3. lubricants

INFOMATICA ACADEMY CONTACT: 9821131002/9029004242
Degree Sem - I Lubricants111
Lubricants
 Syllabus:
 Introduction: Definition, Mechanism of Lubrication, Classification of lubricants,
Solid lubricants (graphite & Molybdenum disulphide), Semisolid lubricants (greases
Na base, Li base, Ca base, Axle greases), Liquid lubricants (blended oils).
 Important properties of lubricants, definition and significance, viscosity, viscosity
index, flash and fire points, cloud and pour points, oiliness, Emulsification, Acid
value and problems, Saponification value and problems.
 Introduction:
 Archaeological investigations reveal that the usefulness of lubrication was known to
the 1400 B.C. Leonardo da Vinci (1452-1519) discovered the fundamental
principles lubrication and described the effects of lubrication on the coefficient of
friction between two rubbing surface.
 The importance of lubrication was widely recognized with the advent of industrial
era and large scale consumption of lubricant in various fields since 1947 such as
automotive applications, industrial machinery and aviation.
 All material surfaces though they appear to be smooth show many irregularities in
the form of peaks (asperities) & troughs (valleys) when examined under a microscope
of high magnification. This is known as surface roughness.
 When two flat surfaces are placed over one another, the asperities of the upper
surface rest on those of the lower surface Thus, the surfaces make contact at these
points only .While over most of the area they are separated.
 Thus the real area of contact is very much smaller than the apparent area of
contact.

 Whenever a machine works, it‘s moving (sliding or rolling) parts rub against each
other with the result a friction is developed.
 Friction is the force of resistance to the relative motion of two solid surfaces in
contact. When the applied force is more than the resistance between two moving
surfaces, damage of the surfaces occur.
 Due to friction large amount of heat is liberated at the rubbing surfaces, which gets
heated up, damaged & even results in the welding of the two surfaces (seizure) and
the efficiency of the machine is also reduced.
 Here, if lubricating substances are applied on the moving surfaces, the friction
reduces and thereby wear also gets minimized. Thus the ill—effects of friction is
minimized.
 The use of lubricants to reduce friction depends upon the type of surfaces in contact
and also upon the working condition of various machines.
 Lubricant and Lubrication:
 Lubricant:
o Lubricant is defined as a substance introduced between two moving surfaces
(sliding/rolling) with a view to reduce the frictional resistance between them.
 Lubrication:
o Lubrication is the process of reducing frictional resistance between moving
surface with the use of lubricants in between them.
o The lubricant keeps the two surfaces apart so that frictional resistance and
consequent destruction of material is minimized.
 Purpose of Lubrication / Function of Lubricants:
1) It covers the surface irregularities thus forming a layer between the rubbing surfaces
t avoiding wear and tear.
2) It reduces loss of energy in the form of heat thereby acting as a coolant.
a. In Transformers:
i. The function of the lubricating oil in an electrical transformer is to insulate the
windings and to carry away the heat generated when the transformer is on
load.
b. For Cutting Tools:
i. Cutting fluids (emulsions) are used as lubricants for machining, cutting
grinding of metals.
ii. In these operations the tool comes in contact with the work piece which high

friction, results in excessive power consumption and liberation of large
amount of heat.
iii. Heat liberated may overheat the tool and may lover its temper. Therefore, the
main function of cutting fluid is to act as a coolant.
3) It reduces expansion of metal due to frictional heat generated, which in turn helps to
maintain shape, size & dimensions of metal part, in contact.
4) Lubricants reduce wastage of energy as the efficiency of machine is enhanced.
5) It avoids Seizure (welding of two surfaces due to heat) of moving surfaces.
6) It reduces the maintenance and running cost of the machine.
7) It acts as a Seal. E.g. Lubricant used between cylindrical wall and piston of an I.C
engine-acts as a Seal preventing the leakage of gases under high pressure from the
cylinder thus minimizing the power loss.
8) The presence of lubricants on machine surface prevents the attack of moisture,
dust. This helps to control corrosion of the moving machine parts.
9) It acts as a Cleaning agent because they have the tendency to wash off solid part;
produced due to combustion or wear (solid particles are transported away from the
sliding surface). This helps to control corrosion of the surfaces.
 Mechanism of Lubrication:
During lubrication of a machine, a film of lubricant is interposed between moving
surface in order to reduce frictional resistance, which minimizes damage of the rubbing
surface. There are three types of mechanism by which lubrication is done:
 Fluid Film or Thick Film or Hydrodynamic Lubrication:
 In this type of mechanism a liquid lubricant is applied in the form of thick film
between moving surface. The film is at-least surface contact & welding of
junctions rarely occurs.
 The resistance to movement is only due to internal resistance between the
particles of the lubricant moving over each other.
 In such system friction is dependent on the viscosity of lubricant, relative velocity
and area of the moving surfaces but is independent of load. Coefficient of friction
(f) is 0 001- 0.03, which is much lower as compared to that for un-lubricant
surfaces (0.5-1.5).
 f=F/W where "F" is the force required to cause motion. "W" is the applied load.
Mechanism of Hydrodynamic lubrication can he better understood by considering
the operation of journal bearing.
 The bearing consists of a shaft rotating at affair speed with moderate load. The
lubricant is applied in annular space.

