C.F. ‘Chubb’ Michaud, CWS-VI

So far in this series, we have discussed the importance of accounting for pressure losses in filtration systems up to the point of entry. It was also pointed out that static pressure measured at a convenient hose bib outside the home is not a true representation of the dynamic or working pressure available to the point of entry under normal use conditions. Small pipe size and long runs reduce available pressure and restrict flow when water is being used downstream. Filters that require pressure and flow for backwashing and regeneration may not have enough water available to perform those functions unless all downstream flow is locked out or a booster pump is installed. In Part 5, we will discuss the designed flow through laterals.

Flow through laterals
An often ignored but critical part of any filter design is flow distribution within the filter. This is particularly true with ion exchange and carbon beds or other adsorptive media beds where capacity is a function of how much water is flowing relative to how much media is used. Distribution gets more critical with larger tank sizes and systems with large ratio differences between the service flow and backwash or regeneration. The feed stream has to contact all of the media to obtain expected capacity. Water entering a tank through a single port at the top and exiting through a single port at the bottom will not contact all of the media. Water will take the path of least resistance; it makes a beeline from one end to the other without stopping to visit all the media.

If we assume that all of the particles stack up in a bed the way you stack canon balls, the angle formed in the matrix is approximately 60 degrees, but because some of the angles are rotated, we will call it 45 degrees. After water contacts the top of the bed, it flows over the particles in all directions, splitting the flow. This flow splits again and again until it reaches the side of the tank. Without an engineered distribution system, flow may reach all the way to the sides of the tank but the flow will not be uniform (plug flow) across the entire bed. It is estimated that in worse cases of poor distribution, up to 50 percent of the flow will pass through a center core of media, representing only about 15 percent of the total and that the outer 50 percent of the media may see only about 20 percent of the flow. Under low-flow conditions, such as during regeneration, flows are skewed even more toward the center. The difference between point-to-point flow and plug flow is illustrated in Figure 4.

In most systems, service flows are much higher than regeneration flows. Most residential systems are actually designed to maximize the service flow. How many gpm can I squeeze out of the system at 15 psi ∆P? As a result, distribution points are wide open to provide as little restriction as possible without allowing resin to escape. Flows will approach 10 gpm/ cu. ft. and higher. The resin bed acts as a very efficient distributor at this flowrate. Water sees most of the bed except for three to four inches at the top edges and bottom, resulting in a capacity loss of about 10 percent. Underbedding can avoid the capacity drop.8 In regeneration, we have a different story. Regeneration flows for ion exchange systems are generally under 1.0 gpm/cu. ft. and often as low as 0.25 gpm/cu. ft. With the flow as low as 1/20th the service flow, ∆P is at 1/400th. There is no reason for the regenerant to distribute itself; it takes the shortest and lowest ∆P path. It drops straight down, only passing through approximately 15 to 20 percent of the bed.

I confirmed this on a 63-inch top-mount softener that had convenient side flange windows. The customer was complaining about short runs and high leakage. I observed the regeneration brine, which entered the top through a center-point metal diffuser and could see the phase separation lines from the entering brine going straight down to the bed in a column no more than 20 inches wide. After a few minutes of brining, a crater formed around the riser pipe (a result of the resin dehydrating in the presence of brine). After the regeneration and rinse cycles were over and the system returned to standby, we drained the vessel and removed the upper window to sample the resin. I took samples from the top six inches of the bed near the window in the outer third of the column. I then took a sample diagonally through the middle third of the bed about 18 inches down and finally, a sample of the inner third about halfway down the bed. A lab analyzed the three samples to determine how much sodium and calcium was on the resin. Crude? Yes. But it was very telling. The outer third of the bed was < 10 percent in sodium form. It was still exhausted. The middle third was about 40 percent sodium form, considered poorly regenerated. It would show some capacity but also very high leakage. The inner core (about 24 inches) was > 85 percent sodium and adequately regenerated. The total brine flow was about 32 gpm through 50 cu. ft. of resin. This is perfectly acceptable. Service flow was up to 200 gpm; this is also perfectly acceptable. The ratio of service to regeneration flow was 6:1. Ideal! Why wasn’t it working? This particular customer had a 0.2-ppm hardness shutoff. That’s right. Not 2 grains. Not 2 ppm but 0.2 ppm. A smaller system (i.e., residential) with less stringent performance (i.e , < 1 gpg) would have gotten by; larger systems need some design help. We then set out to improve the distribution.

Because this was a top-mount system, the upper diffuser had to remain. We blinded off the entire bottom rows of slots, however, and installed a machined PVC baffle to redirect the inlet flow outward and high up in the head space of the tank rather than straight down. This also provided some pressure drop through the top distributor. The control valve was converted to up-flow brining to address the low leakage requirements. The question remained, “How do we make sure that the brine goes all the way out to the sides of the resin bed so it regenerates the entire bed?”

