Steam traps Applications and Recommendations

Role of the steam traps Basically : To get rid of the condensate . To eliminate air and CO2 But also : To prevent steam to escape into the condensate return. And: To make heating systems as efficient as possible . To prevent their corrosion . To get rid of impurities .

Main selections criterions Sub-cooling or not . Air and CO2 elimination . Resistance to water hammers . Resistance to dirt . Freezing resistance . Life time . Energy efficiency

Main lines Ideally ,the steam trap should : React immediately. Not subcool or back up the condensate . Be resistant to water hammer . Be freeze resistant

Calculation of the radiation losses Calculation of pipe diameter Calculation of start-up loads Calculation of the radiation losses Pressure drop in steam lines

Pressure By-pass valve Steam Air D1 D2 H Isolating valve 100 m 100 m Q = 36000 kg/h D1 = 10” Steam pressure : 15 bar Steam temperature : 200°C Ambiant temperature : 0°C Steam front velocity : 1 m/s Flow across BP valve : kg/m*m/s KVS BP valve in critical flow Opening time of BP valve :

Tracing applications Ideally the steam traps should : Work efficiently at low load (< 5 kg/h) . Be dirt resistant . Be freeze resistant . Work with high back pressure .

General purpose of a tracer - To keep process fluids at process temperature - To prevent thickening or solidification of products - To prevent freezing

Options Product Clamp strap Steam tracer - Attached tracers - Welded tracers ( short or continuous welds ) - Heat conducting paste - Wrapped copper tubing

Jacketed pipes Steam Product

Calculation of the tracing system Insulation 8O mm Lowest outside t°= - 10°C Steam pressure : 10 bar Steam t° : 184°C Product line t°= 80°C Tracers

Example : Product t° to be maintained : 80°C Lowest outside t° : - 10°C Product line pipe size 8’’ : 220 mm Pipe thickness 1 8 mm Insulation thickness 2 80 mm Thermal conductivity of pipe wall 1 55 W/m.K Thermal conductivity of insulation 2 0.07W/m.K Heat transfert coefficient product to wall 1 465 W/m².K Heat transfert coefficient wall to air 2 10 W/m².K

Q = S . k . t 1° Let’s calculate the heat loss of the product line : With k = 1 ---------------------- 1 + 1 + 1 + 2 1 2 1 2 W/m².°C

Solution : S =  . D . L = 3.1416 x (0.22 + 0.16 ) x 1 = 1.194 m²/m k = 1 = 0.803 W/m².k ______________________ 1 + 1 + 0.008 + 0.08 465 10 55 0.07 Dt : -10°C + 80°C = 90 °C

Q = 1.194 m²/m . 0.803 W/m².K .90 = 86 W/m = 0.086 kW/m x 3600 kJ/h 2000 kJ/kg = 0.15 kg/h.m of steam

= 0.16 kg/h.m of steam ---> 1 tracer is enough ! What is the amount of heat actually transmitted by a 1/2” steel tracer ? Q = S . k .t With S = 0.047 m²/m for a 1/2” tracer k = 10 W/m².K t = steam t° - air t° = 184°C - (-10°C) Q = 0.047 . 10 . 194 = 91 W / m = 0.091 kW/m . 3600 kJ/h 2120 kJ/kg = 0.16 kg/h.m of steam ---> 1 tracer is enough !

Quick selection table for 1/2” tracers Product line Winterizing Solidification t° Solidification t° 25 to 65°C 65 to 150 °C 2” 1 1 2 3” 1 1 3 4” 1 2 3 6” 2 2 3 8” 2 2 3 10” 2 3 6 12” 2 3 6 14” 2 3 8 16” 2 3 8 18” 2 3 10 20” 2 3 10

Heat exchangers Ideally the steam traps should : Make heating systems highly efficient thus not allowing condensate subcooling . Get rid of non-condensible gases . Be resistant to water hammers . Work with back pressure .

Q = 15850 l/h Temperature regulator / control valve 6 Bar(g) 165°C Heat exchanger 1500 kW 70°C secondary fluid outlet t° 70°C in 2000 kg/h Condensate return 2 Bar(g) back pressure 10°C primary fluid inlet t° Q = 15850 l/h

Frequently encountered problems ... Flooded heat exchangers Temperature swings Cold traps Water hammers Mechanical stress Leaking gaskets Corrosion

V max Non condensibles Condensate Deposits V=O