Practice technologist > Viscosity > Teaching aids > Laboratory
The guide focuses on methods and examples of the calculations made in the implementation of technological operations receiving installation, storage and shipment of raw materials for the preparation of lubricating oils (base oils and additives), cooking oil trade with specified quality. The manual is considered a quality relationship and quantitative relationships between the individual parameters of quality additives and commodity-term products, which resulted in the ability to predict a certain quality indicators oils and evaluation of the results of the laboratory analysis.
Proposed a graphical method of calculation by nomograms:
- The amount of additives needed for the preparation of oil
- The density of base oil and commodity products at different temperatures.
Manual was developed with the involvement of the materials and data processing plant documentation preparation of lubricants, technical standards for raw materials and manufactured products now
The volume, density and mass of liquid
The amount of liquid may be expressed as in the bulk and in terms of the weight (ton kg).
The weight and volume of the liquid are related by
M = d x V (1.1)
where: M - liquid mass m (kg)
V - volume, cubic meters (L), occupied by liquid,
d - the liquid density t / m3 or kg / dm3
Weight accepted into the tank (shipped from the tank) product is calculated by the expression
M = ( n2 – n1 ) х f х d ( 1.2 ),
where: d - the density of the fraction at a temperature of loading;
f - the calibration factor is numerically equal to the volume tank 1 cm height.
n1 and n2 - the initial and final levels cm.
Artwork (n2 - n1) x f, equal to the volume received (shipped) product is also determined by the calibration table as the difference between the volume of the tank at vzlivah n1 and n2. Density is not a constant and depends on the temperature of the liquid. Density at various temperatures are related by
d2 = d1 + k х ( t1 - t2 ) ( 1.3 )
where: d1 - density at temperature t1;
d2 - the density at temperature t2,
k - the temperature coefficient characterizing the change in the density of the liquid at a temperature change of 1 degree;
for lube oil and commodity it is taken equal to 0.000673 and 0.00066, respectively, for the values of the density of 0.700 - 0.799 and 0.800 - 0.890 t / m3.
Determination of density at various temperatures can be carried out according to the schedule, which reflects the change in density depending on the temperature of the product.
Depending using 1.1, 1.2 and 1.3, you can perform almost all the necessary calculations of volume, density and mass for receive operations, internal transfers and shipments of raw materials and commodity products.
Examples of calculations
In reservoir adopted fraktsіya 420-500 ° C with a temperature of 41 ° C. Passport density fraction d20 = 878 kg / dm3. Calculate the density of the fraction in the vessel.
According to 1.3 d41 = d20 + 0,00066 х ( 20 – 41 ) =0,878 – 0,014 = 0,864 kg / dm3
Tanker working volume of 8.72 m3 of oil loaded with a temperature of 32 ° C and a density of 0.885 passport kg / dm3. Calculate the mass of the loaded oil.
According to 1.3 d32 = d20 + k х ( 20 – 32 ) = 0.885 – 0,00066 х 12 = 0,877 – 0,0079 = 0,877 t/м3
According to 1.1 M = d х V = 0,877х 8,72 = 7,647 t.
At 4/2 tank entered the residual fraction at 70 ° C. Initial vzliv is 42 cm, the end - 665 cm, passport density d20 = 0,905. Calculate the mass of the received product.
According to 1.3 d = 0,905 – 0,00066 х 50 = 0,872 t/м3,
According to 1.2 М = ( 665 – 42 ) х 0,449 х 0,872 = 243,92 t.
Determined graphically (see. Annex) oil density D at 36 ° C, if the passport is its density 0.880 t / m3.
Choosing a graphics sheet with the appropriate conditions of temperature and density values. From a point 36 on the x-axis (temperature), draw a vertical line up to intersection with the oblique line, which has 0,880 mark, and hence draw a horizontal line to the intersection with the y-axis (density), where is the value of the density - 0.870 t / m3. The interstices between the parallel slant lines are generated in units of 0.010 density and grid 10 are separated segments. This allows you to operate on values of densities up to 0,001 units of density.
Calculations based on the routing.
Flow chart determines the component composition of oil, expressed in percentages by weight of components in the finished oil. Consequently, calculations based on the technological cards are basically calculating amounts of the components at their percentage content in the oil. Calculations of this type allow to determine the weight quantities of the components needed to prepare a given quantity of oil, as well as the amount of oil which can be prepared from a particular presence of one of the mixed components.
