Does a pump have a maximum flow rate for a given system?

AI Thread Summary
A pump, particularly a centrifugal type, has a maximum flow rate determined by the system it is connected to, where increasing power beyond a certain point does not yield a proportional increase in flow rate due to rising back pressure. As flow rate increases, the resistance in the system also increases, requiring more power to maintain flow. The relationship between power, flow rate, and pressure is defined by specific equations, indicating that flow is proportional to pump speed and pressure is related to flow rate squared. In practical terms, a pump can reach a point where, despite increased power input, the flow rate does not increase linearly due to system constraints. Understanding these dynamics is crucial for efficient pump operation in hydraulic systems.
TimNJ
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Hi everyone,

I am an electrical engineering student and in my free time, I make educational youtube videos about electronics. (youtube.com/SolidStateWorkshop). I'm working on a video right now about transistors. To help explain their operation, I am attempting to use a hydraulic circuit consisting of a pump and a valve. If you'd like to know more about how i plan on doing that, I can elaborate but I don't want to confuse transistors with fluid dynamics at the moment.

I would like to know if a pump (perhaps a centrifugal type) has a maximum flow rate it can produce for a given system. Is there a point at which increasing power to the pump (ie trying to increase its RPM) will NOT result in a greater flow rate? I'm assuming there is some sort of "variable-power controller" for the pump. I'm guessing that by incrementally increasing power, you will not see a linear increase in flow rate. As flow rate increases, resistance (back pressure?) increases and thus more power is required to move the water. If the power is being limited, then the flow rate is also limited. I'm probably like 900% wrong though...

I've attached a graph of what I am after. I am curious if such a relationship exists in fluid dynamics or if there are other factors that will make a hydraulic system behave very differently from an electric one.

Thanks in advance.

Tim
 

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QUOTE="TimNJ, post: 5197049, member: 488196"]I would like to know if a pump (perhaps a centrifugal type) has a maximum flow rate it can produce for a given system.[/QUOTE]In the real world, there would be flow from the outlet at a higher pressure back to the inlet at a lower pressure from the certain features of pump design, such as the clearance of the pump tip blades with its casing. So in the real world a pump has a maximum pressure that it can produce. But at maximum pressure, flow is zero so power output of the pump is also zero. That is not to say power input from the pump shaft is zero. Power input to the pump and power output of the pump are 2 different things, and determine the efficiency at which the pump is operating.

TimNJ said:
Is there a point at which increasing power to the pump (ie trying to increase its RPM) will NOT result in a greater flow rate?
As you know, or should know, an electrical motor can have its rpm increase/ decrease by change in voltage to the motor. One doesn't increase set out to increase power but rather one sets out to change one of the 2 variables that by their product is power. For an electrical system that is Power=EI. For a fluid system that is Power=QΔP, where Q is the fluid flow, P is the pressure differential across the pump.

Affinity Equation #1: Q ≅ RPM, Flow is proportional to the pump speed.

TimNJ said:
As flow rate increases, resistance (back pressure?) increases
The system that the pump is connected to determines the pressure at the outlet of the system. A pump not connected to any piping system will have an output pressure of zero. Once again, at another operating condition, pump power output will be zero, from the equation Power = PQ, where P(pressure) = 0. we can have lots of flow but still no pump power output. Power input from the shaft is not zero though. The pressure is determined by the flow of the fluid to the system, the relationship, that the "back pressure" as you say, related to the velocity of the flow

Affinity Equation #2: ΔP(pressure) ≅ Q(flow rate)2 ≅ V(elocity of the flow)2

TimNJ said:
thus more power is required to move the water
Simply stated, from Power = P(pressure)Q, Pressure is proportional to the velocity of the fluid squared and flow rate is proportional to the velocity of the fluid.
Thus we have,

Affinity Equation #3: P(ower) ≅ the cube of the flow Q.

Hopefully one can make sense of all that.
 
Try googling for sample pump curves and valve curves.
 
256bits said:
In the real world, there would be flow from the outlet at a higher pressure back to the inlet at a lower pressure from the certain features of pump design, such as the clearance of the pump tip blades with its casing. So in the real world a pump has a maximum pressure that it can produce. But at maximum pressure, flow is zero so power output of the pump is also zero. That is not to say power input from the pump shaft is zero. Power input to the pump and power output of the pump are 2 different things, and determine the efficiency at which the pump is operating.

As you know, or should know, an electrical motor can have its rpm increase/ decrease by change in voltage to the motor. One doesn't increase set out to increase power but rather one sets out to change one of the 2 variables that by their product is power. For an electrical system that is Power=EI. For a fluid system that is Power=QΔP, where Q is the fluid flow, P is the pressure differential across the pump.

Affinity Equation #1: Q ≅ RPM, Flow is proportional to the pump speed.The system that the pump is connected to determines the pressure at the outlet of the system. A pump not connected to any piping system will have an output pressure of zero. Once again, at another operating condition, pump power output will be zero, from the equation Power = PQ, where P(pressure) = 0. we can have lots of flow but still no pump power output. Power input from the shaft is not zero though. The pressure is determined by the flow of the fluid to the system, the relationship, that the "back pressure" as you say, related to the velocity of the flow

Affinity Equation #2: ΔP(pressure) ≅ Q(flow rate)2 ≅ V(elocity of the flow)2Simply stated, from Power = P(pressure)Q, Pressure is proportional to the velocity of the fluid squared and flow rate is proportional to the velocity of the fluid.
Thus we have,

Affinity Equation #3: P(ower) ≅ the cube of the flow Q.

Hopefully one can make sense of all that.

Absolutely wonderful response. Thank you so much. I understand that power is a product of two quantities. I'm assuming that the magical power regulator box has a method of varying power. In simplest terms, I take it that pressure is determined by the system the pump is attached to (valves, heat exchanges, pipes). Increasing flow rate subsequently increases pressure. If the pump was unregulated, it would likely consume more and more current (because it needs more power). But if power is the independent variable, then, the flow rate follows a cube root relationship where increasing power by a set amount has less and less effect on the flow rate. Hope that's about right!

Thanks again.
 
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