ARTICLE #13: PARALLEL OPERATION OF CENTRIFUGAL PUMPS and the Importance of Curve Stability
Some time ago, one of our readers asked a question about a parallel operation of pumps. This prompted a discussion which was posted under Q&A Section, Question #25. Later on, another reader inquired about the unstable curves and their effect on pumps parallel operation, which resulted in another Pump Magazine Article (#8). This generated interest from our friends at Pump-Flo company, who asked our permission to reprint the article on their web site. Pump-Flo is a maker of pump hydraulic selection curves program, for the pumps made by several pump manufacturers. Their readers also liked the topic on parallel operation, and more discussion followed, which we posted in Q&A Section, under Question #48.
We have been receiving more questions and comments from our readers, and the latest one (which we show below) convinced us of a need to respond with a formal article (#13), which is what you are reading.
A chemical pump supplier engineer was asked why two of his pumps in parallel appeared to be unstable at certain points on the curve, notwithstanding that the curve was an inherently stable curve with continuous rising gradient to shut off. Both pumps are identical, same diameter impeller, end suction type.
This instability was experienced in a mining plant on a cooling tower system with acidic water at 50 deg C. The instability could be corrected by altering the settings of the discharge gate valve, i.e. changing the head and flow on the pump. The suction head was more than adequate at 10 psig. The NPSHr was less than 0 psig.
His explanation was linked to the suction manifold, which directed liquid in a straight feed to one pump [A] and liquid via 2 x 90 degree elbows to the second pump [B]. There was a very short straight portion of pipe between the 2nd elbow and the suction flange on pump B, approximately 3-4 pipe diameters [pipe dia 10"]
This arrangement it was argued, caused the suction head on the 2nd pump to fluctuate rapidly, both increasing and decreasing the suction head to pump B beyond the smooth flow suction head experienced by pump A. This fluctuation or vortexing could be substantial enough to cause the 2 pumps to fight themselves and result in the instability experienced.
While the argument is plausible and credible, is this a recognized phenomenon in the industry that can and has occurred on a number of other occasions? The solution, if this is the cause of the instability, would be of course to change the manifold to more smoothly direct flow to the 2nd pump, eliminating the 90 degree elbows.
National Process Equipment,
Answer: Dear Hugh,
As we understand it, this is a setup you have:
Note: elbows EL1 and EL2 are shown within the same plane for sketch simplicity, but in reality most likely are in different planes.
The “stable” curve the supplier engineer pointed to is probably a manufacturer curve from the catalog, which probably looks something like this:
The H-Q curve is “continuously rising”, i.e. deemed stable.
Pump Magazine Article #8 talked about unstable (“drooping”) curves:
As was explained in Article #8, unstable curves cause problem for pumps operating in parallel, which would explain Hugh’s problem. However, what is puzzling is why the two pumps, having stable curves, behave as if their curves are unstable?
However, the pumps are not identical. Clearly, the suction approach to Pump B is different from Pump A. While it is generally known that sharp turns, multiple elbows, short pipes, and other obstructions are not a good thing for pumps, and numerous Case Histories have been published to illustrate and warn the pump users not to do that (Hydraulic Institute even issued guidelines on suction approach dimensions, sump configurations, etc. – all to help avoid problems) – questions nevertheless come up each time another problematic parallel operation surfaces.
There are many reasons why “tipsy-turvey” suction could cause instability. Let’s take a look how an apparently “stable” H-Q curve, even as tested by the manufacturer, can become “unstable” when a pump ends up with a “curvey” suction. There is no other way to understand this without having to delve into the heart of the impeller hydraulics. The head generated by the impeller is the difference between the velocity angular momentum between the impeller inlet and exit:
H = [ (U x Vtheta)OD - (U x Vtheta)eye ] / g
The “g” is gravitational constant, U is peripheral rotational velocity, subscript “theta” means “peripheral projection of the absolute flow velocity vector”, and H is impeller head. If fluid velocities are taken in feet per second, and g=32.17 ft/sec2, then head H comes out in feet. There is also hydraulic efficiency involved, but we skip that for simplicity.
Hydraulic designers refer to this in terms of “velocity triangles”, which essentially is a set of fluid velocity vectors are the inlet and the exit of the impeller:
W1 V2 W2
Alpha1 = 90 degrees Alpha2 = 20 degrees
The projection of the absolute velocity “V” onto peripheral velocity “U” is called Vtheta . For most end-suction pumps, the absolute flow angle at the inlet is 90 degrees, which is also referred to as “no-prewhirl” condition. Thus the projection of the “V1” velocity onto “U1” is zero, and thus the product (U x Vtheta)eye = 0
For example, if a Pump A develops 100 feet of head at the BEP point of 1000 gpm, the impeller exit component contributes all 100 feet if there is no pre-whirl, and there is no effect of the impeller inlet – no addition and no subtraction. At lower flow, the exit velocity triangle changes (Vtheta)OD component gets larger, and thus more head is generated. At the shut-off, there is no through-flow, and the (Vtheta)OD essentially overlays directly over the vector “U2”. Thus, at zero flow (shut valve), the head is the highest, - for typical end-suction pumps, such as ANSI, for example, the “rise to shut-off” is approximately 110 - 130%. And in order for the H-Q curve to be “stable” this rise must be “continuous” – in order to avoid instabilities if operated in parallel.
