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The problem of work performance in reciprocating compressors

2018-05-07View Original

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I would like to ask: what is the difference between cranking a reciprocating compressor at 2 MPa and 0.5 MPa with the return valve fully open, in terms of power generation? Personally, I think the difference is minimal. Can the work done by the piston in such cases be expressed as W = ΔP (pressure difference) * S (area) * L (stroke)? Is the increase in pressure, which leads to an increase in gas density, the fundamental reason for the increase in work done? (Waiting online: lol)
Reply #22018-05-07
Just take a look at the current changes and you’ll know.
Reply #32018-05-07
The current did increase; I’m just wondering why it increased :)
Reply #42018-05-08
You are right; under normal operation, the pressure difference determines the current. The higher the pressure, the greater the density, which means the mass of gas per unit volume increases, and thus more work needs to be done on that gas
Reply #52018-05-08
The higher the pressure, the more gas is compressed by the piston each time, which naturally results in a greater electric current

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