If we suppose the flow at the outlet of a water tank to be the discharge of a pump,and if we connect a pump to this outlet it is as if we have two pumps in series.In such arrangement we don't expect the flow rate to be increased specially if the system curve is near to flat.But in practice the flow rate is increased, and to some extent the more the pump size we have more flow rate. what is the reason (or what is wrong with my antecedents)?
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gravity flow as pumping 2
- Thread starter yazguli
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- #3
I mean a water storage tank with a definite height or installed at a definite height and an outlet say a 2" or 4" pipe at it's bottom from which water can flow by gravity at a certain flow rate.
The question is to increase this flow rate by means of a pump.
The question is to increase this flow rate by means of a pump.
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- #4
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As the tank drains, the static pressure available to the pump will decrease unless the tank is being refilled. So the flow rate from the pump will decrease accordingly as the pump follows its curve.
The tank must either be of a size that it will provide the minium duration (30, 60, 90+ minutes) of the system or a reliable source of refilling it must be provided.
The tank must either be of a size that it will provide the minium duration (30, 60, 90+ minutes) of the system or a reliable source of refilling it must be provided.
Sdpaddler50
Mechanical
You don't increase the flow rate by adding a pump. As others have noted you are increasing the pressure only. The booster pump adds energy, not flow.
Sdpaddler50
Mechanical
I do not agree. Conduct a flow test of a city hydrant (static, residual and flow) and then plot the results. If you then add a booster pump to that same city supply, and do the same flow test you will have increased the pressure, not the flow. A booster pump can not magically add flow. It increases the energy of the water only.
- Thread starter
- #11
Yazguli, you are correct about parallel/series pumps.
Sdpaddler50, I think we are saying the same thing, just looking at it differently. You are correct that the pump does not magically add flow. The energy added to a system by a pump is a mix of both potential (pressure) and dynamic (velocity or flow).
I can pump into a bladder tank and store the potential energy. This is commonly done on domestic water systems.
If I need a higher pressure on a fire system, I will add a pump. The only limit is the pressure rating of the pipe/fittings, etc.
If I need a higher flow on a fire system, I will add a pump. The only limit is that i do not create such "negative pressure" on the suction inlet of the pump to produce cavitation.
Typically, this flow limit is indicated by extending the flow curve to where it intersects 20 psig. If the flow curve without the pump cannot achieve a certain flow rate, adding a pump will not achieve that flow rate because of cavitation. But if it can, adding the pump will achieve the flow rate, not because I am adding "flow", but because I am adding enough energy into the system to achieve the flow/pressure I desire.
Sdpaddler50, I think we are saying the same thing, just looking at it differently. You are correct that the pump does not magically add flow. The energy added to a system by a pump is a mix of both potential (pressure) and dynamic (velocity or flow).
I can pump into a bladder tank and store the potential energy. This is commonly done on domestic water systems.
If I need a higher pressure on a fire system, I will add a pump. The only limit is the pressure rating of the pipe/fittings, etc.
If I need a higher flow on a fire system, I will add a pump. The only limit is that i do not create such "negative pressure" on the suction inlet of the pump to produce cavitation.
Typically, this flow limit is indicated by extending the flow curve to where it intersects 20 psig. If the flow curve without the pump cannot achieve a certain flow rate, adding a pump will not achieve that flow rate because of cavitation. But if it can, adding the pump will achieve the flow rate, not because I am adding "flow", but because I am adding enough energy into the system to achieve the flow/pressure I desire.
stookeyfpe
Specifier/Regulator
Q1 = Q2. Spaddler50 is accurately stating the function of a pump. A pump cannot create mass - it is only designed to move it at a rated flow rate and pressure.
if you do find a pump that creates mass, let me know. I know a few desert dwellers that would interested in purchasing the product.
if you do find a pump that creates mass, let me know. I know a few desert dwellers that would interested in purchasing the product.
I agree that a pump doesn't add mass - never said it did.
But when talking about flow, you have to look at dQ/dt.
If I have an atmospheric tank with a drain pipe on the bottom and valve and I open the valve, it will start to empty and have a certain flow rate all dependant on the height of the liquid in the tank, pipe diameter, material, etc. It will take a certain period of time for the tank to empty. Say the tank is 1000 gallons and it takes 1000 minutes to empty. Flow rate is 1 gallon per minute
If I put a pump on the pipe and open the valve and turn on the pump, the pressure in the piping will be greater (depending on the pump curve).