 When journal bearing is stationary' the two surfaces remain in contact, but as
the shaft begins to rotate, it is slightly displaced from the journal axis, drawing a
layer of lubricant.
 It gets squeezed between the surfaces, due to which a pressure is developed in
the lubricant. Which is sufficient to keep the shaft & the bearing apart, the shaft
than floats drawing continuously a layer of lubricant between the surface.
 The efficiency of thick film Lubrication depends upon
a. Viscosity of the Lubricant:
The lubricant chosen should have adequate viscosity &should remain in place
and separate the surfaces under working condition. Highly viscous liquid en
used much energy will be required to overcome internal resistance to keep
them in motion.
Hydrocarbon oils are suitable for the same. To maintain suitable viscosity of
blended with polymers and antioxidants (amino phenols) are added to prevent
[O] of unsaturated compounds present in traces.
b. Design of Bearing:
c. Rate of Rotation:
E.g. Clocks, Guns, Sewing machines, scientific instrument are provided with
this type of lubrication.
 Boundary Lubrication or Thin Film Lubrication:
 Fluid or hydrodynamic is effective only if the lubricants film is 1000A thick. But
when the pressure developed in the lubricant is unable to support the load. the
lubricant film gets squeezed out. This happens
1) When It heavy load is encountered
2) Starting or stopping operation of a machine.
3) When viscosity is low.
4) The machine is to be operated at a comparative low speed.
 Under these circumstances a layer of lubricant adsorbed (i.e. surface attached) by
physical or chemical or both on the surface serves as lubricant.

 The adsorbed layer may not be able to cover all .abnormalities on the surface and
contact between surfaces occur which lead to some wear and tear coefficient.
 Of friction in such case is 0.05 to 0.15. Thus a thin layer of lubricant or
boundary film is maintained between the 2 metal surfaces which carry the load.
This kind of lubrication is called ‗Boundary Lubrication‘.
 Boundary film lubrication is influenced by the oiliness property of the lubricant.
Vegetable and animal oils have higher oiliness than mineral oil.
 It is because of the fatty acid content in animal and vegetable Oil. But they are
not thermally stable.
 Oiliness property of mineral oil is improved by addition of vegetable oil anal
animal of which also impart good thermal stability.
 Presence of fatty acids in of group chemically reacts with COOH the metal
surface fuming a thin film of metallic soap. The soap formed gets adsorbed on the
surface acting as lubricant in boundary lubrication.
 Effectiveness of boundary lubrication depends on the structure and chemical
character of oil. The lubricant molecule should have:
1) Long hydrocarbon chain.
2) Polar groups to promote spreading and orientation over the metallic surface at
high pressure.
3) High Viscosity Index, Resistance to Heat & Oxidation, good Oiliness, low pour
Point.
4) Active atoms or groups to form chemical bonds with the metal.
5) Lateral attraction between the chains.
 In IC-engines, blended mineral oil is used. Other substitutes are graphite or
Molybdenum di sulphide either in dry form or as a suspension in oil is suitable.
 Extreme Pressure Lubrication:
 When the operating condition is severe as with high load and high speed even the
layer of lubricant become ineffective.
 Contact between the surfaces takes place and lead to development high
temperatures and can even cause seizare.

