Relationship between the vacuum pump’s pumping speed and exhaust volume
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Hello, dear teachers! In many samples, the parameters of vacuum pumps that I’ve seen describe the \"gas extraction rate\", but sometimes users request the \"air discharge volume\" parameter instead of the gas extraction rate. I’m thinking that if there’s inhalation, then there must be exhalation; therefore, the rate of air intake and the volume of air expelled should be the same. But I think the displacement should be influenced by the pump chamber structure. Could the teachers also explain the relationship between the two? The user asked whether the “displacement” parameter could be selected using the exhaust rate. Or perhaps the “air displacement volume” mentioned by the user refers to the “air extraction rate””
I understand that the air extraction rate and the exhaust volume are in the same relationship
The pumping speed of a vacuum pump and the mass flow rate of the exhaust gas are the same, but the volume flow rate is different. Assuming that the temperatures at the inlet and outlet are the same, P1V1=P2V2; the values can be determined by converting between the inlet and outlet pressures
Personally, I think that if the customer is already familiar with positive-pressure equipment such as fans and air compressors, then the exhaust volume should refer to the suction volume.
Vacuum pump specifications usually mention the \"suction volume\" or simply the \"air volume\", which is what you refer to as the pumping rate. Of course, the same amount of gas is expelled as is drawn in; the difference lies in the pressure. The principle behind a vacuum pump is to compress the gas that is drawn in and then transport it, with the exhaust pressure being higher than the intake pressure. According to the gas equation P1V1=P2V2, the volume of gas expelled is smaller than the volume of gas drawn in, which means that the exhaust rate is lower than the intake rate.