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Density Effect on Pump Power and Load 5
- Thread starter sheiko
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[ul]
[li]It's part of the formula we use to convert pump head to pressure[/li]
[li]You'll need it to calculate the hydraulic force acting on the impeller when the centrifugal pump is operating off the best efficiency point[/li]
[li]We need specific gravity to calculate the horsepower of the motor we need to operate the pump[/li]
[li]NPSH and cavitation are directly related to specific gravity. The lower the specific gravity the lower the vaporization pressure.[/li]
[li]Low specific gravity fluids cause a number of mechanical seal problems[/li]
[/ul]
[li]It's part of the formula we use to convert pump head to pressure[/li]
[li]You'll need it to calculate the hydraulic force acting on the impeller when the centrifugal pump is operating off the best efficiency point[/li]
[li]We need specific gravity to calculate the horsepower of the motor we need to operate the pump[/li]
[li]NPSH and cavitation are directly related to specific gravity. The lower the specific gravity the lower the vaporization pressure.[/li]
[li]Low specific gravity fluids cause a number of mechanical seal problems[/li]
[/ul]
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See attached, sheiko.
Good on ya,
Goober Dave
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Good on ya,
Goober Dave
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And just to expand on micalbrch's post, sheiko is a valued member in my opinion. Lots of help to others, and a number of purple star votes...
Good on ya,
Goober Dave
Haven't see the forum policies? Do so now: Forum Policies
Good on ya,
Goober Dave
Haven't see the forum policies? Do so now: Forum Policies
- Thread starter
- #11
Sorry guys for not having been specific.
Re-reading my post I realize the questions may seem naive.
Well my questions arose from the quote below (source: Lieberman N. - A working guide to process equipment - 3rd Ed - section 30.1.1 - page 375):
"Cooling the liquid will raise the pump's discharge pressure. If the discharge control valve is now opened, the pump's discharge pressure will drop back as the flow increases. The discharge pressure drops because the pump develops less feet of head at a higher flow rate.
The specific gravity of the liquid has increased. The feet of head of liquid has decreased by the same amount. So deltaP remains constant.
How does this affect the amp load on the motor driver?
The feet of head has gone down, and the specific gravity has gone up by the same amount. Therefore, the horsepower or work per barrel of liquid pumped remains constant. But the volumetric flow has increased. Then the amp load on the motor driver would also increase."
That is what I don't understand: how can the horsepower remain constant and the amp load increase?
Hope the query is clear enough this time.
"We don't believe things because they are true, things are true because we believe them."
Re-reading my post I realize the questions may seem naive.
Well my questions arose from the quote below (source: Lieberman N. - A working guide to process equipment - 3rd Ed - section 30.1.1 - page 375):
"Cooling the liquid will raise the pump's discharge pressure. If the discharge control valve is now opened, the pump's discharge pressure will drop back as the flow increases. The discharge pressure drops because the pump develops less feet of head at a higher flow rate.
The specific gravity of the liquid has increased. The feet of head of liquid has decreased by the same amount. So deltaP remains constant.
How does this affect the amp load on the motor driver?
The feet of head has gone down, and the specific gravity has gone up by the same amount. Therefore, the horsepower or work per barrel of liquid pumped remains constant. But the volumetric flow has increased. Then the amp load on the motor driver would also increase."
That is what I don't understand: how can the horsepower remain constant and the amp load increase?
Hope the query is clear enough this time.
"We don't believe things because they are true, things are true because we believe them."
Try again, but this time please do not mix your remarks and questions within the quotation of the printed text.
I'll try to answer anyway.
"Cooling the liquid will raise the pump's discharge pressure. If the discharge control valve is now opened, the pump's discharge pressure will drop back as the flow increases. The discharge pressure drops because the pump develops less feet of head at a higher flow rate.
Forget about pressure when dealing with pumps. Head is the only meaningful term in pump work. Discharge pressure is simply and only the result of pump discharge HEAD x product density at any given time.
