Although condensate return has nothing to do with combustion control, it does have a significant impact on the efficiency of the boiler. Over the past number of years I have been developing a combustion calculator and every now and again I am being confronted by the percentage of condensate being returned and its influence on the hot well (feed water) temperature. I am referring to a system where condensate is returned from the process to an open (vented) hot well at atmospheric pressure (0 bar (g)), and where it mixes with cold make-up water to replenish steam and condensate losses in the reticulation system and processing plant. The mixture of condensate and make-up is delivered to the boiler via a feed water pump.
Theoretically one can assume that condensate is at 100 ᵒC (assuming no heat loss from the reticulation system) and make-up is at ambient (±25 ᵒC) temperature. Let us further assume that 50% condensate is returned to the hot well and that make-up is 50% of steam supply. Simple arithmetic suggests that the temperature of the water in the hot well should be at 62,5 ᵒC.
With one of our own outsourced boilers I was however often confused by the obvious discrepancy between theory and the practical reality on this matter. Although the percentage condensate returned varied between 40% and 60% the feed water temperature was mostly in the vicinity of 94 ᵒC. Leaking steam traps in the plant? (The maintenance team denied it was the case.) Probably, but not necessarily so. Let us once again look at first principles governing the physics of steam and condensate.
With our outsourced boiler steam is supplied to the plant at 8 bar pressure and then reduced to 5 bar at the point of use. After passing through a process heat exchanger the condensate is returned to the boiler house. From steam tables we know that:
- The enthalpy of saturated steam at 8 bar is 2774 kJ/kg
- The enthalpy of saturated steam at 5 bar is 2757 kJ/kg. Since no heat is exchanged or lost in the PRV, the steam at 5 bar will be slightly superheated. This superheat is absorbed by the process in the heat exchanger.
- The enthalpy of saturated condensate at 5 bar is 671 kJ/kg. This is the energy remaining in the condensate after it has passed through the process heat exchanger, but before it has passed through the steam trap.
- The enthalpy of saturated condensate at atmospheric pressure is 419 kJ/kg. This is the energy contained in the liquid condensate after passing through the steam trap and on arrival at the hot well.
It seems that passing through the steam trap has caused the condensate to give up 252 kJ/kg of its internal energy? Where did it go to? Of course, it was used to create flash steam (some 0,11 kg per kg of condensate) at atmospheric pressure and 100 ᵒC. This mixture of condensate and flash steam still contains the sum total of the original condensate energy at 5 bar, namely 671 kJ/kg. And the total of this energy, when mixed in with make-up water at 25 ᵒC in the hot well, will cause its temperature to rise according to the formula: Energy = 4,19*kg*(T2-T1). One kg of condensate will thus increase the temperature of 2 kg of feed water by ((671+4,19*25)÷(4,19*2)) = 92,6 ᵒC.
We seem to run into a problem here, since water at atmospheric pressure boils at 100 ᵒC, so if we mix too much condensate with make-up water in the hot well, it will start boiling and excess heat will escape as flash steam through the vent. Theoretically (meaning no energy losses taken into consideration) at least it seems to indicate that with an open hot well there is a certain maximum percentage of condensate that can be returned before the hot well starts boiling.
Our attempt at calculating the maximum condensate returned, hinges on the following equation:
Energy of feed water at boiling point (100 ᵒC) = (Energy of condensate + Energy of make-up), remembering that 100% feed water consists of x% condensate and (100-x)% make-up water. Thus, to calculate the percentage (x) of condensate returned at which the water in the hot well starts boiling, the following formula applies:
100%*4,19 kJ/kg*100 ᵒC = [x%*671 kJ/kg]+[(100-x)%*4,19 kJ/kg*25 ᵒC].
Solving this equation for x provides us with a maximum condensate return of 55,5% before the water in the hot well starts boiling.
I am the first to admit that this is a purely theoretical exercise. Depending on condensate system heat losses, more condensate can be returned before boiling takes place. In fact, the more condensate is returned, the less make-up and water treatment will be required. But from an energy point of view alone:
- Incurring cost to recover all of the available steam condensate may not be the most economical solution.
- Aim at returning sufficient condensate to bring the hot well as close to boiling as possible, without causing feed pumps to cavitate.
This post was compiled by René le Roux for Le Roux Combustion, all rights reserved. Do you want to know more about efficiency of combustion or combustion optimization? Please contact us for your professional boiler automation, steam system efficiency and coal characterization needs.
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