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 Inläggsrubrik: Carburettor Jet size
InläggPostat: fre 08-01-25 23:10 
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Hi

1/ Wikipedia
---------------
When reading the excellent topic "E85_in_standard_engines" http://en.wikipedia.org/wiki/E85_in_standard_engines , I see that they recommend to enlarge jet size by some 25-30% and jet diameter by some 13%
sic "An excellent starting point, if one doesn't want to experiment with multiple test trials over the 25% to 30% range, is simply to increase the fuel flow by 27%, which just requires increasing the diameter of the jets by 13%."

I also read at the very last line sic "E85 ... liquid density of approximately 0.77 - 0.79 compared to gasoline ... density of 0.72 - 0.78"

2/ This excellent :P etanol nu forum and website
-------------------------------------------------------
In this forum, international part "burning temp of E85 vs Benzin", http://etanol.nu/forumrecover/viewtopic ... ec39a9de0e post dated 30May07 9:04 Aryan states sic "Ethanol is heavier than gasoline. The mass of a liter of gasoline is ca 737g at 15°C, the density of E85 is 780gr/l at 15°C. This means that we need (737/780) * (14,7/9,765) = ca 1,4 times as much volume of E85 per amount of air. "

Which means that for a carburetted car, jet surface should be increased by some 40% logically leading to jet diameter increased by some 18%. Such a 40% value appears as the de facto value in http://etanol.nu/ombygg-forgasare.php

3/ Question 1
---------------
Why such a huge difference between 2 well recognised E85 web sites ?
Wikipedia related post suggests to increase diameter by some 13% while this website suggests some 18%

4/ Question 2
-----------------
When I make the calculation
(737/780) * (14,7/9,765) = 1.422 and not 1.40

In addition I read very often that at 15°C gasoline density is between 0.72 and 0.78 with 0.75 as an average
E85 is betwen 0.77 and 0.79 with 0.78 as an average
So such values lead to increase fuel by
(750/780)*(14.7/9.765) = 1.447 # 1.45

which is farther from 40%

So
why did you select gasoline average density of 737 and not 750 ?
which is the more appropriate value for fuel increase 40, 42, 45% ?
Best
Bye


Senast redigerad av Ceyal35 lör 08-01-26 13:59, redigerad totalt 4 gånger.

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I agree with what you say. The volume ratio between E85 and gasoline should be about 1,48 according to my calculations. The only practical test of this on my part is my conversion of a 6hp lawnmower where I increased the main jet from 0,67 mm to 0,8 mm which gives a factor of 1,42 in area increse. The engine worked just fine on E85 after that. se here http://etanol.nu/forumrecover/viewtopic.php?p=5810#5810
However, it is not only the jet area which is important. If the fuel viscosity plays a role that could also affect the amount of fuel reaching the engine. I have not investigated that question.

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Hello Ceyal35 and GL,
I have no experience with carburator tuning + Ethanol --> theory only:

@1:
http://www.megamanual.com/flexfuel.htm
AFR_E0 = 14.7:1
AFR_E85 = 9.7:1
Definition of AFR is: mass_air / mass_fuel , the unit of mass is [kg]
The engine (gas pedal) is limited by flown mass of air -->
--> 14.7/9.7 = 1.52 --> you need 52% more mass of E85 than E0 to achieve same lambda than before.
Today we have E5 in most EU countries, when your engine was dsegined ETBE and later MTBE was up to 5% in petrol.


@2:
Aryan considered density, because we think/talk/calculate fuel consumption in volume units.
--> 737kg/m^3 / 780kg/m^3 * 14,7 / 9,7 = 1.43 you need 43% more volume of E85 than E0 to achieve same lambda than before.



@3:
The main difference are the units



@4:
Don't get confused by some % deviation. Petrol is a mixture of a lot of fluids --> datas depends on source, raffinery, quality, MOZ/ROZ... some standrads (ISO, DIN???) set limits within a 'fluid' can be called petrol/gas Eurosuper95 or what ever.
Ethanol is just one stuff (in theory). There is something added to avoid the natural limit of water (Aceotrop) and for taxes reasons. After all units of operation (processes) it isn't pure E100. Some other substances contaminate Ethanol, for example acids, catalytic converters... but of corse just a very minor part --> datas are more narrow (also standardized)




Finally I'd like to confuse you compeltely because of my calculation. :)
In my opinion it's not allowed just to increase area proportional wanted flow.