 Certain additives have been found to be effective under such extreme conditions
to maintain lubrication and prevent damage to machinery parts. These are called
extreme pressure additives'.
 These additives react with the metal surface, which is initiated by prevailing high
temperature forming a durable solid film. These are:
1) Chlorine containing organic compounds (chlorinated esters).
2) Sulphur containing organic compounds (Sulpharised oils).
3) Phosphorous containing organic compounds (Tricresyl phosphate).
 These additives react with Metal to form metallic chloride, sulphide and
phosphides. which serve a lubricant layer between the surfaces).
 Melting point of these compounds is high and hence they retain their properties
even at high temperature.
 In continuous running operations at lower temperature ordinary lubricant with
adequate viscosity can maintain lubrication.
 These additives develop only under high temperature conditions.
 Extreme pressure additives are added is cutting fluid, in wire drawing, etc.
 Comparisons:
1)Fluid film & Extreme:
Sr.
No.
Fluid film lubrication or Thick
film / Hydrodynamic
Lubrication
Extreme Pressure Lubrication
1.
This mechanism for lubrication
is useful in machines where
load is low and speed is not very
high.
This mechanism for lubrication is
useful In machines where load & speed
both are high & also machine is under
pressure & temperature.
2.
In this liquid lubricant with
high viscosity is applied as thick
film on machine surfaces in
contact. Thickness of film is
1000A
In this mineral oils with special
additives are used. Additives improve
specific property of lubricant as per
requirement of working condition of
machines.

3.
The co-efficient of friction is
lowered to about 0.001 to 0.03
(un-lubricated machine parts
0.5 1.5to  )
The blending can be done to have
almost nil friction of machine parts,
became on surface, metal forms
temporary products with additives. E.g.
chlorides, sulphates etc.
4.
Lubricant must have adequate
viscosity and high viscosity
index.
In this, the lubricant with additives
remains on surface of metal for long
time as there is adherence due to
chemical reaction.
5.
It shows great impact of
seasons, as temperature affects
viscosity.
Seasonal impact is almost eliminated.
6.
E.g. sewing machines, watches,
clocks etc.
E.g. cutting tools, rock crushing
machines, wire drawing machines etc.
2)Fluid film & Boundary
Sr.
No.
Fluid film lubrication or Thick
film / Hydrodynamic
lubrication
Boundary lubrication
1.
This mechanism for lubrication is
useful in machines where load is
low and speed is not very high.
This mechanism for lubrication is useful
In machines where speed is low, load is
high and a journal / shaft starts moving
at a fixed interval.
2.
In this liquid lubricant with high
viscosity is applied as thick film
on machine surfaces in contact.
Thickness of film is 1000A
In this a lubricant capable of getting
adsorbed on lubricating surface is used;
as a thin film introduced in the clearing
spaces of moving surfaces.
3.
The co-efficient of friction is
lowered to about 0.001 to 0.03
(un-lubricated machine parts
0.5 1.5to  )
The co-efficient of friction is lowered to
extent of 0.05 to 0.15.

4.
Lubricant must have adequate
viscosity and high viscosity index.
Lubricant must have tendency to develop
adherence with the surface, high V. I,
high resistance to oxidation, low pour
point, adequate oiliness & high stability
at elevated temperature.
5.
It shows great impact of seasons,
as temperature affects viscosity.
Seasonal impact is minimized due to thin
film and also adherence to the surface.
6.
E.g. sewing machines, watches,
clocks etc.
E.g. Gears, rail axle, boxes, tractors,
rollers etc.
 Classification of Lubrication:
On the basis of physical state lubricants may be classified into following type:
1) Liquid Lubricants / Lubricating Oils.
2) Semi Solid Lubricants.
3) Solid Lubricants.
1) Liquid Lubricants:
Lubricating oils reduce friction & wear between two moving / sliding metallic surface
by providing a continuous fluid film in between them.
Good lubricating oil must possess the following properties:
a. Adequate viscosity for particular service condition.
b. Low freezing point.
c. High boiling point.
d. Good sticking characteristics.
e. High oxidation characteristics.
f. Heat stability.
g. Non-corrosive properties.
h. Stability to decomposition at the operating temperature.
i. Liquid lubricants are further classified on the basis of their origin into:

