Does a carburetor have constant fuel mass flow, constant liquid volume flow, or constant molecule count flow?

Question

I have a generator with carburetor. It seems to consume a bit more fuel than I expected, 0.44 liters per hour for 386 watt production, or 1.14 liters per kWh. I'm also bit more annoyed by the exhaust fumes than I expected to be for a 4-stroke engine.

One factor that might affect the fuel consumption and exhaust fumes is the type of gasoline used. I understand that today nearly every location on this planet is using E10 gasoline (10% ethanol, 90% fossil fuel). However, I'm not using that and I'm not using E0 either; I'm using "small engine gasoline" that consists of only simple chain hydrocarbons and contains no aromatic hydrocarbons. The small engine gasoline has been specifically created to stay useful for 3-5 years in storage, to never block carburetors even if aged, and to evaporate completely without leaving any residue.

The small engine gasoline has bit different properties that normal gasoline:

• 95 E0 without oxygen: 32.7 MJ/l, 755 kg/m³, H/C 1.78
• 95 E10: 31.1 MJ/l, 743 kg/m³, H/C 1.89; 3.2% O
• small engine gasoline: 30.8 MJ/l, 685 kg/m³, H/C 2.17

I'm trying to understand how a carburetor works. If I have a carburetor and one cubic meter of air passes through it, it picks up some amount of fuel. I understand that the fuel is in the liquid state, so the carburetor isn't pulling evaporated gas but rather liquid that forms droplets and evaporates in the cylinder.

But how is the amount of fuel picked up by the carburetor determined?

Does it pick a certain volume of liquid fuel, such as always one liter of fuel per one cubic meter of air (just an example, not to suggest that would be correct air/fuel ratio)?

Or does it pick a certain mass of liquid fuel, such as always one kilogram of fuel per one cubic meter of air?

Or does it pick a certain number of fuel molecules, such as 4*10²⁴ molecules per one cubic meter of air?

Of course the air pressure affects the operation of the carburetor, so let's assume the altitude is fixed, and the position of the choke valve is fixed.

I quickly calculated that one liter of both 95 E10 and small engine gasoline require almost the same number of external oxygen atoms to burn (only 0.5% difference), but that one kilogram of small engine gasoline requires 8% more oxygen atoms to burn than one kilogram of 95 E10. So if the carburetor is "picking kilograms" rather than "picking liters", then its adjustment might be off for the small engine gasoline, which might explain the annoying exhaust and little bit higher fuel consumption than what I expected (however, the little bit higher fuel consumption can also be explained by 6% lower energy density of small engine gasoline when compared with oxygenless 95 E0 -- it's possible the manufacturer has stated the fuel consumption with oxygenless 95 E0).

Answer 1

It's not constant mass or constant volume, but more complicated - for an ideal free jet, mass flow is area * square root of (2 * density * pressure difference) - so a higher density increases the mass flow and decreases the volumetric flow, both by a ratio of square root (new density / old density)

And if the pressure loss in the jet internal passages is significant, the viscosity of the fuel could change things as well

So really adjusting based on maths can't help you much here, just tweak things until it's not obviously rich or lean

Answer 2

Carburetors basically work by aspiration, as air flows through a venturi, it draws an amount of gasoline depending on velocity of the air. The devil is in the details. There will be a main orfice that governs flow of gasoline to the venturi. Also the float level will determine the liquid head pressure of the gasoline. There will be a separate idle system for low speed ( low air velocity). Automobile carbs likely have a system that adjusts gasoline flow depending on intake manifold vacuum. I have tried to modify without success; I think the best you can do it make sure it is operating as designed, no restricted passageways. There are a lot of details so the devil has a lot to do.

Answer 3

Ideally a carburetor would pick up enough fuel by mass so that you end up with a near desired stoichiometric ratio of air to gas. E10 has a stoichiometric ratio of about 14.1:1. 14.1 kilograms of air to 1 kilogram of fuel. Note that from the mass of the fuel or air, you can compute the number of molecules.

https://en.wikipedia.org/wiki/Air%E2%80%93fuel_ratio

I liked this discussion including a short table with stoichiometric air fuel ratios:

https://www.jalopyjournal.com/forum/threads/afr-and-e10.1052108/

Carburetors, especially small engine carburetors, are not that accurate. Plus small engines tend to have more incomplete combustion just because of their simple design.

