Flammability of a water + oil mixture

[orbs]

Hello
I had fun at the following experience:
in a metal bowl I poured 10 ml of water, 1 ml of dishwashing detergent, I stirred well, I added 10 cc of petroleum, I emulsified well.
with a blowtorch I put to a boil, the emulsion ignited with a pilot, then was self maintained until exhaustion of the liquid.
the bowl was empty.
it can be assumed that the water emulsified by the surfactants was vaporized during the combustion of the oil while preserving the temperature sufficient for the ignition of the oil.

in parallel I proceeded under the same conditions to the combustion of petroleum alone resulting in a longer flame, more yellowish red, t a lot of black smoke, and the characteristic smell of burnt petroleum

the combustion of the emulsion produced a shorter, longer yellow light flame

I conclude nothing because too many parameters are random;
however I wanted to point it out knowing that this is nothing new and that this process is known.
but what a pleasure I wanted to share with you

[elephant]

Very interesting experience, orbs. In fact, it would have been interesting to do a little calorimetry, for example, by placing a small saucepan above the flame.

So:

with the same amount of water in the pan

test 1: combustion of X cc of petroleum: delta T °?

test 2: combustion of the same quantity of oil + water: delta T °

It is of course necessary to adapt the quantity of oil and water in the pan so that the temperature rise is significant, for example for test 1, passage from 15 to 80 ° c to avoid being in the errors of measurement .

The following experiment is to repeat test No. 2 with different proportions of water.

[dedeleco]

Not surprising, since the calorific value of petroleum is around 38000KJ / liter:
http://fr.wikipedia.org/wiki/Pouvoir_calorifique
and the energy to bring the water to 500 ° C is much lower less than 3000KJ / liter
http://en.wikipedia.org/wiki/Latent_heat
and therefore there is energy to burn the whole, less violently.

Alcohol at 50% burns worse, because it only makes 21000KJ / liter.
http://fr.wikipedia.org/wiki/Pouvoir_calorifique

[chatelot16]

calorimetry? putting a pan over the flame is not really calorimetry ...

how do you know what proportion of the heat will heat the pan or be lost?

it is not even worth measuring: the combustion of oil produces the energy it must produce, the vaporization of water consumes the energy provided by the vaporization energy, as combustion energy is largely superior to the energy of vaporization of water it can continue to burn

but this result is interesting: with wood for example, green wood which contains less water than your mixture burns badly: the humidity of the wood does not only consume a little energy, it completely sabotages the combustion and a most of the fuel goes up in smoke without burning

your oil therefore has the quality of burning properly mixed with water thanks to the detergent

[Capt_Maloche]

Hello
I had fun at the following experience:
in a metal bowl I poured 10 ml of water, 1 ml of dishwashing detergent, I stirred well, I added 10 cc of petroleum, I emulsified well.
...
but what a pleasure I wanted to share with you

Hello,

you're right it's a pleasure to share this type of manipulation

Too bad you didn't make a little video, or at least one or two photos, it lacks a little "cheerful" all that

and why dose the water + detergent in ml and the oil in cc?
what is a cc? a cm3 = 1ml; AH, apparently
according to my converter, 1cc would be equal to 1ml (to be confirmed, I don't know where this unit comes from ...)

so basically you made a mixture of 50% water + 50% petroleum

let's see, the PCI of Petroleum Lampant is about 40 KJ / Kg
The latent heat of vaporization is 2550 KJ / kg + (4.18x80) = 334KJ / Kg to rise from 20 to 100 ° C = 2884 KJ / Kg

that is to say that in theory a maximum mass ratio of 40 / 000 = 2884 could totally disappear

in summary: 1 volume of oil for 10 volumes of water
but what interest?

We know the reactions that allow water vapor to greatly improve combustion

it would be interesting to test different dosages on this principle

this is an interesting lab, would you like to do it with measurements, photos and video?

[Obamot]

Absolutely Capt Maloche. What should be known basically, is if there is a calorific contribution by the water vapor, or a contribution in some way "synergistic", is that it? What I would like to know is if there is a calculated gain for a Pantone engine, and of what order?

Yes Orbs, why did you take two measurement scales? By the way it was petroleum type, but what type?

For the mixture, it looks like the proportions were rather the following:

1l of water = 1 kg
1l of petroleum = 0,8 kg

So it would seem that 10 cc is the equivalent of 40 parts, VS the same water capacity at 50 parts, + 10 parts for the dishwashing liquid .... it would take a little bit to make a round count of 100 parts (But I could be wrong).

[elephant]

For our bottom of garage experiments, the volume ratio seems to me the easiest to handle. All it takes is a graduated cylinder. Nothing prevents us from making a volume / mass conversion table afterwards.

[dedeleco]

If we measure by calorimetry, we will find less heat and no contribution by water vapor. To be measured with a calorimeter.

With alcohol at 50 °, we do the same, for flambé pancakes, with less heat !!!