Examples of calculations
Determine the number of components for the preparation of 400 tonnes of oil.
As calculated by the general formula
mi = M x ni
where M - amount of oil is prepared, that is;
ni - component content in oil,%
mi - the weight of the component m.
According to the documentation contains 51.0% oil base oil Fr.420-500, Fr.> 500 - 43.2% additive Infintum D-1180 - 5.7% and 0.1% depressant.
Calculating:
m1 = 400 x 51 = 204 m - Fr.420-500
m2 = 400 x 43.2 = 172.8 m - fr.> 500
m3 = 400 x 5.7 = 22.8 m - D-1180
m4 = 400 x 0.1 = 0,4t - depressant additive
In the presence of 0.6 tons Dizopak additives. Determine the amount of oil A, which can be prepared using a number of additives and the required number of the remaining components.
In accordance with the routing oil contains D: fr.> 500 - 30.0%, the fraction Fr.420-500 - 66.9%, additives Dizopak - 3.0% , depressant - 0.1%.
Calculating:
M = 0.6 0.03 m = 20 (amount of oil)
m1 = 20 x 0.669 = 13.38 m (fr. 420 - 500 °C)
m2 = 20 x 0.300 = 6.0 m (fr. >500 °C)
m3 = 20 x 0.001 = 0.02 m (the number of depressant)
In some cases the calculations are complemented by technological calculations cards, which use specific quality parameters of the mixed components and the final product - the base number, density, viscosity, and other active elements. The most commonly used as a calculation factor TBN.
Base number in the calculation of cooking oils (TBN)
The calculations for the preparation of oils using as a settlement factor TBN supposed equality
M х BN = m1 х BN1 + m2 х BN2 + … + mI х BNIi
where: M - mass is prepared oil, t;
BN (base namber) - oil TBN, mg KOH per 1 g;
m1, m2, ... mi - mass of mixed components;
BN1, BN2, ... BNi - alkaline components of oil.
Examples of calculations
Determine the amount of additive Dizopak (TBN 150 mg KOH per 1 g) needed for the preparation of 150 tonnes of oil A with base number of 10.5 mg KOH per 1 g
Alkalinity oil is produced only by the additive, so
M х BN = m(add) х BN(add)
m(add) = M х BN / BN(add)
m (add) = 10.5 x 150 /150 = 10.5 t - quantity of additive.
Determine the amount of additive Infineum D-1180 (alkalinity. Number of 170 mg KOH per 1 g of oil) required to previously prepared 600 t oil (base number of 6.0 mg KOH per 1 g) to bring the oil to the condition X TBN 10 5 mg KOH per 1 g.
(m1 + m2 ) х BN = m1 х BN1 + m2 х BN2
Is There: m1 = 600; BN1 = 6,0; BN2 = 170; BN = 10,5;
m2 – required.
(600 + m2) х 10,5 = 600 х 6,5 + 170 х m2
6300 + 10,5 m2 = 2400 + 170 m2
159,5 m2 = 2400
m2 = 15,04 t - amount of additives.
Calculate the number of alkaline mixture, resulting in random association 200 t oil №1 (base number of 5.0 mg KOH per 1 g), and 90 tons of oil №2 alkali number of 9.0 mg KOH per 1 g.
M х BN = m1 х BN1 + m2 х BN2
BN = ( m1 BN1 + m2 BN2 ) / M
Is There: m1 = 200 ; m2 = 90 ; BN1 = 5,0 ; BN2 = 9,0 ; M = m1 + m2 = 290;
BN = ( 200 х 5 + 90 х 9 ) / 290 = 6,24 мг КОН на 1 g - TBN oil mixture.
At 0.6 m available. Additive Infineum D-1180 at base number 170 mg KOH per 1 g Calculate the amount of oil which can be prepared from this amount of additive and the amount necessary for the preparation fractions 420-500 °C and fractions > 500°C. According to the routing oil additive contains 3.6% D-1180. M = Amount of oil 0.6 3.6 x 100 = 16.7 t. In accordance with the documentation is contained in the fraction >500°C the 48.0% and 48.4% of a fraction of 420 - 500.
Calculate:
m2 = 16,7 х 48,0 : 100 = 8,02 т - amount fraction >500°C
m3 = 16,7х 48,4 : 100 = 8,08 т - amount fraction 420 – 500°C.
Practice technologist > Viscosity > Teaching aids > Laboratory