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- How does vacuum pump heat rejection vary with flow rate/load?

I'm working on a project to replace a couple of vacuum pump skids for a lab campus. The existing pumps are water-cooled, liquid-ring pumps. Since they are water cooled, there is effectively no heat rejection into the mechanical room(s) except through un-insulated exhaust pipes. The new pumps are rotary vane, air-cooled, so they will reject heat to the space. So, how much?

The vendor told me to figure 2,500 BTU/hr/hp, which is just another way of saying "all of it". But I don't think that's true. A couple of issues:

1. The vacuum exhaust is ducted (piped) out of the building. Now, since this volume is small and the piping isn't insulated, I'm comfortable saying it still gets rejected into the room. But it would still be nice to know what fraction goes into the airstream. But there may not be an easy way to find this out.

2. I'm pretty sure the heat rejection (and pump horsepower) varies with load, but I'm not certain of that or exactly how. I'm more familiar with compressors, so I'll use an analogy/comparison, to make sure I understand, and I'm looking for confirmation/correction:

Picture a reciprocating compressor (I assume the same thermodynamics applies to a rotary compressor/pump). It's just a piston in a cylinder. As the piston squeezes the air, the air heats up. A certain fraction of the work is held in the compressed air and the rest is released as heat. The amount of lost heat is significant, but it isn't 100%. In addition, the heat rejection changes with load. Consider the firsts stroke of the compressor to pump-up a surge tank/system. The gauge pressure is zero and the pressure rise/compression is pretty close to zero. The compression stroke has nearly zero force behind it, so the work done is near zero and the heat of compression is near zero. As the system pumps up, the pressure rises, force on the piston rises, and work done with each stroke rises. The work done and heat rejection increases as the system is pumped-up to operating pressure. Maximum work is done on the last stroke, and the system if operating at continuous maximum flow operates at maximum power and maximum heat rejection.

Assuming the above is correct:

[As I type this, I realize this is going in a different direction than I thought...]

The vacuum pump kind of works in reverse, but not exactly. The first stroke also does near zero work. But after that, we have two competing parameters:

1. Pressure difference increases with each stroke.

2. Input mass decreases with each stroke.

From this, I conclude: Peak work (power) for a vacuum pump system occurs somewhere in the middle of pump-down, when there is a significant pressure difference, but still a significant mass flow rate. Heat rejection increases as a fraction of power input as the system is pumped down. But once the system is pumped-down, the work done on the last stroke is effectively zero; There is a large compression ratio, but effectively zero volumetric flow rate. And even at maximum working flow rate, the power is still effectively zero.

Thoughts?

The vendor told me to figure 2,500 BTU/hr/hp, which is just another way of saying "all of it". But I don't think that's true. A couple of issues:

1. The vacuum exhaust is ducted (piped) out of the building. Now, since this volume is small and the piping isn't insulated, I'm comfortable saying it still gets rejected into the room. But it would still be nice to know what fraction goes into the airstream. But there may not be an easy way to find this out.

2. I'm pretty sure the heat rejection (and pump horsepower) varies with load, but I'm not certain of that or exactly how. I'm more familiar with compressors, so I'll use an analogy/comparison, to make sure I understand, and I'm looking for confirmation/correction:

Picture a reciprocating compressor (I assume the same thermodynamics applies to a rotary compressor/pump). It's just a piston in a cylinder. As the piston squeezes the air, the air heats up. A certain fraction of the work is held in the compressed air and the rest is released as heat. The amount of lost heat is significant, but it isn't 100%. In addition, the heat rejection changes with load. Consider the firsts stroke of the compressor to pump-up a surge tank/system. The gauge pressure is zero and the pressure rise/compression is pretty close to zero. The compression stroke has nearly zero force behind it, so the work done is near zero and the heat of compression is near zero. As the system pumps up, the pressure rises, force on the piston rises, and work done with each stroke rises. The work done and heat rejection increases as the system is pumped-up to operating pressure. Maximum work is done on the last stroke, and the system if operating at continuous maximum flow operates at maximum power and maximum heat rejection.

Assuming the above is correct:

[As I type this, I realize this is going in a different direction than I thought...]

The vacuum pump kind of works in reverse, but not exactly. The first stroke also does near zero work. But after that, we have two competing parameters:

1. Pressure difference increases with each stroke.

2. Input mass decreases with each stroke.

From this, I conclude: Peak work (power) for a vacuum pump system occurs somewhere in the middle of pump-down, when there is a significant pressure difference, but still a significant mass flow rate. Heat rejection increases as a fraction of power input as the system is pumped down. But once the system is pumped-down, the work done on the last stroke is effectively zero; There is a large compression ratio, but effectively zero volumetric flow rate. And even at maximum working flow rate, the power is still effectively zero.

Thoughts?