 Vegetable / Animal Oil:
 Vegetable / Animal oil are extracted from plants or animals respectively. These
are triglycerides of higher fatty acids.
 Commonly used vegetable oils are palm oil coconut oil, castor oil, groundnut oil,
and olive oil.
 Source of animal oil are fatty tissues.
 E.g.: Lard oil, (pigs). Tallow oil (cattle) & Neat foot oil (from cattle feet).
 Advantages:
a. They have very good Oiliness property (sticking of oil to the machine even
under high temperature & heavy load.
 Disadvantages:
a. They are very costly.
b. They tend to undergo oxidation in presence of oxygen forming gummy
products. hardened or thickened in presence of air.
c. They undergo hydrolysis in presence of moisture forming acidic products,
which cause corrosion of metal surfaces.
Now a days, they are used as blending agent with other oils to produce desired
effects.
 Mineral Oil:
 Mineral Oil is obtained by fractional distillation of petroleum or crude oil above
composed of long chain hydrocarbons of 12 — 50 C atoms.
 Advantages:
a. Cheap.
b. Abundantly available.
c. Have better thermal stability.
d. Available in wide range of viscosities.
 Disadvantages:
Poor oiliness as compared to vegetable and animal oils. To overcome this, mineral
oils are made into blends of suitable oiliness by adding vegetable / animal oils.
Lubricants
(Origin)
Vegetable/Anim
Oil
Mineral Oil Blended Oils Synthetic Oils

 Blended Oils / Compound Oils:
 The mineral oils obtained from crude oil by fractional distillation do not possess
all the desirable properties. The property is improved specific additives. Such
lubricants are called blended oils.
Additive Chemicals used Function
Oiliness
Carriers
Vegetable & Animal oils,
fatty acid like palmitic
acid & Oleic acid.
Forms a strong durable film on the
metal surface.
Viscosity
Index
Polymers like PS
Polyesters, Hexanol.
Reduces rate of change of Viscosity with
temperature.
Pour Point
Depressants
Phenol, certain
condensation products
of chlorinated wax with
naphthalene.
Lowers the Pour point separation or wax
from at thereby inhibiting the lubricant
from becoming solid
Extreme
Pressure
Additives
Organic materials that
contain Sulphur.
Phosphorous. Chlorine.
These additives are adsorbed on the
metal surface or react chemically with
metal producing a surface layer thereby
preventing tearing up of the metal under
high speed & load.
Anti-Oxidants
Phenolic or amino
compounds
Retard oxidation of oil by getting
themselves preferentially oxidized. Eg: In
IC engines, Turbines where (o) is a
serious problem.
Corrosion
Inhibitors
Organic compounds of
Phosphorous
Protect the metal from corrosion by
preventing contact between the metal
surfaces & the corrosive substance.
 Synthetic Lubricants:
 Blended oil though satisfies most of the requirements of a suitable lubricant but
for certain machines which work under severe extreme conditions it is found to
be unsuitable.
 E.g. in DIE casting fire resistant hydraulic fluids are desirable.
 Hot running bearings & Hot rolling mills where extreme temperature and
pressure are encountered.
 In Aircraft engines - when an aircraft in a cold region is to be taken off, the oil

has to be pumped at about 0° to 50°C depending upon the temperature of the
region.
 On the contrary when the same engine is landing in other country where the
temperature is high the engine nay get heated to very high temperature.
 These types of working conditions evolve the need of a lubricant that can be
useful between very wide ranges of temperature such as -50°C to 250°C.
 Synthetic lubricants should possess the following properties.
1) V. I. flash & Fire point should be high.
2) Low pour point.
3) Oiliness and Thermal stability at high temperature.
4) Resistance to corrosion and oxidation high.
 E.g.:
a. Chloroflouro H‘s - Chlorinated HC"S E.g. chlorinated diphenyl compounds.
b. Polyalkere Glycols: Low pour point, high Flash & fire point, V.I can be used
up to300 C .
 2 2 2 2n
OH H C CH OCH CH OH  
c. Silicones: High V.l. low volatility, good oxidation stability, moisture resistant,
can be used up to 230 C e.g. Aircraft's, Missile.
d. Fluorinated HC‘S e.g. Fluorinated tert Amines, Ethers, esters etc.
 Disadvantage:
High cost therefore used only for special machines such as in Aircraft and
refrigeration Industry.
2) Semi - Solid Lubricants:
 The most widely used semisolid lubricants are grease and Vaseline. Semisolid
lubricants are used under working condition where:
1) Where oil cannot remain in place due to high load, low speed intermittent
operation of machine parts, sudden jerks etc due to higher frictional
resistance of grease than oils. Rail axle boxes.
2) In situations where dripping or spurting of oil is undesirable, because unlike
oils, grease if used do not splash or drip over articles being prepared by the
machine. E.g. Machineries used in textile mills, paper industry, and food
product manufacturing unit.
3) In places where the bearing has to be sealed against entry of dirt, water & grit
because grease is less liable to contamination by these.
4) In bearing and gears that work at high temperature.
 Lubricating grease is the dispersion of soap in a liquid lubricating oil which may