One of the problems with using volume measurements is the volumetric density of fuel changes with temperature. On a hot day, one liter of gasoline will contain fewer molecules, fewer moles, of gasoline than on a cold day. This is why modern jet aircraft will display number of pounds of fuel, not gallons.

https://blog.aopa.org/aopa/2017/07/19/get-ready-to-use-pounds-not-gallons-by-chip-wright/

The amount of air that an engine is pulling in depends on temperature and pressure. Generally, on colder days with higher pressure you can make more power because you can get more molecules of air into the engine. Boyles law or the Ideal Gas Law is an explanation:

https://en.wikipedia.org/wiki/Ideal_gas_law

PV = nRT

Pressure * Volume = number_of_moles * Reynolds_number * temperature

At wide open throttle, a small engine should be able to pull in 75% or more of its displacement volume every two cycles for a four stroke.

https://hpwizard.com/volumetric-efficiency.html

With the atomospheric pressure and the temperature you can estimate the number of moles of air that the engine gets into the combustion chamber.

From your description, the engine sounds like its running rich, high fuel consumption and smoke. My first reaction would be to put in a clean air filter, plug and replace the main jet on the carburetor, possibly jetting down one or two sizes, if it is replaceable. In my motocross years, I found that the labels on the jets, the amount of fuel they should flow, can be off multiple sizes. I would find this by timing the amount of fuel that the jet could flow through a burette. Some were good, some were not. Also, if you are at a higher altitude or hotter temperature, you will need to jet down.

It would be interesting to see a data sheet for your "small engine gasoline".

Answer 4

I looked at the mathematics of carburetors and discovered some interesting features.

A simple carburetor has essentially these components, starting from air filter side (1) and ending to engine valve side (4):

1. Choke valve. This is used to create a negative pressure in the main jet when the engine is cold, because fuel doesn't vaporize well when cold and thus more fuel is needed.
2. A thin throat, where the main fuel jet is located. The main fuel jet is simply a small tube for fuel that ends in a nozzle spraying the fuel into the airflow.
3. Throttle valve. This is the mechanism for limiting airflow into the engine.
4. Idle jet. This is after the throttle valve so it works using the large negative pressure created by throttling the engine and the purpose is to offer enough fuel when the engine is throttled (I'll explain later why this is necessary).

The basic equation explaining carburetor operation is Bernoulli's principle. Using the interesting parts of it, the negative pressure of airflow is proportional to air_density * air_velocity². This explains why the carburetor has a thin throat. By making the throat thin, to get enough air the velocity of the air rises to a large value. This increases the negative pressure of the airflow a lot.

However, note that negative pressure is proportional to air velocity squared. This has two problems:

1. If the engine is operating at low RPM but at full throttle, the airflow is lower too than it would be at high RPM, and a second power relationship means the engine gets far less fuel than it would need. So you can expect a carbureted engine to run lean at low RPM, and to run rich at high RPM, when wide open throttle.
2. If the engine is throttled, the airflow reduces markedly. This reduces the negative pressure so much that the engine gets practically no fuel -- in fact, there can be so little fuel the engine would not run without idle jet. To mitigate this issue, the idle jet is necessary. It operates not by negative pressure of airflow, but by negative pressure caused by throttling the engine.

These two jets, main jet and idle jet, give a rather poor approximation for the correct air/fuel ratio. You can't expect a carbureted engine to have as precise air fuel ratio than a fuel injected engine has. No wonder modern cars use fuel injection!

Ok, now we know that negative pressure is proportional to airflow squared, and that to mitigate the problems this causes when throttling the engine, a secondary idle jet is used. But how does the fuel flow relate to the negative pressure?

We can use the Darcy-Weisbach equation. According to it,

pressure/length = friction_factor * fuel_density/2 * fuel_velocity^2 / hydraulic_diameter

At first, it looks like fuel flow squared would be proportional to pressure (which is proportional to airflow squared), so the two squares would cancel out each other. However, not so! We know fuel density, length of the jet, pressure (from Bernoulli's principle) and hydraulic diameter of the nozzle. But we don't know the friction factor. Let's take a look at it first.

In laminar regime (probably the regime in most carburetors), the friction factor is:

friction_factor = 64/reynolds_number

And the Reynolds number is:

reynolds_number = fuel_velocity*characteristic_linear_dimension / kinematic_viscosity
  = fuel_velocity*characteristic_linear_dimension*fuel_density / dynamic_viscosity

In this case, the characteristic linear dimension describes the size of the nozzle (and is the same as hydraulic diameter if it's circular nozzle). So Reynolds number is proportional to fuel velocity, which means Darcy friction factor is inversely proportional to fuel velocity, and this changes the fuel velocity in the final formula to be a linear relationship and not a square relationship.

So pressure (proportional to airflow squared) is proportonal to fuel flow (not squared here). Twice as much air, 4x as much fuel, unless the increased airflow is caused by reducing throttling (in which case the idle jet fuel flow reduces).

Another interesting observation about this is that fuel flow is affected by fuel viscosity. Different fuels have different viscosities.

If we put the dynamic viscosity form of Reynolds number in the formula, we get the result that fuel density cancels out, and fuel velocity is dependent on nozzle characteristics and dynamic viscosity. So dynamic viscosity dictates volume flow.

