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Pump Cavitation in a Transfer Line

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Pump Cavitation in a Transfer Line

A centrifugal pump transfers water at 25°C from an open storage tank to a process vessel. The pump is located 3 m above the tank liquid level. The suction line is 10 m long, with a 4-inch schedule 40 pipe (internal diameter 102 mm), two 90° elbows (K=0.75 each), and a fully open gate valve (K=0.2). The pump curve provided by the manufacturer shows a required NPSH (NPSHr) of 4.0 m at the design flow rate of 50 m³/h. The water vapor pressure at 25°C is 3.17 kPa, and the atmospheric pressure is 101.3 kPa. The density of water is 997 kg/m³. The pump is currently operating at 50 m³/h, but operators report noise and vibration, and the flow rate is fluctuating. The suction line has a strainer (K=2.0) that is partially clogged, increasing its resistance. The pump inlet pressure gauge reads 20 kPa (gauge). The discharge pressure is stable.

1. Calculate the available NPSH (NPSHa) for the current operation, assuming the strainer is clean (K=2.0). Use the Darcy-Weisbach equation with a friction factor f=0.02 for the pipe. Determine if cavitation is likely.

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First, calculate the flow velocity in the suction line: volumetric flow rate Q = 50 m³/h = 50/3600 = 0.01389 m³/s. Pipe internal diameter D = 102 mm = 0.102 m. Cross-sectional area A = πD²/4 = π*(0.102)²/4 = 0.00817 m². Velocity v = Q/A = 0.01389/0.00817 = 1.70 m/s. Next, calculate the head loss due to friction and fittings in the suction line. The suction line includes: 10 m of pipe, two 90° elbows (K=0.75 each), one gate valve (K=0.2), and a strainer (K=2.0). Total K for fittings = 20.75 + 0.2 + 2.0 = 3.7. Friction factor f=0.02. Head loss hL = [f(L/D) + ΣK] * v²/(2g). L/D = 10/0.102 = 98.04. f*(L/D) = 0.0298.04 = 1.961. Sum = 1.961 + 3.7 = 5.661. v²/(2g) = (1.70)²/(29.81) = 2.89/19.62 = 0.1473 m. hL = 5.661 * 0.1473 = 0.834 m. Now, NPSHa = (P_atm - P_v)/(ρg) + z - hL, where z is the static head relative to pump centerline. The pump is 3 m above the tank liquid level, so z = -3 m (since the pump is above the liquid level, the static suction head is negative). P_atm = 101.3 kPa, P_v = 3.17 kPa. (P_atm - P_v)/(ρg) = (101.3 - 3.17)1000 / (9979.81) = 98,130 / 9780.57 = 10.03 m. So NPSHa = 10.03 - 3 - 0.834 = 6.196 m. Since NPSHa (6.2 m) > NPSHr (4.0 m), cavitation should not occur under clean strainer conditions. However, the observed noise and vibration suggest otherwise, so the strainer may be clogged, increasing hL and reducing NPSHa.

2. The pump inlet pressure gauge reads 20 kPa (gauge). Calculate the NPSHa from this gauge reading and compare it to the calculated value from the previous task. What does this indicate about the strainer condition?

