Stirlingengineforum.com

I have been reading about mechanical drives for beta engines at http://mac6.ma.psu.edu/stirling/drives/index.html

I am having trouble visualising the quantities VAR (Volume Amplitude Ratio) and VPL (Volume Phase Lag)

Quoting from the source

The volume amplitude ratio, VAR , is defined as the change of the volume of the compression space divided by the change in volume of the expansion space during a complete revolution. VAR = 1 is suggested by many researchers for this quantity.

Now, in my simple mind, in a beta configuration engine a fixed mass of working fluid (air in this case) is alternately compressed and expanded as the working volume cyclically increases and decreases during a single rotation of the crank. OK, consider the power piston at top dead centre, just about to begin
compression. Does the the change of the volume of the compression space not equal the swept volume of the cylinder as the piston heads towards BDC?

An now on the way back to TDC the change in volume of the expansion space must be the same swept volume. So how can VAR be anything but 1?

Refering to VPL the source states the volumetric phase lag , VPL , refers to the angular offset between the volume of the compression space as function of crank angle and that of the expansion space. VPL = 90° is an often cited optimal value with the compression space lagging behind the expansions space. For strictly sinusoidal function this definition of phase lag is unique for others one might look at the phase lag of the maxima and the minima of compression and expansion space.

Now reading this definition I am none the wiser. I am sure that what this really amounts to is the swept volumes of the power piston and the displacer and the phase lag between the position of the power piston and the position of the displacer.

Would I be correct in saying that VAR equals the ratio of the power piston swept volume to the displacer swept volume? If that is the case VAR could have any physical value, but experience suggests a value of 1.5 (that is the swept volume of the displacer should be 1.5 x the swept volume of the power piston, with larger values necessary in some cases, for example Low Delta T engines.?

Bob

Dear BobC

Over the last thirty two years I have wasted more time, money and materials than I care to admit to, building SEs in accordance with Academics (invariably unproven) theories. If you have very good heat exchangers and are prepared to run the heater bright red hot and precharge to 100 barg on Hydrogen you may well find a displacement ratio (Lambda) of 1:1 gives good power output. For ordinary mortals like myself the best I can target are reasonable heat exchangers, a few barg of Air and a just red heater so it doesn't burnout in a few hours because it's made of 304 SS. The big problem is that whereas the heater may be at 600c the average temperature of the air will probably be around half this temperature and as a concequence will have expanded too little to maintain a pressure difference across the compression piston during the power stroke.

If you have enough patience, some of this is covered in more detail in my thread 'Is anybody on this forum building Stirling cycle engines'. I also have an article I wrote for the UK SES which, if you email me, I would be happy to send to you.

Should you wish to check my credentials, in the gallery there is a picture of my recently developed Alpha and a link to a youtube video.

In answer to your original question, the displacement ratio will need to be 0.5 to 0.75 depending on the quality of your heat exchangers and the temperature at which you wish to run your engine. Be cautious when measuring the displacement ratio, on a Beta it is not the ratio of the swept volumes as the compression volume is invariably less than its swept volume.

Hope this helps.

GeoffV