Power required for compression

[joemde]

Hello

I have been trying to calculate the theoretical power required for compression of a small portable 2 stage air compressor with the following spec.

Volume flow rate Q = 2x10^-5 m³/s
Inlet pressure P₁ = 1 Bar (10⁵Pa)
Outlet pressure = 200 Bar
interstage pressure P_x = 14 Bar
k = 1.4
N = number of stages
Motor power = 400 w

Using the equation for power during adiabatic compression I get a theoretical power of 16 watts. Why the huge discrepancy between the theoretical and the actual power required? Is my calculation correct? I am aware there are mechanical losses and inefficiency's in the pump but are they so great as to require 25 x the power?
Thanks very much for any help.

 

[zdas04]

Give me a break. You really think you are doing 200 compression ratios in a "small portable 2 stage compressor"? Not happening on this planet. Try again.

David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.

 

[joemde]

Why not? You can buy hand pumps that do it..

 

[pmover]

like zdas04 stated, double-check your input parameters.

secondly, a pump and compressor are not the same device and suggest you not confuse the two.

thirdly, homework assignments are not permitted in this website. please read the site policies.

good luck!
-pmover

 

[dcasto]

Lets see, you are building a heating source to ignite a methane flame?
the discharge temps on your homework are exceeding 1000F

 

[tomato5]

joemde sir, if you havent found a reaonable answer, I will do my best to help you dont worry about smartalec answers

 

[BigInch]

tomato, Go for it.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[zdas04]

Yeah, Tomato explain to us how you can do 200 compression ratios in two stages in a "small" (and therefore unlikely to be intercooled) 2-stage recip. The three people who provided useful albeit off the cuff responses that you so calmly call "smartalec answers" have something like 110 years of compression experience among them, maybe you have more. Seems like I would have come across a wise, helpful 130 year old Engineer sometime in my brief 35 years of messing with compressors, but maybe not.

I sit with bated breath awaiting your pearls of wisdom without price.

David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.

 

[joemde]

Hi tomato5, thanks for the offer of help, and everyone else for the responses. It turns out i was mistaken in the original post. Technically the compressor is three stage as the inlet pressure is '7 Bar'.

Each stage is intercooled. I was just trying to figure out the power required assuming adiabatic compression, given the flow rate stated.

Thanks again.

 

[BigInch]

Still a way to go.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[rakuday]

Joemde,

What is the application of this compressor? The flow rate is negligible for any real compressor that can be built. As others have mentioned, check your input parameters.

The problem statement clearly indicates it to be school homework. You first talk of compressor and then hand pumps for such high pressures, get your fundamentals right even to do well in your school.

 

[tomato5]

zdas04 now he has an answer, and does not need more help but if he did. i woud provide it to him to the best of my ability whatever you have said in your response, could have been very nicely and politely said in about 1/3rd the number of words sarcasm does not befit us just creates friction be honest say what your technical mind says and on your way to new challenges

 

[iainuts]

Hi joemde,
The machine you have sounds a bit like the Rix Microboost. Inlet pressure 0.2 to 1.7 barg, discharge 150 barg, 3 stages, 0.37 kW, 2x10^-5 m³/s which I'm assuming is the 1'st stage volumetric displacement.

http://compressors.rixindustries.co...50?&plpver=10&origin=keyword&by=prod&filter=0

Given your requirements of 7 barg inlet, 200 barg outlet, 3 stages and I'm assuming air or similar diatomic gas, the power required is around 0.1 kW. Even the original 1 barg inlet, 200 barg discharge with 2 stages isn't completely ridiculous. Most of the machines I work with have compression ratios of 8:1. They include metal diaphragm machines and hydraulic intensifier designs such as the Hydro Pac. I also design cryogenic reciprocating machinery in the 100 hp range with 2 stages that can easily do 8:1 compression ratios and often slightly higher. I'm not so surprised by the high compression ratio.

To determine power required, I would suggest using a thermophysical properties database such as Refprop. You can then pull enthalpy values for your gas directly from the database and use those to determine power required for each stage.

To answer your question, "Why the huge discrepancy between the theoretical and the actual power required?" I don't think there's such a large discrepancy. Perhaps your volumetric displacement is off? I'm assuming you don't mean standard cubic meters per second which would be an even smaller machine. I've assumed the first stage displacement is 2x10^-5 m³/s.

You'll find that manufacturers who make reciprocating machines that are as small as this one (and even much larger) have to consider a trade between motor size and adding relatively large flywheels since most of the power required for a recip is only needed for a short period of time at the very end of a stroke when the gas pressure has risen close to the discharge pressure and must be expelled from the cylinder. On larger machines, manufacturers will install flywheels, but for smaller machines those flywheels aren't as cost effective as simply increasing motor size by a factor of 3 or more. It's very common to find small, reciprocating machines like this with relatively large motors for that reason.

 

[joemde]

Iainuts, thank you so much for your answer! Having recalculated the power required, and taking into consideration what you said about increasing motor size in the absence of a flywheel, i get a much more reasonable value.

Thanks again for the helpful post and happy new year!