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The NPSHa can also be calculated from the pump suction pressure: NPSHa = (P_suction - P_v)/(ρg) + v²/(2g), where P_suction is the absolute pressure at the pump inlet. The gauge reads 20 kPa, so absolute pressure = 20 + 101.3 = 121.3 kPa. P_v = 3.17 kPa. (P_suction - P_v)/(ρg) = (121.3 - 3.17)1000 / (9979.81) = 118,130 / 9780.57 = 12.08 m. The velocity head v²/(2g) = 0.1473 m (from previous). So NPSHa = 12.08 + 0.147 = 12.23 m. This is much higher than the 6.2 m calculated from the tank level and friction losses. The discrepancy is large. The gauge reading suggests a higher NPSHa, which would not cause cavitation. However, the gauge may be located at a point where the pressure is not representative, or there may be an error. Alternatively, the gauge could be reading incorrectly. But if we trust the gauge, NPSHa is 12.2 m, which is well above NPSHr, so cavitation is not expected. Yet the symptoms indicate cavitation. This inconsistency suggests that the gauge reading is unreliable or that the cavitation is occurring due to other reasons, such as air entrainment or the pump operating far from its design point. However, the problem states the flow rate is fluctuating, which could be due to cavitation. A more likely explanation: the gauge is located at the pump inlet, but the pressure there may be lower than the gauge reads due to a clogged strainer causing a pressure drop upstream that is not reflected in the gauge? Actually, if the strainer is clogged, the pressure at the pump inlet would be lower, so the gauge should read lower, not higher. The gauge reads 20 kPa, which is positive, so perhaps the gauge is correct and the NPSHa is indeed high. But then why cavitation? Maybe the NPSHr is higher than stated, or the fluid is hotter than 25°C. Given the information, the most likely conclusion is that the strainer is not clogged, but there is another issue. However, the problem asks to compare and indicate strainer condition. The calculated NPSHa from tank level (6.2 m) is lower than from gauge (12.2 m). The difference is 6 m, which is large. This suggests that the friction loss is much higher than assumed, meaning the strainer is clogged. If the strainer were clean, the gauge reading should correspond to NPSHa of about 6.2 m, which would be a gauge pressure of: P_suction = P_v + ρg*(NPSHa - v²/(2g)) = 3.17 + 9979.81(6.2 - 0.147)/1000 = 3.17 + 9.78*(6.053) = 3.17 + 59.2 = 62.37 kPa absolute, or 62.37 - 101.3 = -38.9 kPa gauge. But the gauge reads +20 kPa, which is much higher. This indicates that the actual pressure at the pump inlet is higher than expected, which would mean the friction loss is lower, not higher. That contradicts the clogged strainer hypothesis. Therefore, the gauge reading is inconsistent with the system. Possibly the gauge is faulty or located incorrectly. Given the symptoms, the most plausible cause is that the strainer is clogged, causing a higher pressure drop, which would reduce NPSHa. But the gauge reading does not support that. Hence, the gauge is unreliable. The calculated NPSHa from the tank level (6.2 m) is still above NPSHr, so cavitation should not occur. However, if the strainer is partially clogged, the additional pressure drop could reduce NPSHa below 4.0 m. For NPSHa to be 4.0 m, the total head loss would need to be 10.03 - 3 - 4.0 = 3.03 m. The current hL is 0.834 m, so additional loss of 2.2 m is needed. The strainer K would need to increase by ΔK = (2.2 * 2g)/v² = (2.2*19.62)/2.89 = 14.9. So total K for strainer would be 2.0+14.9=16.9, which is a 8.5 times increase. This is possible if the strainer is significantly clogged. Therefore, the strainer is likely clogged, and the gauge reading is either inaccurate or taken at a different location. The recommended action is to clean the strainer and verify the gauge.

3. Propose three operational or design changes to prevent cavitation in this system, and explain the trade-offs of each.

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  1. Lower the pump or raise the tank: Reduce the static suction lift by moving the pump closer to the tank or elevating the tank. This increases NPSHa by the reduction in z. Trade-off: May require structural changes and additional space; if the pump is already installed, relocation can be costly and disruptive.

  2. Increase suction pipe diameter: Replace the 4-inch line with a larger diameter (e.g., 6-inch) to reduce velocity and friction losses. This reduces hL and increases NPSHa. Trade-off: Higher capital cost for larger pipe and fittings; may require rework of existing piping and supports.

  3. Install a booster pump or use a low-NPSH pump: Add a small booster pump upstream to increase suction pressure, or replace the existing pump with one having a lower NPSHr. Trade-off: Additional equipment and energy consumption; booster pump adds complexity and maintenance. Alternatively, a low-NPSH pump may be more expensive and less efficient.

Other options: Use a strainer with a larger open area or install a differential pressure indicator to monitor clogging; but the three above are primary. Each measure should be evaluated based on cost, ease of implementation, and reliability.