Condensate flashing to the low-pressure header
Steam systems · flash recovery
A plate exchanger on the 150 psig main drains through a float-and-thermostatic trap that passes about 3,000 lb/h of condensate. Downstream of the trap the line drops into a flash vessel held at 15 psig by the low-pressure header it feeds, and the tank vent above it breathes a white plume all shift — which is the clue that starts this calculation. Condensate leaving a trap is saturated water at the trap's own pressure, so at 150 psig it is sitting at 185.5 °C, and the moment it lands in a 15 psig vessel it is 68 degrees hotter than water at 15 psig is allowed to be. Something has to give, and what gives is that part of the water boils itself. The steam tables supply the four numbers this needs, each read at the pressure it belongs to. At 150 psig (164.7 psia) the saturated liquid enthalpy is h_f1 = 787.5 kJ/kg and the saturation temperature 185.5 °C. At 15 psig (29.7 psia) the same column reads h_f2 = 507.8 kJ/kg at 121.0 °C, with a latent heat of h_fg2 = 2,199.5 kJ/kg.
Every number in this problem is editable — change any value below and the whole chain recalculates.
- ṁ_c = 3,000 lb/h — Condensate through the trap
- p₁ = 150 psig — Trap pressure (164.7 psia, T_sat 185.5 °C)
- p₂ = 15 psig — Flash vessel pressure (29.7 psia, T_sat 121.0 °C)
- h_f1 = 787.5 kJ/kg — Liquid enthalpy at 150 psig, from the tables
- h_f2 = 507.8 kJ/kg — Liquid enthalpy at 15 psig, from the tables
- h_fg2 = 2,199.5 kJ/kg — Latent heat at 15 psig, from the tables
- (a)the percentage of the condensate that flashes off in the vessel
- (b)the flash steam that percentage releases, in pounds per hour
- (c)what the recovered steam is worth against the boiler's own rating
Read the top line as an energy balance and the whole thing falls out. Each kilogram arrives carrying 787.5 kJ and is only allowed to keep 507.8 kJ as liquid at the new pressure, so 279.7 kJ is surplus sensible heat with nowhere to go. It buys latent heat instead, at 2,199.5 kJ for every kilogram boiled — that ratio IS the flash fraction. Note which pressure each enthalpy is read at: h_f1 upstream, h_f2 and h_fg2 both downstream. Reading h_fg at the HIGH pressure is the standard wreck, and it understates the flash by about 10 %.
Carried onward at full precision, not this rounded figure.
A percentage does not size a vent, a header or a receiver — pounds per hour do. The share from part (a) against the trap's 3,000 lb/h gives the flash rate directly. Worth keeping in view: the OTHER 87.3 % stays liquid and still has to leave the bottom of the vessel, so the vessel needs a trap of its own sized for roughly 2,620 lb/h, and the flash outlet needs to be big enough that 381 lb/h of low-density steam does not carry water over into the header.
Carried onward at full precision, not this rounded figure.
Steam is steam, whoever made it, so the flash can be read against the same 34.5 lb/h-per-horsepower yardstick the boiler is sold by. Careful with what the answer means: boiler horsepower is defined from and at 212 °F, where the latent heat is 2,256.5 kJ/kg, and this steam is leaving at 15 psig where it is 2,199.5. So the flash truly carries 105.7 kW of latent heat while the horsepower yardstick values it at 108.5 kW — 2.5 % apart, and that 2.5 % is the whole difference between a nameplate unit and a thermodynamic one.
Carried onward at full precision, not this rounded figure.
Therefore 12.7 % of the condensate flashes as it enters the vessel, releasing 381 lb/h of 15 psig steam into the low-pressure header — 105.7 kW of latent heat, or 11.1 boiler horsepower the burner never has to fire, out of a stream that was on its way to a receiver vent.
Why this order
The order here is the order of the physics, and it only runs one way. The flash fraction is a property of the two PRESSURES and nothing else — change the trap's flow from 3,000 lb/h to 30,000 and part (a)'s 12.7 % does not move a digit — so it is settled first, from four table readings, before any flow rate is allowed into the problem. Only then does part (b) scale it. That separation is worth internalizing because it is what makes the flash chart on a trap manufacturer's wall possible at all: one curve per letdown, good for every plant. The trade carries a serviceable mental version of it — about one percent of flash for every 10 psi of letdown in this range — and 135 psi of letdown predicting 13.5 % against the tables' 12.7 % is close enough to catch a decimal-point error on a ladder.
What the calculation actually teaches is that sensible heat and latent heat are the same currency in two denominations. The 279.7 kJ/kg of surplus is not lost when the pressure drops; it is spent, and what it buys is a phase change in a small part of the stream. Check the surplus a second way and the tables agree with themselves: saturated water cooling from 185.5 °C to 121.0 °C is a 64.5 K slide, and at about 4.34 kJ/(kg·K) for water at those temperatures that is 280 kJ/kg — the same figure the h_f column gave, reached without ever opening the enthalpy page. The consequences are all in the pipework. Flash is why a condensate line two-thirds full of water still hammers, because 12.7 % by mass at 15 psig is over 96 % by VOLUME and that vapour is moving; it is why a receiver vent plumes; and it is why a flash vessel is one of the few pieces of steam equipment that pays for itself out of a stream everybody else treats as waste. Turn the vessel pressure down to 5 psig and rerun the worksheet: the flash rises past 15 %, but the header it can feed gets less useful, which is the trade-off every flash-recovery design is really arguing about.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.