E61 Preinfusion  Page 2
 Jake_G
 Team HB
Bernoulli would not agree.
Sorry.
Bernoulli's equation assumes zero frictional losses. The flow coefficient of valves and orifices describes these losses well. You are referring to the venturi effect, which is also occurring and is independent of friction. Bernoulli states that as a fluid speeds up it exchanges static pressure for velocity pressure. This is recoverable. Friction losses are not, and these are the losses that occur across a gicleur any time there is flow. The frictional loss is embodied as an irreversible pressure drop that is proportional to the square of the flow rate. Double the flow rate, quadruple the pressure drop.
Sorry.
Bernoulli's equation assumes zero frictional losses. The flow coefficient of valves and orifices describes these losses well. You are referring to the venturi effect, which is also occurring and is independent of friction. Bernoulli states that as a fluid speeds up it exchanges static pressure for velocity pressure. This is recoverable. Friction losses are not, and these are the losses that occur across a gicleur any time there is flow. The frictional loss is embodied as an irreversible pressure drop that is proportional to the square of the flow rate. Double the flow rate, quadruple the pressure drop.
Sorry. No.bluesman wrote:But as long as the flow rate at the outlet is unrestricted at working pressure, that pressure drop across the system (i.e between pump and "shower head" in the group) will remain constant. The force exerted by the water will change with the area of the outlet, but the pressure (ie the force per square area) will be the same at both ends minus only the slight differential required to force the water through the puck (which, as you point out, is why it flows at all).
while not a "hydraulics" engineer, I am a mechanical engineer with a strong background in fluid dynamics and hydraulics. How can I help?bluesman wrote:It would be great if an HB participant turns out to be an hydraulics engineer and will weigh in
LMWDP #704

 Supporter ♡
I'd love to hear your take on using bottom papers under a puck and why the flow seems to be so different and seemingly increases extraction around the edges of the puck.Jake_G wrote:
while not a "hydraulics" engineer, I am a mechanical engineer with a strong background in fluid dynamics and hydraulics. How can I help?
You know, since you're offering.
 bluesman
I really appreciate your knowledge, skil and experience, Jake. I've enjoyed and learned from your posts over the years. But Bernoulli's equation is applied with validity and utility every day to real world systems in which there is some friction. This discussion of pressure and flow in brew paths is one such illustration. When there's no resistance at all to flow at the outlet because the end orifice is so much larger in area than any other part of the path, there's a large pressure drop across the gicleur because it's too small to pass a flow volume sufficient to generate any back pressure downstream. The outflow velocity is essentially that driven by gravity.Jake_G wrote:Bernoulli would not agree. Sorry. Bernoulli's equation assumes zero frictional losses.
But as the resistance at the puck increases and pressure builds in the path, there's a slight drop across the gicleur because of the increased pressure throughout the flow path. Look at the measured pressures across the gicleur when the shot is coming to its end
Here's a useful description (from the BU physics department) of the phenomenon to which I refer:
"Bernoulli's equation has some surprising implications. For our first look at the equation, consider a fluid flowing through a horizontal pipe. The pipe is narrower at one spot than along the rest of the pipe. By applying the continuity equation, the velocity of the fluid is greater in the narrow section. Is the pressure higher or lower in the narrow section, where the velocity increases?
Your first inclination might be to say that where the velocity is greatest, the pressure is greatest, because if you stuck your hand in the flow where it's going fastest you'd feel a big force. The force does not come from the pressure there, however; it comes from your hand taking momentum away from the fluid.
The pipe is horizontal, so both points are at the same height. Bernoulli's equation can be simplified in this case to .
The kinetic energy term on the right is larger than the kinetic energy term on the left, so for the equation to balance the pressure on the right must be smaller than the pressure on the left. It is this pressure difference, in fact, that causes the fluid to flow faster at the place where the pipe narrows."
 Jake_G
 Team HB
Thanks for the kind words, David!bluesman wrote:I really appreciate your knowledge, skil and experience, Jake. I've enjoyed and learned from your posts over the years. But Bernoulli's equation is applied with validity and utility every day to real world systems in which there is some friction.