When the valve is opened, the pump's HEAD will decrease as flow increases. That is due to the pump's head acting on the system curve and presumedly increasing the energy acting on the system, resulting in an increase in flow through the system. The new pump discharge head will settle at a new point of intersection between the pump's head curve and the system curve. That flowrate may also vary with the change in product density, because any change in product density will most likely change the system curve, especially if the viscosity changed too. The pump curve DID NOT change.
Once the new operation point is established, you can calculate discharge pressure if you like. It will be head * density.
The CHANGE IN DISCHARGE HEAD BETWEEN THE TWO OPERATING POINTS IS NOT DEPENDENT ON THE CHANGE IN PRODUCT DENSITY IN ANY MANNER, OTHER THAN HOW THAT MIGHT HAVE AFFECTED THE SYSTEM CURVE. The statement, "The feet of head of liquid has decreased by the same amount" IS NOT NECESSARILY TRUE. The change in head was caused by the change in flowrate in the system. Any change in PRESSURE will be the result of head change between operating points and the new product density x the new discharge head. DELTA P DOES NOT HAVE TO REMAIN CONSTANT.
"The feet of head has gone down, and the specific gravity has goen up by the same amount." ALSO NOT TRUE.
"Therefore, the horsepower or work per barrel of liquid pumped remains constant." Again NOT TRUE. Power consumed is DENSITY X FLOWRATE X HEAD at any given time. You have varied the flow and the density and the head, none of which MUST be equal to any product of the other two.
Calculate your power required and convert that to electric power, take your voltage and calculate the amps.
The following explaines in practical terms why it is advantagous to forget about pressure, until you need to know it, usually for determining the wall thickness of the pipe you will need to connect to the pump discharge.
If you pump diesel with a constant speed centrifugal pump discharging at a head of 1000 feet and your flowrate is 500 cfs, then switch to gasoline, your flowrate will go up and up as the gasoline pushes the diesel down the pipeline. Your discharge head will stay at the same 1000 feet. Your discharge pressure, if you want to know, will drop, because 1000 feet x gasoline density is less pressure. Your gasoline flowrate will increase due to less viscosity then diesel and the pump will eventually move to the new gasoline flowrate as the diesel clears the pipeline. The final discharge pressure will be the density of gasoline x new final head when gasoline has completely displaced the diesel from the pipeline. Power consumption will be whatever it is. That may be less because you now have the same 1000 feet discharge head x a lighter density product, but flowrate has also changed, probably gone up, as the reduced viscosity of gasoline may have really increased the flow quite a lot, so you can't say for sure that power is now less, remained the same, or gone up.
If you controlled the flowrate of the new lighter density gasoline product to the same flowrate of the previous diesel product, then you could say for sure that power required to pump that flow controlled gasoline decreased from where it was when diesel was being pumped, because flowrate stayed the same, density decreased and head stayed the same.
"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek
I'll try to answer anyway.
"Cooling the liquid will raise the pump's discharge pressure. If the discharge control valve is now opened, the pump's discharge pressure will drop back as the flow increases. The discharge pressure drops because the pump develops less feet of head at a higher flow rate.
Forget about pressure when dealing with pumps. Head is the only meaningful term in pump work. Discharge pressure is simply and only the result of pump discharge HEAD x product density at any given time.
When the valve is opened, the pump's HEAD will decrease as flow increases. That is due to the pump's head acting on the system curve and presumedly increasing the energy acting on the system, resulting in an increase in flow through the system. The new pump discharge head will settle at a new point of intersection between the pump's head curve and the system curve. That flowrate may also vary with the change in product density, because any change in product density will most likely change the system curve, especially if the viscosity changed too. The pump curve DID NOT change.
Once the new operation point is established, you can calculate discharge pressure if you like. It will be head * density.
