Energy to pressurize. 2

[John31415]

Sorry if this is a stupid question:
What is the energy cost as a function of pressure. I.e., how much more energy does it take to pressure to Y bar as opposed to Y/2 bar. It seems like it should be a linear function but there are no doubt losses..

 

[moltenmetal]

Your question is too open-ended to give a simple answer. The flow, service, duty cycle and head difference required determine the type of pump used, which in turn determines the mechanical efficiency you can achieve. For a given fluid at a given flow, the type of pump may change, or may not change, depending on what x and x/2 are in terms of head. You may be able to do x/2 with a centrifugal but x may require a regenerative turbine, multi-stage centrifugal or a positive displacement pump for instance. The mechanical efficiency amongst those choices can vary greatly and may not be the overriding criterion for the best selection.

If this is a question about compressing gases, you're in the wrong forum.

 

[BigInch]

I assume the fluid is a liquid, since you posted in the pump forum. Liquids being relatively incompressible, I assume you also mean more like the work required to double the discharge pressure Pd of the same given flowrate, each time beginning from the same suction pressure Ps and that is linear. Pd₂-Ps =
2*(Pd₁-Ps). Assuming you would reach those pressures with drivers and pumps designed to do so efficiently in each case, and each process therefore could be done at the same net efficiency in producing each head, the work needed is indeed linear.

What would you be doing, if you knew that you could not fail?

 

[John31415]

Fair enough I'm looking to get some intuition for various pressures of sea water. It seems that the higher the pressure the higher expected losses should be. Would you have a (rough) estimate of the losses for 10,50,100 bar? Obviously, the pump cost varies immensely however I'm not sure how much the energy cost varies in practice.
Thanks so much!

 

[BigInch]

It's closely linear.

What would you be doing, if you knew that you could not fail?

 

[bimr]

The pump power and electrical cost can be calculated with the following equations:

P = QH / 3960 η


P ~ Power, HP
Q ~ Flow Rate, GPM
H ~ Total Dynamic Head, Feet-Water
η ~ Efficiency, % (Use 62% as typical)

http://www.icalcul8.com/pump_power.php

EP = WP x 0.746 x 1/eff Equation 29

where:

EP = Electrical power (kW)

WP = Water power (hp)

eff = Overall efficiency of pump and motor system (decimal value, 0 to 1) Use 0.8

https://sites.google.com/a/lbl.gov/...nsumption/well-pump-energy-calculation-method

 

[electricpete]

Would you have a (rough) estimate of the losses for 10,50,100 bar?

Are you interested in losses or fluid power. These are two different (mutually exclusive) types of power. BigInch was discussing fluid power

=====================================
(2B)+(2B)' ?

 

[John31415]

I'm not sure that they are mutually exclusive. Don't losses and power add up to 1? The system has mechanical losses due to efficiency problems so you have to subtract them from the amount of fluid power realistically available.

 

[bimr]

The pump power and electrical cost is directly proportional to the pressure (head). The pump efficiency for a centrifugal pump is approximately 62%, but will vary slightly depending on the BEP of the pump and the duty point.

P = QH / 3960 η

 

[BigInch]

LINEAR!
10, 50, 100 Bar pressures, given the same flowrate and same pump efficiency will be 1E, 5E, 10E. Where E is the energy required to reach 10 Bar.

Assuming you are talking about different pumps (and their drivers) that are each designed to reach their BEP delivery heads at similar values of η (which perhaps could be somewhat doubtful technically given the range of 10x, but still within presumptive theory). Attempting that range with the same (one) centrifugal pump/driver combo probably won't be able to reach such a range of heads with the same efficiency and at the same flowrate. With one pump I would expect some variation of efficiency, but aside from that possible efficiency variation, STILL VERY CLOSELY LINEAR.

What would you be doing, if you knew that you could not fail?

 

[77JQX]

For a constant flow rate, the power delivered to a fluid is exactly double for a doubled pressure rise. Depending on how you deliver the power, the efficiency of the machine may vary, and the power supplied to the machine may deviate from exactly linear.

