Lev Nelik answers a reader's question on thermodynamic sub-processes that are highly time dependent
by Lev Nelik, Ph.D., P.E., Pumping Machinery, LLC, P&S Editorial Advisory Board
May 28, 2015

Letter from a Reader

The following comments relate to scenarios described in Parts 1 and 2 of “Can Deaerators Create Pump Trips?” (Pumps & Systems, March and April 2015), which discuss handling power plant transients.

Steady flows to and from the deaerator (DA) and gradual isentropic thermodynamic changes from the high to low temperature/pressure conditions are assumed. A gradual power plant cool down period after a shutdown event may mitigate general or local flashing.

From Part 1 at Point A:

  • DA total tank volume V = 20,000 gallons = 2,674 cubic feet (ft3).
  • Tank liquid (water) volume (Vliq) = 10,000 gallons = 1,337 ft3.
  • Mass of tank liquid
    equation
  • Mass of tank vapor (mvap) = 0 lbm (all mass is saturated liquid at vf).
  • Total mass of tank contents (M) = mliq + mvap = 76,564 lbm.
  • Using the above values, the calculated specific volume (“v”) should agree with the temperature-specific volume (T-v) diagram at Point A. However, calculated
    equation
  • This value is greater than vf = 0.01746 ft3/lbm from the diagram. Since the vapor volume Vvap = mvap x vg = 0 ft3, the analysis might be conducted by assuming that the total tank volume revV = Vliq + Vvap = 1,337 ft3. In other words, assume the total DA tank volume is 10,000 gallons and completely full of liquid (water) at Point A. Then, at Point A on the diagram,
    equation

Using revV and “v” for calculations at Point B, X = 0.003%, vapor mass mvap = X x M = 2.3 lbm, vapor volume Vvap = mvap x vg = 24.5 ft3, liquid mass mliq = (1-X) x M = 76,562 lbm, liquid volume Vliq = mliq x vf = 1,312.3 ft3. Therefore, the total volume and mass of the tank contents remain the same. However, the vapor volume has increased by the same amount that the liquid volume has decreased. The less dense vapor must occupy a larger space than the reduction in liquid volume. Because there is no additional tank space to occupy, some liquid may be expelled from the 10,000-gallon DA or excess vapor pressure might activate a relief valve.

Expelled tank liquid may imply that the total mass of the DA tank contents decreases at Point B: Specific volume at Point B is greater than at Point A. Using “vb” = 0.018 ft3/lbm at Point B, the calculated vapor volume is now about 66 ft3 at Point B compared with 24.5 ft3 if “v” = 0.01746 ft3/lbm. Since the new liquid volume is now 1,312 ft3, about 41 ft3 must be expelled from the tank at Point B: 66 + 1,312 – 41 ft3 = 1,337 ft3 = the tank volume.

However, I would not expect a DA tank to be either half full of liquid without any vapor or completely full of liquid. A more likely location for Point A might be to the right of the saturated liquid state where a two-phase liquid and vapor state exists—Point A1.

The originally calculated specific volume, “v” = 0.03492 ft3/lbm, represents a liquid and vapor state at 302 F. If this value is chosen, a new set of values can be determined at Point A1 and Point B1: X = 0.17% at B1, ∆Vvap = +23 ft3, and ∆Vliq = -23 ft3. In this case, the vapor has enough room for expansion going from a calculated 1,341 ft3 at A1 to 1,363 ft3 at B1. However, while the calculated liquid volume has decreased at B1, the corresponding liquid mass has increased. This seems unlikely.

After some trials, it was determined that reasonable changes in volume and mass occur if the specific volume is greater than about 0.29 ft3/lbm for the given data in this article. If another specific volume such as “v” = 0.5 ft3/lbm is chosen, another set of liquid and vapor state values can be determined at Points A2 and B2: X = 4.52% at B2, ∆Vvap = -1.33 ft3, and ∆Vliq = +1.33 ft3. In this case, the vapor volume has contracted going from a calculated 2,587 ft3 at A2 to 2,586 ft3 at B2. Now, the calculated liquid volume and mass have both increased at B2, while the calculated vapor volume and mass have decreased at B2. Vapor condensation, from state A2 to B2, is now the result. This seems to be a realistic outcome after slowly cooling. Although it should be noted that, with “v” = 0.5 ft3/lbm, there is a substantial difference in liquid and vapor mass and volume values when compared with the original calculations from above.

After an emergency trip situation, there may be ongoing automatic and/or manual adjustments as various plant elements (DAs, pumps, etc.) react to changing system conditions. If the liquid+vapor mass (M) within a fixed-volume DA tank can decrease or increase in response to some of these changes, maybe the specific volume is not necessarily constant at different thermodynamic states.

Besides the dynamics occurring inside the DA tank, I would expect that any major net positive suction head (NPSH) problem at the pump suction might be attributed to entrained air, not water vapor, being transferred from the DA liquid to the boiler feed pump. Before being shut down, the injected pegging steam might have removed most of this air.

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