1. Steam traps Applications and Recommendations

2. 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 .

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

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

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

6. Pressure
By-pass valve
Steam
Air D1 D2 H
Isolating valve
100 m
100 m
Q = 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 :

7. 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 .

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

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

10. Jacketed pipes
Steam
Product

11. 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

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

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

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

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

16. = 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 = m²/m for a 1/2” tracer
k = 10 W/m².K
t = steam t° - air t° = 184°C - (-10°C)
Q = = 91 W / m
= kW/m kJ/h
2120 kJ/kg
= 0.16 kg/h.m of steam ---> 1 tracer is enough !

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

18. 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 .

19. 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 = l/h

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

21. V max
Non condensibles
Condensate
Deposits
V=O