Double-suction pumps, for example, are different. Their casing forms a “toilet-bowl-shaped” suction, which – just as a toilet! – creates a prewhirl:
Alpha1 = 45-60 degrees
Now the product (U x Vtheta)eye is not zero! If this product amounts to, say, 30 feet, the impeller would generate less head: 100-30 = 70 feet. However, hydraulic designers would compensate for that by altering the impeller exit geometry – width, blade angles, etc. – so that the exit portion would produce extra head, say 130 feet – and the net result would remain the same 130-30 = 100 feet.
Therefore, by changing impeller geometry, it possible to change the curve shape, including the rise to shut-off. Likewise, by changing the pre-rotation (pre-whirl), it is also possible to change the rise-to-shut-off. In fact, the change could go either way – the head at the BEP can either increase, or decrease – depending whether the pre-rotation is positive (as shown above), or negative. The shut-off head, however, does not depend on prerotation component.
When a designer accounts for a known pre-rotation in the casing (such as for double suction pumps example), he or she does it in a way to preserve the flow dynamics, and maintain smooth flow and good efficiency. The designer has not control, however, if the pre-rotation is created artificially, such as bad suction piping, multiple elbows, and so on, - including the direction of the prerotation. If the “deck gets stacked the wrong way”, the flow in the suction piping may end up with negative rotation, - and double-turning elbows are known to be nasty on flow patterns.
The resulting velocity inlet triangle may thus look like this:
Alpha1 = 140 degrees, i.e. negative Vtheta results.
This could create a negative inlet term - for example, (U x Vtheta)eye = - 40 feet (minus). Since the exit is fixed, the net head would become 100 – (-40) = +140 feet, - which is obviously greater then the shut off head, - i.e. an unstable curve got created:
As Hugh rightly noted, the solution, if this is the cause of the instability, would indeed be to change the manifold to a more conventional and appropriate inlet piping configuration. Even though there could always be other reasons involved, a good piping is not a guarantee, but a beginning to a trouble-free pump operation.
Next, let’s look at the parallel operation in more detail, starting with stable curves.
Situation 1: Single Pump
The pump must generate pressure (head) to overcome hydraulic losses in the system. These losses consist of losses in a valves, bends, elbows, heat exchanger and pipe friction. From basic hydraulics, a friction loss is proportional to the square of flow:
HLOSS = k x Q2
The summation of individual losses is equal to a total system hydraulic loss:
HLOSS = KVALVE X Q2 + KELBOW X Q2 + KHE x Q2 +… = (KVALVE + KELBOW + KHE +…) x Q2 = kSYSTEM x Q2
We were able to do the math above and add the individual hydraulic coefficient due to the fact that the flow is the same throughout the system. We could not do that so easily if the system had branches, with flows branching out all over. This is analogous to the electric circuit, where the same current flowing from one component to the next result in different voltage drops across the individual resistors. In hydraulics, the pressure drop due to fluid flow is similar to the voltage drop due to electric current.
Once the kSYSTEM is calculated, a system curve can be plotted: hLOSS
If the pump curve and a system curve are plotted together, an intersection thus determines the pump operating point, because it belongs to both curves:
Situation 2: (2) pumps in parallel, but with a common resistance element (valve)
Flows Q1 and Q2 combine, and that is what flows through the valve: Qvalve = Q1+Q2
The valve curve is similar parabola as shown in the previous example:
h = k x Qvalve2 = k x (Q1+Q2)2
Mathematically, a pump curve is a Head as a function of Flow. Or, it can be also expressed as Flow as a function of Head, i.e.:
Q1 = f(H1), and Q2 = f(H2)
In general, the pumps do not have to be identical.
Then, Qtotal = Q1+Q2 = f(H1) + f(H2)
Since Qtotal = Qvalve, the combined pump curve and a valve (i.e. system) curves can be plotted and the intersection would determine the head (pressure) where the pump/system will operate:
1 pump 2 pumps (Qtotal = Q1 x 2)
500 600 Flow, gpm
(for simplicity both pumps here were assumed to be identical)
Note an interesting thing: when only one pump was operating (point “500 gpm”), the flow was 500 gpm. But when the second pump also came on line, they together, did not produce 500x2 = 1000 gpm! Instead, only 600 gpm flows through the system! Clearly, the answer to that is in the shape of the system resistance curve. However, if the pumps were pumping against mainly a static head (such as pressurized tank) – then the flows would be additive! Why? Because the static head curve is not a parabola, but a straight line, independent of flow, and thus parallel to the flow axis:
100 psi (system)
500 1000 Flow, gpm
500 gpm 1000 gpm
Note: in general, a combination of friction as well as static heads may be present, although usually either one or the other scenario is more predominant.