The CHANGE IN DISCHARGE HEAD BETWEEN THE TWO OPERATING POINTS IS NOT DEPENDENT ON THE CHANGE IN PRODUCT DENSITY IN ANY MANNER, OTHER THAN HOW THAT MIGHT HAVE AFFECTED THE SYSTEM CURVE. The statement, "The feet of head of liquid has decreased by the same amount" IS NOT NECESSARILY TRUE. The change in head was caused by the change in flowrate in the system. Any change in PRESSURE will be the result of head change between operating points and the new product density x the new discharge head. DELTA P DOES NOT HAVE TO REMAIN CONSTANT.
"The feet of head has gone down, and the specific gravity has goen up by the same amount." ALSO NOT TRUE.
"Therefore, the horsepower or work per barrel of liquid pumped remains constant." Again NOT TRUE. Power consumed is DENSITY X FLOWRATE X HEAD at any given time. You have varied the flow and the density and the head, none of which MUST be equal to any product of the other two.
Calculate your power required and convert that to electric power, take your voltage and calculate the amps.
The following explaines in practical terms why it is advantagous to forget about pressure, until you need to know it, usually for determining the wall thickness of the pipe you will need to connect to the pump discharge.
If you pump diesel with a constant speed centrifugal pump discharging at a head of 1000 feet and your flowrate is 500 cfs, then switch to gasoline, your flowrate will go up and up as the gasoline pushes the diesel down the pipeline. Your discharge head will stay at the same 1000 feet. Your discharge pressure, if you want to know, will drop, because 1000 feet x gasoline density is less pressure. Your gasoline flowrate will increase due to less viscosity then diesel and the pump will eventually move to the new gasoline flowrate as the diesel clears the pipeline. The final discharge pressure will be the density of gasoline x new final head when gasoline has completely displaced the diesel from the pipeline. Power consumption will be whatever it is. That may be less because you now have the same 1000 feet discharge head x a lighter density product, but flowrate has also changed, probably gone up, as the reduced viscosity of gasoline may have really increased the flow quite a lot, so you can't say for sure that power is now less, remained the same, or gone up.
If you controlled the flowrate of the new lighter density gasoline product to the same flowrate of the previous diesel product, then you could say for sure that power required to pump that flow controlled gasoline decreased from where it was when diesel was being pumped, because flowrate stayed the same, density decreased and head stayed the same.
"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek
- Thread starter
- #13
Thanks BigInch. Especially for the last example. I indeed tend to forget the effet of liquid viscosity on volumetric flowrate.
As for the use of pressure, I believe it was useful in the quote to explain and distinguish the following effects:
1/ effect of specific gravity on pump discharge pressure (as the pump head in independant of density). A higher density means a higher pump differential pressure and thus a higher discharge pressure.
2/ effect of valve opening on discharge pressure. Indeed, when the valve opens, the valve pressure drop decreases and, as a result the pump discharge pressure also decreases.
3/ effect of discharge pressure on the pump head. At the new (higher) density, the decrease of the pump discharge pressure causes the decrease of the discharge head and, as a result the pump differential pressure and head decrease.
Last, trust me everything between " " is from the quote!
"We don't believe things because they are true, things are true because we believe them."
As for the use of pressure, I believe it was useful in the quote to explain and distinguish the following effects:
1/ effect of specific gravity on pump discharge pressure (as the pump head in independant of density). A higher density means a higher pump differential pressure and thus a higher discharge pressure.
2/ effect of valve opening on discharge pressure. Indeed, when the valve opens, the valve pressure drop decreases and, as a result the pump discharge pressure also decreases.
3/ effect of discharge pressure on the pump head. At the new (higher) density, the decrease of the pump discharge pressure causes the decrease of the discharge head and, as a result the pump differential pressure and head decrease.
Last, trust me everything between " " is from the quote!
"We don't believe things because they are true, things are true because we believe them."
- Thread starter
- #14
Please replace 2/ and 3/ for clarity.
2/ effect of valve opening on pump differential pressure. Indeed, when the valve opens, the valve pressure drop decreases and, as a result the pump differential pressure decreases (intersection of system and pump curve).
3/ Effect of pump differential pressure on pump head. At the new (higher) density, the lower pump differential, the lower the pump head and the higher the flow.
"We don't believe things because they are true, things are true because we believe them."