I think we are essentially saying the same thing, but focusing in at a different scale.
Also, nice article by Michael Teahan, but he makes an inaccurate conclusion.
I'll start my friendly rebuttal/clarification with another reference:
Calculation of Flow through Nozzles and Orifices
The calculations start with Bernoulli, but they are augmented to account for frictional and viscous losses, which are nonrecoverable. It's those little eddy currents after the orifice that mess up everything Bernoulli figured out.
The pressure at the Vena Contracta is accurately predicted using Bernoulli's equation, but ignoring the viscous friction losses, you'd assume that the pressure would fully recover (and Bernoulli calculates that it will), but it doesn't. How much pressure is lost is a function of the flow rate. At zero flow rate, there is no pressure drop at all. And at average espresso flow rates, the pressure drop is less than 1 bar. But the problem is that espresso does not flow at an average rate. It starts out nearly choked, and then speeds up as solids are extracted from the puck and IRA resistance is reduced. As such, the pressure drop across the gicleur starts off very low, but depending on your grinder and the style of shot you are pulling, you could conceivably have a pressure drop of 1 bar or more across the gicleur by the end of your shot.
A typical shot for me might have 6s of dwell before any drops. Then comes a rather linear increase from 7s to 26s where 36g appear in the cup. So 20s for 36g of liquid. That's an average flow rate of 1.8g/s but since it is always speeding up and starts at zero, the actual final flow rate is roughly twice that, or 3.6g/s.
Reviewing this chart below:
You can see that this theoretical shot goes from basically zero flow and full pressure (assuming the pump was set to 9 bar) and ends at a spot on the curve where you've got around 7.5 bar at the puck.
What strikes me as odd is that in Michael's article, and the image you posted, he shows the puck pressure increasing at completion and the effective orifice diameter decreasing. This makes no sense. If solids are being extracted from the puck, it will offer less resistance to flow and the drop across the gicleur will be greater, not less.
All the info you linked to regarding wetting and preinfusion is spot on. You start with your water debit as the flow rate and the pump pressure as the pressure drop and as the puck begins resisting flow, the drop across the gicleur gets lower and lower, but never zero.
Aside from the PI chamber and spring, the gicleur is solely responsible for what the ramp up rate to the final brew pressure looks like. The thought that the puck "takes over" as the primary resistance to flow is also accurate, but as the puck gives up solids, the flow rate increases and the gicleur contributes more and more to the overall pressure drop.
Again. I think most people say "Doesn't matter" or "Basically zero", and I say in my nerdiest of nerd voices "Actually, at a flow rate of 2.673ml/s, the pressure drop across a 0.5mm gicleur is 0.876 bar, sir."
Cheers!
 Jake
LMWDP #704
 bluesman
As an old auto racer who's built many cars, I've always been interested in the effects of shape on flow. Everything from carb intake horns and jets to fuel lines and fittings to combustion chambers and exhaust headers to parts protruding into the airstream can be optimally contoured for maximum flow with appropriate resistance (which varies from minimal to an ideal value eg for exhaust back pressure).
I've played with chamfering, rounding, and polishing the orifices on gicleurs, but only informally when cleaning them. I've always wondered if the actual bore shape and the contour of the openings could be optimized for best results in the cup. The turbulence you describe is a real phenomenon that is usually best minimized wherever encountered  it's one cause of cardiac problems when a heart valve is leaky or rough. I suspect that it causes more local scaling on the end(s) of the gicleur and elsewhere, just as it causes or aggravates bacterial deposition in the heart (the disorder known as bacterial endocarditis).
Maybe a gicleur bore shaped like a trumpet bell inside would give smoother preinfusion and pressure dynamics so the puck doesn't get slammed as hard. I haven't seen anyone try this in a measured, controlled way. I used countersinks and tiny files to clean and reshape old ones when playing with Oscar 1  but I never evaluated the effect (if any) in the cup beyond casual drinking and unrecorded observations. If I were so inclined, a proclivity much in decline once you pass 70, I'd test that theory properly. But having downsized to an apartment and sold my shop, I'll leave it to others to pursue.
