Boiler Efficiency Test and the Cold-End Limit on an Economiser
A textile plant runs a packaged three-pass fire-tube boiler rated at 10 t/h of saturated steam at 10.5 bar gauge. A four-hour test was carried out at steady load. Data: steam output 10.0 t/h at dryness fraction 0.98; feedwater temperature 105 °C; furnace oil firing rate 720 kg/h with a net calorific value of 41.5 MJ/kg; flue gas oxygen 6.5% dry and flue gas temperature 240 °C leaving the boiler; ambient air 30 °C; blowdown 4% of steam flow; radiation and convection losses 2.0% of fuel input. The plant engineer proposes fitting an economiser to heat the feedwater, quoting an installed cost of LKR 18 million. The boiler runs 6000 h per year and furnace oil costs LKR 280/kg. The fuel contains 2.5% sulphur.
1. Determine the boiler efficiency by both the direct and indirect methods, and identify the largest recoverable loss.
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Steam properties. At 11.5 bar absolute, h_f = 787 kJ/kg, h_fg = 1994 kJ/kg, h_g = 2781 kJ/kg. With dryness 0.98: h_steam = 787 + 0.98 × 1994 = 2741 kJ/kg. Feedwater at 105 °C: h = 4.18 × 105 = 439 kJ/kg.
Useful heat.
- Steam: 10 000 × (2741 − 439) = 23.02 GJ/h
- Blowdown: 400 × (787 − 439) = 0.14 GJ/h
- Total = 23.16 GJ/h
Fuel input. 720 × 41.5 = 29.88 GJ/h.
Direct efficiency = 23.16/29.88 = 77.5%.
Indirect check. Excess air from O₂ = 6.5/(21 − 6.5) = 44.8%, so total air ≈ 1.448 × 14.0 = 20.3 kg/kg fuel and flue gas ≈ 21.3 kg/kg fuel. Dry gas loss = 21.3 × 1.05 × (240 − 30)/41 500 = 11.3%. Radiation 2.0%. Blowdown 0.14/29.88 = 0.5%. Sum of accounted losses and output = 91.3%, leaving about 8.7% for water vapour from fuel hydrogen and moisture, unburnt losses and casing leakage - a credible closure for a short test.
Judgement. The dry flue gas loss at 11.3% is the dominant and most easily recovered item, and it is driven by two things: the 240 °C stack temperature and the 44.8% excess air. Reducing excess air to 4% O₂ would cut the loss by roughly 1.5 percentage points at almost no capital cost - but with 2.5% sulphur in the fuel, lower excess air raises SO₃ and the acid dew point, so the two measures interact and must be judged together.
2. Size an economiser that cools the flue gas to 180 °C. Estimate the feedwater temperature rise, the annual fuel saving and the simple payback.
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Recovered heat. Cooling 21.3 kg of gas per kg of fuel through 60 K: 21.3 × 1.05 × 60 = 1342 kJ/kg fuel.
Duty. 720 kg/h × 1342 kJ/kg = 0.966 GJ/h = 268 kW.
Feedwater rise. Mass through the economiser is 10 400 kg/h (steam plus blowdown). ΔT = 0.966 × 10⁶/(10 400 × 4.18) = 22.2 K, so the feedwater enters the boiler at about 127 °C instead of 105 °C. This is comfortably below saturation at 11.5 bar, so no steaming occurs in the economiser.
Fuel saving. 1342/41 500 = 3.23% of firing rate, i.e. 0.0323 × 720 = 23 kg/h. Annual: 23 × 6000 = 139 t/yr, worth 139 000 × 280 = LKR 39.0 million per year.
Payback. 18/39.0 = 0.46 year, about 5.5 months on a simple basis.
Caveats to state explicitly. These are gross figures. Deduct the additional flue-gas-side draught loss (typically 0.5–1.5 mbar, which may need a larger ID fan or a taller stack), the water-side pressure drop and pumping power, periodic tube cleaning, and a maintenance allowance of perhaps 5% of the saving. Even on the net figure the payback is under a year. Note also that cooling to 160 °C instead of 180 °C would recover a further 20 K, worth about 0.9 percentage points of efficiency - but see the acid dew point constraint.
3. The fuel contains 2.5% sulphur and the feedwater enters the economiser at 105 °C. Assess the cold-end corrosion risk and state what you would actually specify. What other low-cost measures would you check first?
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The cold-end problem. In a counterflow economiser the flue gas leaves at the water inlet end. With gas at 180 °C and water at 105 °C, the tube-wall temperature at that end is roughly 125–135 °C. For a fuel with 2.5% sulphur, 1–3% of the sulphur converts to SO₃ and the resulting acid dew point is typically 140–160 °C, depending on the SO₃ concentration and the excess air level. The wall is therefore below the dew point, and the cold end will run wet with sulphuric acid. Condensate is not merely a corrosion nuisance: it captures soot, blocks the gas passages and can cause a stack fire. Cooling to 160 °C would make this considerably worse, with wall temperatures below 120 °C.
What to specify.
- Limit the gas outlet to 180 °C and confirm the predicted wall temperature by calculation, not by rule of thumb.
- Provide a feedwater bypass so the inlet water temperature to the economiser can be held at 120–130 °C at low load, when the gas flow and temperature both fall and the cold end is most at risk.
- Use a corrosion-resistant cold-end section - enamel-coated or PTFE-coated tubes, or a cast-iron gilled section in the first row - and arrange the economiser for easy replacement of that row.
- Fit a stack temperature thermocouple and a condensate drain with a pH check so the operator can see the problem developing.
Cheaper measures to check before spending capital. First, reduce excess air from 6.5% to about 4–4.5% O₂ with a proper oxygen trim; this alone is worth roughly 1.0–1.5 percentage points of efficiency and a fraction of the economiser cost. Second, check blowdown control against measured TDS rather than a fixed 4% valve setting, and consider heat recovery from the blowdown to the feedwater. Third, verify the dryness fraction: 0.98 indicates carryover, which wastes fuel and threatens the superheater and turbine, if any, and is a chemistry problem masquerading as an efficiency one.