However, if you're asking about the energy required to pressurize a closed container with a fluid, then it isn't linear. It's been far too long since I've had any thermo to answer this, but I wanted to point out that everyone has been answering questions about power and flow, rather than the stored energy in a closed vessel at one pressure versus another pressure.

 

[BigInch]

Now you are complicating the problem considerably. No container has been mentioned up until now. Even if it had, container or not, work, or power when flowrates are constant, required by the pump (pressurizing an essentially incompressible fluid) is still linear. Are you introducing new conditions. 1.) There is perhaps some compressible fluid initially present in the container (air?) 2.) And perhaps the expansion of the container due to internal pressure must now be considered???

If you begin with a gas in the container, compressiblity varies the density with pressure, so even with an ideal gas and holding actual volumetric flowrate constant, at double the container pressure you would be pumping double the density and power is now 4 times what you needed before, so all concepts of linearity fly out the window.

Pumping an incompressible fluid into a perfectly rigid container, initially with a vacuum, will still require essentially double the work at double the pressure. Since there were no time constraints ever mentioned in the OP, or clarification posts, all we can really talk about here is the amount of work anyway. Power is the rate at which work is done and no reference time frame was ever stated in which to discuss that. Pumping an incompressible fluid into a rigid container, initially with a vacuum, will see no pressure rise, except for vapor pressure above the fluid line, and static pressure of the fluid below, until the container is full. When you reach 5 bar, the work being done at the moment is exactly 1/2 that done when you reach 10 bar, and 1/10 that at 50 bar. If you happen to be at the same flowrate in each case, then power will increase in the same linear proportion.

Most of us assumed that this never was a question about energy. The OP said it was a stupid question and talking about energy and power without a time frame reference is the main reason that it was.

What would you be doing, if you knew that you could not fail?

 

[77JQX]

I brought up the closed container because the OP doesn't mention flow, or power, or efficiency. When I read the OP's question, I saw the words energy and pressure and I assumed he meant the energy required to bring a container to a given pressure. Maybe it's because of the thread where the guy wanted to know how many car batteries he needed to bring for his pump to pressure test his pipeline. Thus I was surprised to see all the responders giving answers to a pipe flow problem. Since the OP hasn't corrected anyone, he's given tacit agreement that this is indeed a flow problem.

OP aside, for those that stumble upon this thread, it is my assertion that work - or energy - required to bring a vessel from 0 to 2P is not simply twice the work required to bring it from 0 to P. So I disagree with BigInch, and I had to think carefully before posting, since he usually provides good advice.

I agree with BigInch that, for an incompressible fluid and a constant fill rate, the power at 5 bar is 1/2 the power at 10 bar, and 1/10 the power at 50 bar. But energy (work) is the integral of power over time. If power varies linerally with pressure, then the work performed varies exponentially.

And if you're dealing with a compressible fluid, then things get really complicated.

 

[BigInch]

Hi 77JQX,
I don't think we disagree on energy. Maybe we disagree on work.
You say, "But energy (work) is the integral of power over time."
Yes, Energy is, but work is not. Work is the integral of force * distance F*D.
Power, the first derivative of FD with respect to time.

Energy is the integral of Power over time.
Time takes us back to the reason we are talking about flowrates; the OP didn't give us any time frame reference to discuss energy, so we had to change the question, introduce time via flowrates so we could talk about power instead. He seemed OK with that. Still you, me, we, none of us can tell how much energy will be consumed in pressurizing a vessel to any given pressure. We would need a time frame reference to talk about that and we still don't have one.

But, If we want to assume a time frame reference, I guess we can. Let's assume it takes time T to fill a vessel. This is a pumping problem, we're in that forum, so can we also assume common type, plain, centrifugal pumps, all pumps run at same speed, just produce different discharge heads. Built for purpose, so all have same efficiency at their given discharge heads. Vessels contain the same volumes. That should be enough to get us 2 * pressure = 2 * work = two 2 x power. The pumps are directly connected to the vessels, so can we simplify and say pump pressure = vessel pressure.