I notice that most diagrams like the ones we show depict the gicleur as a hole in a a thin barrier plug. I wonder of the length of the bore could affect the product. Maybe a longer, wider bore of proper shape would improve flow characteristics enough to taste. If the bore were longer, it could be wider and more carefully shaped to create improved pressure gradients. The added length would lend some more friction to counteract the wider bore. And surface finish with very fine microroughness would facilitate a boundary layer and make laminar flow smoother.
In the immortal words of Shel Silverstein, I'll just have myself another espresso
I've played with chamfering, rounding, and polishing the orifices on gicleurs, but only informally when cleaning them. I've always wondered if the actual bore shape and the contour of the openings could be optimized for best results in the cup. The turbulence you describe is a real phenomenon that is usually best minimized wherever encountered  it's one cause of cardiac problems when a heart valve is leaky or rough. I suspect that it causes more local scaling on the end(s) of the gicleur and elsewhere, just as it causes or aggravates bacterial deposition in the heart (the disorder known as bacterial endocarditis).
Maybe a gicleur bore shaped like a trumpet bell inside would give smoother preinfusion and pressure dynamics so the puck doesn't get slammed as hard. I haven't seen anyone try this in a measured, controlled way. I used countersinks and tiny files to clean and reshape old ones when playing with Oscar 1  but I never evaluated the effect (if any) in the cup beyond casual drinking and unrecorded observations. If I were so inclined, a proclivity much in decline once you pass 70, I'd test that theory properly. But having downsized to an apartment and sold my shop, I'll leave it to others to pursue.
I notice that most diagrams like the ones we show depict the gicleur as a hole in a a thin barrier plug. I wonder of the length of the bore could affect the product. Maybe a longer, wider bore of proper shape would improve flow characteristics enough to taste. If the bore were longer, it could be wider and more carefully shaped to create improved pressure gradients. The added length would lend some more friction to counteract the wider bore. And surface finish with very fine microroughness would facilitate a boundary layer and make laminar flow smoother.
In the immortal words of Shel Silverstein, I'll just have myself another espresso
 Jake_G
 Team HB
My guess is that a perfect, bellshaped gicleur would be transparent to the flow up until you hit choked flow, which is where the velocity through the bore hits sonic speeds. At that point, the gicleur becomes a "speed limit" and does not allow any more flow through, regardless of the inlet pressure.
Below this speed limit, Bernoulli has it right and it is as if the gicleur isn't even there.
Assuming one designed a bellshaped jet to limit the flow rate to 4ml/s, you would have a 4ml/s water debit and fill rate to begin wetting and preinfusion with zero pressure at the puck, as you would get with a typical gicleur. Once the puck begins generating back pressure, the pressure drop across the jet would shift, so long as the flow rate is not less than 4ml/s. Whatever pressure the puck is generating, the jet would bear the other portion, up to whatever pressure the pump bypass/OPV is set to. So, if the puck is absorbing water at 4ml/s and generating 2 bar of pressure whilst doing so, the jet would be choking the flow from the pump and the pump would hum along at 9 bar, with a 7 bar drop across the jet.
Now, the instant the puck begins to choke the flow down to less than 4ml/s, the jet would become invisible to the flow, since it would no longer be choking the flow. So at that moment, the puck pressure would hit 9 bar, rather abruptly. This of course assumes the puck provides enough resistance to limit the flow to less than the choked flow of your jet, but that's a reasonable assumption.
It would indeed be interesting to do some modeling and experimentation to determine what this transition looks like and if it is a good or bad thing for the puck, but my guess is that the transition behavior that those eddy currents give us in a traditional gicleur is probably better.
Cheers!
 Jake
Below this speed limit, Bernoulli has it right and it is as if the gicleur isn't even there.
