In my work we design a system in power plant . It has a pump and multi-branches, now we use a software to calculate the exact head and flow rate of pump. The system in the start up will need pressure of 6 bars and flow rate of 86 m3/h and in normal operation it need a pressure of 3 bars and flow rate of 86 m3/h . Now we will just buy one pump with one speed so one curve only ( more expert engineers told me that), that covers the p=6bars and flow rate of 86 ,,now when the pump works in the normal operation it will gives p=3bars but much higher flow rate than we needed , so in one branch w put a control valve , they told me to measure the pressure drop in the control valve that needs to create more resistance on the pump so the pump head increases and flow rate dcreases again to 86 ,,,,, the control valve will be installed in one branch which needs flow rate of 60 m3/h ,,,, so my question is does installing the cv and the pressure drop that is caused will affects the flow rate or velocity in that branch ? in my case when we install the cv the pump will produce flowrate of 86 and 60 of them will pass throught the cv ,, so does these flowrate of 60 will decrease further as cv cause throttling or what ?
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control valves
- Thread starter anber
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Hi anber,
Could you explain how does the system demand changes from start-up to normal operation?
I understand that you have a pump feeding a manifold and there are many pipes branching out. Do you know the system curve of each branch? If it is more than 3 branches, you could alleviate some iterations using your software (is it Fathom or something similar?) including a Flow Control Valve and finding its K-factor for the desired flow.
Depending on your configuration, when you increase the resistance on one branch the pressure in the manifold increases and the flow in the other branches tends to increase How much? You'll have to find again the combined system curve and intersect with your pump-curve.
To your question, if you adjust the CV so that the pressure in the pump is 6bar, only then the flow will be of 86 m3/h and it will be split inversely proportional to the resistance of each branch.
If you throttle down a branch, the system curve will go up, the pressure will increase, the flow reduces, and the new flow will be again split inversely proportional to the resistance of each branch.
Hope it helps.
Could you explain how does the system demand changes from start-up to normal operation?
I understand that you have a pump feeding a manifold and there are many pipes branching out. Do you know the system curve of each branch? If it is more than 3 branches, you could alleviate some iterations using your software (is it Fathom or something similar?) including a Flow Control Valve and finding its K-factor for the desired flow.
Depending on your configuration, when you increase the resistance on one branch the pressure in the manifold increases and the flow in the other branches tends to increase How much? You'll have to find again the combined system curve and intersect with your pump-curve.
To your question, if you adjust the CV so that the pressure in the pump is 6bar, only then the flow will be of 86 m3/h and it will be split inversely proportional to the resistance of each branch.
If you throttle down a branch, the system curve will go up, the pressure will increase, the flow reduces, and the new flow will be again split inversely proportional to the resistance of each branch.
Hope it helps.
- Thread starter
- #3
hi, eloyRD
ok after the system curve goes up and the pressure increases and the flow reduces and the flow will split in each branch and the branch that contains the cv will have flowrate of 60 ,, so does this flowrate will decreases as it passes through the cv or will remain the same ? i am using a software called pipenet but i need to know what will happen in real life
ok after the system curve goes up and the pressure increases and the flow reduces and the flow will split in each branch and the branch that contains the cv will have flowrate of 60 ,, so does this flowrate will decreases as it passes through the cv or will remain the same ? i am using a software called pipenet but i need to know what will happen in real life
Sorry, now I think that I misunderstood your doubt.
Every time you change a parameter (like throttling up or down a valve) you'll be in transient state until you reach steady state (that is the situation that is calculated by EPANET).
The system will stay there until a parameter change. If you have a variable discharge pressure or something like that, the flow will keep changing. If that is your situation and your require constant flow, you will require a flow control valve.
Every time you change a parameter (like throttling up or down a valve) you'll be in transient state until you reach steady state (that is the situation that is calculated by EPANET).
The system will stay there until a parameter change. If you have a variable discharge pressure or something like that, the flow will keep changing. If that is your situation and your require constant flow, you will require a flow control valve.
- Thread starter
- #5
mmmm i still don't get it ,,if the cv is throttled to a certain value , so the system resistance increases , the pump head increases and the flow rate decreases . Now assume that at this certain valve opening , the pump produces p=6 bars and flow rate of 80 m3/h,,, now this flow rate is divided into branches according to the resistance in each branch ,, now there is a branch that contains the above cv that we adjust it to a certain value ( cv we mentioned above) and this branch will have flow rate of 60 m3/h ,, does this flow rate will reamain constant in that branch downstream of the cv ( we now have a steady state and we didn't change the valve opening and the head and flow rate from the pump is constant )
So that means that the valve affects upstream and downstream flow in an equal manner. That is because it affects flow, regardless of if the flow is an upstream or a downstream flow, because as you have already found out, upstream flow must be the same as downstream flow. The valve only affects the energy needed to drive that flow. A nearly closed valve has high resistance and will consume lots of the available energy as flow passes, therefore the flow that passes must be relatively low.
Find what you like to do, earn a living at it, and then make your lifestyle fit your income. — Chuck Yeager
Find what you like to do, earn a living at it, and then make your lifestyle fit your income. — Chuck Yeager
- Thread starter
- #8
Head is not distance, it is potential energy. Pumps provide potential energy to a fluid which can then be used to drive flow.
Flow uses energy to overcome friction created by the fluid as it moves against the pipe wall and around obstacles in the system. Energy is consumed when accelerating a fluid's velocity. Energy is again consumed when raising a fluid to a higher elevation.
The energy needed to raise a certain amount of fluid 50 meters higher, is very conveniently measured in head as simply 50 meters of head. That is exactly what a pump that has a 50 meter head capacity can do. A pump that has a 100 m head capacity at a flowrate of 2 m3/s can raise 2 m3 of fluid by an elevation of 100 meters every second.
