Computing brew pressure of direct lever espresso machines

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rpavlis
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#1: Post by rpavlis »

It is fairly straight forward to compute the pressure with good accuracy that develops during a pull in direct lever machines. The following calculation utilises La Pavoni dimensions. The handle length may vary slightly from model to model, though most must be close to 27-28 cm.

Here is the computation:


This is a computation of the handle force required to produce 9 bar gauge pressure on coffee sample in portafilter of La Pavoni manual espresso machines.


Piston diameter: 44mm
Distance between lever pins 25mm
Length from fulcrum pin to end of handle: 275mm

9 bars = 9e5 Pascals
(1Pa = 1 N/m²)

For piston

A=πr²

A=π*2.2*2.2 cm²=15.205 cm² = 1.5205e-3 m²

Mechanical advantage of lever is 275/25.0=11.0

****


Thus the force required on piston is 9e5*1.5205e-3 or 1368 N

Because of 11.0 mechanical advantage, force required on end of handle is 1368/11 = 124.4 N

On Earth g=9.8 m/sec²

Gravitational force F=mg The downward force on a 1kg mass is thus 9.8N

Corresponds to force on 12.7 kg.

Hanging a mass of 12.7 kg on the end of the La Pavoni machine's handle should produce 9 bar gauge pressure provided the handle is exactly horizontal.


One can, of course, do similar computations for any direct lever machines.

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TomC
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#2: Post by TomC »

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yakster
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#3: Post by yakster »

That works out to be about 28 pounds. I guess if you have a feel for a 30 pound tamp, you'd be in the ballpark.
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number9
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#4: Post by number9 »

"Hanging a mass of 12.7 kg on the end of the La Pavoni machine's handle should produce 9 bar gauge pressure provided the handle is exactly horizontal. "

<s>

Note that the handle will only be horizontal for one instantaneous moment. At all other times it is not horizontal, and you will need to draw a free body diagram on the end of the handle and calculate the two component forces on the handle. One could, of course, integrate over the arc length of the motion of the handle and get an average force that you need on the handle and use that, instead of constantly changing your force on the handle in order to hold a constant force on the piston.

</s>

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dominico
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#5: Post by dominico »

number9 wrote: Note that the handle will only be horizontal for one instantaneous moment. At all other times it is not horizontal, and you will need to draw a free body diagram on the end of the handle and calculate the two component forces on the handle.
The effects of moment and angle for the ultimate brew pressure on a Pavoni are discussed here: Pressure Profiling a La Pavoni

Basically at the "effective angles" of the lever a constant 30lb of force will put you in the 8 bar to 9 bar ballpark.
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mgrayson
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#6: Post by mgrayson »

I'm a mathematician, not a physicist, so I could be confused, but if the rod connecting the piston to the lever is close to vertical throughout the pull, then the vertical force multiplier will stay constant even when the lever is close to vertical. Sure, less force is in the tangential direction, but the angle between the lever and the connecting rod increases the downward force by the same factor.

A downward force of f yields a tangential force of f*sin(theta), where theta is the angle from the vertical.
But a rotation by l*dtheta of the arm moves the piston down by (l/10)*sin(theta)*dtheta. The sin(theta)'s cancel out and the net vertical force on the piston is just 10*f. Of course, it isn't a perfect cancellation if the connecting rod isn't vertical, but it's not as bad as it looks at first glance.

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OldNuc
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#7: Post by OldNuc »

As posted above the practical result is the pressure varies from about 8 bar to max of a bit over 9 bar for that magic 30lb force. The angle the handle moves trough is not that great so you end up with a Dromedary shaped pressure hump.

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rpavlis (original poster)
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#8: Post by rpavlis (original poster) »

If one look carefully at the handle arrangement on these machines, one can see that if the force be applied tangential to the handle, the force needs to be increased by the secant of the angle of the handle off horizontal to keep a constant force on the piston rod. With constant tangential force, the force will otherwise fall by the cosine of the angle off horizontal.

However, if the force be applied vertically, the vertical component of the force applied to the piston remains at a constant ratio without regard to angle. The ratio of the horizontal distance between the pins and the horizontal distance between the fulcrum and the end of the handle is constant.

OldNuc
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#9: Post by OldNuc »

Most operators do not consistently apply that force vertically. It can be done easily if you think about it though. This might be the cause of the shot to shot inconsistency experienced by some people.

donn
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#10: Post by donn »

It seems to require fewer terms in English, where 1 "bar" is 14.5 PSI.

If the piston is 44 mm, that's 1.73 inches, area 2.36 in².
9 bar is 130.5 lbs/in², over 2.36 in² = 307.46 lbs.
Mechanical advantage 11 = 28 lbs., 12.7 kg.

I'm assuming that the pressure is relative, so 0 PSI would be ambient pressure, not a vacuum.

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