A performance test is being conducted on a double pipe counter flow heat exchanger that carries engine oil and water at a flow rate of \(2.5 \mathrm{~kg} / \mathrm{s}\) and \(1.75 \mathrm{~kg} / \mathrm{s}\), respectively. Since the heat exchanger has been in service over a long period of time it is suspected that the fouling might have developed inside the heat exchanger that might have affected the overall heat transfer coefficient. The test to be carried out is such that, for a designed value of the overall heat transfer coefficient of \(450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and a surface area of \(7.5 \mathrm{~m}^{2}\), the oil must be heated from \(25^{\circ} \mathrm{C}\) to \(55^{\circ} \mathrm{C}\) by passing hot water at \(100^{\circ} \mathrm{C}\left(c_{p}=4206 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at the flow rates mentioned above. Determine if the fouling has affected the overall heat transfer coefficient. If yes, then what is the magnitude of the fouling resistance?

To calculate the heat transfer from oil, we will use the formula: \(\dot{Q}=\dot{m} \cdot c_{p} \cdot \Delta T\) Where \(\dot{Q}\) is the heat transfer, \(\dot{m}\) is the mass flow rate, \(c_{p}\) is the specific heat, and \(\Delta T\) is the temperature difference. For the engine oil, we have: \(\dot{m}_{oil} = 2.5 \thinspace \mathrm{kg/s}\), \(\Delta T_{oil} = (55-25)^{\circ} \mathrm{C}=30^{\circ}\mathrm{C}\), Assuming the specific heat of oil, \(c_{p_{oil}} = 2000 \thinspace \mathrm{J/kg \cdot K}\) (typical value), we can calculate the heat transfer from oil: \(\dot{Q}_{oil} = 2.5 \mathrm{\thinspace kg/s} \times 2000 \mathrm{~J/kg \cdot K} \times 30 \mathrm{K} = 150000 \mathrm{~W}\)

Next, we'll use the heat transfer from oil to determine the temperature difference for the water: \(\dot{Q}_{water} = \dot{m}_{water} \cdot c_{p_{water}} \cdot \Delta T_{water}\) Where \(\dot{m}_{water} = 1.75 \thinspace \mathrm{kg/s}\) and \(c_{p_{water}} = 4206 \thinspace \mathrm{J/kg \cdot K}\). We can rearrange the equation to solve for the temperature difference for the water: \(\Delta T_{water} = \frac{\dot{Q}_{water}}{\dot{m}_{water} \cdot c_{p_{water}}} = \frac{150000 \mathrm{~W}}{1.75 \mathrm{\thinspace kg/s} \times 4206 \mathrm{~J/kg \cdot K}} \approx 20.09 \mathrm{K}\) Since water enters the heat exchanger at \(100^{\circ}\mathrm{C}\), its exit temperature will be approximately \(79.91^{\circ}\mathrm{C}\).

To determine the LMTD, we will use the following formula: \(LMTD=\frac{\Delta T_{1}-\Delta T_{2}}{\ln \frac{\Delta T_{1}}{\Delta T_{2}}}\) Where \(\Delta T_{1}\) represents the temperature difference between the hot and cold fluids at one end of the heat exchanger and \(\Delta T_{2}\) represents the temperature difference at the other end. For our heat exchanger, we have \(\Delta T_{1} = 100^{\circ}\mathrm{C} - 25^{\circ}\mathrm{C} = 75^{\circ}\mathrm{C}\) and \(\Delta T_{2}= 79.91^{\circ}\mathrm{C} - 55^{\circ}\mathrm{C} = 24.91^{\circ}\mathrm{C}\). Thus, we can calculate the LMTD: \(LMTD=\frac{75 \mathrm{K}-24.91 \mathrm{K}}{\ln \frac{75 \mathrm{K}}{24.91 \mathrm{K}}} \approx 46.59 \mathrm{K}\)

Now that we have the LMTD, we can calculate the surface area required to transfer the heat using the designed overall heat transfer coefficient, \(U = 450 \mathrm{\thinspace W/m^2 \cdot K}\): \(A = \frac{\dot{Q}_{oil}}{U \cdot LMTD} = \frac{150000 \mathrm{~W}}{450 \mathrm{\thinspace W/m^2 \cdot K} \times 46.59 \mathrm{K}} \approx 7.03 \mathrm{\thinspace m^2}\)

Since the calculated surface area (7.03 \(\mathrm{m^2}\)) is less than the given surface area (7.5 \(\mathrm{m^2}\)), fouling is likely present in the heat exchanger. To determine the magnitude of the fouling resistance, we'll use the following equation: \(R_{f} = \frac{1}{U_{actual}\cdot A} - \frac{1}{U_{design}\cdot A}\) Where \(U_{actual}\) is the actual overall heat transfer coefficient after fouling and \(A\) is the given surface area. First, we need to find the actual overall heat transfer coefficient: \(\dot{Q} = U_{actual} \cdot A \cdot LMTD\) \(U_{actual} = \frac{\dot{Q}}{A \cdot LMTD} = \frac{150000 \mathrm{~W}}{7.5 \mathrm{\thinspace m^2} \times 46.59 \mathrm{K}} \approx 431.20 \mathrm{\thinspace W/m^2 \cdot K}\) Now, we can calculate the fouling resistance: \(R_{f} = \frac{1}{431.20 \mathrm{\thinspace W/m^2 \cdot K} \cdot 7.5 \mathrm{\thinspace m^2}} - \frac{1}{450 \mathrm{\thinspace W/m^2 \cdot K} \cdot 7.5 \mathrm{\thinspace m^2}} \approx 0.00049 \mathrm{\thinspace m^2 \cdot K/W}\) So, fouling has affected the overall heat transfer coefficient, and the magnitude of the fouling resistance is approximately \(0.00049 \mathrm{\thinspace m^2 \cdot K/W}\).