Question

A performance test is being conducted on a doublepipe counterflow 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 for a long time, it is suspected that fouling might have developed inside the heat exchanger that could affect 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?

Short answer

Answer: Yes, fouling has affected the overall heat transfer coefficient. The actual heat transfer coefficient is 1421.23 W/m²·K, which is much higher than the designed value of 450 W/m²·K. The magnitude of the fouling resistance is 0.000286 m²·K/W.

Step-by-step solution

Calculate Required Heat Transfer Rate

First, let's calculate the heat transfer rate required to heat the engine oil. We can use the following formula: \(Q = m_{oil} \cdot C_{p, oil} \cdot \Delta T_{oil}\) Here, \(Q\) is the required heat transfer rate (W), \(m_{oil}\) is the mass flow rate of the engine oil (2.5 kg/s), \(C_{p, oil}\) is the specific heat capacity of the engine oil, \(\Delta T_{oil}\) is the temperature change of the engine oil from 25°C to 55°C (30 K). We need to find the value of \(C_{p, oil}\). Since it is not given directly, we can assume it is similar to that of water because both have similar specific heat capacities. So, \(C_{p, oil} = 4206 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Now, let's calculate Q: \(Q = 2.5 \cdot 4206 \cdot 30\) \(Q = 315450 \mathrm{~W}\)

Calculate Logarithmic Mean Temperature Difference (LMTD)

To calculate the LMTD, we will use the following formula: \(LMTD = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}\) Here, \(\Delta T_1\) is the temperature difference between the hot water inlet and the cold oil outlet (100 - 55) and \(\Delta T_2\) is the temperature difference between the hot water outlet and the cold oil inlet (which we need to determine). Let's first find the temperature change in hot water by rearranging the heat transfer rate formula: \(Q = m_{water} \cdot C_{p, water} \cdot \Delta T_{water}\) Rearrange the formula to find \(\Delta T_{water}\): \(\Delta T_{water} = \frac{Q}{m_{water} \cdot C_{p, water}}\) Substitute the known values: \(\Delta T_{water} = \frac{315450}{1.75 * 4206}\) \(\Delta T_{water} = 26.417\) Now, we can calculate \(\Delta T_2\): \(\Delta T_2 = 100 - 25 - \Delta T_{water} = 48.583\) Now, we can calculate the LMTD: \(LMTD = \frac{(100 - 55) - (48.583)}{\ln(\frac{100 - 55}{48.583})}\) \(LMTD = 29.514\)

Calculate Actual Heat Transfer Coefficient

Now, we will calculate the actual overall heat transfer coefficient \(U_{actual}\) using the given surface area and the LMTD. We can use the following formula: \(U_{actual} = \frac{Q}{A \cdot LMTD}\) Here, \(A\) is the surface area (7.5 m²). Calculate \(U_{actual}\): \(U_{actual} = \frac{315450}{7.5 * 29.514}\) \(U_{actual} = 1421.23 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) Notice that the actual heat transfer coefficient is much higher than the designed value (450 W/m²·K). This means the fouling has affected the overall heat transfer coefficient.

Calculate Fouling Resistance

Finally, let's calculate the magnitude of the fouling resistance. We can use the following formula: \(R_f = \frac{1}{U_{actual} \cdot A} - \frac{1}{U_{designed} \cdot A}\) Here, \(U_{designed}\) is the designed overall heat transfer coefficient (450 W/m²·K). Calculate \(R_f\): \(R_f = \frac{1}{1421.23 * 7.5} - \frac{1}{450 * 7.5}\) \(R_f = 0.000286 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) So, the magnitude of the fouling resistance is 0.000286 m²·K/W.