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Axial Thrust and Viscosity (Pumps)
- Thread starter Vic123
- Start date
Axial load is set by the pressure differential across the pump. If the dP is the same for the viscous fluid, I would expect the axial load to be the same as well.
Axial position should be the same regardless, once the axial load reaches a certain point the rotor will be hard against the active thrust bearing, and then the axial position should not change unless the thrust bearing is being damaged.
-The future's so bright I gotta wear shades!
Axial position should be the same regardless, once the axial load reaches a certain point the rotor will be hard against the active thrust bearing, and then the axial position should not change unless the thrust bearing is being damaged.
-The future's so bright I gotta wear shades!
With increased viscosity you get,
Diff head (diff pressure) will be less, hence also axial thrust.
Flow capacity will decrease.
Efficiency will decrease.
Horsepower required will increase.
Some axial pump designs are pressure balanced, thus thrust loads are not affected one way or the other.
Diff head (diff pressure) will be less, hence also axial thrust.
Flow capacity will decrease.
Efficiency will decrease.
Horsepower required will increase.
Some axial pump designs are pressure balanced, thus thrust loads are not affected one way or the other.
electricpete
Electrical
"With increased viscosity you get,
Diff head (diff pressure) will be less"
Looking from the system viewpoint if viscosity increases and system doesn't change (no control valve), the DP would normally go up, right?
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Diff head (diff pressure) will be less"
Looking from the system viewpoint if viscosity increases and system doesn't change (no control valve), the DP would normally go up, right?
=====================================
Eng-tips forums: The best place on the web for engineering discussions.
electricpete,
I suppose I should have said I was describing the effect of increasing viscosity on the pump, but I thought that part was obvious.
First the pump,
If you introduce a fluid into a pump with a higher viscosity than the previous fluid, the required power to maintain the previous flowrate increases. If that power increase is not supplied by the driver, let's say it was at max power output with the previous fluid, the pump's response to the higher viscosity fluid is to decrease its differential head output. That directly leads to a decrease of discharge pressure. To be technically correct; this is only if the densities of both fluids were also the same, since it is head x density = pressure. That said, less head and pressure would be delivered to the pipeline inlet.
Now the pipeline,
A higher viscosity fluid in the pipeline needs a higher inlet pressure, hence also inlet head, to continue at the same flowrate of the previous fluid. But we note that the pump's differential head and discharge pressure has just been decreased, as the more viscous fluid started to be pumped into the pipeline. The combination of more pressure required at the pipeline inlet to continue at the same flowrate plus the lower pressure coming from the pump with the more viscous fluid causes the system as a whole to start decreasing the net flowrate. It eventually arrives at a new and lower stable flowrate, at the point where the new pump curve (modified for the higher visc fluid) intersects the new pipeline curve (also modified for the higher visc fluid).
As you may gather, this can get more complicated if the density of the fluid also changes. In multiple product pipelines it is common to have several products in the same pipeline, lined up one after another. The hydraulic performance of one of those types of pipelines with several pump stations, control valves and a couple of high mountain passes, and different receiving stations or delivery terminals along the way can get rather complicated. That's not to mention the last little problem of making sure that the different products have an available tank to put them in at whatever station(s) is/are going to be their final destination. Well what can I say, it keeps me moving too.
I suppose I should have said I was describing the effect of increasing viscosity on the pump, but I thought that part was obvious.
First the pump,
If you introduce a fluid into a pump with a higher viscosity than the previous fluid, the required power to maintain the previous flowrate increases. If that power increase is not supplied by the driver, let's say it was at max power output with the previous fluid, the pump's response to the higher viscosity fluid is to decrease its differential head output. That directly leads to a decrease of discharge pressure. To be technically correct; this is only if the densities of both fluids were also the same, since it is head x density = pressure. That said, less head and pressure would be delivered to the pipeline inlet.
