cdot K\)), \(RF_i\) is the fouling factor on the inner side of the tube (\(m^2\cdot K/W\)), \(h_o\) is the convection heat transfer coefficient on the outer side of the tube (\(W/m^2\cdot K\)), and \(RF_o\) is the fouling factor on the outer side of the tube (\(m^2\cdot K/W\)).

Find the inlet temperature of the hot fluid.

The inlet temperature of the hot fluid, \(T_{h, in}\), is given as \(150^{\circ} \mathrm{C}\).

Calculate the exit temperature of the hot fluid.

To find the exit temperature of the hot fluid, \(T_{h, out}\), we can apply energy conservation. The heat transfer rate calculated in Step 1 for the cold fluid, \(Q_{c}\), should be equal to the heat transfer rate for the hot fluid. \(Q_h = m_hC_{p,h}(T_{h, in} - T_{h, out})\), where: \(Q_h\) is the heat transfer rate for the hot fluid (\(W\)), \(m_h\) is the mass flow rate of the hot fluid (\(kg/s\)), and \(C_{p,h}\) is the specific heat capacity of the hot fluid (\(J/kg\cdot K\)). Set \(Q_c = Q_h\) and solve for \(T_{h, out}\).

Calculate the log mean temperature difference (LMTD).

Using the calculated values in step 3 and 4, we can find the log mean temperature difference (LMTD) given by: \(LMTD = \frac{(T_{h, in} - T_{c, out}) - (T_{h, out} - T_{c, in})}{\ln(\frac{T_{h, in} - T_{c, out}}{T_{h, out} - T_{c, in}})}\)

Calculate the heat exchanger surface area.

To calculate the surface area of the heat exchanger, \(A\), we can use the equation: \(A = \frac{Q_c}{U_o\cdot LMTD}\) The calculated surface area gives the size of the heat exchanger required for the given heat transfer specifications.