be a mineral oil, synthetic oil, blended oil. To improve heat resistance of grease,
inorganic solid thickening agents like finely divided clay, colloidal silica, carbon
black etc are added.
3) Solid Lubricants:
 Solid lubricants are used
1) Where a liquid or semisolid lubricant film cannot be maintained or their
presence is undesirable. E.g. in commutator blades of electric motors &
Generators.
2) The operating temperature or load is too high for even a semisolid lubricant to
remain in position.
3) Comustible lubricant must be avoided.
4) Where parts to be lubricated are not easily accessible.
5) When heavy machinery is to be operated at high speed & moderate load or at
very high load & low speed.
 Commonly used solid lubricants are Graphite, Molybdenum disulphide, mica,
talc etc. solid lubricants. If not manufactured properly contain grain particles
which may damage delicate parts/ surface of machinery.
 Hence solid lubricants are to be checked carefully for fineness. Solid lubricants
are used in the form of fine powder or as suspension in water or oil.
A. Graphite:
a. Structure:
i. It consists of mesh of hexagonal carbon rings arranged in flat parallel
layers & each carbon atom is linked to three other carbon atoms by a
covalent bond.
ii. But the distance from the fourth election is more than double. Hence this
forth valence bond is not fixed.
iii. But moves about & hence there is no strong bonding. Further, the planer
surface does not have covalent links & are.
iv. Free to move relative to one another which gives slippery & hence lubricant
property to graphite.

b. Properties:
i. High strength & chemical stability at elevated temperature.
ii. It is soft & has lubricating property.
iii. High electrical conductivity.
c. Uses:
i. As lubricant
ii. Making electrodes
iii. Resistors
iv. Making batteries etc.
B. Molybdenum Disulphide  2MoS
a. Structure:
i. It has sandwich like structure in which a layer of Mo atom is between the 2
layers of sulphur atoms.
ii. The layers are parallel to each other & inter-layer distance = 3.13A.
iii. The layers are held together by weak Vander Wall‘s force. So they can slide
over other easily due to softness of  2MoS
b. Properties:
i. Stable in air upto 400 C
ii. Can also work in vacuum.
c. Uses:
i. Space crafts
ii. Automatic & truck-chessis.

 Theory Questions:
1) What are lubricants? Give their important functions.
2) How are lubricants classified? Explain with examples.
3) Define grease. Under which situation it is used as a lubricant?
4) Mention the essential qualities of a good lubricant.
5) What is extreme pressure lubrication? Discuss with suitable examples.
6) What are the different mechanisms of lubrication? Explain the boundary or thin film
lubrication.
7) Define & Explain the importance of the following :
a. Oiliness
b. Viscosity & Viscosity index
c. Cloud and pour point
d. Fire and Flash Point
e. Acid Value and Saponification Value
f. Emulsification
8) Write short note on: Liquid Lubricants
 Properties of Lubricants:
 Viscosity Index (V.I.):
 Viscosity Index is defined as the rate of change of viscosity of oil with respect to
temperature. On heating the oil becomes thin. i.e. their viscosity decrease
 If the decrease in viscosity is rapid, the oil is said to have a ―Low Viscosity Index‖.
 If the viscosity of the oil is slightly affected on raising temperature it is said to
have ―High Viscosity Index‖.
 An ideal lubricant should have HIGH VISCOSITY INDEX and appropriate
viscosity.
. 100
L U
V I
L H

 

Where, U = Viscosity at 100°F of the test oil.
L = Viscosity at 100°F of the low-viscosity standard oils (L-oils) having
V.I = 0 & having the same viscosity at 210°F as the test oil.
H = Viscosity at 100°F of the high viscosity standard oils (H-oils)
having
V.I 100 & have the same viscosity at 210°F as the test oil.
o Significance:
Viscosity is the main determinant of the operating characteristics of the
lubricants.