Now the pipeline,
A higher viscosity fluid in the pipeline needs a higher inlet pressure, hence also inlet head, to continue at the same flowrate of the previous fluid. But we note that the pump's differential head and discharge pressure has just been decreased, as the more viscous fluid started to be pumped into the pipeline. The combination of more pressure required at the pipeline inlet to continue at the same flowrate plus the lower pressure coming from the pump with the more viscous fluid causes the system as a whole to start decreasing the net flowrate. It eventually arrives at a new and lower stable flowrate, at the point where the new pump curve (modified for the higher visc fluid) intersects the new pipeline curve (also modified for the higher visc fluid).
As you may gather, this can get more complicated if the density of the fluid also changes. In multiple product pipelines it is common to have several products in the same pipeline, lined up one after another. The hydraulic performance of one of those types of pipelines with several pump stations, control valves and a couple of high mountain passes, and different receiving stations or delivery terminals along the way can get rather complicated. That's not to mention the last little problem of making sure that the different products have an available tank to put them in at whatever station(s) is/are going to be their final destination. Well what can I say, it keeps me moving too.
Normally, if the viscosity were to increase the result would be as described by BigInch
"With increased viscosity you get,
Diff head (diff pressure) will be less, hence also axial thrust.
Flow capacity will decrease.
Efficiency will decrease.
Horsepower required will increase."
However, the statement in a later posting is not clear,
"If that power increase is not supplied by the driver, let's say it was at max power output with the previous fluid, the pump's response to the higher viscosity fluid is to decrease its differential head output."
If the input power increases due to a change of viscosity and you are already at the maximum output of the driver, you are now in an overload situation, unless of course the shift in flow and head is sufficient to counteract the increased power; this is unlikely.
A pump running at a constant speed can not, by itself produce less product to match the input power available. Some type of control device would be needed to limit pump output and reduce power input.
Naresuan University
Phitsanulok
Thailand
"With increased viscosity you get,
Diff head (diff pressure) will be less, hence also axial thrust.
Flow capacity will decrease.
Efficiency will decrease.
Horsepower required will increase."
However, the statement in a later posting is not clear,
"If that power increase is not supplied by the driver, let's say it was at max power output with the previous fluid, the pump's response to the higher viscosity fluid is to decrease its differential head output."
If the input power increases due to a change of viscosity and you are already at the maximum output of the driver, you are now in an overload situation, unless of course the shift in flow and head is sufficient to counteract the increased power; this is unlikely.
A pump running at a constant speed can not, by itself produce less product to match the input power available. Some type of control device would be needed to limit pump output and reduce power input.
Naresuan University
Phitsanulok
Thailand
I assumed that driver power was not infinite and was already running at full power output with the previous fluid, so if a higher viscosity fluid was introduced, the head must decrease, because heads always decrease with a higher viscosity fluid. You would only be in an overload condition if you tried to continue at the same flowrate with the new higher viscosity fluid. Since head decreased, differential pressure decreased and flow capacity decreased. Nobody said a constant speed pump would change flowrate by itself; you just got through adding a more viscous fluid. That changed the flowrate due to hydraulic changes simultaneously occurring inside the pump with the introduction of the hi visc fluid. Net result is flow decrease from the pump nozzle, even without considering the rest of the system curve which will also increase and tend to reduce system flows.
The one "control device" you get whenever you install and operate a real pump system, is the maximum driver power output to the pump shaft. When the driver is at max, its at max and the pump can't get anymore power.
The one "control device" you get whenever you install and operate a real pump system, is the maximum driver power output to the pump shaft. When the driver is at max, its at max and the pump can't get anymore power.
"When the driver is at max, its at max and the pump can't get anymore power"
So very true, that's why properly protected motors on a system where the pump is capable of running out on its curve beyond the point of maximum power available will shut down from overload if and when the pump operates in this region, whereas, poorly protected motor capable operating in overload will fail from "burn-out", unless of course you have a non-overloading motor fitted,
A motor can only be a controlling device if it shuts down from overload and then it is really the motor protection that is the control.
