How does the specific gravity affect the pressure output from a pump? 2

[ChEMatt]

If I have a pump that can put out, deadhead, 100' of water, and I change to another liquid that has the same viscosity but has a 0.7 specific gravity, how would that affect the discharge pressure at the outlet flange of the pump?

Had a discussion with a co-worker on the subject and we seemed to agree that the column of liquid produced by the pump would INCREASE by a certain percentage due to the drop in density, but the outlet pressure of the pump would stay the same because of the equal viscosities. But I am uncertain that this is correct. Is it?

Thanks for your help!

 

[JJPellin]

The head will be the same. The pressure will be lower by the ratio of the specific gravities.

Johnny Pellin

 

[Compositepro]

This is a key concept that many new engineers did not grasp in school. Head and pressure are similar but not identical terms.
A centrifugal pump will be rated in feet of head, not feet of water or psi. If you are pumping water, then the two are the same. Centrifugal pumps impart velocity (momentum) to a fluid by spinning it with the impeller. Any mass will rise to the same height, given the same initial upward velocity.

 

[zdas04]

But the energy imparted by the impeller is a function of the density of the fluid--denser fluids take more energy per unit hp than less dense fluids.

David Simpson, PE
MuleShoe Engineering

Law is the common force organized to act as an obstacle of injustice Frédéric Bastiat

 

[ChEMatt]

I always had trouble understanding head. I associate it with pump pressure as well as line losses.

I would argue (conceptually so) with David that the premise was that the viscosity is the same for both liquids, so would this not imply that the density would not adversely affect the capability of the pump? This was one of the sticking points that my co-worker and I were having trouble with was resolving the density difference, taking into account the equation to convert head to pressure and vice versa. It makes sense that a more dense fluid would resist flow more, but is that resistance not due to viscosity? I'm not trying to split hairs here, just trying to understand as best I can.

Thanks,

-Matt

 

[Compositepro]

First keep in mind that most centrifugal pumps are powered by induction motors so the rotational speed is pretty much constant regardless of the load on the motor. If you close the outlet of the pump so there is no flow, the load on the motor will drop very significantly. All the impeller does is spin the same water around inside the pump. Many people intuitively think that the load on the motor will go up if you block-in a pump. Wrong. When you open the valve to allow flow through the pump, water enters the center of the impeller at close to zero velocity, and leaves the O.D. of the impeller at tip velocity. Accelating a mass requires energy. Higher density means more mass and more energy.

When you mention resistance, you are not thinking like an engineer. How exactly do you define resistance? A one ton mass has greater resistance to moving than a one pound mass. This is due to inertia. There is also rolling or sliding friction. This is equivalent to viscosity. Viscosity is normally neglected when discussing the theory of centrifugal pumps.

 

[zdas04]

My post should have been "per unit volume" not "per unit hp". If I'm taking one cc of 1000 kg/m[sup]3[/sup] fluid, then it will take some amount of energy to accelerate that mass. If the fluid is 800 kg/m[sup]3[/sup] then it takes less hp to accelerate that volume. If the fluid is 1200 kg/m[sup]3[/sup] then it takes more hp to accelerate that volume. Viscosity is (mostly) irrelevant to this calculation. Work is a function of how much you are lifting and how high your are lifting it and a centrifugal pump at constant speed is a constant volume device. If pump head is constant, then the only wiggle room left is energy required to accelerate the fluid.

David Simpson, PE
MuleShoe Engineering

Law is the common force organized to act as an obstacle of injustice Frédéric Bastiat

 

[25362]

There are cases when power consumption goes up with reduced flow rates. This happens with axial pumps having high specific speeds (high flow rates low heads). See Thread407-27055

 

[Compositepro]

That is why an axial flow pump should not be called a centrifugal, although they they sometimes are. They are not the same. They are both rotodynamic pumps.

 

[25362]

IMO, it is important to point out that the power P, depends not only on the head, H, the flow rate Q, and the specific gravity ρ, but also on the pump efficiency η as in:

P ∝ HQρ/η​

At flow rates lower than the BEP, η may drop proportionally more than the increase of the product HQ (assuming constant specific gravity), actually increasing P.