I made a calculation to estimate average jet speed:
Given:
Fuel consumption: V_mpg_E0 = 6.66666l/100km = 4l/60km = 4000ccm/60km
car speed: v_car_average = 60km/h
jet Ø: Ø_jet_E0 = 1.2mm = 0.12cm

Calculated:
Fuel flow: V_dot_E0 = V_mpg * v_car_average = 4000ccm/60km * 60km/h
V_dot_E0 = 4000ccm/h = 1.11111ccm/s

Jet area: A_jet_E0 = Ø_jet_E0^2 * pi / 4 = 0.12cm^2 * pi / 4
A_jet_E0 = 0.01131cm^2

Fuel flow through jet: V_dot_E0 = v_jet_E0 * A_jet_E0 -->
v_jet_E0 = V_dot_E0 / A_jet_E0 = 1.111111ccm/s / 0.01131cm^2
v_jet_E0 = 98.2cm/s ~ 1m/s

Reynolds, a characteristic number to compare flow/streaming under different, but similar conditions:
Reynolds: Re = v_jet * Ø_jet / ny

ny, cinematic viscosity [mm^2/s], don't mix with dynamic viscosity eta [Pa·s][kg/s/m] (eta = ny * density)
ny_E100 = 1.52E-6m^2/s
ny_E0 = 0.6E-6m^2/s (petrol)
ny_E85_estimated = [0.85*1.52 + 0.15*0.6]E-6m^2/s = 1.382E-6m^2/s


Re_E0 = v_jet-E0 * Ø_jet_E0 / ny_E0
Re_E0 = 1m/s * 1.2E-3m / 1.52E-6m^2/s
Re_E0 = 790


In my opinion Re_E0 = E_85 should be fulfilled - that's the key of my idea:

Re_E85 = v_jet_E85 * Ø_jet_E85 / ny_E85
v_jet_E85 = V_dot_E85 / A_jet_E85
V_dot_E85 = V_dot_E0 * 1.27 (1.25...1.30)
A_jet_E85 = Ø_jet_E85^2 * pi / 4

Re_E85 = [V_dot_E85 / ( Ø_jet_E85^2 * pi / 4) * Ø_jet_E85] / ny_E85
Re_E85 = [4 * V_dot_E85 / Ø_jet_E85 / pi] / ny_E85
Re_E85 = 4 * V_dot_E85 / Ø_jet_E85 / pi / ny_E85
Ø_jet_E85 = 4 * V_dot_E85 / Re_E85 / pi / ny_E85
Ø_jet_E85 = 4 * V_dot_E0 * 1.27 / Re_E0 / pi / ny_E85
Ø_jet_E85 = 4 * 1.11111E-6m^3/s * 1.27 / 790 / pi / 1.382E-6m^2/s
Ø_jet_E85 = 0.0016m = 1.6mm

Ø_jet_E85 / Ø_jet_E0 = 1.6mm / 1.2mm = 1.333333
Ø of E85 jet should be 33% bigger than previous.


If you consider that Ethanol reduces engine temprature you could lean out a little bit.

Never mind of all read numbers. I'm convinced your carburated engine does not provide YOUR wanted fuel map. Carbs approximate just some requirements: idle...tested at MOT (IMHO stupid, worthless), WOT....enriched to get max power, avoid to damage the engine, partial load....depends on luck how well your carb is tuned!

A 2CV friend of mine tried to find out best 'map' with a changeable jet + NB O2 sensor. He could just find compromises --> EFI is the way to go if you're interested in a serious mixture.

»Horst

PS: I doublechecked my calcs but of corse anything could be wrong - no waranty!

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I checked density specifications from OKQ8 which is Swedens largest oil company and they specify a density for
petrol class 1 with 5% ethanol to 745kg/m3
http://www.okq8.se/data/internal/data/1 ... in_MK1.pdf

E85 has according tothe same supplier a deisity of 780 kg/m3
http://www.okq8.se/data/internal/data/1 ... %20E85.pdf

(E85 "Vinter"= winther quality is essencially E75 and has a density of 770 kg/m3)

So a propper calculation would be (745 / 780) * (14,7 / 9,76500) = 1,43782741

However a petrol car is not really designed for E05 but for a purer petrol quality with a higher density, which could have been as low as 680 - 720 kg/m3.

Which would result in
(680 / 780) * (14,7 / 9,76500) = 1,31237938
(720 / 780) * (14,7 / 9,76500) = 1,38957816

So the "truth" is somewhere between 1,4-1,3 like also Horst calculations indicate. However there are many things that can compromise a theoretical calculation, for example many carburetted cars run rich at WOT and lean a halv throttle so one might have to experiment a little.