If we put the kinematic viscosity form of Reynolds number in the formula, we get the result that fuel density doesn't cancel out, and fuel mass flow rate (density times velocity) is dependent on nozzle characteristics and kinematic viscosity.

So, the answer to the main question I asked is:

1. If there are two fuels with same dynamic viscosity, their volume flow is the same
2. If there are two fuels with same kinematic viscosity, their mass flow is the same
3. If there are two fuels with different viscosity, you can't say anything about their relative flow rates without knowing their viscosities.

Obviously, viscosity is dependent on temperature too, and air cooled engines usually don't have a stable temperature.

The alkylate fuel I used is almost pure isooctane produced by alkylation units in oil refineries (in fact, it has research octane rating of 98). If you know how octane rating is defined, isooctane is the "good stuff" (octane rating 100) and n-heptane is the "bad stuff" (octane rating 0). According to https://www.chemiway.co.jp/en/product/data/i_data01.html, the dynamic viscosity of isooctane is 0.503 cP at 20 degrees Celsius.

For regular gasoline, I found several sources (for 20 degrees Celsius):

• http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/viscosity.html says 0.6 cP
• https://www.engineersedge.com/heat_transfer/dynamic_viscosity_13955.htm says 0.29 cP (but this is so far away from the other sources it probably isn't true)
• https://pubs.rsc.org/en/content/articlehtml/2017/ra/c7ra00357a says non-ethanol gasoline has 0.529 mm²/s = 0.397 cP viscosity (750 kg/m³ density) and E10 has 0.546 mm²/s = 0.412 cP viscosity (754 kg/m³ density)

So apparently we can conclude that the viscosity of gasoline can vary a lot.

Probably the way the small engine manufacturer decided to calibrate the fuel flow is that all gasoline variants (with viscosity 0.4 - 0.6 cP) have enough fuel flow to produce the maximum power the engine is capable of producing. If somebody buys 1900 W generator and finds out it's capable of producing only 1600 W power with a certain type of gasoline, the owner wouldn't be happy. So probably the manufacturer errs on the side of rich mixture if the viscosity happens to be less than 0.6 cP.

The viscosity of isooctane / alkylate gasoline is 0.503 cP, about the middle of the range of gasoline viscosities. So it's expected to have slightly rich mixture because normal gasoline can have a slightly higher viscosity in some cases.

Also, by observing that viscosity affects fuel flow, we can explain why carburetor engines are so sensitive to stale gasoline. Stale gasoline has higher viscosity, so it doesn't flow well enough into the engine, leading to all kinds of running problems. On the other hand, a fuel injected engine will just see from the oxygen sensor that fuel isn't flowing very well, and it automatically compensates by having the injectors open for a longer amount of time.

Edit: one of the answers suggested using Bernoulli's principle for the fuel flow too (mass_flow = area * square root of (2 * density * pressure difference)). Let's investigate this.

Here's a GNU Octave / Matlab program determining whether Bernoulli's negative pressure or fuel viscosity loss is larger.

% Parameters begin here
rho = 735 % gasoline density in kg/m^3
mu_cP = 0.6 % gasoline viscosity in cP
fuel_line_diameter = 0.6e-3 % fuel jet diameter in meters
fuel_line_length = 20e-3 % fuel jet length in meters
fuel_liters_per_hour = 1.5 % how many liters per hour gas the engine uses

% Code begins here
kinvisc = mu_cP*1e-3/rho
v = fuel_liters_per_hour*1e-3/3600/(pi*(fuel_line_diameter/2)^2)
Re = v*fuel_line_diameter/kinvisc
if Re > 2300
    error("Flow no longer perfectly laminar")
end
f_D = 64/Re
p_viscosity = fuel_line_length * f_D * rho/2 * v^2/fuel_line_diameter
p_flow = 0.5*rho*v^2

For 735 kg/m³ gasoline density, 0.6 cP gasoline viscosity, 1.5 liters/hour fuel consumption and 20 mm long fuel jet, regardless of the fuel jet diameter, viscosity pressure loss is about twice as significant as Bernoulli flow pressure loss is. Fuel jet diameter doesn't affect this relative relationship at all, it just affects the negative pressure that needs to be caused by the carburetor throat.

It would require 10 mm long fuel jet (shorter than my assumption of 20 mm), or 3.0 liters per hour fuel consumption (unrealistically high for this engine, about twice as much as this engine can consume) or 0.3 cP viscosity (very low for gasoline, although as the gasoline tank heats up this could be realistic) to have as significant Bernoulli flow pressure loss as the viscosity pressure loss is.

So it appears that the main equation determining fuel flow is the one for viscous flow in a tube, not the Bernoulli's principle applied for fuel flow too (you still apply Bernoulli's principle for airflow, however).