Reducing the power required to drive a slider crank

Hi folks,

I'm currently working on the design of a triplex plunger pump. I have conducted an investigation into the loading of the pump at maximum operating pressure using an analytical method (free body diagrams) and using rigid body dynamics to validate.

In both analyses the pump has been treated like a slider crank mechanism (except with three sliders 120° apart). My project requires that I use industry standard dimensions for the initial design so I have based the dimensions on the Weir 2500 TWS Destiny frac pump. The pump has a 10'' stroke and 4'' plungers, I have rounded these figures to 250mm and 100mm for the sake of working in metric units. Weir claim in their brochure that the pump can achieve 150MPa output with a flow rate of 170 GPM (12.881*10-3m3s-1) whilst being run from a 2500hp external source. Weir rate their maximum rod load at 273000lbf.

My design has a 250mm stroke, 100mm plunger diameter and I am striving to meet the same output potential as the case study pump. I have calculated maximum torque at 157.63kNm and in turn calculated the power requirement at around 2905hp, the analytical results and computer generated ones match each other. I have not even taken friction into account and my power requirement is considerably higher. My rod load is even 10000lbf lower than their maximum rating.

I'm struggling to see how I can reduce the power requirement of the design, I cannot change the flow rate required so logic would suggest I need to reduce the torque somehow. However, I cannot reduce the stroke of the piston so I cannot reduce the diameter of the crank, I cannot reduce the operating pressure so I cannot reduce the load on each piston. There is a gearbox between the input and pump however, with the exception of any parasitic losses this should not affect the power requirement, it seems more logical to attempt to reduce torque at the crank.

My normal area of focus isn't mechanical so I'm not really sure how to proceed here in order to reduce the maximum net torque. If anyone has any suggestions on things I could look into please help!

Cameron
 

jrmichler

Science Advisor
925
823
First, a basic calculation of power required at 100% efficiency: 150 MPa X 145 PSI/MPa X 170 GPM / 1714 = 2157 hp at 100% efficiency. That nicely matches the reported 2500 hp driver because the efficiency of piston pumps is quite high.

You need that much power to move that much liquid at that pressure. You can reduce power only by pumping less fluid, pumping at lower pressure, or by increasing pump efficiency. There is no escaping the laws of physics.

You can make a smaller pump and pump the same amount of fluid by running at higher RPM (and lower torque), in which case the power will be the same. Be aware that smaller pumps running faster create other problems, such as shorter life and reduced reliability.
 
First, a basic calculation of power required at 100% efficiency: 150 MPa X 145 PSI/MPa X 170 GPM / 1714 = 2157 hp at 100% efficiency. That nicely matches the reported 2500 hp driver because the efficiency of piston pumps is quite high.

You need that much power to move that much liquid at that pressure. You can reduce power only by pumping less fluid, pumping at lower pressure, or by increasing pump efficiency. There is no escaping the laws of physics.

You can make a smaller pump and pump the same amount of fluid by running at higher RPM (and lower torque), in which case the power will be the same. Be aware that smaller pumps running faster create other problems, such as shorter life and reduced reliability.
Thanks for the reply jrmichler. So the error in the situation is that I am calculating power in the wrong way and that's why it's too high. I must have glossed over the fluid power formula as my instinct was to calculate the angular velocity required to meet the flow rate and then use the formula P=T*ω. Is it wrong of me to expect the methods to be interchangeable? Could you explain where the 1714 comes from? I would imagine it is a different constant in a metric equivalent of the formula?
 

russ_watters

Mentor
18,990
5,143
Their system is 86% efficient, which is quite good.

2905/2500 is 116%, which is not a lot

Your piston is 5% larger.

You didn't say your rpm.

To me, this feels like a fool's errand trying to duplicate the performance of a commercially available piece of equipment that a company spent decades perfecting. But if I tried and came within 10-20% of their performance I'd throw myself a parade.
 

russ_watters

Mentor
18,990
5,143
Thanks for the reply jrmichler. So the error in the situation is that I am calculating power in the wrong way and that's why it's too high. I must have glossed over the fluid power formula as my instinct was to calculate the angular velocity required to meet the flow rate and then use the formula P=T*ω. Is it wrong of me to expect the methods to be interchangeable?
For a piston, the torque on the lever arm is not constant, so that seems a more difficult method.
Could you explain where the 1714 comes from? I would imagine it is a different constant in a metric equivalent of the formula?
I don't recognize it per se, but it is a rolled-up constant of unit conversions. The formula for power is just pressure times flow rate. In English units it's gpm×ft head/3960=bhp
 
Their system is 86% efficient, which is quite good.

2905/2500 is 116%, which is not a lot

Your piston is 5% larger.

You didn't say your rpm.

To me, this feels like a fool's errand trying to duplicate the performance of a commercially available piece of equipment that a company spent decades perfecting. But if I tried and came within 10-20% of their performance I'd throw myself a parade.
For a piston, the torque on the lever arm is not constant, so that seems a more difficult method.

I don't recognize it per se, but it is a rolled-up constant of unit conversions. The formula for power is just pressure times flow rate. In English units it's gpm×ft head/3960=bhp
Thanks for the replies russ_watters. To be quite frank I'm chasing my own tail. Honours year projects were predetermined and then allocated with a like it or lump it attitude. Ballpark figures at this stage are more than I could have hoped for at the start but I'm interested in exploring different methods and comparing the results. I calculated RPM at approximately 131 RPM (at the crank, so maybe 700-800 input considering their gear ratio). The reason I opted initially for the torque calculation was that I had already constructed a spreadsheet of calculations for the instantaneous torque at 1° intervals throughout the crank rotation so this was just a couple of columns extra and made it easy to graph the power throughout the cycle. Thanks for the formula, I'm digging through some books now in search of an SI equivalent.
 