That is also why I made it possible on the caculation page http://etanol.nu/ombygg-forgasare.php
to enter your own percentage.

/Aryan

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ami8i skrev:
Hello Ceyal35 and GL,
I have no experience with carburator tuning + Ethanol --> theory only:

---------
Reynolds, a characteristic number to compare flow/streaming under different, but similar conditions:
Reynolds: Re = v_jet * Ø_jet / ny

ny, cinematic viscosity [mm^2/s], don't mix with dynamic viscosity eta [Pa·s][kg/s/m] (eta = ny * density)
ny_E100 = 1.52E-6m^2/s
ny_E0 = 0.6E-6m^2/s (petrol)
ny_E85_estimated = [0.85*1.52 + 0.15*0.6]E-6m^2/s = 1.382E-6m^2/s


Re_E0 = v_jet-E0 * Ø_jet_E0 / ny_E0
Re_E0 = 1m/s * 1.2E-3m / 1.52E-6m^2/s
Re_E0 = 790


---------
»Horst

PS: I doublechecked my calcs but of corse anything could be wrong - no waranty!


Horst, you used the etanol figure for ny_E0. Using ny_E0 = 0.6E-6m^2/s (petrol) instead gives Re_E0 = 2000. Does that affect your further calculations?

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InläggPostat: mån 08-01-28 23:06 
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Hi
1/ Aryan: thanks very much for the explanation
In http://en.wikipedia.org/wiki/Gasoline : The weight-density of gasoline is given as 737.22 kg/m³

I fully understand the calculations based on stoechiometric AFR (calculated in volume instead of mass)

So from such calculation the enrichment value should be in the 40% in volume and not between 25-30% as given by Wikipedia "E85 in standrd engines"

2/ GL : tanks for your testimony
As far as your last comment is concerned, I crosscheck the value of cinematic viscosity of gasoline in various publications

In (sorry in French) http://fr.wikipedia.org/wiki/Viscosit%C3%A9 : dynamic viscosity is 1.2 E-3 for ethanol and 0.65 E-3 for Petrol at 20°C

In the english counterpart http://en.wikipedia.org/wiki/Viscosity at 25°C : dynamic viscosity is 1.074 E-3 for Ethanol .. not given for gasoline

In (sorry in French) http://www.econologie.com/articles.php? ... =646&prt=2 Gasoline Viscosity: 0.5 up to 0.75mm²/s i.e 0.5 up to 0.75 E-3 at 20°C

so Values used by Ami8 are the correct ones

We see from the above that both dynamic and cinematic viscosities of Ethanol are roughly twice as musch Gasoline viscosity ..


3/ Ami8 Thanks also for the explanation and calculation
I understand
- that you take into account cinematic viscosity which is an interesting and brilliant idea
- that the key element is V_dot_E85 = V_dot_E0 * 1.27 (1.25...1.30)
- the other key element being that the two Reynolds number should be equal Re_E0 = Re_E85

Where do you draw the 1.27 factor ? from the amount of Ethanol to produce the same heat as Gasoline ?

I do not see any relationship with steochiometric AFR for both Ethanol and Gasoline ? Is that normal ?

I need to further think about your interesting and brilliant idea before being able to firther comment
Best
Bye


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Ceyal35 skrev:
Hi
1/ Aryan: thanks very much for the explanation
In http://en.wikipedia.org/wiki/Gasoline : The weight-density of gasoline is given as 737.22 kg/m³

I fully understand the calculations based on stoechiometric AFR (calculated in volume instead of mass)

So from such calculation the enrichment value should be in the 40% in volume and not between 25-30% as given by Wikipedia "E85 in standrd engines"


Yes that is right. I think the 25-30% on wiki is based on the fact that the avarage fuel consumption usually increases 25-30%. This is because the engine gains some efficiency on E85.

On the other hand if you do not need max power and stoch mixture E85 has the possibility to run in a much wider range of both weaker and richer mixtures.

I leave the other question to Horst :-)

/Aryan

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Hello scientists,
good to know that people follow my thoughts - thank you!

@viscosity:
My source was http://www.geiger.cc/kraftstoffe.php#Stoffwerte quotet content (table) of (blue) "Kraftfahrtechnischen Taschenbuch" and "Autoelektrik / Autoelektronik" of Robert BOSCH GmbH.