The bottom distributors were the double-stacked bayonet design with 16 laterals radiating out to the sides from a central hub. These laterals had a slotted outer sleeve that slipped over a molded pipe with lots and lots of 5/16th holes distributed evenly along the entire length. The total surface area of all of the holes in all of the laterals added up to roughly the equivalent of a six-inch pipe (rated at > 800 gpm flow). During regeneration at 32 gpm, ∆P was essentially zero. There was no reason for the brine to go to the ends of the laterals, even in up-flow mode (see Figure 5). The media and underbedding had to be removed and the laterals had to be rebuilt (which meant someone had to crawl down to the bottom of a wet pit and empty the tanks).

Figure 5 shows flow distribution through a wide-open distributor of the type used in the above example. The bulk of the flow exits near the hub, giving an uneven backwash and poor distribution in service. The designed lateral tightens the flow by applying the ∆P right at the orifices rather than the hub, the slotted sleeves or the media bed. This forces the flow to distribute to all of the outlets. Placing a flow restrictor downstream and external to the tank does absolutely nothing for internal distribution.

The hydraulic equation
There is a hydraulic equation that helps to solve for ∆P and orifice size for restricting fluid flow. If you plug in the flow-perorifice hole and the desired ∆P, it will tell you the size of the hole you need (see Equation 1). Keep in mind that with a huband-lateral design, the holes are not uniformly spaced. They are spaced to handle equal flows or drain areas of the tank and they get closer together as they extend out the lateral. This is illustrated in Figure 5 for a hub-and-lateral design.

Equation 2. Hydraulic formula for determining orifice size and pressure drop9
Q= K(k)d2h1/2
Where:
Q = flowrate in gpm
K is a conversion constant = 19.636
k is an efficiency factor for a straight drilled hole (not beveled) = 0.61
d is the diameter of the orifice in inches (squared)
h is the pressure drop in feet of head. 1 psi = 2.31 feet of head.

Note: h1/2 means the square root of h. Convert the pressure to feet of head and take the square root.
Here is an example: What size orifice is required to produce a ∆P of 1.0 psi at a flow of 1.0 gpm? Solving for d2, we have: d2 = Q/(K)(k)(h)1/2
d2 = 1.0/(19.636)(0.61)(2.31)1/2
d2 = (1.0)/(19.636)(0.61)((1.5199) = 1.0/18.2 = 0.549
d = 0.2344 inches

The error introduced by using a 0.25-inch hole is that the ∆P drops to 0.77 psi; in larger systems, numbered drills should be used to achieve the tolerance needed. The spacing between the holes is illustrated in Figure 6 for a hub-and-lateral design. The exact placement of the hole is calculated as being halfway into the zone area. In other words, half the area of the zone is inside the radius of the hole and half is to the outside. The hole placement is the ratio of 21/2/2, the square root of two divided by 2 (or 1.414/2 = 0.71) approximately 70 percent of the distance between the zone boundaries.

Figure 6 illustrates the concept of the equal-area zones, which are numbered from 1 to 5. For our 63-inch tank with an ID of 60 inches, the holes are located at 9.5, 16.4, 21.2, 25.1 and 28.5 inches from the center of the hub. So what size orifice and why? Using the above example of a 63-inch softener vessel with 16 laterals off a central hub, I did a little trial-and-error math and decided that five drain zones would suffice. If you divide the radius of the tank by six you will get a good design approximation for the number of zones. That’s a total of (5 x 16 =) 80 holes. I used a service flow of 200 gpm and a regeneration flow of 32 gpm. I plugged these values into my hydraulic equation and determined (from experience) that I needed at least 0.15 psi ∆P during regeneration to force brine all the way to the ends of the laterals. With a flowrate of only 0.4 gpm/hole, (32/80 = 0.4), my orifice size worked out to 0.2368 inches. Then, calculating ∆P at full service flow of 200 gpm or 2.5 gpm/hole, I determined that ∆P in service would be 6.0 psi (just from the laterals) and at 10 gpm/sq. ft. of bed, another 2 psi from the resin, for a total ∆P across the bed of 8 psi. This was acceptable to all. Before modifying the laterals, the ∆P across the bed was measured at about 3 psi (so mostly from the resin bed and not the laterals). To make the changes in design, we first had to sleeve the laterals to block all the existing holes and then redrill with the appropriate number drill to make the changes. The measured ∆P on the reinstalled system was noted at just under 9 psi at a flow of 180 gpm. This calculates to 11.1 psi at 200 gpm. We never said this was easy…only doable!

Header/lateral design
There are limitations in tank size for the hub-and-lateral design simply because the tips of the laterals get farther apart as the tanks get larger. If we pick a distance, say 12 inches apart at the tips, we can use a 16-lateral hub and get a tank circumference of 192 inches. The diameter of this tank would be 61 inches. For tanks above a 60-inch diameter, we opt to use a header/lateral rather than a hub-and-lateral to limit the distance between drain points. Figure 7 illustrates a typical header/lateral design for a 60-inch tank with seven laterals off a header. The laterals are spaced nine inches apart with a diagonal distance between holes of 10.5 inches. The red line grid placed over the cross-sectional representation of the tank outlines the drain areas for each hole. A physical count shows we will need 40 holes. The total surface area of the tank is 2,826 in2; each hole will drain 71 in2 or about the equivalent of a 9.5-inch-diameter tank. Hole placement is staggered as shown.

