Cavitating Boiler Feedwater Control Valve
A 100 mm single-seat globe valve controls 45 m3/h of boiler feedwater from a deaerator to a boiler. Upstream pressure is 1,200 kPa(g) at 160°C; downstream pressure is 700 kPa(g). At 160°C the vapor pressure of water is about 618 kPa absolute. The valve has a rated Kv of 70 and normally operated near 60% travel. After a pump impeller was trimmed, the same flow is now controlled at about 35% travel, and the valve emits a sharp crackling noise. The downstream pipe shows pitting, and the plug and seat are eroded after six months. The loop is otherwise stable. A spare valve with the same Kv and standard trim is available, and maintenance proposes replacing the eroded trim during the next shutdown. The process cannot tolerate a long outage, and the boiler feedwater system has a high-energy bypass that must remain protected.
1. Using the pressures given, determine whether the valve is flashing or cavitating. Calculate the cavitation index sigma and explain what the result means.
Read the worked answer
Convert gauge pressures to absolute by adding about 101.3 kPa: P1 = 1,200 + 101.3 = 1,301.3 kPa abs P2 = 700 + 101.3 = 801.3 kPa abs Pv = 618 kPa abs
The cavitation index is: sigma = (P2 - Pv) / (P1 - Pv) = (801.3 - 618) / (1,301.3 - 618) = 183.3 / 683.3 = 0.268
Because P2 is greater than Pv, the liquid leaving the valve is still above vapor pressure; this is not flashing at the outlet. However, the vena contracta inside the valve can fall below Pv, causing vapor bubbles that collapse downstream. That is cavitation, and it fits the crackling noise, downstream pitting, and eroded trim.
A sigma of 0.268 is low. Compare it with the valve manufacturer's incipient cavitation index; if the valve's index is around 0.3–0.5, the valve is operating in a damaging cavitation region. Flashing would require P2 below Pv, which is not the case here.
2. Calculate the required Kv at the design flow and compare it with the rated Kv. What does this tell you about the valve selection and the proposed trim replacement?
Read the worked answer
Use Kv = Q × sqrt(SG / ΔP), with Q in m3/h and ΔP in bar. At 160°C the feedwater SG is about 0.90. ΔP = 1,200 - 700 = 500 kPa = 5.0 bar Kv = 45 × sqrt(0.90 / 5.0) = 45 × 0.424 = 19.1
The valve's rated Kv is 70, so the required Kv is only about 27% of the rating. The valve is grossly oversized, which explains why it runs at 35% travel and why the pressure drop is concentrated at a small opening. That low opening promotes cavitation and trim damage.
Replacing the eroded parts with the same Kv and standard trim would repeat the problem. The correct fix is to reduce the valve capacity or use a purpose-designed anti-cavitation trim, and to verify the actuator and positioner for the new opening range.
3. Recommend a corrective action plan, including immediate and shutdown work, and explain how you would verify improvement.
Read the worked answer
Immediate actions: monitor noise, vibration, and downstream wall thickness; avoid extended operation at very low travel; and do not use the high-energy bypass as a permanent cure unless it is engineered and approved. Confirm the actual flow and pressure drop, because the impeller trim changed the operating point.
Shutdown work: replace the oversized valve or trim with a correctly sized anti-cavitation design, such as a cage trim, angle valve, or hardened multistage trim. Consider a downstream restriction or other means to raise P2 and reduce ΔP, but check the effect on the deaerator, pump, and boiler. Inspect downstream piping for thinning and replace if required. Verify fail-safe action and positioner calibration.
Verification: calculate the new sigma at normal and maximum flow, measure noise after restart, confirm the valve operates in a reasonable travel range, and inspect the trim after a short run-in period. Document the new Kv, flow, ΔP, and cavitation margin. Use LOTO, depressurization, high-temperature PPE, and a proper pressure test during the outage.