Naresuan University
Phitsanulok
Thailand
So very true, that's why properly protected motors on a system where the pump is capable of running out on its curve beyond the point of maximum power available will shut down from overload if and when the pump operates in this region, whereas, poorly protected motor capable operating in overload will fail from "burn-out", unless of course you have a non-overloading motor fitted,
A motor can only be a controlling device if it shuts down from overload and then it is really the motor protection that is the control.
Naresuan University
Phitsanulok
Thailand
Interesting concept. You seem to imply that infinite power draw is a possibility.
Torque (ft-lbs) is what produces rotation, not power. Power equivalent is in units of ft-lbs/sec, or torque expended per time unit. Torque outputs of engines and motors decrease with rotational speed increase. At maximum conditions, torque output ==== 0.0000 When torque output of a motor or engine is 0, further angular acceleration is prohibited. When angular acceleration =0, speed does not increase.
Or, do you think Newton was and I am wrong about this? When I run my electric golf cart at max rpm and we go up a steep hill, I slow down, regardless of the power draw. What you see is the motor requiring more power as it tries to produce that tiny tiny tiny extra little bit of torque it needs, but even infinite power draw won't do it. Is your golf cart different?
Torque (ft-lbs) is what produces rotation, not power. Power equivalent is in units of ft-lbs/sec, or torque expended per time unit. Torque outputs of engines and motors decrease with rotational speed increase. At maximum conditions, torque output ==== 0.0000 When torque output of a motor or engine is 0, further angular acceleration is prohibited. When angular acceleration =0, speed does not increase.
Or, do you think Newton was and I am wrong about this? When I run my electric golf cart at max rpm and we go up a steep hill, I slow down, regardless of the power draw. What you see is the motor requiring more power as it tries to produce that tiny tiny tiny extra little bit of torque it needs, but even infinite power draw won't do it. Is your golf cart different?
electricpete
Electrical
Artisi and 25632 are right.
BI - since you mentioned motors, the most common industrial application is a squirrel cage induction motor supplied with constant frequency power, which acts close to constant speed (speed droop typically 1% from no load to full load). To suggest that it acts like a constant power control device is ridiculous.
With a fixed speed drive and no changes to the control system, looking at dp vs flow curves, increase in viscosity will cause system curve to shift up and pump curve to shift down (assuming no automatic flow or pressure control system response). The operating point at the intersection will surely shift to the left (lower flow), but whether it shifts up or down (higher or lower dp) would seem indeterminate without specific curves.
There may be a variety of control applied as well. My point was simply to request a clarification of the conditions being compared.
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BI - since you mentioned motors, the most common industrial application is a squirrel cage induction motor supplied with constant frequency power, which acts close to constant speed (speed droop typically 1% from no load to full load). To suggest that it acts like a constant power control device is ridiculous.
With a fixed speed drive and no changes to the control system, looking at dp vs flow curves, increase in viscosity will cause system curve to shift up and pump curve to shift down (assuming no automatic flow or pressure control system response). The operating point at the intersection will surely shift to the left (lower flow), but whether it shifts up or down (higher or lower dp) would seem indeterminate without specific curves.
There may be a variety of control applied as well. My point was simply to request a clarification of the conditions being compared.
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I don't think I said or implied anything about constant or variable speed. I said torque output is responsible for angular acceleration and the resultant power draw to infinity is the result of the motor attempting to make an infitesimal amount of more torque. If torque is not produced, there can be no acceleration or resultant speed change. It will continue to run at that speed, practically speaking.
As for your previous question, I repeat, "Looking from the system viewpoint if viscosity increases and system doesn't change (no control valve), the DP would normally go up, right? ".
RIGHT!
From the "pipe" point of view, the pressure differential at the pipe inlet to the pipe outlet must go up to move the same flowrate and you are indeed 100% right.
From the "pump" point of view, the differential head decreases, given that the density of both of the fluids remain equal and that the pump suction pressure is held constant, so the discharge pressure of the pump goes down.