If this filter is designed to flow at 200 gpm, we will have a flow of 5 gpm/hole. As a carbon filter, it will backwash at 200 gpm; we would design our ∆P for the best distribution at 200 gpm with about 2 to 3 psi ∆P. As an ion exchange bed, however, we have to design for the regeneration flowrate. For cation, this might be 20 to 30 gpm and for anion, it could be 15 to 20 gpm. What size will the orifices be? Using the Hydraulic Equation Q = K(k)d2h1/2, we can solve for the diameter of the holes as:

d2 = Q/(K)(k)(h)1/2

We will assume a minimum pressure drop during regeneration is 0.1 psi through the laterals. Due to the density of the regenerant solutions, the actual value will be higher. Let’s look at the minimum flows and check the resulting service ∆P. For the anion at 15 gpm, we find the orifice size is 0.2552 in. at 0.1 psi but the resulting service flow ∆P is 17.8 psi, which is far too high. We increase the regeneration flow (to narrow the ratio between service flow and regeneration flow) to 20 gpm and find an orifice size of 0.2947 in. and a service ∆P of 10 psi. This is better but still high. We increase the flow again to 25 gpm and our orifice goes to 0.3295 in. and service ∆P drops to 6.4 psi. This is considered acceptable. We can use the same values for the cation laterals since it will have the same service rate. If we regenerate the cation at 30 gpm, however, we need to size the orifices at 0.3609 in. and our ∆P in service drops to 4.5 psi.

Orifice design
The shape of the orifice will have a huge impact on the resulting pressure drop. The value for ‘k’ in the above examples is for a straight drilled hole with square edges. The value is 0.61. If we taper the holes or counter-sink them, the value rises to 0.98. This greatly reduces the ∆P at any given flow. In our example above, tapering or beveling the holes would necessitate increasing the regeneration flows to 47 gpm to maintain a 0.1 psi ∆P and service ∆P would drop to 1.72 psi. For the most consistent performance from an engineered lateral, drill the holes straight through with square edges.

Backwashing a bed through a designed bottom distributor is generally not a problem for ion exchange since the backwash flow will be intermediate to the regeneration and service values. The same laterals will work. For GAC filters, backwash rates may be equal to service rates. Here, ∆P in the bottom distributor can be designed at 2 to 3 psi and will work well for both functions. High-performance media filters intended for particulate removal may have backwash rates that are 6 to 7 times the service rates. It is suggested that for these systems, design for a 1 psi ∆P in service. Backwash ∆P will be about 5 psi, which should be no problem.

Separate regeneration laterals
There will be occasions when the ratio between service flow and regeneration flow becomes too high. This is particularly true when designing mixed beds because the regeneration flowrates are based on only a portion of the bed (cation or anion) and service flow is based on the whole bed (cation plus anion). This can also occur when the resin bed is used to remove a particular contaminant that is very slow to elute and longer regeneration times are needed. In such cases, flow distribution may be designed to go through a separate lateral utilized only for the regeneration. This way, service flow and regeneration flow are independent of one another. It may even include those situations where a separate lateral is used for the introduction of regenerant chemical and the discharge of the spent regenerant.

Summary
Water flowing through a filter bed will take the path of least resistance, the straightest line from point A to point B, unless it is forced to do otherwise. This is key in building larger systems. You need to design the laterals to optimize the flow distribution within the tank. The small price paid in higher ∆P is well worth the extra capacity and efficiency gained. In systems that demand fractional ppm leakages, there is no other way to achieve that goal other than through sound design. Hydraulics not only plays a role in getting water to and from a filter but determines just how the water flows through the filter itself. Systems should be designed to provide ‘plug’ flow; i.e., a uniform flow that moves through the filter like a piston in an engine. Good hydraulics equals good filtration efficiency as well as the media utility and must be designed into the system.

References

  1. Michaud, C. F., “Factors Affecting the Brine Efficiency of Softeners,” WC&P Internationsl, February, 2002.
  2. Cameron Hydraulic Data, Ingersoll-Rand Corp., New York, NY, 14th Edition, 1968. pg. 67.

About the author
C.F. ’Chubb’ Michaud is the Technical Director and CEO of Systematix Company of Buena Park, CA, which he founded in 1982. He has served as chair of several sections, committees and task forces with WQA, is a Past Director and Governor of WQA and currently serves on the PWQA Board, chairing the Technical and Education Committees. Michaud is a past recipient of the WQA Award of Merit, PWQA Robert Gans Award and a member of the PWQA Hall of Fame. He can be reached at (714) 522-5453 or via email at AskChubb@aol.com

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