My thoughts about the fuel E85 compared to other ethanol products which not had gone through an expensive chemical dehydration process, have led me into this experiment. The aim is to ascertain how much water it can absorbed without gasoline starts secede. Furthermore, this has developed into a self-composed method, for from an arbitrary sample of E85 determine its potential water content.

In order to illustrate which quantities of water this is, in the case of E85 and its absorptive capacity, have I done three trials. One with E85 and two with commercial alcohols (cleaning and washer fluid). Just these three samples do not account for MTBE content in the unleaded petrol 98 RON. According to a vague indication contains unleaded 98 RON about 10% MTBE. Furthermore, is the water absorption capacity, greater for unleaded 98 RON than for unleaded 95 RON. 95 RON also contain MTBE but the content is up to 20%! Shuffled 15% 95 RON with 85% ethanol, will the MTBE content fall to 3%. The level of gasoline for the three samples is about 1.5 ml lower then. MTBE is an octane enhancing fossil fuel, but is soluble in water/alcohol.

For the samples are distilled water added in 100 ml ethanol fuel. The importation of water is interrupted when the sample shows signs of turbidity. Muddiness indicates that phase separation has begun. Phase separation is the same as the stratification of gasoline and ethanol/water element and depends on the temperature and the proportions of the components. A major component of ethanol provides large water absorption, while a small provides less absorption.

T-red | T-blue | E85 |

85 ml T-red and 15 ml U98 | 85 ml T-blue and 15 ml U98 | 98 ml E85 and 2 ml U98 |

Amount of water = 25 ml | Amount of water = 17 ml | Amount of water = 22 ml |

Percentage of water = 24% | Percentage of water = 18% | Percentage of water = 22% |

The percentage is the weight of the water quantity, the weight in proportion to the total mass of fuel. The calculations are made on the assumption that the density of ethanol is 0.79 gram/ml and 0.74 gram/ml of gasoline.

Since these are large amounts of water became the change of the proportions of ethanol and gasoline rather drastically. These will therefore be smaller than they were before the water was added.

To investigate how well the absorption of the water-saturated sample was I have to do this temperature test:

It had to endure -18 Celcius (zero Fahrenheit)and also heated to the boiling point.

It showed no signs of separation.

Weight per cent water in the ethanol portion of each sample:

T-red | T-blue | E85 |

73% ethanol & 27% water | 80% ethanol & 20% water | 76% ethanol & 24% water |

According to Statoil's information sheet (no longer exist) is/was the density of E85 784 kg/m

100 ml E85 (Statoil) therefore weighs 78.3 grams and can be divided into these components:

- 3 % is MTBE => 2.35 grams
- 85 % is ethanol => 66.56 grams
- 12 % is gasoline => 9.40 grams

When E85 is contaminated of water change the proportions. We have a water component which occupies a certain percentage, how much we do not know. A layer sample shows not the actual content of gasoline and ethanol, if there received water after it was produced. If E85 is diluted with water will the proportion of gasoline and the ethanol get lower than it was from the beginning.

The analytical result for the water content will always be based on the algorithm for V-Power/98ron.

Imagine now that a layer samples showed E90, although the sample contains water - for example, 10 % water in ethanol (10 % of water not is equivalent to 10 g of water here). Since the sample contained 10 % water, was it E89. Another example: We have E65, it contains 10 % water - thus it is E63. We have E75 and it contains 10 % water - it is E73, etc. One can then conclude that the difference is marginal in comparison with possible measurement errors. The percentage deviation of the E63 and E65 is 0.6 % - slightly more than half a per cent and for E89/E90 - 1.3 %. But always count on with an error around ± 0.5 % for the lower and ± 1 % for the higher quality (if the content of water is up to 10 %).

We get to work. Measure out 100 ml of E85 and carefully add water while stirring (at room temperature). Stop when the sample shows signs of turbidity. When this happens begin what is called phase separation - the gasoline/alcohol start to separate. Note the amount of water added in millilitres. I assume you have already done a layer sample, if not - determine the quality of a layer sample. For example, you could add 10 ml of water in the E88. - What percentage is it? 10 ml is 10 grams and the weight of ethanol is 88 x 0.783 = 68.9 g.

The addition of water in this sample was: 100 x 10 / (10 + 68.9) = 12.7 %

Algorithm for max-ceiling to V-Power: (88 - 60)

Algorithm for max-ceiling to E5: (88 - 55)

The sample contains water because only 12.7 % of water could be added (not 28 or 30). The proportion of water in the alcohol for E88 is (with V-Power as the reference): 30.0 - 12.7 = 17.3 %

To compensate for the error mentioned above, the result should be this (17.3 - 2): 15.3 %