Lockout After a Service on a Pressure-Jet Burner
A pressure-jet burner on a 35 kW boiler has just been serviced: nozzle replaced with the same size and angle, electrodes cleaned and re-set, photocell wiped, pump strainer cleaned, and the oil filter changed. On restart the burner fires, runs for about 90 seconds, then locks out. The engineer resets it; it fires again and locks out after a similar period. The flame looks steady through the inspection port while it is running. The oil tank is a plastic bunded tank 8 m away with a top outlet; there is a de-aerator at the boiler. The engineer suspects the photocell.
1. Evaluate the engineer's suspicion and list the more likely causes in order of probability.
Read the worked answer
A photocell fault usually causes a lockout at the end of the safety time (typically 5-15 seconds after the burner lights), not after 90 seconds of steady running. A 90-second run-then-lockout points to a flame that is lost or a control that loses its flame signal after the burner has warmed up, or to an oil supply problem that develops as demand continues.
More likely causes, in order:
- Oil supply/air: a partially blocked filter or strainer, a de-aerator that is not venting properly, or a suction line drawing air that accumulates in the pump until the flame fails. The 8 m top-outlet line and de-aerator are prime suspects.
- Photocell seeing the flame then losing it: a photocell that is marginal or a flame that moves as the burner heats can drop the signal; a dirty or misaligned photocell, or one affected by heat, is possible but less likely than oil.
- Solenoid valve or pump: a solenoid that drops out when hot, or a pump coupling that slips.
- Control box: a control that times out because the flame signal is weak (e.g. photocell resistance drifting with temperature).
- Electrode/HT lead: arcing that worsens as components warm, though this usually causes a failed light-up rather than a run-then-lockout.
The engineer should not replace the photocell on suspicion; the flame signal should be measured and the oil supply checked first.
2. Describe the diagnostic sequence you would follow, and explain what the 90-second timing tells you.
Read the worked answer
What the timing tells you: the burner lights and proves flame, so ignition, electrodes, and the photocell can see a flame at start. The lockout comes later, so the fault develops during running. The control's lockout after a proven flame is a flame-failure lockout, which means the flame signal dropped below the threshold or was lost.
Diagnostic sequence:
- Observe the flame and photocell signal: watch the flame through the port until lockout; note whether the flame shrinks, lifts, or becomes unstable before lockout. Measure the photocell current/flame signal with a meter if the control allows, and compare with the manufacturer's minimum.
- Check the photocell: clean the lens, confirm it is correctly seated and not affected by stray light or heat, and test/replace only if the signal is marginal.
- Check the oil supply: pump pressure and vacuum at the pump; a rising vacuum indicates a restricted line or filter. Bleed the de-aerator and check for air. Inspect the tank outlet filter and pump strainer.
- Check the solenoid and pump: confirm the solenoid stays energised and the pump maintains pressure when hot; check the coupling.
- Check the control box and wiring: look for a loose connection or a control that is timing out; substitute a known-good control only after the above are eliminated.
- Re-test: run for at least 15 minutes and confirm stable flame signal and no lockout before signing off.