Situation 3: (2) pumps in parallel, but NOT with common resistance
Now, that is getting interesting. In the previous example, the change of the valve setting (changing valve “k” coefficient) affects both pumps the same way. But what if we have a system like this:
Common valve, V0
Let’s assume, for simplicity, two identical pumps, and the valves are identical also. Assume also that each valve is throttled the same percentage (i.e. having the same coefficient “k”):
Head,ft PUMP CURVE hsys
(Hso= 120 ft)
HBEP = 100 ft VALVE CURVE
250 500 (BEP)
Digitize several points form the valve curve into a Table below to make it easier to work with, as we go:
We know the following:
Assume there are no other losses except for the (2) valves. In other words, all other losses have been combined into the (2) “equivalent” valves V0 and V2
Valve V0 sees the combined flows: from Pump 1 plus Pump 2
The head generated by the Pump 1 is equal to the pressure drop (head loss) across the Valve V0, because of the pressure continuity – the nodal point at the exit of the Pump 1 is the same as the entry point into the Valve V0
Valve V2 sees flow from Pump 2 only
System must be in balance, to satisfy individual pump H-Q curves, and individual valve h-Q curves.
The solution is iterative:
a) Guess at the flow from Pump 1, Q1=250 gpm.
b) Per H-Q curve of Pump 1, head H1=118 feet
c) Head loss across valve V0 is equal to head generated by Pump 1, i.e. valve h1=H1=118 ft
d) From the valve V0 curve, Q0 = 543 (we actually “cheated” a bit here – got the equation for the parabola, h0 = Q02/2500, and then just plugged-in the numbers)
e) Total pumps flow is equal to what is flowing through valve V0, i.e. Qtotal=Q0=543
f) Flow through Pump 2 is the difference: Q2 = Q0 – Q1 = 543-250 = 293 gpm
g) From the Pump 2 curve (we used identical pumps in this case), H2 = 116 ft (approximately, as it is more difficult to derive the polynomial equation for the pumps. For the valve it is easy - parabola)
h) A pressure drop across the valve V2 is the difference between the pressures at its exit and inlet. The exit pressure is equal to what Pump 1 generated, and the inlet is what Pump 2 generated, i.e. h2 = 116-118 = -2 ft. This is impossible, i.e. our first guessed flow from Pump 1 was wrong. Let’s re-guess and go around again:
a) New guess, Q1=300 gpm
b) H1 = 115 ft
c) h0 = H1 = 115 ft
d) Q0 = 536 gpm
e) Qtotal = Q0 = 536 gpm
f) Q2 = Q0 – Q1 = 536 – 300 = 236 gpm
g) H2 = 119 ft from pump curve
h) h2 = H2 – H1 = 119-115 = 4 ft
i) From valve curve (or interpolating from the Table), Qvalve2=100 gpm
j) Q2 = Qvalve2 = 100 gpm (from the flow continuity – what flows to Valve 2 has to come out from Pump 2). This is not equal to Pump 2 flow as derived at step (f), so we need to guess again. (But we are getting closer).
a) Next guess, Q1 = 400 gpm
b) H1 = 113 ft
c) h0 = H1 = 113 ft
d) Q0 = 532 gpm
e) Qtotal = Q0 = 532 gpm
f) Q2 = Q0 – Q1 = 532-400 = 132 gpm
g) H2 = 119.5 ft from pump curve
h) h2 = H2 – H1 = 119.5-113 = 6.5 ft
i) Qvalve2 = 127 gpm
j) Q2 = Qvalve2 = 127 gpm
We now reached a reasonable convergence. Thus the final answer is:
Pump 1: flow = 400 gpm, head = 113 ft
Pump 2: flow = 132 gpm, head = 119.5 ft
The above solution has a numeric (analytical) counterpart, and could be solved by writing a set of simultaneous equations and solving them:
h2 = k2 x Q22
h0 = k0 x (Q1+Q2)2 – we could even have different valves or valve settings (different k0 and k2)
H1 = f(Q1) or Q1 = f(H1)
H2 = f(Q2) or Q2 = f(H2) – pumps also could be different
H1 = h0
h2 = H1 – H2
(6) equations and (6) unknowns: Q1, Q2, H1, H2, h1, h2
Clearly, even a simple case of just two pumps and two valves can be a rather time consuming task to solve. Clearly, for a real pumping system, with many pumps and components, a manual method is impractical. This is when computers come handy, as Ray Hardee P.E., Engineered Software, Inc, explained in the discussion for Question #48. Obviously a more refined math and methods can be employed – but let’s leave that for the programmers and mathematicians! For now, and from the practical point of view, - if we have a piping network program like that and a clear manual of how to use it – a solution of even a complex system would become an easy job.
One of such companies is Sunrise Systems, and you can get more information about them and their product via their web site at www.sunrise-sys.com
Note that we have only touched on systems with stable curves. What happens if the pump curves are not stable? Well, mathematically, we get multiple solutions. In other words, just as was explained in Article #8, a pump would be technically able to operate at two flows – for the same given head. How do programs like the ones mentioned solve this?! That, however, is another issue. In the meantime – stay away from the unstable curves, and, for that matter – from bad piping!
Keep on pumping!
Dr. Lev Nelik, P.E., Apics
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