Update after some calculator button mashing:

Power (W) = Pressure (Pa) * Flow Rate (m3s-1)

W->hp = 0.001341
Pa->psi = 0.000145
m3s-1->gpm = 15850.32

0.000145*15850.32/0.001341 = 1713.87

Thanks for the replies guys, I'll press forth.
 

russ_watters

Mentor
18,990
5,143
Thanks for the replies russ_watters. To be quite frank I'm chasing my own tail. Honours year projects ...
Oh, I assumed this was real world. School projects are mental exercises, so it's not a fool's errand (there is generally no expectation of significant success).

Why do you feel you are chasing your tail? What exactly are the requirements of the project? To me, if you succeed in designing a system that comes within 20% of a mature commercial product, that's a big win. So I d not see a problem here (except that your understanding of the engineering issues isn't good enough yet).
 
Oh, I assumed this was real world. School projects are mental exercises, so it's not a fool's errand (there is generally no expectation of significant success).

Why do you feel you are chasing your tail? What exactly are the requirements of the project? To me, if you succeed in designing a system that comes within 20% of a mature commercial product, that's a big win. So I d not see a problem here (except that your understanding of the engineering issues isn't good enough yet).
Apologies, I should have said that at the start!

As far as chasing my tail, I'm in Scotland, the story goes like this: I started out on a systems eng programme that allowed exit at any year with a qualification (over 4 years certificate, diploma, bachelors degree, honours). Through lack of students interested in the honours year they decided to scrap it. My options were to stay and graduate with a bachelors (students get government funding here so employers expect you to graduate with honours) or, transfer to a different university and a different programme. I've spent the past 2 years on a computer aided design programme at a different uni. Without an application the subject is very broad, there is little in the way of fundamental principles in favour of 100% tutorial based learning: "this is how you model xyz: step 1...". The mechanical students learn more FEA than we do and the aircraft students learn CFD (we don't even touch on it). Projects are a mixed bunch they can be anything from designing coffee mugs to designing something to do with detecting gamma rays in space. I really tried to get the coffee mug project purely for the testing scenarios I was imagining, the project coordinator wasn't sold. My project isn't so bad though. I just find myself doing the same section 2 or 3 times over before returning any meaningful results, that's more to do with how I learn than anything else really, I have to make the mistakes first otherwise it never burns in!

It was my decision to commit to the course so I'll see it through the last few months until the end. In September I return to my original university for direct entry into the final year of mech eng with a focus on renewable energy I'll have to pay the tuition fees because I'll no longer be eligible for the government funding however, I'll end up in the field that I wanted at first. :smile:
 
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You state a preference for SI units, but then your discussion is an odd mix of SI and Imperial units (power in hp, rod load in lbf, for example). Why is this? The system description will be exactly the same in either system of units, but consistency is important to avoid errors.
 
You state a preference for SI units, but then your discussion is an odd mix of SI and Imperial units (power in hp, rod load in lbf, for example). Why is this? The system description will be exactly the same in either system of units, but consistency is important to avoid errors.
Thanks for the reply Dr. D. That's a valid concern and I appreciate you highlighting it. The reason for the mixture of units is that the case study pump has all of it's technical documentation in imperial units, this is due to Weir's SPM sub division being based in Texas. I would generally be happy to work in either standard however, being at a UK university our research materials primarily use SI units. Values from the case study pump are set out at the start of the write-up as a guideline of what can ordinarily be expected of this type of pump. Values that have been generated using analytical methods and simulations are given in SI units however, selected figures are converted later at a comparison stage.

Would you suggest that it would be more appropriate to give the case study figures already converted to metric values? It wouldn't be any trouble to run through the write up and amend the figures used.
 
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Since I am originally from Texas and have actually worked with triplex pumps a bit, I would tend to favor US Customary units (nearly the same as Imperial units, but not quite) .... but that is really immaterial.

Over the years, I've developed a few general rules that I try to follow:
1) When asked to solve a problem, solve it in the units given;
2) If mixed units are involved in a problem statement, make a choice, usually the one that involves the fewest conversions;
3) If there are any electric machines involved (motors, generators, solenoids), always opt for SIs

For your university work, since you are in the UK, I would suggest that you convert to SI the US Customary units for the comparison machine for all parameters, with minimal rounding; state these once parameters once in both sets of units. There after, do all work and comparisons in SI.
 
Since I am originally from Texas and have actually worked with triplex pumps a bit, I would tend to favor US Customary units (nearly the same as Imperial units, but not quite) .... but that is really immaterial.

Over the years, I've developed a few general rules that I try to follow:
1) When asked to solve a problem, solve it in the units given;
2) If mixed units are involved in a problem statement, make a choice, usually the one that involves the fewest conversions;
3) If there are any electric machines involved (motors, generators, solenoids), always opt for SIs

For your university work, since you are in the UK, I would suggest that you convert to SI the US Customary units for the comparison machine for all parameters, with minimal rounding; state these once parameters once in both sets of units. There after, do all work and comparisons in SI.
Thank you for the advice Dr.D, that seems like a logical way of presenting the numbers. Just out of curiosity would you say generally that there is still a lot of employment in this industry in the US?
 
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I am pretty far removed from this industry now (my work with triplex pumps was 40 years ago), but I know that fracking is on the increase. I would imagine that there is considerable interest in such things right now.
 

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