1) If the viscosity of oil is too low, a liquid film cannot be maintained between
moving consequently excessive wear takes place.
2) If the viscosity is too high, excessive friction will result.
 Flash Point anew Fire Point:
 Flash Point is defined as the lowest temperature at which the lubricating oil
gives off enough vapors when heated at a specified rate that ignite for a moment
when a flame is brought near the surface of the oil.
 Fire Point of the oil is defined as the lowest temperature at which the vapours of
the oil burn continuously for at-least 5 seconds when a flame is brought near the
surface of the oil which heated in the apparatus at the specified rate.
In most cases, the fire points are 5' to 40°Higher than its flash point.
 Significance:
Flash & Fire points are of importance when oil is exposed to high temperature
service. A good lubricant should have flash point at-least above the temperature
to which it is exposed.
These safeguards against risk of fire during storage, transport and use of the oil.
 Measurement:
It is determined either by Open Cup apparatus (Cleaveland Apparatus) or by C
apparatus. (Abel's and Pensky-Martin apparatus)
Closed cup apparatus gives more reproducible results.
 Cloud Point & Pour Point:
 Cloud Point: When oil is cooled slowly is at a specified rate, the temperature at
which it becomes cloudy or hazy in appearance is called its ―Cloud point‖.
 Pour Point: When oil is cooled slowly at a specified rate, the temperature at
which the oil ceases to flow or pour is called its ―pour point‖.
 Significance:
1) Cloud & Pour points indicate the suitability of lubricants in cold conditions.
2) They help w to know the lowest temperature up to is which the oil can be
suitable as a liquid lubricant.
3) Lubricant used in machine working at low temperate should possess low pour
point.
4) Otherwise solidification of lubricant will cause jamming of the machine. It has
been found that presence of wax in the lubricating oil raise the pour point.
 Measurement:
The test is carried out using Cloud & Pour point apparatus.

 Acid Value Or Neutralization Number:
 It is defined as the number of milligram of KOH required to neutralize free acids
present in 1 gram of the oil sample.
 Source:
When a mineral is exposed to contr6rTis rise and fall of temperatures during
working conditions or on prolonged exposure to oxygen from air undergo cleavage
in the long hydrocarbon chain of which the oil is composed forming tree organic
acids.
If vegetable or animal oil are used as lubricant or for blending (glycerides of
higher fatty; easily oxidisable unsaturated sites in tatty acids tends to absorb
oxygen on exposure acids.
|
|
C C  
Unsaturated parts of Vegetable / animal oil
2O
COOH COOH 
 Effect:
Presence of carboxylic acids affects the quantity of oil increasing the acid value.
Lubricating oil should possess acid value less than 0.1.
Value greater than 0.1 indicates that oil has been oxidized and will lead to
corrosion of metal surfaces besides gum and sludge formation. Free mineral acids
are not present in the lubricants unless refined in a faulty manner.
 Determination:
The oil sample weighed (1 gm ) & mixed with alcohol (50 rid for l gm of oil
sample). The mixture is warmed for 10-15 mts on water bath.
The free fatty acid separate out from oil. The Mixture is then titrated against 0.1
B KOH solution Phenolphthalein Indicators. The volume of KOH consumed is
noted down & is substituted in the formula.
KOHVolume of KOH soln run down N 56
Acid Value
Weight of oil in grams
 

(Long Hydrocarbon chain)
Effects of Heat or Oxygen from air
(Carboxylic acid)

End point is colorless to pink.
Unit of acid value is mg of KOH / gm of oil
 Significance:
Acid value helps us to know the suitability of oil.
Higher the acid value more the corrosion of machine surfaces which cause
increased friction and wear. A good lubricant should possess low acid value.
 Saponification Value:
 Saponification is the hydrolysis of an ester with alkali (NaOH / KOH) to give
alcohol and sodium or potassium salt of the acid (soap). Saponification value is
the characteristic property of vegetable or animal oils and not of mineral or
synthetic oil.
 Mineral oil being mixture of hydrocarbons do not react with KOH and so are not
saponifiable. Vegetable / animal oils are triglycerides of mixed fatty acids. The
fatty acids normally found present in these oils are from 4 20toC C & many oils
have even higher fatty acids.
 The saturated as well as unsaturated nature. The general empirical formula of
these acids is 2 2n nC H O : Oleic  18C , Palmitic  16C , & Stearic  17C . Vegetable
and animal oils are triglycerides of fatty acids. Alkali saponifies them.
Where, R, 1 2,R R are the alkali groups
 During saponification, on treatment of oil with aqueous alkali, the soaps are
formed and glycerol is released. They require large amount of alkali to get
hydrolyzed. Their saponification values are high and each fatty oil has its own
characteristic value. E.g.:
Vegetable oils
Groundnut oil =190 —196
Castor oil =183-186
Coconut oil — 246-260
Animal oils
Lard oil =195
Neatfoot oil =194 to 199

