Heat Exchanger Fouling and Performance

First, calculate the heat duty from the hot stream: Q_hot = m_hot * Cp_hot * (T_hot_in - T_hot_out) = 10 kg/s * 2.5 kJ/(kg·K) * (150 - 100) K = 10 * 2.5 * 50 = 1250 kW. Check with cold stream: Q_cold = m_cold * Cp_cold * (T_cold_out - T_cold_in) = 15 * 4.18 * (45 - 25) = 15 * 4.18 * 20 = 1254 kW. Close enough (rounding). So Q ≈ 1250 kW. Now, determine the log mean temperature difference (LMTD) for countercurrent flow (since it's a 1-2 exchanger, we need the correction factor F). For a 1-2 exchanger, the LMTD correction factor F depends on P and R. P = (T_cold_out - T_cold_in)/(T_hot_in - T_cold_in) = (45-25)/(150-25) = 20/125 = 0.16. R = (T_hot_in - T_hot_out)/(T_cold_out - T_cold_in) = (150-100)/(45-25) = 50/20 = 2.5. Using the formula for F for 1-2 exchanger: F = (√(R²+1) * ln[(1-P)/(1-PR)]) / [(R-1) * ln((2-P(R+1-√(R²+1)))/(2-P(R+1+√(R²+1))))]. This is complex; alternatively, use charts. For P=0.16, R=2.5, F is approximately 0.95 (from typical charts). Let's compute LMTD for countercurrent: ΔT1 = T_hot_in - T_cold_out = 150 - 45 = 105°C; ΔT2 = T_hot_out - T_cold_in = 100 - 25 = 75°C. LMTD_cc = (105 - 75)/ln(105/75) = 30/ln(1.4) = 30/0.3365 = 89.15°C. Then corrected ΔT = F * LMTD_cc = 0.95 * 89.15 = 84.7°C. Now, U_current = Q / (A * ΔT) = 1250 kW / (50 m² * 84.7 K) = 1250,000 W / (50 * 84.7) = 1250,000 / 4235 = 295.2 W/(m²·K). So U_current ≈ 295 W/(m²·K). The design U_clean = 500 W/(m²·K). The fouling factor Rf = 1/U_current - 1/U_clean = 1/295 - 1/500 = 0.00339 - 0.002 = 0.00139 m²·K/W. This is a significant fouling resistance.

The fouling resistance Rf = 0.00139 m²·K/W. If the fouling layer has thermal conductivity k_f = 0.5 W/(m·K), then the thickness t_f = Rf * k_f = 0.00139 * 0.5 = 0.000695 m = 0.695 mm. This is a thin layer, less than 1 mm. Now, consider the pressure drop increase: shell-side pressure drop increased from 30 kPa to 50 kPa, a 67% increase. For flow across tube banks, pressure drop is roughly proportional to the square of velocity and inversely proportional to some power of the flow area. If fouling reduces the flow area, the velocity increases. However, a 0.7 mm fouling layer on tubes (tube OD 25 mm) would reduce the shell-side flow area slightly. The shell-side cross-sectional area is complex, but if fouling is uniform on tubes, the effective diameter increases by 2*0.7 = 1.4 mm, from 25 mm to 26.4 mm, which is a 5.6% increase in diameter. This would reduce the flow area between tubes, increasing velocity by roughly (25/26.4)² = 0.897, so velocity increases by about 11%. Pressure drop would increase by about 1.11² = 1.23, i.e., 23% increase, not 67%. Therefore, a uniform 0.7 mm fouling layer does not fully explain the 67% pressure drop increase. The pressure drop increase may be due to localized severe fouling, such as blockage in the shell-side baffle windows or inlet/outlet nozzles, or fouling that is not uniform. Alternatively, the fouling could be on the shell side but also causing channeling or increased roughness. Thus, the estimated fouling thickness is plausible for the heat transfer degradation but insufficient to explain the pressure drop rise; further inspection is needed.

Given the significant fouling (Rf=0.00139 m²·K/W) and increased pressure drop, cleaning is necessary. Options include:

1. Mechanical cleaning: Hydroblasting or rodding of the shell side. This requires shutting down the exchanger, draining, and opening the shell. The production impact depends on whether a spare exchanger exists or if the process can be bypassed. If no spare, schedule cleaning during a planned shutdown. Trade-off: Production loss during downtime; however, mechanical cleaning is effective for hard deposits.

2. Chemical cleaning: Circulate a cleaning solution (e.g., inhibited acid or caustic) to dissolve deposits. This can be done in situ without opening the exchanger, reducing downtime. Trade-off: Requires compatibility of cleaning chemicals with materials; may not remove all deposits; disposal of cleaning solution.

3. Operational changes: Reduce the fouling rate by adjusting process conditions. For example, lower the hot stream inlet temperature if possible, or increase cooling water flow to reduce wall temperature (if fouling is temperature-driven). Install a filter upstream to remove particulates. Trade-off: Changes may affect process efficiency or require additional equipment.

4. Design modification: If fouling is chronic, consider replacing with a larger exchanger or one with enhanced tubes (e.g., finned tubes) to provide more area and lower fouling rate. Trade-off: Capital cost.

Immediate action: Plan for cleaning during the next available opportunity. In the meantime, monitor U and pressure drop to avoid exceeding limits. If production critical, consider installing a spare exchanger. Also, investigate the root cause of fouling (e.g., suspended solids, scaling) and address it.