Bernoulli Law Formula for Coffee Roasting

Bernoulli Law Formula for Coffee Roasting

This topic came up in the old “Huky Form” which is currently closed but still open for search results. https://www.hukyforum.com/index.php?topic=1091.0

If the maximum pressure is 3kPa does 1.5kPa produce 1/2 the heat output? Does 2kPa produce twice the heat of 1kPa? Maybe those assumptions are not entirely correct?

ROR, projected roast progress, trial and error are more likely predominate factors in choosing gas setting. If you worked out Artisan graphic values based on simple arithmetic like I did, they may not be accurate.

Here’s the full discussion and at the end of this post is an excel file using Eugenm’s formula with user input. I ended up downloading the excel file to Google Drive then opening in Google Sheets.

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AuthorTopic: Huky heat power output setting  (Read 815 times)

Offline eugenm

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Huky heat power output setting
« on: June 12, 2016, 10:03:16 AM »

First of all I would like to say hello to all members of this beautiful Huky community as this is my first post !

I would like to share a bit of knowledge gained in my line of profession that would help understand how gas flow varies with the total pressure of the system.

I guess at one point everybody asked the question how much should the pressure of the gas has to be lowered in order to lower the heat power by a certain amount. This can be easily calculated using Bernoulli’s hydrodynamic law.

Let us assume that at the beginning of the roast, when we drop the green beans in the drum the initial gas pressure is P1 (kPa or mbar or psi). This pressure corresponds to a gas flow F1 (m3/h or l/s or cfm) that generates a total heat power which we will denote Q1 (kW or BTU/h). The total heat power is directly proportional with the gas flow meaning Q2/Q2=F2/F1, where F2 is a new value for the gas flow which generates a total heat power Q2.
Bernoulli will help us understand that the ratio between the a new value of gas pressure and the initial gas pressure value is equal with the square of the ratio between the new gas flow value and the initial gas flow value : P2/P1 = (F2/F1)^2

Let us analyze a simple scenario : we have an initial gas pressure at which we charge the drum with green coffee beans of P1=2kPa. This pressure will corresponds to a gas flow F1 that generates a heat power Q1. We want now to reduce the heat power by 50% and we want to learn what the total gas pressure P2 would have to be :
We want to reduce the heat power by half which means Q2/Q1=0.5 ; That means the ratio between the new gas flow and the initial gas flow is also 0.5 : F2/F1 = 0.5
At what total gas pressure we will have the F2 gas flow ?
P2/P1=(F2/F1)^2=0.5^2=0.25 which means P2=P1*0.25=2*0.25=0.5kPa

The conclusion is that in order to reduce by half the initial heat power we need to reduce by 4 times the initial gas pressure !!!

Hope this would be of some help,
Eugen

 

Offline LTB

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Re: Huky heat power output setting
« Reply #1 on: June 12, 2016, 12:18:28 PM »

Waw ! Extremly intersting !

Welcome to you and thanks to share that bit of knowledge ! I already thought that the heat power between , lets say, 2 kPa and 2.5 kPa did not seem to be huge, but now I can understand why :)

 

Offline eugenm

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Re: Huky heat power output setting
« Reply #2 on: June 12, 2016, 12:37:11 PM »
Up to a point this can be seen on the color and size of the flame, which seems not to change when we apply small adjustments on the gas flow with the needle valve. We actually have to turn a couple of times the blue knob to feel a difference.
Indeed if we lower the pressure from 2.5 kPa to 2.0 kPa we will only reduce the power with approx. 10%

 

Offline day

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Re: Huky heat power output setting
« Reply #3 on: June 12, 2016, 11:12:53 PM »
Ha, great news. I recently purchased a Extech HD755, which has accuracy of approx +-.04 or .05″WC. I was a little worried and considered getting a %R scale, but in the end decided it should be solid, but now I can rest easy!  :)
« Last Edit: June 13, 2016, 10:38:45 PM by day »

 

Offline LTB

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Re: Huky heat power output setting
« Reply #4 on: June 13, 2016, 08:09:39 AM »

Indeed if we lower the pressure from 2.5 kPa to 2.0 kPa we will only reduce the power with approx. 10%

Hum… finally i think i did’nt get it at all…..

 

Offline edtbjon

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Re: Huky heat power output setting
« Reply #5 on: June 13, 2016, 11:21:20 AM »

Up to a point this can be seen on the color and size of the flame, which seems not to change when we apply small adjustments on the gas flow with the needle valve. We actually have to turn a couple of times the blue knob to feel a difference.
Indeed if we lower the pressure from 2.5 kPa to 2.0 kPa we will only reduce the power with approx. 10%

Even though it’s “only 10%” that minute change can drastically change the profile. E.g for a 400g charge, going through the middle phase at 2.5 vs. 2.0 kPa will differ by at least a minute, maybe even more. It is indeed a balancing act, which is also why it’s a good idea to start learning the Huky (and any drum roaster) with big charges, as they are “easier” to balance.

Now, dare I roast even lighter?

 

Offline eugenm

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Re: Huky heat power output setting
« Reply #6 on: June 13, 2016, 11:36:41 AM »
Hey LTB,
Let us denote the initial pressure P1=2.5kPa and the final pressure P2=2kPa. At P1 there will be a gas flow F1 that corresponds to a heat power Q1, at P2 there will be a gas flow F2 that corresponds to a heat power Q2. We want to calculate the final heat power Q2 related to the initial one Q1. Here is a picture with the equation :

So Q2 will be a little less then 90% of Q1, another way of saying is that from P1 to P2 the heat power goes down a bit more then 10%@edtbjon,
You are perfectly right, small changes of the total heat power will affect the roasting profile, no question about it. I was just thinking that a small calculation tool can help have an idea on how the total heating power varies with the total gas pressure.

 

Jump to:
=> Roasting

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I asked Apple our family mathematician if she could create an excel spreadsheet based on Eugenm’s formula. You can find the file below and hopefully get it working.

It’s very interesting at least and I’m wondering how the general roasting community, especially those with a scientific background make of it. Unfortunately Bernoulli’s law does not take into account the rate of heat absorption/radiation green coffee exhibits during the coffee roasting process.

coffee roasting bernoulli

Here’s the link to the Dropbox file, while I try to embed the excel file.

https://www.dropbox.com/s/7fsh16kd5cvrtph/coffee%20roasting%20bernoulli.xlsx?dl=0

 

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