Both of those, from a "system" point of view, decrease flow.
As for motors, just take a look at the motor torque output vs speed, and realize,
1.) That angular acceleration is a product of torque which is the product of two real things in this world, force and distance,
2.) Torque is not the result of power applied. Power is not a real force in this world. Energy is real. Power is the rate that energy is spent and, as such, is only a concept of the human mind. If you didn't have a watch, you would have no way to measure power or even a concept of power is. As far as I know, Mother Nature does not own a watch and She only lives in the present. She has no memory of what happened during the last second that passed, and neither does energy.
3.) Power is the measurement of how fast torque is applied.
Realizing these, you will understand my answer.
As for your previous question, I repeat, "Looking from the system viewpoint if viscosity increases and system doesn't change (no control valve), the DP would normally go up, right? ".
RIGHT!
From the "pipe" point of view, the pressure differential at the pipe inlet to the pipe outlet must go up to move the same flowrate and you are indeed 100% right.
From the "pump" point of view, the differential head decreases, given that the density of both of the fluids remain equal and that the pump suction pressure is held constant, so the discharge pressure of the pump goes down.
Both of those, from a "system" point of view, decrease flow.
As for motors, just take a look at the motor torque output vs speed, and realize,
1.) That angular acceleration is a product of torque which is the product of two real things in this world, force and distance,
2.) Torque is not the result of power applied. Power is not a real force in this world. Energy is real. Power is the rate that energy is spent and, as such, is only a concept of the human mind. If you didn't have a watch, you would have no way to measure power or even a concept of power is. As far as I know, Mother Nature does not own a watch and She only lives in the present. She has no memory of what happened during the last second that passed, and neither does energy.
3.) Power is the measurement of how fast torque is applied.
Realizing these, you will understand my answer.
Huh? All finally quiet on the Western Front, or is this a lull in the battle?
electricpete
Electrical
“I don't think I said or implied anything about constant or variable speed.”
You have been assuming constant power. I think this may be a reasonable approximation for some drivers such as turbine with fixed governor valve position. But constant power is not consistent with constant speed (like sync motor or approximately like induction motor) when discussing variations in torque-vs-speed curve.
P = 2*PI*T*N
If speed N is constant (induction motor assumption) and we change the torque curve which changes T, we cannot possibly keep P constant.
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You have been assuming constant power. I think this may be a reasonable approximation for some drivers such as turbine with fixed governor valve position. But constant power is not consistent with constant speed (like sync motor or approximately like induction motor) when discussing variations in torque-vs-speed curve.
P = 2*PI*T*N
If speed N is constant (induction motor assumption) and we change the torque curve which changes T, we cannot possibly keep P constant.
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Finally. Thanks for the expl. I can agree I was assuming constant power, because where changes in viscosity are concerned, there is (usually) not much power change to speak of in relation to the most important variables in pump flow power consumption, which are Head, Flow and Density. The considerations of power variation due to a fluid with a different viscosity that could still be pumped in a similar pump, were not IMO significant to the original question to even require mentioning, but technically true, so I listed it.
I see where in this bit I said, "I assumed that driver power was not infinite and was already running at full power output", and I agree that to be 100% correct, perhaps I should have included the word "rated", as in, "full >rated< power". I agree that it would be smoked at some point on its way to infinity, but, given the above relationship to H,Q & Den, the real difference in power and torque did not really become relavent unitl the 5 or 6th comment. (or so I thought)
A couple of other observations: Even speed is not necessarily constant these days with the number of VFDs being sold for IMs, and I don't recall saying anything about IMs in my original explanation. Artisi, injected motors into the discussion without any real reason to do so. Up until that point, you guys just assumed that some kind of an electric driver, also without reason. IYAM, diesels is the way to go. Motors are just sometimes necessary evils... I suppose. IMO
I see where in this bit I said, "I assumed that driver power was not infinite and was already running at full power output", and I agree that to be 100% correct, perhaps I should have included the word "rated", as in, "full >rated< power". I agree that it would be smoked at some point on its way to infinity, but, given the above relationship to H,Q & Den, the real difference in power and torque did not really become relavent unitl the 5 or 6th comment. (or so I thought)
A couple of other observations: Even speed is not necessarily constant these days with the number of VFDs being sold for IMs, and I don't recall saying anything about IMs in my original explanation. Artisi, injected motors into the discussion without any real reason to do so. Up until that point, you guys just assumed that some kind of an electric driver, also without reason. IYAM, diesels is the way to go. Motors are just sometimes necessary evils... I suppose. IMO
electricpete
Electrical
ok, now everyone's assumptions are clear.
