Steam Coil That Only Heats on Mild Days
An air handler serving a paint booth uses a 15 psig saturated steam coil to heat 12,000 scfm of outdoor air from 40°F to 90°F. The site is near sea level. Condensate drains through a thermostatic trap into a return header held at 5 psig; the trap sits 2 ft below the coil and must lift condensate 8 ft, with 30 ft of 1 in. return piping. Since a trap was replaced last winter, the coil reaches setpoint only on mild days, the trap cycles loudly, and banging is heard at the coil during morning warmup. The control valve modulates steam to hold discharge temperature; on cold mornings it is 100% open with 78°F discharge. A survey found the new trap's published capacity is 900 lb/h at a 15 psi differential. The steam main holds a steady 15 psig. Maintenance believes the coil is simply undersized and wants to add a second coil.
1. Compute the full-load condensate load from the coil duty and compare it with the trap's capacity at the differential actually available.
Read the worked answer
Coil duty: the sensible heat equation at sea level is Q = 1.08 × cfm × ΔT = 1.08 × 12,000 × (90 − 40) = 648,000 Btu/h.
Condensate load: at 15 psig, saturated temperature is 250°F and h_fg ≈ 945 Btu/lb. Adding the small sensible subcooling allowance for condensate leaving at saturation, use about 945 Btu/lb: 648,000/945 = 686 lb/h at full load.
Available differential across the trap:
- Trap inlet: 15 psig
- Return header: 5 psig
- Static lift: 8 ft ÷ 2.31 ft/psi = 3.5 psi
- Friction and flashing in 30 ft of 1 in. return at 686 lb/h: roughly 1 psi
- Total backpressure ≈ 5 + 3.5 + 1 = 9.5 psig
- Available ΔP = 15 − 9.5 = 5.5 psi
Trap capacity at 5.5 psi: an orifice trap's capacity varies approximately with the square root of differential pressure, so 900 × √(5.5/15) = 900 × 0.606 = 545 lb/h - below the 686 lb/h required. At full load the trap is the constraint, not the coil, which is consistent with the discharge flattening at 78°F rather than the coil simply running out of surface. Adding a second coil would make the shortfall worse by raising the condensate load on the same trap.
2. Explain what happens at low and moderate load, including the morning banging, and why a thermostatic trap is a poor choice for this arrangement.
Read the worked answer
The control valve modulates steam pressure to hold 90°F discharge. As outdoor temperature rises, the required coil temperature - and therefore the required steam pressure - falls. The trap can only discharge when the coil pressure exceeds the total backpressure of about 9.5 psig (5 psig header + 3.5 psi lift + about 1 psi line loss). Once the valve throttles coil pressure below roughly 9.5 psig, the trap cannot open at all: this is stall.
Consequences:
- Condensate backs up into the coil and floods the lower rows, reducing effective surface. The coil then delivers less heat than the surface should, the valve opens further, and the unit hunts between flooded and cleared states.
- The coil becomes a water-logged slug. When the valve finally opens with a pressure rise sufficient to blow the trap, the slug is pushed into the return line, producing the reported banging and the loud trap cycling.
- Because the lift adds backpressure, the stall point is high; with a gravity drain and no lift, stall would occur at a much lower pressure and the coil would behave far better.
A thermostatic trap is a poor match because its element follows temperature, not the need to pass large condensate loads at low differential, and it tends to close on subcooled condensate while remaining shut during stall. Even at full load it is undersized at 5.5 psi. The problem is not coil surface and not steam supply pressure.
3. Recommend a corrected drain arrangement and the checks needed to confirm it, with the tradeoffs.
Read the worked answer
Root cause: the combination of an 8 ft lift, a 5 psig return header, undersized trap capacity at the real differential, and a modulating load that drives the coil into stall.
Recommended arrangement (in order of preference):
- Drain by gravity to a vented condensate receiver at the low point, then pump from the receiver to the 5 psig header. This removes the lift from the trap's duty entirely; the trap then sees about 15 − 1 = 14 psig at full load and still functions at low load because the receiver is at atmospheric pressure. It is the cleanest fix and eliminates stall.
- If a receiver is impractical, use a condensate pump-trap (electric or pressure-powered) at the coil outlet. These are designed for lifts into pressurized returns and for stall conditions.
- If neither is possible, use a float-and-thermostatic trap with a lifting fitting and vacuum breaker, size the return line at 1½ in. to cut line loss, and keep the lift under about 4 ft. Even then, accept that stall will occur below roughly 10 psig coil pressure.
Sizing: for 686 lb/h at full load, select the trap at the actual differential with at least a 2:1 safety factor - about 1,400 lb/h at 5.5 psi if a lift is retained, or about 1,400 lb/h at 14 psi with a receiver.
Checks before ordering: verify steam main pressure at the coil inlet during a cold-morning run (not just at the header), confirm the control valve actually strokes fully, confirm the coil is not air-bound (install a thermostatic air vent at the coil high point), inspect the 1 in. return for flashing erosion, and temperature-check the coil face for cold rows. After the change, trend discharge temperature and trap cycles across the full outdoor temperature range to prove stall is gone. Do not add coil surface; the coil is not the limiting item.