Pump Power = Q * den * Head / eff
Power pump 1 = Q * den * Head 1 / eff
Power pump 2 = Q * den * Head 2 / eff, flowrates the same so Q2 = Q1.
When Head 2 = 2 * Head 1, then Power pump 2 = 2 x Power pump 1.
It seems simple to see now that, if both pumps run for time T, P * T = Energy, so pump 2, pressurizing at 2 x Head 1, consumes 2 x Power over the same time T to fill each vessel, so Energy used by pump 2 is 2 * Energy consumed by pump 1.... Or not???
I think so.

What would you be doing, if you knew that you could not fail?

 

[77JQX]

BigInch - If both vessels are the same size, and both pumps run for time T, and Q1=Q2, then how does Head2 get to 2*Head1?

The heart of my argument is that Head is a function of time, and thus power is a function of time:

Pump Power(t) = Q * Head(t) * den / eff

(I've applied liberal doses of assumptions about constant Q and constant efficiency to come up with the preceeding. We could, of course, write Q(t) and eff(t) into the above, but that doesn't seem to me to be especially relevant).

We've agreed energy is the integral of power wrt time. We can't simply say Energy = Power * Time, since power varies with time. In one case, we integrate power from nearly zero to P1 (Q*H1*den/eff). In the second case is the first integral, plus the integral of power from P1 (Q*H1*den/eff) to P2 (Q*H2*den/eff). The only way the energy could be linear is if the second integral is equal to the first. I don't think it is.

Maybe I'm engaging in some sloppy thinking - I can't stop thinking that pressurizing a vessel is analogous to compressing a spring. From Hooke's Law, the work (W2) to compress to F2=2*F1 is 4*W1. That, and I've never seen the difference between work and energy as a nit worth picking.

 

[BigInch]

The heads are different because the pumps have different BEP discharge heads, different curves.

Why should head be a function of time. you keep on trying to change the question with additional conditions. If thats the rules, then I might start adding conditions too. Head produced by the pump is a function of flowrate and pump curve specifications. If I hold Q to a constant rate, like we said, the pump curves say flow Q is discharged at head H1 for pump1 and H2 for pump 2. Each pump is lifting Qo to its discharge head associated with Qo, no matter what is going on in the attached vessel, at least up to the time where vessel pressure equals pump dischrge pressure and all flow stops; the pump lifts flow Q to head H and the Power consumed in the process is = Q * den * H/eff. Power2 is 2 * Power1.

If you must have variable heads to make your theory work, then at least be fair and make the head of each pump vary over time in the same proportion to its final discharge head H1 and H2 where H2 = 2 H1; something like at any given time discharge head H2 = 2 * H1. now head is not constant over time and Still power2 = 2 x power 1.

What would you be doing, if you knew that you could not fail?

 

[77JQX]

Why would head vary? I figure that if you start with a container at one pressure, and end with it at another pressure, that implied changing head during the pressurization process.

Anyway, this document answers my questions:

http://www.hse.gov.uk/research/crr_pdf/1998/crr98168.pdf

For water in a pipe, the answer is that stored energy is closely linear with pressure. I ran values for water to 10 bar in HDPE pipe (8” DR9) and Steel pipe to 100 bar (8” SCH 60). While the energy stored in pipe strain isn’t linear, it’s overwhelmed by the energy stored in the compression of water.

I learned something – which is why I come here. Thanks, everyone!

 

[BigInch]

I ignored the container, because the energy used by the pump is constant and that is actually what you pay for. Differential head is a function of pump flowrate. It isn't important to the question what the container is doing, or how much energy is stored there, or how much is not. Only thing that matters to cost is how much the pump lifted and how high it lifted it.

Anyway, yes it is linear.

What would you be doing, if you knew that you could not fail?

 

[electricpete]

Are you interested in losses or fluid power. These are two different (mutually exclusive) types of power. BigInch was discussing fluid power

I'm not sure that they are mutually exclusive. Don't losses and power add up to 1?

We are in agreement, just terminology. The losses and fluid power add up to the input power. Said another way, we take the input power and split it into two buckets: losses and fluid power. What goes in one bucket can not also be in the other bucket... it is only in one or the other (mutually exclusive). That is what I meant when I said these are "different (mutually exclusive)".


=====================================
(2B)+(2B)' ?