Please review the basics of fluid mechanics and flow, particularly as expressed by Bernoulli. This addresses energy conservation, loss, gain and conversion of energy to head (elevation), velocity and pressure of static and flowing fluids.
Find what you like to do, earn a living at it, and then make your lifestyle fit your income. — Chuck Yeager
Flow uses energy to overcome friction created by the fluid as it moves against the pipe wall and around obstacles in the system. Energy is consumed when accelerating a fluid's velocity. Energy is again consumed when raising a fluid to a higher elevation.
The energy needed to raise a certain amount of fluid 50 meters higher, is very conveniently measured in head as simply 50 meters of head. That is exactly what a pump that has a 50 meter head capacity can do. A pump that has a 100 m head capacity at a flowrate of 2 m3/s can raise 2 m3 of fluid by an elevation of 100 meters every second.
Please review the basics of fluid mechanics and flow, particularly as expressed by Bernoulli. This addresses energy conservation, loss, gain and conversion of energy to head (elevation), velocity and pressure of static and flowing fluids.
Find what you like to do, earn a living at it, and then make your lifestyle fit your income. — Chuck Yeager
LittleInch
Petroleum
anbar,
You appear to trying to model a complex network system without understanding fluid dynamics. This will not end well and posting random questions on an internet forum might end up confusing you further.
You need to do some reading up and understand in your own mind how fluids work in a network.
Sometimes it is easier to understand if you think of the system as an electrical circuit with wires instead of pipes, amps instead of current, etc
In terms of this question you haven't explained how the system needs 6 bar at the start up but only 3 bar during normal operation for the same flow.
Running a 6 bar pump continuously at 3 bar is just a waste of energy.
Network analysis is complex and change in one location or branch will impact on another location. Changes in size, flows and pressures are all part of the analysis to get what you need.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
You appear to trying to model a complex network system without understanding fluid dynamics. This will not end well and posting random questions on an internet forum might end up confusing you further.
You need to do some reading up and understand in your own mind how fluids work in a network.
Sometimes it is easier to understand if you think of the system as an electrical circuit with wires instead of pipes, amps instead of current, etc
In terms of this question you haven't explained how the system needs 6 bar at the start up but only 3 bar during normal operation for the same flow.
Running a 6 bar pump continuously at 3 bar is just a waste of energy.
Network analysis is complex and change in one location or branch will impact on another location. Changes in size, flows and pressures are all part of the analysis to get what you need.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
LittleInch
Petroleum
Volts,
Joules is energy, same as fluid, variable resistors are control valves, generators or batteries are pumps / fluid supply...
It falls down at times (it is an analogy after all), but sometimes it can help people understand what's going on in a system / network
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Joules is energy, same as fluid, variable resistors are control valves, generators or batteries are pumps / fluid supply...
It falls down at times (it is an analogy after all), but sometimes it can help people understand what's going on in a system / network
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
No. Volts corresponds to fluid Pressure.
I would always try to think in terms of how electricity works like water.
Fluid head is energy as in ft-lbs/unit mass of fluid = ft-lbs/lb = ft, so has to be electrically represented by Joules = NM/some kind of unit mass of electron.
The difficulties in the analogy come when trying on one hand to represent fluid pressure head (pgh) by Volts and then trying to make that work for energy in Joules, when it should actually (I think) be Joules/unit mass. Just as head can be converted to pressure, pgh or to velocity, v^2/2/g
Find what you like to do, earn a living at it, and then make your lifestyle fit your income. — Chuck Yeager
I would always try to think in terms of how electricity works like water.
Fluid head is energy as in ft-lbs/unit mass of fluid = ft-lbs/lb = ft, so has to be electrically represented by Joules = NM/some kind of unit mass of electron.
The difficulties in the analogy come when trying on one hand to represent fluid pressure head (pgh) by Volts and then trying to make that work for energy in Joules, when it should actually (I think) be Joules/unit mass. Just as head can be converted to pressure, pgh or to velocity, v^2/2/g
Find what you like to do, earn a living at it, and then make your lifestyle fit your income. — Chuck Yeager
Maybe I can simplify the problem, if I understand it correctly. There is one constant speed pump, and the vendor has provided a Q ( gpm) vs H (ft ) curve.
1) I would first convert the gpm vs ft curve to a W(lb/hr) vs psi curve, knowing the pump's suction pressure and fluid density.
2) for each selected process flow, I would use this curve to obtain the pump discharge pressure
3) for each of these process flows, I would also tabulate the requuired or known pressure downstream of the the valve
4) the required valve Cv ( simplified) would be from the relation of W(lb/hr)= 63.4*Cv*SQRT( DP/sv), where sv =specific volume, assuming no flashing or cavitation occurring. DP= (Pi-Po)
Not so tough. With luck the range of required Cv does not exceed a 15:1 ratio, if it does ,then it may be necessary to use 2 valves ( 20+80) in parallel.
"...when logic, and proportion, have fallen, sloppy dead..." Grace Slick
1) I would first convert the gpm vs ft curve to a W(lb/hr) vs psi curve, knowing the pump's suction pressure and fluid density.
2) for each selected process flow, I would use this curve to obtain the pump discharge pressure
3) for each of these process flows, I would also tabulate the requuired or known pressure downstream of the the valve
4) the required valve Cv ( simplified) would be from the relation of W(lb/hr)= 63.4*Cv*SQRT( DP/sv), where sv =specific volume, assuming no flashing or cavitation occurring. DP= (Pi-Po)
Not so tough. With luck the range of required Cv does not exceed a 15:1 ratio, if it does ,then it may be necessary to use 2 valves ( 20+80) in parallel.
"...when logic, and proportion, have fallen, sloppy dead..." Grace Slick
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