Pump Efficiency, Head, Flow, and related technical info

[chem geek]

Mark,

I also made a spreadsheet which is how I came up with the formula, and I carefully measured the graphs with a ruler to get reasonable accurate numbers. My data, from both the full IntelliFlow and the IntelliFlow 4x160 graphs, is as follows:
(I'm continuing to edit this to straighten out the columns)

GPM  Head  Out. Power  In. Power  Eff.  (RPM/350)^2-(GPM^2)/470
     (ft.)                                           (RPM/322)^3+80
3450 RPM
  0   96      0          1310     0.0%  97.180164    1310.277154
 20   94    353.7502     1620    21.8%  96.32910017
 40   93    699.9738     1930    36.3%  93.77590868
 60   89   1004.8011     2210    45.5%  89.52058953
 80   84   1264.4688     2510    50.4%  83.56314272
100   77   1448.8705     2780    52.1%  75.90356826
120   67   1512.8466     2980    50.8%  66.54186613
140   56   1475.2136     3010    49.0%  55.47803634

3110 RPM
  0   80      0                         78.95591837  980.9766836
 20   79    297.3007                    78.10485454
 40   76    572.0216                    75.55166305
 60   71    801.5829                    71.2963439
 80   65    978.458                     65.33889709
100   57   1072.5405                    57.67932262
120   48   1083.8304                    48.3176205
						
2350 RPM
  0   46      0                         45.08163265  468.7191816
 20   45    169.3485                    44.23056882
 40   42    316.1172                    41.67737733
 60   38    429.0162                    37.42205818
 80   31    466.6492                    31.46461138
				
2070 RPM
  0   35      0           330     0.0%  34.97877551  345.6705539
 20   34    127.9522      450    28.4%  34.12771168
 40   32    240.8512      550    43.8%  31.57452019
 60   28    316.1172      630    50.2%  27.31920104
 80   22    331.1704      690    48.0%  21.36175423

1500 RPM
  0   19      0                         18.36734694  181.0895264
 20   18     67.7394                    17.51628311
 40   15    112.899                    14.96309162
 60   11    124.1889                    10.70777247

950 RPM  chart said 750, but was probably wrong
  0   7       0                         7.367346939  105.6804838
 20   6      22.5798                    6.516283109

690 RPM
  0   4       0           95     0.0%  3.886530612   89.83965015
 20   3      11.2899      95    11.9%  3.035466782

I do not get the same data you do in your table for actual GPM. For example, at 3450 RPM and a head of 80 feet, the graph looks like this is 90 GPM, not the 97 "Actual GPM" shown in your table. I understand the pump affinity equations that are valid in regions of equivalent efficiency. I just was curiously looking for a general solution. I believe what I said before to be true, that the formula I came up with is within 1-2 feet of head in accuracy which probably means that it is accurately reflecting what is going on -- a static head based on RPM^2 minus a dynamic head based on GPM^2. The "static head" is really a balanced dynamic head between the motor pushing water one way and it leaking at the same rate the other way. The GPM^2 term appears to just be the dynamic head of the net amount of flow through the pump so while the RPM^2 "push" amount remains constant (since we are looking at a curve at constant RPM), the "leakage" amount decreases as the head decreases so that a net GPM increases. Or another way of looking at it is that the pump RPM is pushing water at a GPM determined from (RPM/350)^2-(GPM^2)/470 = 0; GPM = sqrt(470*(RPM/350)^2) = RPM/16.1 so at 3450 this is 214 GPM. So the pump's RPM (and impeller size) cause 214 GPM to move when there is no head. Any resistance (head) results in a backward "leakage" which causes the net GPM to drop (since this backward movement meets a lot of frictional resistance), but 214 GPM is always being moved "forward" by the pump even when the net GPM is 0 (meaning that the "leakage" is also 214 GPM so no net movement is seen). At least this is my best understanding of how this works. I don't have the pump curves (for this pump) going down towards 0 head, but I have seen other pumps' curves that do appear to have the parabolic shape (see this link, for example).

Regardless of which formula is used, I also wanted to estimate the electrical efficiency and that has been harder to determine. Unfortunately, only 3 RPMs showed electrical power curves on the graph so my best-guess formula isn't validated very well, but it does appear to roughly be a combination of some small fixed losses (resistance, probably) plus an amount of "work" that is related to the cube of the RPM. This makes sense since the velocity of water is proportional to the RPM as is the GPM (as shown with the formula in the previous paragraph) while the frictional losses (i.e. head) are roughly proportional to the GPM^2 so the power is expected to be the product of the flow rate times head so proportional to GPM^3 and since GPM and RPM are directly related, this is proportional to RPM^3. What is harder to figure out is how the electrical power doesn't track the output power or the amount of head as might be expected. It instead seems to have a linear increase with net GPM and then slows down its increase as it approaches around 50% efficiency. I would presume that eventually as output power dropped at lower GPM, that the electrical power would start to drop as well. The electrical power should be directly related to the pressure resistence against the impellers (I'm assuming a constant RPM) which is a difference in pressure on one side of the impeller compared to the opposite side. So normally one would expect the electrical power to be proportional to head alone, but apparently the "leakage" causes some of the pressure on the opposite side of the impeller to be higher so that the net difference isn't the full amount of head (so there is less net pressure and therefore less "work"). As the head declines, the leakage also declines and apparently reduces the pressure on the opposite side of the head faster than the overall head drop so that there is a net electrical power increase. Near the point of optimal efficiency, these two effects act so that the amount of electrical power tracks the output power (near 50% efficiency) which essentially means that the "leakage" reduces the back pressure faster than the drop in overall head, but at a rate that is similar to the output power. I suspect that in the region at lower head and high GPM that the electrical power drops in rough proportion to the head since the "leakage" effect would be much smaller at that point. So, at very low head, the pump curve should be near vertical instead of the parabolic formula that I came up with (assuming that the leakage becomes minimal at low head). It seems like I'm missing something here, at least for figuring electrical power -- either that or it truly is as complicated as it sounds and not easily computed.

Richard