The same mechanism can have opposite effects:
brutal spraying of water, disperses around it,
if very finely divided, it homogenizes the fuel by dispersing it in very fine drops, increasing its surface area to evaporate and its ignition capacity and therefore improves the quality of combustion.
if not very divided, the explosion of large drops very violent disperses large drops of fuel far, towards the cold zones preventing it from burning properly.
Very moist wood shows this type of incomplete combustion.

[dedeleco]

chatelot16 writes:

calorimetry? putting a pan over the flame is not really calorimetry ...
how do you know what proportion of the heat will heat the pan or be lost?

a calorimeter is not at all a pan on the flame !!
http://en.wikipedia.org/wiki/Calorimeter
http://fr.wikipedia.org/wiki/Calorim%C3%A8tre
We recover all the heat and all the products !!

Bomb calorimeters
Bomb calorimeter

A bomb calorimeter is a type of constant-volume calorimeter used in measuring the heat of combustion of a particular reaction. Bomb calorimeters have to withstand the large pressure within the calorimeter as the reaction is being measured. Electrical energy is used to ignite the fuel; as the fuel is burning, it will heat up the surrounding air, which expands and escapes through a tube that leads the air out of the calorimeter. When the air is escaping through the copper tube it will also heat up the water outside the tube. The temperature of the water allows for calculating calorie content of the fuel.

In more recent calorimeter designs, the whole bomb, pressurized with excess pure oxygen (typically at 30atm) and containing a known mass of sample (typically 1-1.5 g) and a small fixed amount of water (to absorb produced acid gases), is submerged under a known volume of water (ca. 2000 ml) before the charge is (again electrically) ignited. The bomb, with sample and oxygen, forms a closed system - no air escapes during the reaction. The energy released by the combustion raises the temperature of the steel bomb, its contents, and the surrounding water jacket. The temperature change in the water is then accurately measured. This temperature rise, along with a bomb factor (which is dependent on the heat capacity of the metal bomb parts) is used to calculate the energy given out by the sample burn. A small correction is made to account for the electrical energy input, the burning fuse, and acid production (by titration of the residual liquid). After the temperature rise has been measured, the excess pressure in the bomb is released.

Basically, a bomb calorimeter consists of a small cup to contain the sample, oxygen, a stainless steel bomb, water, a stirrer, a thermometer, the dewar (to prevent heat flow from the calorimeter to the surroundings) and ignition circuit connected to the bomb.

Since there is no heat exchange between the calorimeter and surroundings → Q = 0 (adiabatic); no work performed → W = 0 Thus, the total internal energy change ΔU (total) = Q + W = 0

Also, total internal energy change ΔU (total) = ΔU (system) + ΔU (surroundings) = 0 → ΔU (system) = - ΔU (surroundings) = -Cv ΔT (constant volume → dV = 0)

where Cv = heat capacity of the bomb

Before the bomb can be used to determine heat of combustion of any compound, it must be calibrated. The value of Cv can be estimated by Cv (calorimeter) = m (water). CV (water) + m (steel). Resume (steel)

m (water) and m (steel) can be measured;

Cv (water) = 1 cal / gK

Cv (steel) = 0.1 cal / gK

In laboratory, Cv is determined by running a compound with known heat of combustion value: Cv = Hc / ΔT

Common compounds are benzoic acid (Hc = 6318 cal / g), p-methyl benzoic acid (Hc = 6957 cal / g)

Temperature (T) is recorded every minute and ΔT = T (final) - T (initial)

A small factor contributes to the correction of the total heat of combustion is the fuse wire. Nickle fuse wire is often used and has heat of combustion = 0.9813 kcal / g

In order to calibrate the bomb, a small amount (~ 1 g) of benzoic acid, or p-methyl benzoic acid is weighed. A length of Nickle fuse wire (~ 10 cm) is weighed both before and after the combustion process. Mass of fuse wire burned Δm = m (before) - m (after)

The combustion of sample (benzoic acid) inside the bomb ΔHc = ΔHc (benzoic acid) xm (benzoic aicd) + ΔHc (Ni fuse wire) x Δm (Ni fuse wire)

ΔHc = Cv. ΔT → Cv = ΔHc / ΔT

Once Cv value of the bomb is determined, the bomb is ready to use to calculate heat of combustion of any compounds by ΔHc = Cv. ΔT

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[Alain G]

calorimetry? putting a pan over the flame is not really calorimetry ...

how do you know what proportion of the heat will heat the pan or be lost?

it is not even worth measuring: the combustion of oil produces the energy it must produce, the vaporization of water consumes the energy provided by the vaporization energy, as combustion energy is largely superior to the energy of vaporization of water it can continue to burn

but this result is interesting: with wood for example, green wood which contains less water than your mixture burns badly: the humidity of the wood does not only consume a little energy, it completely sabotages the combustion and a most of the fuel goes up in smoke without burning

your oil therefore has the quality of burning properly mixed with water thanks to the detergent

Chatelot Hi!

Disagree with you to wet wood! I have long wood heated with a wood stove and I can assure you that if the stove is able to reach a high temperature smoldering, semi moist wood gives better energy yields a very dry wood .

I remind you that a good stove has a sheet that forces the flame and smoke to come together to get better combustion because it forces the smoke to pass through the flame!