The ny datas were given in [mm^2/s] at 20°C:
Benzine~0,6 (gasoline)
Dieselkraftstoff~4
Ethanol~1,5
Methanol~0,75
I converted gasoline and Ethanol into SI unit [m^2/s] - just comparing other website with given ny of Ethanol http://de.wikipedia.org/wiki/Alkohol
--> ny_E100 = 1.52E-6m^2/s

GL, you're absolutely right I took the wrong number! --> I continue with corrected numbers:

Sidenote: I remember that we had to use one of the vicosities to consider temperature. One of them - cinematic or dynamic - is rather temperature stable and the other one depends of temperature. 'Of corse' I forgot which one was the stable on. If you're really really interested in I'll blow some dust of the shelf to find it out. Maybe this rule was used for gases only....


@Fuel Density:
Density of gasoline depends also on temperature. Aryans quoted 780kg/m^3 for E85 is given at 15°C.
http://www.aral.de/aral/faq.do?category ... ntPage=2#2 says standardized Otto fuel (gasoline/gas/petrol) is 720...775 kg/m^3 at 15°C, average 745kg/m^3.
10°C = 749,5kg/m^3
15°C = 745kg/m^3
20°C = 740.5kg/m^3
30°C = 731.5kg/m^3
The factor is about 0.9kg/m^3 each °C.

I don't know how density of Ethanol depends on temperature.



@1.27 factor
I took the median 1.27 of 1.25...1.30 for V_dot_E85 = V_dot_E0 * 1.27 (1.25...1.30) without thinking anything - just copied your quoted site http://en.wikipedia.org/wiki/E85_in_standard_engines ... to enlarge jet size by some 25-30% ...
I had to take my own brain and remember what I typed some lines earlier:
737kg/m^3 / 780kg/m^3 * 14,7 / 9,7 = 1.43
--> V_dot_E85 = V_dot_E0 * 1.43
To consider Aryans result of
(720 / 780) * (14,7 / 9,76500) = 1,38957816
I continue with factor 1.40.
I don't believe that the engine was designed for pure petrol. As written MTBE and/or ETBE was added which is similar to Ethanol.

EDIT:
aryan skrev:
Yes that is right. I think the 25-30% on wiki is based on the fact that the avarage fuel consumption usually increases 25-30%. This is because the engine gains some efficiency on E85.
Aryan, very good argument that increased efficiency could cause an error of calculated 1.40. --> with factor 1.25...1.30 the jet would be even smaller, see below.



@New calculation:
Re_E0 = v_jet-E0 * Ø_jet_E0 / ny_E0
Re_E0 = 1m/s * 1.2E-3m / 0.6E-6m^2/s
Re_E0 = 2000

2000 is not much away from Re_critical~2300 which marks the limit of laminar(Re<Re_critical>Re_critical) --> at WOT the jet is flown turbulent!


Re_E0 = Re_E85:

Re_E85 = v_jet_E85 * Ø_jet_E85 / ny_E85
v_jet_E85 = V_dot_E85 / A_jet_E85
V_dot_E85 = V_dot_E0 * 1.40
A_jet_E85 = Ø_jet_E85^2 * pi / 4

Re_E85 = [V_dot_E85 / ( Ø_jet_E85^2 * pi / 4) * Ø_jet_E85] / ny_E85
Re_E85 = [4 * V_dot_E85 / Ø_jet_E85 / pi] / ny_E85
Re_E85 = 4 * V_dot_E85 / Ø_jet_E85 / pi / ny_E85
Ø_jet_E85 = 4 * V_dot_E85 / Re_E85 / pi / ny_E85
Ø_jet_E85 = 4 * V_dot_E0 * 1.40 / Re_E0 / pi / ny_E85
Ø_jet_E85 = 4 * 1.11111E-6m^3/s * 1.40 / 2000 / pi / 1.382E-6m^2/s
Ø_jet_E85 = 0.00072m = 0.72mm

Ø_jet_E85 / Ø_jet_E0 = (1.2mm - 0.72mm) / 1.2mm = 0.40
Ø of E85 jet should be 40% smaller than previous with E0!

Since I haven't read ever that anyone has to use a smaller jet my idea/calculation is wrong. If I would use 1.27 instead of 1.40 I'd get an Ø of 0.65mm!

Reynolds would say that the influence of viscosity is more important than simple contnuity. But I guess it isn't. Maybe the influence of temperature is the problem?

»Horst

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Senast redigerad av ami8i tis 08-01-29 01:50, redigerad totalt 1 gång.