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I for one don't consider this a "battle" but rather a discussion to which we are possibly talking about or looking at from a different view point or understanding.
Yes, I introduced the subject of "electric motors" on the assumption we are talking about electric motor drives and in conjunction with your comment below;
And I quote-
"If that power increase is not supplied by the driver, let's say it was at max power output with the previous fluid, the pump's response to the higher viscosity fluid is to decrease its differential head output."
My understanding of this is, rightly or wrongly, if the driver cannot produce the extra power required by the increase in vis. then the pump will somehow produce less head (output).
I now understand that you are talking about diesel drive units - and if your above statement is still applied to the discussion - this doesn't change much, as demanding extra power from a diesel engine already operating a full load, or as you have put full rated power, is still overload. The saving grace being you can reduce the engine speed slightly to get back to the continuous engine rating.
As for your golf cart - I guess it is a DC motor drive and not an AC drive.
Naresuan University
Phitsanulok
Thailand
Yes, I introduced the subject of "electric motors" on the assumption we are talking about electric motor drives and in conjunction with your comment below;
And I quote-
"If that power increase is not supplied by the driver, let's say it was at max power output with the previous fluid, the pump's response to the higher viscosity fluid is to decrease its differential head output."
My understanding of this is, rightly or wrongly, if the driver cannot produce the extra power required by the increase in vis. then the pump will somehow produce less head (output).
I now understand that you are talking about diesel drive units - and if your above statement is still applied to the discussion - this doesn't change much, as demanding extra power from a diesel engine already operating a full load, or as you have put full rated power, is still overload. The saving grace being you can reduce the engine speed slightly to get back to the continuous engine rating.
As for your golf cart - I guess it is a DC motor drive and not an AC drive.
Naresuan University
Phitsanulok
Thailand
If electricpete is happy, I'm happy.
Its probably DC then, I know its not diesel.
Its probably DC then, I know its not diesel.
I like this explanation even better than mine.
To understand the differences in power characteristics between an electric motor and internal-combustion engine, we'll first examine characteristics of a [COLOR=white red] standard 3-phase electric motor. [/color] Figure 2 shows the torque-speed relationship of a 20 hp, 1800 rpm, NEMA Design B motor. Upon receiving power, the motor develops an initial, locked-rotor torque, and the rotor turns. As the rotor accelerates, torque decreases slightly, then begins to increase as the rotor accelerates beyond about 400 rpm. This dip in the torque curve generally is referred to as the pull-up torque. Torque eventually reaches a maximum value at around 1500 rpm, which is the motor's break-down torque. As rotor speed increases beyond this point, torque applied to the rotor decreases sharply. This is known as the running torque, which becomes the full-load torque when the motor is running at its rated full-load speed - usually 1725 or 1750 rpm.
The torque-speed curve for a 3600-rpm motor would look almost identical to that of the 1800-rpm motor. The difference would be that speed values would be doubled and torque values would be halved.
Common practice is to ensure that torque required from the motor will always be less than breakdown torque. [COLOR=white red] Applying torque equal to or greater than breakdown torque will cause the motor's speed to drop suddenly and severely [color], which will tend to stall the motor and most likely burn it out. If the motor is already running, it is possible to momentarily load a motor to near its breakdown torque. But for simplicity of discussion, assume the electric motor is selected based on full-load torque.
