Steam coil on an air handler: duty and condensate load
Air handler · steam coil duty and condensate load
An air handler on a Canadian plant roof draws 8,000 cfm of outdoor air through a steam preheat coil before anything else touches it. On the design day the outdoor sensor reads 5 °F and the leaving-air controller holds 55 °F downstream of the coil, so the coil is being asked for a 50 F° lift across the full airflow. The coil is fed from a 5 psig branch off the low-pressure header, through a modulating valve, and it drains through a float-and-thermostatic trap into a return line. Two numbers come from the steam tables at that coil pressure. At 5 psig (19.7 psia) the saturation temperature is 108.4 °C, or 227 °F — the temperature the tube surface sits at whenever the valve is open — and the latent heat on the same row is h_fg = 2,234.1 kJ/kg, which is 960 BTU for every pound. Everything the coil does is steam condensing on the inside of those tubes and giving up exactly that latent heat, which means the pounds of steam it swallows per hour and the pounds of condensate the trap must clear per hour are the same number.
Every number in this problem is editable — change any value below and the whole chain recalculates.
- V̇ = 8,000 cfm — Airflow across the coil
- T_ea = 5 °F — Entering air, design day
- T_la = 55 °F — Leaving air, controller setpoint
- ΔT = 50 F° — Lift the coil must produce
- p = 5 psig — Coil steam pressure (19.7 psia, T_sat 227 °F)
- h_fg = 2,234.1 kJ/kg — Latent heat at 5 psig, from the tables
- (a)the sensible load the coil has to put into the airstream
- (b)the steam that duty condenses, which is also the trap's condensate load
- (c)the coil read against a boiler's own rating
This is the 1.08 rule in SI clothing: the constant is the density of standard air times its specific heat, 1.2014 kg/m³ × 1,004.8 J/(kg·K) = 1,207.2 J/(m³·K), and it comes out as 1.08 BTU per hour per cfm per Fahrenheit degree once the units are pushed through. Sensible only — this coil changes the air's temperature and not its moisture content, so no latent term belongs here. The one honest caveat is the word STANDARD: 5 °F air at any altitude is denser than the 0.075 lb/ft³ the rule assumes, so a fan holding 8,000 cfm is moving more mass than this, and the answer is a little conservative.
Carried onward at full precision, not this rounded figure.
ṁ = Q̇/h_fg is the whole steam side of the coil: the duty divided by what one kilogram gives up when it condenses, with h_fg read at the COIL's pressure and not at the header's, since the modulating valve is what sets the tube pressure. Read the answer twice. It is the steam the valve must pass, and it is the condensate the trap must clear, because in a coil those are the same pounds a few seconds apart. It is also the RUNNING load — a trap gets selected on two or three times it, for the morning start with cold metal and cold air, when condensate forms faster than the steady state ever makes it.
Carried onward at full precision, not this rounded figure.
One air handler, in the units the boiler room speaks. Thirteen boiler horsepower is not a trivial branch: on a 150 BHP plant this single coil is nearly a tenth of the whole boiler at design conditions, which is worth knowing before the header, the PRV station and the condensate pump are sized, and worth remembering on the morning after a shutdown when every coil in the building is calling at once.
Carried onward at full precision, not this rounded figure.
Therefore the coil must put 126.6 kW — 432,000 BTU/h — into the airstream, condensing about 450 lb/h of 5 psig steam, which is exactly the condensate load the trap has to clear and the equivalent of 13.0 boiler horsepower on the plant behind it.
Why this order
Two sides of one energy balance, and the coil is the equals sign. The air side is sensible heat: mass flow times specific heat times the dry-bulb rise, packaged by the trade into 1.08 × cfm × ΔT so it can be done on a clipboard. The steam side is pure latent heat: at 5 psig the condensing surface stays at 227 °F from one end of the coil to the other, so no sensible term appears there at all, and the whole duty is paid for by phase change. The order runs air-first for a practical reason — the airflow and the temperatures are the design decision, and the steam is the consequence. Reverse the order and you have the other everyday version of this problem: a trap of known capacity, a coil that will not hold setpoint, and the question of how much air the coil could actually heat.
The number to keep in your head is around 1,000 BTU per pound of steam, near enough over the whole range of pressures a coil is ever fed at, which makes the mental arithmetic almost free: divide the BTU/h load by a thousand and you have the pounds of condensate per hour. Here that shortcut says 432 lb/h against the 450 the tables give, close enough for a trap catalogue and wrong in the safe direction for a valve. The trade detail worth the most on a Canadian roof is elsewhere, though: 5 °F entering air against a 227 °F tube is a freeze-up waiting to happen, because a modulating valve throttled down at part load leaves condensate lying in the bottom tubes with subfreezing air blowing across them. The fixes are all about never modulating the steam — face-and-bypass dampers or an integral face-and-bypass coil, with the steam valve wide open whenever it is open, so the tubes are either full and hot or shut off entirely. A vacuum breaker on the coil and a trap that can drain by gravity into a non-pressurized return matter for the same reason: a coil that cannot drain will not survive its first cold night.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.