But the flow will be greater as well because it will take less time for the tank to empty. Same 1000 gallons, but with the pump, it only takes 10 minutes to empty. Flow rate is 100 gallons per minute.
But when talking about flow, you have to look at dQ/dt.
If I have an atmospheric tank with a drain pipe on the bottom and valve and I open the valve, it will start to empty and have a certain flow rate all dependant on the height of the liquid in the tank, pipe diameter, material, etc. It will take a certain period of time for the tank to empty. Say the tank is 1000 gallons and it takes 1000 minutes to empty. Flow rate is 1 gallon per minute
If I put a pump on the pipe and open the valve and turn on the pump, the pressure in the piping will be greater (depending on the pump curve).
But the flow will be greater as well because it will take less time for the tank to empty. Same 1000 gallons, but with the pump, it only takes 10 minutes to empty. Flow rate is 100 gallons per minute.
Sdpaddler50
Mechanical
Check out the attached. I have a simple graph showing a public supply, a booster pump, and the combined ; public plus booster supply. Take any point. For our purposes, lets look at 750 gpm. With just the public supply, we only get 53 psi from the public main at 750 gpm. The booster pump now gives us about 96 psi at the 750 gpm flow rate. Bottom line - flow rate has not increased. We have only increased the pressure (energy) of the water. Same principle as voltage and current in electrical systems.
If a contractor runs calculations and finds they are deficient for a particular density, one thing they should check is the available water supply. IE, what is the the available flow at that location. Then, they can determine if the water supply can support the booster pump. Sometimes it cant.
Adding a separate pump and tank, or additional lead in to the building from a public main, that is different. Now, we are increasing non only pressure, but the flow as well.
If a contractor runs calculations and finds they are deficient for a particular density, one thing they should check is the available water supply. IE, what is the the available flow at that location. Then, they can determine if the water supply can support the booster pump. Sometimes it cant.
Adding a separate pump and tank, or additional lead in to the building from a public main, that is different. Now, we are increasing non only pressure, but the flow as well.
- Thread starter
- #16
The fluid (water) flows in a system of pipes and fittings, though a short pipe and a few fittings .As you have friction losses the system curve is parabolic.If in Sdpaddler50's graph you plot such a curve, the intersection point of system curve and the combined curve will indicate that the pressure is less than what we expected and the flow is more.
Sdpaddler50
Mechanical
Q (flow) is not increased by a booster pump - period. I think I have adequately explained, and illustrated that with a mathematical graphical representation which I previously attached. But, I surrender. No more posts from on this subject.
Spadler50,
I ask your indulgence for one additional point - I think we are saying the same things, just focusing on different aspects.
According to Bernoulli, pressure, static, velocity(flow rate) as well as other things are all interrelated and the sum equals a constant value. If I hold velocity constant, then either the pressure or the static have to increase. If i hold pressure and static constant, then velocity (flow) has to increase. If i hold static constant (our case), then pressure and velocity will change accordingly.
I have taken your graph and marked it up to indicate my point. I drew a continuation of the curve and made some observations.
The magenta line indicates holding velocity (flow) constant and it shows there would be increased pressure.
The green line indicates holding pressure constant and it shows there would be increased flow.
With a pump, I get increased pressure for a certain flow or I get increased flow for a certain pressure.
The curve will continue until it reaches zero pressure (which may not actually occur).
At some flow rate, the suction of the pump will exceed the vapor pressure of the water and the pump will cavitate.
Again, I think we are saying the same thing, just interpreting the results differently.
I ask your indulgence for one additional point - I think we are saying the same things, just focusing on different aspects.
According to Bernoulli, pressure, static, velocity(flow rate) as well as other things are all interrelated and the sum equals a constant value. If I hold velocity constant, then either the pressure or the static have to increase. If i hold pressure and static constant, then velocity (flow) has to increase. If i hold static constant (our case), then pressure and velocity will change accordingly.
I have taken your graph and marked it up to indicate my point. I drew a continuation of the curve and made some observations.
The magenta line indicates holding velocity (flow) constant and it shows there would be increased pressure.
The green line indicates holding pressure constant and it shows there would be increased flow.