Note that Figure 2 shows a [COLOR=white red] temporary [/color] large torque excess that can provide additional muscle to drive the hydraulic pump through [COLOR=white red]momentary [/color]load increases. These types of electric motors also can be run indefinitely at their rated hp plus an additional percentage based on their service factor - generally 1.15 to 1.25 (at altitudes to 3300 ft).
Catalog ratings for electric motors list their usable power at a rated speed. [COLOR=white red] If the load increases, motor speed will decrease [/color] and torque will increase to a value higher than full-load torque (but less than break-down torque). So when operating the pump at 1800 rpm, the electric motor has more than enough torque in reserve to drive the pump.
Entire text is here,
To understand the differences in power characteristics between an electric motor and internal-combustion engine, we'll first examine characteristics of a [COLOR=white red] standard 3-phase electric motor. [/color] Figure 2 shows the torque-speed relationship of a 20 hp, 1800 rpm, NEMA Design B motor. Upon receiving power, the motor develops an initial, locked-rotor torque, and the rotor turns. As the rotor accelerates, torque decreases slightly, then begins to increase as the rotor accelerates beyond about 400 rpm. This dip in the torque curve generally is referred to as the pull-up torque. Torque eventually reaches a maximum value at around 1500 rpm, which is the motor's break-down torque. As rotor speed increases beyond this point, torque applied to the rotor decreases sharply. This is known as the running torque, which becomes the full-load torque when the motor is running at its rated full-load speed - usually 1725 or 1750 rpm.
The torque-speed curve for a 3600-rpm motor would look almost identical to that of the 1800-rpm motor. The difference would be that speed values would be doubled and torque values would be halved.
Common practice is to ensure that torque required from the motor will always be less than breakdown torque. [COLOR=white red] Applying torque equal to or greater than breakdown torque will cause the motor's speed to drop suddenly and severely [color], which will tend to stall the motor and most likely burn it out. If the motor is already running, it is possible to momentarily load a motor to near its breakdown torque. But for simplicity of discussion, assume the electric motor is selected based on full-load torque.
Note that Figure 2 shows a [COLOR=white red] temporary [/color] large torque excess that can provide additional muscle to drive the hydraulic pump through [COLOR=white red]momentary [/color]load increases. These types of electric motors also can be run indefinitely at their rated hp plus an additional percentage based on their service factor - generally 1.15 to 1.25 (at altitudes to 3300 ft).
Catalog ratings for electric motors list their usable power at a rated speed. [COLOR=white red] If the load increases, motor speed will decrease [/color] and torque will increase to a value higher than full-load torque (but less than break-down torque). So when operating the pump at 1800 rpm, the electric motor has more than enough torque in reserve to drive the pump.
Entire text is here,
Hi BigInch.
The main point is an electric motor runs at its design speed, a direct function of the power supply frequency.
If you load it, it essentially does not change speed but rather increases its current draw. As soon as this current draw exceeds FLA(Full Load Amps) the motor is running on borrowed time. This time is borrowed from the motor's overload protection or the time for the windings to reach insulation breakdown temperature.
From an operational point of view; [COLOR=white red]If the load increases, motor speed will decrease[/color] is a non-starter. If the motor slows enough to actually matter to a process it's in trouble.
Keith Cress
Flamin Systems, Inc.-
The main point is an electric motor runs at its design speed, a direct function of the power supply frequency.
If you load it, it essentially does not change speed but rather increases its current draw. As soon as this current draw exceeds FLA(Full Load Amps) the motor is running on borrowed time. This time is borrowed from the motor's overload protection or the time for the windings to reach insulation breakdown temperature.
From an operational point of view; [COLOR=white red]If the load increases, motor speed will decrease[/color] is a non-starter. If the motor slows enough to actually matter to a process it's in trouble.
Keith Cress
Flamin Systems, Inc.-
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