Energy to pressurize.
Energy to pressurize.
(OP)
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..
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..





RE: Energy to pressurize.
If this is a question about compressing gases, you're in the wrong forum.
RE: Energy to pressurize.
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?
RE: Energy to pressurize.
Thanks so much!
RE: Energy to pressurize.
What would you be doing, if you knew that you could not fail?
RE: Energy to pressurize.
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/hes-public/calculation-methodology/calculation-of-energy-consumption/miscellaneous-equipment-energy-consumption/well-pump-energy-calculation-method
RE: Energy to pressurize.
=====================================
(2B)+(2B)' ?
RE: Energy to pressurize.
RE: Energy to pressurize.
P = QH / 3960 η
RE: Energy to pressurize.
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?
RE: Energy to pressurize.
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.
RE: Energy to pressurize.
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?
RE: Energy to pressurize.
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.
RE: Energy to pressurize.
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?
RE: Energy to pressurize.
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.
RE: Energy to pressurize.
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?
RE: Energy to pressurize.
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!
RE: Energy to pressurize.
Anyway, yes it is linear.
What would you be doing, if you knew that you could not fail?
RE: Energy to pressurize.
=====================================
(2B)+(2B)' ?