Fig. 1
Fig. 1 shows the critical elevation diagram and distances between key points. The specific information was summarized in other documents.
Fig. 2
Fig. 2 shows a simple ASPEN process simulator PFD that allows looking at various scenarios assuming no second well and ignoring water supplied to Tank 3 from the spring. The pipe roughness is assumed to be 0.0015 mm.
Facts and factoids about the systems:
1) The 60-HP, 175 GPM nominal booster pump consumed used about 48 HP at 280-psig discharge.
2) The measured flow of the submersible pump was 218 GPM. At the time of the test, 41 min, the level in the tank seemed constant => booster pump flow close to submersible’s?
3) It took 4 min. 40 sec. for the water to “reach” Tank 1. At 218 GPM the water would the Cruzero in 1.3 hours => system was already full of water?
4) After shutting down the booster pump and closing the discharge valve the liquid head read 165 psig. This is equivalent to an elevation of 381’, above the Cruzero but below any of the tanks => check valves present at this elevation?
Table 1
Table 1 summarizes results obtained from the ASPEN model. Case 1 assumes the existing piping geometry with the measured pump discharge pressure of 280 psig. This is the attempt to establish the system’s characteristics. The analysis indicated that both tanks can be reached with the total flow of 146 GPM, 84% of design. Otherwise, at least one of the two legs would choke. About 45% of the flow goes to Tank 3. This result (about 50-50 split) intuitively makes sense as piping to Tank 3 is smaller (2.5’’) but Tank 1, with 4’’ and 3’’ piping, is about 100’ higher than Tank 3. However the flow seems low (see below).
t1 and t3 are times required to fill up Tanks 1 and 3, respectively. At some point one tank will be full and the natural distribution model is no longer valid as one leg of the pipe has to be closed or one tank will have to remain only partially filled.
Case 2 assumes that the 3’’ stretch of piping to in reality 4’’. Now the total flow goes to 186 GPM, 107% of design.
Case 3 assumes the same piping geometry as Case 1 but Tank 1 at the same lower elevation as Tank 3. Now the total flow is 191 GPM.
Case 4 assumes both tank at lower elevation, Tank 3 50’ below Tank 1. Now the total flow is 200 GPM.
Case 5 assumes the original geometry, total flow of 218 GPM. The booster pump discharge pressure would have to be 346.3 psia (331.6 psig). Case 6 assumes 200 GPM. The discharge pressure would have to be 331.6 psia (316.9 psig).
Typical accuracy of such meters is 1% of the range, in this case 1% of about 900 psi or 9 psi.
Let’s analyze the pump performance.
Table 2
Table 2 shows power calculation at observed flow of 218 GPM and discharge pressure of 280 psia. The HP is 47, almost exactly the same as measured. The assumed efficiency, however, is on the high side (85%). If the efficiency was a more typical 70-80% the flow or discharge pressure would have to be lower.
Considering Case 1, the difference between 146 GPM and 218 GPM is 72 GPM => 2952 gal would accumulate in 41 min (no accumulation observed).
Are the pipes actually larger than assumed (like in hypothetical Case 2)? Is the 3’’ stretch of pipe to Tank 3 shorter than assumed? Is the discharge pressure higher (pressure gauge incorrect)? Is the flow meter incorrect? Are one or both tanks at elevation lower than measured by GPS (Cases 3 and 4)?
Two pieces of evidence (no accumulation, higher measured power) indicate that the flow rate is higher than predicted by the model (Case 1). If the piping information is correct this would indicate that at least one tank is at an elevation lower than measured (Cases 3 and 4).
To summarize:
With 280-psi discharge P and flow of 200-218 GPM the water would not be physically able to reach any one of the two tanks unless:
1) The discharge P is greater than 280 psig (the gauge is incorrect)
and/or
2) At least some of the piping is larger bore
and/or
3) The flow is lower than 200-218 GPM (contradicted by power readings and lack of accumulation)
and/or
4) At least one of the tanks is at a lower elevation than predicted by GPS.
and/or
5) The ASPEN model assumptions, such as the PVC pining absolute roughness, were off. The roughness was obtained from multiple sources and assumed to be on the low side of the range.
It could be a combination of 2 or more factors. For example, the gauge may be off, by, say, 20 psi (2%) not 37 psi, and the tanks may be located a slightly lower elevation, and the pipe roughness may be a little lower. More work is required.
Let’s assume that (1) is true. Now we can reconstruct the system curve and look at the potential benefit of having a VFD (variable frequency drive or a variable speed drive).
Fig. 3
Fig. 3 shows the normal operating curve (upper black curve), the system curve (red curve) and the new operating curve with a VFD (lower black curve). When the flow (w) is reduced, proportional to speed, the new operating point (op’) lowers frictional head loss (DHfr).
Table 3
Table 3 shows Case 6 one more time. If we reduce the flow by 25% (Case 7) the pump discharge pressure will reduce to 297 psia. If we reduce it by 50% (Case 8) the pump discharge pressure will reduce to 271 psia. Cases 9 and 10 further reduce the flow. At 50 GPM (Case 10) 97% of the flow goes to Tank 3, the one at lower elevation. The discharge pressure is 258 psia which corresponds to 595 feet of head, a little over the static head to Tank 1.This effectively reproduces the system curve.
The calculations has been made to obtain work (KWhrs) required to fill both tanks. The number is fictitious as once one tank is full it cannot be filled any further. However, it is useful for comparison purposes. Once can save 11% of energy by reducing the flow by 25% and 19% by reducing the flow by 50%, for example by using a VFD. This assumes the same pump efficiency while in reality the efficiency may change. It may even be higher at 150 GPM (the design flow is 175 GPM). However power savings don’t change much by further reducing the flow (Cases 9 and 10). This is because the frictional losses were mostly eliminated. An interesting observation is that if we reduce flow by about 25% then, with the existing system, we will be able to simultaneously fill both tanks at the same time (about 6.5 hrs) without using any balancing valves (that cost capital and power). Reducing the flow by 50% still gives similar filling time of about 10 hrs (a valve on line to Tank 3 may balance the flow).
Fig. 4
Shows simulated system curve (assuming underestimated pressure reading).
Table 4
Table 4 shows simplified economic analysis of installing a VFD. We may or may not have to install a new motor. If we run the pump at night (colder ambient) we may be able to use the old motor without overheating. The new motor cost is about $100/HP. The VFD cost is $50/HP to $240/HP. The power NPV is assumed to be $2000/KW. The VFD loss is 3-5% (assumed 3%). If we get a dirt-cheap VFD, and reduce the flow by 25%, we only save only by keeping the old motor. If we reduce flow by 50% we save with and without a new motor. If we assume an average-priced VFD we only save by reducing the flow by 50% and keeping the old motor. Again, this assumes constant efficiency which may not be the case. There is some VFD cost where, even including the motor cost, the cost of retrofit breaks even with the power savings.
Recommendation:
Reduce flow to about 75-100 GPM (by 50%) while saving 19-22% of power (16% assuming 3% VFD loss) and filling the two tanks for about 10 hrs. It may or may not require a balancing valve on line to Tank 3.
Another way to reduce the flow is to change pump’s wheel, change its geometry, or buy a new pump. The old pump can be sold or traded in.