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ami8i skrev:
@Fuel Density:
Density of gasoline depends also on temperature. Aryans quoted 780kg/m^3 for E85 is given at 15°C.
http://www.aral.de/aral/faq.do?category ... ntPage=2#2 says standardized Otto fuel (gasoline/gas/petrol) is 720...775 kg/m^3 at 15°C, average 745kg/m^3.
10°C = 749,5kg/m^3
15°C = 745kg/m^3
20°C = 740.5kg/m^3
30°C = 731.5kg/m^3
The factor is about 0.9kg/m^3 each °C.


Is this without ethanol, no 5% like we have in Sweden?

ami8i skrev:
I don't believe that the engine was designed for pure petrol. As written MTBE and/or ETBE was added which is similar to Ethanol.


No maybe not, hard to know what fuel a car was designed fo. Modern cars fuel consumption and emmisions are tested with a very pure (& expensive) standardized test petrol. This is one of the reasons why fuel consumption specifications tends to be lower as most people can achieve in reality. Can't find its properties but I don't think it contains MTBE, benzene or other contaminations, it would be intresting to compaire the data on the test fuel. If you look at some very pure petrol called "Alkylat bensin" in Sweden (for small engines like lawn mowers), http://www.okq8.se/data/internal/data/1 ... Bensin.pdf a density of 690 kg/m^3 @ 15°C is specified.

ami8i skrev:
EDIT:
aryan skrev:
Yes that is right. I think the 25-30% on wiki is based on the fact that the avarage fuel consumption usually increases 25-30%. This is because the engine gains some efficiency on E85.


Aryan, very good argument that increased efficiency could cause an error of calculated 1.40. --> with factor 1.25...1.30 the jet would be even smaller, see below.


I do not think you should use a factor 1,25-1,30 in your calculations, the fuel consumtion as expected is caused by a higher efficiency so you can close the throttle more to keep the same speed (less air + less fuel). With the same throttle position and air speed in the carburettor 40% more fuel is not a bad assumption. At WOT and a litle higher top speed the fuel consumption increases even more of course.

greetings Aryan

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Hello Aryan,

@standard petrol:
The ARAL link refers to EN 228. A Google serch would give you the answer: Up to 5% Ethanol are allowed without special branding. EN228 fuels require a mindest level of various properties (density, boiling, vapour, MOZ...) but not the content of substances except some critical like Benol(Benzen?), Sulfur, alcohle, ether.


@test petrol:
Good point to consider. I guess the engines are deeloped for EN228 but the exhaust and mpg tests/exams are made with your quoted test petrol (limits are more naarow). I couldn't find datas too, becassue I don't know the name.


@mpg with E5, Exx, E85:
Hmmm, would be a nice new calculation to consider AFR and energy of each fuel to achieve a given power/torque.
Knowing this you could say the more open the throttle is (more air) the better is efficiency of the engine (compare any shelf diagramme).

»Horst

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Hi
Well I tried the Poiseuille formula to calculate the jet size ... Unfortunately, it does not give appropriate result because as mentionned by Ami8 above, with a Reynolds number of 2000, flow is not purely laminar and is somehow turbulent
In addition, the Poiseuille formula is for an horizontal tube ... not exactly the case of a jet ... which is in the middle of a venturi

So to know why Wikipedia, E85 in standard engine suggests a jet radius increased by 13%, I will try to get in touch which the author of this wikipedia topic
Thanks all for your efforts
Bye

Jet Radius using Poiseuille calculation
---------------------------------------------
As given by http://en.wikipedia.org/wiki/Poiseuille%27s_law , laminar flow PHI (a volume per second) of an uncompressible viscous liquid (whose dynamic viscosity in ETA) thru a circular tube (radius =R, length = L), pressure being P2 and P1 at both end of the tube is

PHI = Pi * R4 * (P2-P1) / 8* ETA* L
with R4 meaning R*R*R*R

Assuming :
- pressures in the intake manifold and the engine being the same for E85 and E0,
- length L of the jets being also the same,
- K= Pi*(P2-P1) / 8*L
we can simplify the above mentioned formula, writing PHI =K*R4 / ETA

Assuming
R = initial radius of the jet as used with gasoline
(R+r) = extended radius of the jet for E85

PHI_E0 = K * R4 / ETA_E0
PHI_ E85 = K * (R+r)4 /ETA_E85

Volume of E85 should be established as such to maintain a stoechiometric AFR … that is to say 1.438 the volume of E0 based on Aryan's calculation
PHI_E85 = PHI_E0 * 1.438

Then
(R+r)4/ETA_E85 = 1.438*R4/ETA_E0
(R+r)/R = Fourth root of (1.438 * ETA_E85/ETA_E0)


At 20°C ... again not really the value in the carb
ETA_E0 = 0.65 * 10-3
ETA_E100 = 1.20*10-3
ETA_E85 = 85% ETA_E100 + 15% ETA_E0 = 1.12*10-3
Then
(R+r)/R = Fourth root of (1.438*1.12/0.65) = Fourth root of 2.45
(R+r)/R = 1.25

so jet diameter should be enlarged by 25%
which does not comply at all with wikipedia 13% of http://en.wikipedia.org/wiki/E85_in_standard_engines .


