Lev Nelik answers a reader's question on thermodynamic sub-processes that are highly time dependent
by Lev Nelik

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
  • 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
  • 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,

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.

In Part 2, I would not expect a significant problem at Pump 2 from an interaction at the junction of its header and the hotter liquid in Header 1. If the liquid in Header 2 is gradually cooling and flowing past the junction, the temperature difference at a junction may be too gradual for a flashing reaction. However, for Pump 1, if it is again started with its header full of 302 F liquid and only 39 psia available at the junction, there might be an insufficient NPSH condition as pointed out in the article. Perhaps Pump 1 would be able to pass the 50-ft header full of hotter liquid quick enough to avoid any significant damage or pump trip.

Of course, the above analysis is based on simplified ideal conditions with gradual thermodynamic changes. Feedback from readers with experience in power plants and the effects of transients on pumps would be insightful.

Lee Ruiz
Oceanside, California

Nelik’s Response

Thank you for your comments. These are very diligent and methodical calculations. I have a few notes to your comments.

Your assumption of the process being gradual and isentropic cannot be assumed, because it is highly transient. This is one of the reasons it is not easy to calculate. Several thermodynamic sub-processes that are highly time dependent are taking place: convection from hotter liquid to colder, phase transformation and cooling of the vapor that is evolving back into liquid phase as it tries to rise through the colder liquid layer. Perhaps the only safe assumption for the process is that it is adiabatic, assuming no heat loss through the pipe to the surroundings occurs as a result of thick insulation.

My assumption that Point A is liquid that occupies half of the tank space with zero vapor above it is problematic. While the pressure is significant, a complete absence of vapor in the space above the liquid is impossible, as it would imply full vacuum. Because the fluid in the DA is at the saturation curve or in the subcooled region, something must be present above the liquid to maintain pressure.

To explain what keeps the void above the liquid at pressure in this case, we would need to understand more precisely the specifics of the DA design.

Your note is well-placed; if we instead make a slight adjustment to starting Point A to be slightly into a two-phase region, this difficulty immediately disappears. The entire space above the liquid would be vapor, the combined volume in the tank could be treated as the entire tank (20,000 gallons) instead of half of it (10,000 gallons), and the total combined specific volume would be total tank volume divided by total mass (which at Point A is mostly liquid). That would make the total combined volume double that if we used only half the tank. It does not affect Point A but makes things easier when going to point B, as we now have a total volume of the entire 20,000 gallons filled with a mixture—some liquid and some vapor. From there, vapor mass fraction “x” and volume can be calculated.

My assumption that the entire system is a constant volume is valid, because otherwise, it is impossible to make any other reasonable assumptions. For example, we cannot assume constant pressure at Point B (As an example, imagine that the tank had a free heavy lid that would apply constant pressure and that the expanding volume of vapor would move the lid up).

The tank described in these articles does not have moveable boundaries. The only openings are to the condensate piping and through the running pumps, but those are essentially closed (by running water).

As explained in Part 2, vapor wants to expand according to a new pressure to which the surface of the DA is now suddenly at (Point B). But it is not a steady state process, and things are very transient. As vapor tries to form and float up, it encounters a colder layer above it and condenses back to liquid almost instantaneously—at many orders of magnitude faster than it takes for vapor to travel through the suction header of the running (colder) pump to reach its entrance. Instead, what likely is happening is convective cooling of the hotter leg by the colder leg via this attempted vaporization and re-condensation.

Visualization of vaporization and re-condensationFigure 1. Visualization of vaporization and re-condensation (Courtesy of the author)

Imagine that you have two layers of fluid of equal volume, separated by a very thin membrane. Initially, 69 psi is enough to keep both liquids subcooled (the lower layer is subcooled but not as subcooled as the upper layer). Suddenly, pressure at the surface drops to 50 psi, at which the top layer is still subcooled, but the bottom layer enters the vapor phase. Will it immediately flash out to many times more vapor volume? If so, where will it go?

Assume that the membrane is thin and raptures as vapor forms. The bubbles trying to form at the lower layer would have to meet the colder layer and return to liquid form. This would go on for some time and eventually both fluids would mix at the average temperature (266+302)/2. If 50 psi is in a vapor region, the fluids will turn to vapor and fill the entire can. If the can is very tall, there will be more space for vapor to form (if it corresponds to vapor condition). Initially, however, all was in two liquid layers. After the drop of pressure (and if the new condition is vapor phase), however, the vapor will expand to take up the complete volume.

Part 2 shows that, for the conditions used in the example and at the resultant average temperature, no two-phase situation—and thus no vapor—exists. This means the system experiences no cavitation or NPSH problem. However, in reality, no vapor would reach the running pump because the colder column is not static and mixing with the hotter liquid. Instead, while vapor wants to form and transmit its energy to heat up cooler liquid, that liquid is actually in motion, taking a heated chunk and moving it toward the pump.

By the time the chunk flows through the entire length of pipe, the resultant rise in temperature of the flowing liquid is minuscule. As a result, it is not even close to heating the cooler leg substantially. In other words, it is the DA temperature that would dictate the temperature of fluid reaching the running pump.