Portable compressor power calculation

AI Thread Summary
The discussion focuses on calculating the theoretical power needed for a small portable two-stage compressor, revealing a significant discrepancy between the calculated 16 watts and the actual motor power of 400 watts. The theoretical power is derived from an adiabatic compression equation, but the low flow rate contributes to this lower estimate. Factors such as mechanical losses, inefficiencies, and heat dissipation during operation are highlighted as reasons for the increased power requirement. Additionally, an alternative calculation based on isentropic enthalpy change suggests a power requirement of about 30 watts, with a high final temperature of 1363 K indicating substantial energy loss. Overall, the conversation emphasizes the challenges in accurately predicting compressor power needs due to inherent inefficiencies.
joe609
Messages
1
Reaction score
0
Hello

I have been trying to calculate the theoretical power required for compression of a small portable 2 stage 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 Px = 14 Bar
k = 1.4
N = number of stages
Motor power = 400 w

Using the equation for power during adiabatic compression

\large \large p=\frac{k}{k-1}P_{1}QN \begin{bmatrix} \left ( \frac{P_{x}}{P_{1}} \right )^{\left ( \frac{k-1}{k} \right )} & -1 \end{bmatrix}

\large \large p=\frac{1.4}{1.4-1}10^{5} \times 2\times 10^{-5} \times 2 \begin{bmatrix} 14^{0.3}-1 & \end{bmatrix}


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.
 
Engineering news on Phys.org
What is the source of that equation? you can't even run a typical light bulb on 16 watts
 
The flow rate of the compressor is very low, keeping the theoretical power requirements low. Calculating the power required based on isentropic enthalpy change between the initial and final states results in about 30 watts, and a final temperature of 1363 K (assuming the input temperature is 300 K). The efficiency of such a compressor would be low because a lot of energy is lost to heat; cooling the output air from 1363 K back down to 300K is lost energy.
 
Posted June 2024 - 15 years after starting this class. I have learned a whole lot. To get to the short course on making your stock car, late model, hobby stock E-mod handle, look at the index below. Read all posts on Roll Center, Jacking effect and Why does car drive straight to the wall when I gas it? Also read You really have two race cars. This will cover 90% of problems you have. Simply put, the car pushes going in and is loose coming out. You do not have enuff downforce on the right...
I'm trying to decide what size and type of galvanized steel I need for 2 cantilever extensions. The cantilever is 5 ft. The space between the two cantilever arms is a 17 ft Gap the center 7 ft of the 17 ft Gap we'll need to Bear approximately 17,000 lb spread evenly from the front of the cantilever to the back of the cantilever over 5 ft. I will put support beams across these cantilever arms to support the load evenly
Thread 'What's the most likely cause for this carbon seal crack?'
We have a molded carbon graphite seal that is used in an inline axial piston, variable displacement hydraulic pump. One of our customers reported that, when using the “A” parts in the past, they only needed to replace them due to normal wear. However, after switching to our parts, the replacement cycle seems to be much shorter due to “broken” or “cracked” failures. This issue was identified after hydraulic fluid leakage was observed. According to their records, the same problem has occurred...
Back
Top