With a pump, I get increased pressure for a certain flow or I get increased flow for a certain pressure.
The curve will continue until it reaches zero pressure (which may not actually occur).
At some flow rate, the suction of the pump will exceed the vapor pressure of the water and the pump will cavitate.
Again, I think we are saying the same thing, just interpreting the results differently.
- Thread starter
- #19
PEDARRIN2,
Sdpaddler50 is not against you when you are speaking about "the more the flow,the less the pressure and vice versa." It is true about any pump and pump curve .
But look at his graph.At any given point on the combined curve the pressure is sum of the pressures, but the flow is equal to the flow of any of them.
I think it is hard to know how the pumps work if we don't draw the system curve.
Look at my attachment.If the system curve is something like what I have added to Sdpaddler50's graph , the intersection points of this curve with his three curves,will indicate that when pump and hydrant are combined the flow is more than that of the hydrant or the pump and the pressure is less than the sum of their pressures.
But if the system curve is nearly flat, e.g the there is a negligible friction loss ,we will not observe such a phenomenon , just as Sdpaddler50 remarks.
What looks contradictory to me is that even when there is a simple system e.g a tank outlet+a pump,and the system curve may be considered as a flat curve,we have increased flow, and this is what Sdpaddler50 is against to according to his graph, which he says is based on experience.
Sdpaddler50 is not against you when you are speaking about "the more the flow,the less the pressure and vice versa." It is true about any pump and pump curve .
But look at his graph.At any given point on the combined curve the pressure is sum of the pressures, but the flow is equal to the flow of any of them.
I think it is hard to know how the pumps work if we don't draw the system curve.
Look at my attachment.If the system curve is something like what I have added to Sdpaddler50's graph , the intersection points of this curve with his three curves,will indicate that when pump and hydrant are combined the flow is more than that of the hydrant or the pump and the pressure is less than the sum of their pressures.
But if the system curve is nearly flat, e.g the there is a negligible friction loss ,we will not observe such a phenomenon , just as Sdpaddler50 remarks.
What looks contradictory to me is that even when there is a simple system e.g a tank outlet+a pump,and the system curve may be considered as a flat curve,we have increased flow, and this is what Sdpaddler50 is against to according to his graph, which he says is based on experience.
yazguli,
A flat system curve would not be realistic. It is my understanding that a system curve represents the various static/friction losses for the piping/valves/accessories. It starts at zero flow, which would represent the elevation difference of the beginning and end points. It progresses up due to increased friction losses for greater flow rates. if there is minimal friction, the curve would be relatively flat, but would trend upward. Since the pump curve starts at a higher pressure (at churn) and trends downward, eventually they have to intersect, which give the design point of the pump (flow and TDH).
The public curve would be what the city main would provide. The intersection of the two curves indicates what the flow in the sytem would actually be (6 flow units).
If the pump is not connected to the public main, then the flow in the system would be about 5.8 flow units.
With the Pump + Public curve (the public would provide suction pressure to the pump), the flow rate of the system would be ~7.1 units.
So without the pump, you only have 6 flow units occuring in the system, but with the pump, you have 7.1 flow units. The system is operating at a higher flow and pressure with the pump than without it.
So, in adding energy to the sytem the pump adds both flow and pressure, which I thought was what I was trying to state.
Am I still missing something?
A flat system curve would not be realistic. It is my understanding that a system curve represents the various static/friction losses for the piping/valves/accessories. It starts at zero flow, which would represent the elevation difference of the beginning and end points. It progresses up due to increased friction losses for greater flow rates. if there is minimal friction, the curve would be relatively flat, but would trend upward. Since the pump curve starts at a higher pressure (at churn) and trends downward, eventually they have to intersect, which give the design point of the pump (flow and TDH).
The public curve would be what the city main would provide. The intersection of the two curves indicates what the flow in the sytem would actually be (6 flow units).
If the pump is not connected to the public main, then the flow in the system would be about 5.8 flow units.
With the Pump + Public curve (the public would provide suction pressure to the pump), the flow rate of the system would be ~7.1 units.
So without the pump, you only have 6 flow units occuring in the system, but with the pump, you have 7.1 flow units. The system is operating at a higher flow and pressure with the pump than without it.
So, in adding energy to the sytem the pump adds both flow and pressure, which I thought was what I was trying to state.
Am I still missing something?
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