 Definition:
Saponification value of the oil is defined as the number of milligram of potassium
hydroxide required to saponifies one gram of vegetable or animal oil.
 Determination:
A known quantity of oil is mixed with known excess of alcoholic standard KOH
soln. say 0.5 N. The mixture is shaken vigorously and allowed to stand for nearly
24 hrs at room temperature alternatively the mixture is refluxed for about 2 hrs,
on water bath using condenser.
Reflux
Oil Alcholic Soap glycerol unreactedKOH KOH   
The quantity of unreacted KOH is found out by titration against 0.5N HCl soln.
using phenolphthalein indicator. End point is pink to colourless. Let the burette
reading be 2V ml .
Volume 01 KOH consumed = Volume in ml of KOH soln added to oil Back
titration reading.
Volume of KOH consumed = Volume in ml of 0.5 NHCI run down in the blank
expt. - Vol in ml of. HCI run down in test experiment.
   2 1 2 10.5 0.5V V ml of N HCI V V ml of N KOH  
Saponification value =
. 0.5 56KOHVol of N KOH consumed N
Weight of oil in grams
 
Note: 1000 ml of I NKOH solution contain 56 gram of KOH
Unit: mg of KOH/gm of oil.
 Significance:
1) Determination of Sap value helps it, ascertain the presence of animal, mineral
and vegetable oil in a lubricant.
2) From the Sap value of oil it can be understood whether the oil is adulterated
or not. It to know the % of adulteration.
% of oil added in pure oil =
changed sap value
x 100
Sap value of pure oil
3) Vegetable or animal oil are used as a blending agent in mineral oil to improve
Oiliness property from the sap value of the blend the % of vegetable oil present
in a blend can be identified.
% of veg. Oil in blend =
Sap value of blend
x 100
Sap value of pure oil
4) Pure mineral oils are mixture of hydrocarbons & not esters. Therefore a pure
mineral oil has sap. value zero.

5) Knowledge of sap. value helps to know the stability of oil in aqueous or
alkaline medium. if in case machine parts face any such conditions.
 Emulsification:
 The ability of oil to get intimately mixed with water forming an emulsion is called
emulsification.
 Emulsion have a tendency to collect dirt, grit, foreign matter etc. thereby causing
& wearing out of the lubricated parts.
 Hence it is desirable that the lubricating oil should form an emulsion with water
which breaks of readily.
 This is called Demulsibity.
 The tendency of lubricant water emulsion to break is determined by ASTM test.
 In this steam at 100 C is bubbled through a test tube containing 20 ml of oil till
the temperature increases to90 C . The test tube is place in a both at90 C . This
time in seconds is noted when the oil & water separate out in distinct layers. It is
called steam emulsion number (SEN) or demulsification number clearly a good
lubricant possesses a low SEN.
 Oiliness:
 It is a measure of lubricants capacity to stick on the surface of machine parts,
under heavy pressure or load. It is important property for extreme pressure
lubrication. Lubricants with good oiliness on subjecting to high pressure stay in
between the lubricating surface. When lubricating oil of poor oiliness is subjected
to high pressure, it has tendency to be squeezed out of the lubricated parts
thereby stopping the lubricating action.
 Mineral oils have poor oiliness whereas vegetable oils have good oiliness.
Vegetables oils are added to mineral oils to improve their oiliness.
 No direct test is available for measuring oiliness.
 Theory Question:
1) Define & Explain the importance of the following :
(a) Oiliness (b) Viscosity & Viscosity index (c) Cloud and pour point
(d) Fire and flash point (e) Acid value and Saponification value (f) Emulsification

 Solved Problems:
There are two types of problems as follows:
Important Formulae
Same formula for both types:
S.V. or A.V. = KOH KOH KOH
oil
V M N
W
 