Assuming one designed a bellshaped jet to limit the flow rate to 4ml/s, you would have a 4ml/s water debit and fill rate to begin wetting and preinfusion with zero pressure at the puck, as you would get with a typical gicleur. Once the puck begins generating back pressure, the pressure drop across the jet would shift, so long as the flow rate is not less than 4ml/s. Whatever pressure the puck is generating, the jet would bear the other portion, up to whatever pressure the pump bypass/OPV is set to. So, if the puck is absorbing water at 4ml/s and generating 2 bar of pressure whilst doing so, the jet would be choking the flow from the pump and the pump would hum along at 9 bar, with a 7 bar drop across the jet.
Now, the instant the puck begins to choke the flow down to less than 4ml/s, the jet would become invisible to the flow, since it would no longer be choking the flow. So at that moment, the puck pressure would hit 9 bar, rather abruptly. This of course assumes the puck provides enough resistance to limit the flow to less than the choked flow of your jet, but that's a reasonable assumption.
It would indeed be interesting to do some modeling and experimentation to determine what this transition looks like and if it is a good or bad thing for the puck, but my guess is that the transition behavior that those eddy currents give us in a traditional gicleur is probably better.
Cheers!
 Jake
LMWDP #704

 Team HB
There will always be a pressure drop across the gicleur if water is flowing, but at some point while making an espresso, after the group is full and the coffee is saturated and pressure has built, it's likely the the effective pressure drop across the gicleur will be effectively zero, or at least so low compared to the drop across the puck as to be ignorable. Or that's what I believe to be true.
 Jake_G
 Team HB
It depends 100% on the flow rate.
The gicleur does not care, nor really experience anything else. The flow rate will always dictate the pressure drop. The easiest way to figure out the relationship is by measuring the water debit and observing the pump pressure. Then you calculate a static C_{v} as equal to the flow divided by the square root of the pressure. That value will be constant and you can calculate the pressure drop by dividing the flow by the C_{v} and then squaring the result.
So, imagine a machine with a water debit of 4ml/s @ 9 bar.
C_{v} = 4/3 = 1.33
What's the pressure drop at 1ml/s?
ΔP=(1/1.33)^{2} = 0.56 bar
2ml/s?
ΔP=(2/1.33)^{2} = 2.25 bar!
Now, it turns out that the gicleur needed to give such a low flow rate is quite small, but still...
Take my GS/3.
Water debit is 7ml/s @ 9 bar
C_{v} = 7/3 = 2.33
Pressure drop @ 1 ml/s
ΔP=(1/2.33)^{2} = 0.18 bar
Basically nothing.
Pressure drop @ 2 ml/s
ΔP=(2/2.33)^{2} = 0.73 bar
Still pretty low.
Pressure drop @ 4 ml/s
ΔP=(4/2.33)^{2} = 2.9 bar
Now the puck only sees 6 bar, even though the panel gauge still shows 9...
The gicleur does not care, nor really experience anything else. The flow rate will always dictate the pressure drop. The easiest way to figure out the relationship is by measuring the water debit and observing the pump pressure. Then you calculate a static C_{v} as equal to the flow divided by the square root of the pressure. That value will be constant and you can calculate the pressure drop by dividing the flow by the C_{v} and then squaring the result.
So, imagine a machine with a water debit of 4ml/s @ 9 bar.
C_{v} = 4/3 = 1.33
What's the pressure drop at 1ml/s?
ΔP=(1/1.33)^{2} = 0.56 bar
2ml/s?
ΔP=(2/1.33)^{2} = 2.25 bar!
Now, it turns out that the gicleur needed to give such a low flow rate is quite small, but still...
Take my GS/3.
Water debit is 7ml/s @ 9 bar
C_{v} = 7/3 = 2.33
Pressure drop @ 1 ml/s
ΔP=(1/2.33)^{2} = 0.18 bar
Basically nothing.
Pressure drop @ 2 ml/s
ΔP=(2/2.33)^{2} = 0.73 bar
Still pretty low.
Pressure drop @ 4 ml/s
ΔP=(4/2.33)^{2} = 2.9 bar
Now the puck only sees 6 bar, even though the panel gauge still shows 9...
LMWDP #704
 Jeff
 Team HB
4 g/s is pretty common for "turbo" shots and not uncommon for extraction phases of profiles with a moderate or long hold.