2/ effect of valve opening on pump differential pressure. Indeed, when the valve opens, the valve pressure drop decreases and, as a result the pump differential pressure decreases (intersection of system and pump curve).
3/ Effect of pump differential pressure on pump head. At the new (higher) density, the lower pump differential, the lower the pump head and the higher the flow.
"We don't believe things because they are true, things are true because we believe them."
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1/ I think you're not listening. Maybe it is difficult for process guys to think in terms of head??? Forget about pressure until you need to know what it is. Work with HEAD. If you work with varying densities, you will soon tire of all the different pump discharge pressure curves you will have to make as soon as the density changes again and again. You use head curves to avoid that. Head is a measure of potential energy applied to the system. That is totally independent of whatever the system happens to contain at the moment.
2/ You're not seeing the big picture. The valve is part of the system curve. When the valve opens, the system curve changes, its heads becoming less for any given flowrate. The initial high head from the pump causes the fluid in the pipe to accelerate, now that the system resistance is lower. That is what causes flow to increase in the piping and only after the fluid starts moving faster down the pipe will the pump will drop its head. It will not drop its head before the fluid starts moving down the piping, so you can see what causes what. It is certainly not a dropping head from the pump that causes flow in the pipe.
3/ This is the last time I will say this. The discharge pressure HAS NOTHING to do with pump head!!! Any pump textbook will tell you that a constant speed centrifugal pump outputs a constant head no matter what the product, or density happens to be. Discharge pressure is whatever it is when whatever fluid happens to be coming out of the pump at any given moment. Detach the pump from the system and you will see that the pump's output head for a given speed is independent of density of product and the product's flowrate. Head = tangential velocity^2/2/g Notice that density of product is not seen in that equation. Pressure therefore does not = head.
"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek
2/ You're not seeing the big picture. The valve is part of the system curve. When the valve opens, the system curve changes, its heads becoming less for any given flowrate. The initial high head from the pump causes the fluid in the pipe to accelerate, now that the system resistance is lower. That is what causes flow to increase in the piping and only after the fluid starts moving faster down the pipe will the pump will drop its head. It will not drop its head before the fluid starts moving down the piping, so you can see what causes what. It is certainly not a dropping head from the pump that causes flow in the pipe.
3/ This is the last time I will say this. The discharge pressure HAS NOTHING to do with pump head!!! Any pump textbook will tell you that a constant speed centrifugal pump outputs a constant head no matter what the product, or density happens to be. Discharge pressure is whatever it is when whatever fluid happens to be coming out of the pump at any given moment. Detach the pump from the system and you will see that the pump's output head for a given speed is independent of density of product and the product's flowrate. Head = tangential velocity^2/2/g Notice that density of product is not seen in that equation. Pressure therefore does not = head.
"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek
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- #16
Good old Camerons. I was just about to suggest Camerons to the OP as I read BI's last post and direct him/her to that page and then I read your post Artisi and opened the attachment. This illustration says it about the best I can think of graphically speaking. I still recommend reading the whole section 1 of Camerons if you can get your hands on a copy, sheiko.
Yep good old Camerons, if you have anything to do with pumps it should be on your shelf.
It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)
It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)
- Thread starter
- #19
BigInch,
1/ I was trying to explain why the author (Norm Lieberman - which is a worldclass engineer) was talking in tetms of pressure. I do agree with all you said.
2/ It is true. Flow decreases before the pump drops its head.
3/ I agree. I was just saying that at a fixed density, a decrease in pump differential causes a decrease of pump head. DeltaP = SG*Head/10.2.
"We don't believe things because they are true, things are true because we believe them."
1/ I was trying to explain why the author (Norm Lieberman - which is a worldclass engineer) was talking in tetms of pressure. I do agree with all you said.
2/ It is true. Flow decreases before the pump drops its head.
3/ I agree. I was just saying that at a fixed density, a decrease in pump differential causes a decrease of pump head. DeltaP = SG*Head/10.2.
"We don't believe things because they are true, things are true because we believe them."
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