 

[Xalii]

For a given system, if you only change the density of the fluid, but not viscosity, the flow will almost be the same (a change might come from the specification of your motor, but not from your pump). I will try a simple explanation: if you don't change the geometry, nor the losses, you will keep the same bucket to do the same work, thus the flow would be the same but the energy required might be bigger or smaller. For a centrifugal pump coupled with an electrical motor, the electrical motor will try to keep a constant velocity and provide the power requested by the pump, thus roughly it can be estimated that if you decrease/increase the density of the fluid, the power needed by the motor to keep the speed the will decrease/increase accordingly, but the head/flow characteristic of the pump will be the same.
A limitation is obviously the maximum capacity of your electrical motor, if you increase the density, you have to be sure that the motor will not burn.
A small change in the pump specification might be seen due to the slipping of the motor. It is assumed that a motor is turning with a constant speed whatever is the load, what is not exact, and a small difference can be noticed. It is thus this small difference that will increase the pump velocity thus change the HQ specification of your pump if your reduce the density of your flow (thus increase slightly the flow).

 

[LittleInch]

My take on this is as follows.

It is difficult to sometime comprehend head and pressure, but when you finally get it a lot of pump systems become clearer.

Centrifugal pumps, as has been noted above give out a fixed head for a given flow at a fixed rotational speed regardless of the liquid inside them. Only when you've fully understood and accepted that will the rest become understandable.

For ease of testing and calculation, these generic head / flow charts are commonly quoted in metres or feet of water. when you order a pump the supplier will use the density of the liquid you've specified to obtain power and pressure figures, but the head would be same.

When you change only the specific gravity of the fluid being pumped in a centrifugal pump two things change:
1) The outlet pressure changes in proportion to the change of density, i.e. lighter fluids create less pressure.
2) The amount of power required to keep the pump spinning at the same speed is also proportional to the density of the fluid when comparing one fluid to another, i.e. a heavier fluid will consume more power for the same flow rate.

Viscosity will affect the downstream system flow resistance and hence the flowrate for a system without flow control will change, but this has very little to do with the pump.

Usually the issue is when someone changes a pump designed for one density of liquid and puts a higher density of liquid in it. Then both pressure and power increase if it is a fixed speed pump which can either exceed the pressure rating or overload the motor. Hence the heaviest density fluid that the pump can see should be used for design and if you're changing duties, be careful to examine the impact of the higher density fluid on the pump and motor.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[Artisi]

Come on guy's this is pumps 101, can't see what the problem is.

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.)

 

[1gibson]

Just to keep the discussion interesting, consider that due to the slip in the motor, a pump never operates exactly per the curve. The performance test curve is corrected to one speed, but if you look at raw test data you will see the actual speed varies based on flow (based on HP.) For low-mid specific speed pumps, you have low HP / higher RPM at shutoff. Depending on nominal speed and size of motor, it might be 5 RPM or so, but the effect is there. Also important to remember that the "rated speed" shown on the pump curve may or may not be the speed that the pump operates at in the field. That could be off by +/- 10 rpm, so watch out for that.

 

[ChEMatt]

Excellent discussion, folks. I'm going to pull out my old unit ops text and take a look at the equations governing pumps. Perhaps that might provide some insight into understanding what's going on. But so far it makes sense. LittleInch's post was especially helpful, I think, but I appreciate everyone's replies.

Sort of off topic, but related to the viscosity thing. I think Mythbusters had a guy swim through corn starch and through water. His lap time was essentially the same. Perhaps it's a somewhat odd demonstration of the fact that the viscosity doesn't figure in to pump performance?

 

[Artisi]

ChEMatt:
Are you saying that viscosity has no effect on pump performance?

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.)

 

[1gibson]

There are well established viscosity correction factors for head, flow, and efficiency.

Don't take this the wrong way, but these days there is no excuse for speculation like this even in general conversation, much less on a technical forum. Take 30 seconds to google it.