Senast redigerad av Ceyal35 tis 08-02-05 13:11, redigerad totalt 2 gånger.

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Blev medlem: tis 06-12-19 15:54
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Ceyal35 skrev:
Hi
---------
Then
(R+r)4/ETA_E85 = 1.438*R4/ETA_E0
(R+r)/R = Fourth root of 1.438 * ETA_E85/ETA_E0


At 20°C ... again not really the value in the carb
ETA_E0 = 0.65 * 10-3
ETA_E100 = 1.20*10-3
ETA_E85 = 85% ETA_E100 + 15% ETA_E0 = 1.12*10-3
Then
(R+r)/R = Fourth Square of 1.438*1.12/0.65 = Fourth root of 2.45
(R+r)/R = 1.25
--------------------------
.

The fourth root should not include the quotient 1.12/0.65 so the (R+r)/R instead should be (fourth root of 1.438)*1.12/0.65 = 1.89 which is obviously not correct either. So back to the hydrodynamics books again!

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Blev medlem: mån 07-09-24 00:23
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Hello Ceyal35,
@Poiseuille:
IMO Hagen-Poiseuille law (as we German soken people call it) is not suitable for describing the flow through a very very short pipe.

That's a simple model of the nozzle, seperated into 4 sections.
The last section 'pressure drop/loss' is very important becasue it is huge - and an important factor of calculation. Don't forget that fuel starts to vapourize here caused on destroyed hydrodynamic flow.
Kod:
*                   *
  *               *
    *           *
      *********


- . -- . -- . -- . -- . --



      *********
    *           *
  *               *
*                   *
jet.....pipe....diffusor....drop



@Alternatives:
Since I don't like just to say 'sorry that's wrong' I ask a 2CV friend who should be familar with damned hydrodynamics. He's educated students in this subject!

Does anyone know this http://en.wikipedia.org/wiki/Darcy-Weisbach_equation ? It works for laminar and turbulent and there is a D/L realationship.


@Wikipedia author:
Good idea to contact him. I bet the number is either poor estimated (coninuity) or result of experiments. I expect no theory, but wozld be hapy to hear/read one. :)

@Manifold pressure:
Pressure in Venturi bottleneck should be leass than in manifold (principle of carburator)

»Horst, happy not to deal with carbs anymore :)

_________________
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Blev medlem: lör 07-11-10 22:05
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GL skrev:
The fourth root should not include the quotient 1.12/0.65 so the (R+r)/R instead should be (fourth root of 1.438)*1.12/0.65 = 1.89 which is obviously not correct either. So back to the hydrodynamics books again!

Hi
Sorry my notation was misleading but is is really fourth root of (1.438*1.12/0.65).. now corrected

Thanks Ami8 .. I need to more deeply study this new idea
Thanks all
Bye


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Blev medlem: tis 06-12-19 15:54
Inlägg: 1244
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Ceyal35 skrev:
GL skrev:
The fourth root should not include the quotient 1.12/0.65 so the (R+r)/R instead should be (fourth root of 1.438)*1.12/0.65 = 1.89 which is obviously not correct either. So back to the hydrodynamics books again!

Hi
Sorry my notation was misleading but is is really fourth root of (1.438*1.12/0.65).. now corrected

Thanks Ami8 .. I need to more deeply study this new idea
Thanks all
Bye

Yes you are right! But going back to the original question "what diameter increase should be used for E85?" I did some more search among the many equations swirling around us. Pouseilles, Darcy-Weissbach and others. I returned to old Bernoulli. Using the values for gasoline used here before, jet diameter 1,2 mm, jetflow speed 1 m/s (assuming a jet length of 1,2 mm) and a gasoline density of 740 kg/m3 the pressure drop according to Bernoulli is 370 Pa and according to Darcy-Weissbach 12 Pa. It is thus apparent that resistance of the main jet is not much influenced by viscosity, only by the density which is involved in the Bernoulli equation.

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