1) 1.55 gm of the oil is saponified with 0.5N alcoholic potassium hydroxide soln. After
refluxing the mixture, it required 15 ml 0.5 NHCL Soln. Find the S.V. of the oil.
 
 
oil
20 15 56 0.5
. . 90.32 mgs of KOH Ans.
1.55
KOH KOH KOHV M N
S V
W
   
  
2) 2.5 g of vegetable oil was mixed with 50 ml of KOH soln. & heated for 1 hr. The
mixture required 26.4 ml of 0.4 N HCl. The blank titration reading was 49 ml. Find
the S.V. of oil.
 
 
oil
49 26.4 56 0.4
. . 202.496 mg/g of KOH Ans.
2.5
KOH KOH KOHV M N
S V
W
   
  
[Note: “50 ml of KOH” is extra confusing data which must be ignored]
3) 5 gms of vegetable oil was saponified using excess of alcoholic KOH (0.5N). The
mixture required 15.0 ml of 0.5 N HCl while blank titration 45.0 of same HCl. Find
the Saponification value of the oil.
Solution:
Given: Weight of oil = 5 gms.
Blank titration reading = 45 ml 0.5 N HCl
Back titration reading = 15 ml 0.5 N HCl.
Thus, volume of 0.5 N KOH required by the oil for Saponification in terms of
0.5 N HCl =45 – 15 = 30 ml.
Types of Lubrication Problems
1) Saponification
Value (S.V.)
2) Acid Value (A.V.)

56Volume of KOH consumed Normality of KOH
Weight of Oil gms
 
30 0.5 56
5
 

= 168 mg. KOH
Ans.: Saponification value of oil = 168 mgs KOH
4) An oil sample of saponification value 180 mgs. KOH, was saponified using 0.4 N
alcoholic KOH solution. The blank titration reading was 50 ml, of 0.4 N HCl solution.
Find the quantity of alcoholic by the oil per gram.
Solution:
Given: Saponification value of oil = 180 mg KOH
Normality of alcoholic KOH = 0.4 N
Normality of HCl = 0.4 N
Blank titration reading = 50 ml
Weight of oil = 1 gm
To find:
Quantity of 0.4 alcoholic KOH required by 1 gm of oil.
56KOHVolume of alcoholic KOH N
Weight of Oil gms
 
56
180 KOHVolume of alcoholic KOH N
Weight of Oil gms
 

  56
180 KOHBlank Back N
Weight of Oil
 

 50 0.4 56
180
1
 

 180 50 22.4 
22.4 1120 180 940   
940
22.4
 
41.96
42 ml  of alcoholic KOH.
Ans.: Quantity of alcoholic KOH required per gm = 42 ml.
5) 5 gms of an oil was saponified with 50 ml 0.5 N alcoholic KOH. After refluxing fo r2
hours, the mixture was titrated by 15 ml of 0.5 N HCl. Find the Saponification value
of oil.

Solution:
Given: Weight of oil = 5 gms
Quantity of 0.5 N alcoholic KOH = 50 ml
Quantity of 0.5 N HCl required (Back titration) = 15 ml
To find: Saponification value of oil.
Now,
Weight of Oil gms
 
= [Volume of KOH added to oil – Back titration reading ]
56KOHN
Weight of Oil
 
 50 15 0.5 56
. .
5
S V
 

35 0.5 56
. . 196
5
S V mg KOH
 
 
Ans.: Saponification value of oil = 196 mg KOH.
6) 2.5 gms of an oil were saponified by alcoholic KOH (0.25 N). The blank reading with
0.5 N HCl was 40 ml while back was 20 ml of same HCl. Find Saponification value of
oil.
Solution:
Given: Weight of oil = 2.5 gms
Back titration reading = 20 ml 0.5 N HCl
Normality of KOH = 0.25 N.
Strength of alcoholic KOH & HCl are given as, 0.25 N & 0.5 N respectively.
Point of equivalence in titration would reach as,
2 ml 0.25 N KOH = 1 ml 0.5 N HCl.
Thus quantity of KOH added to oil (blank) = 80 ml.
And quantity of KOH unreacted (back) = 40 ml.
Weight of Oil gms
 
  56
. . KOHBlank Back N
S V
Weight of Oil
  

 80 40 0.25 56
. .
2.5
S V
 


40 0.25 56
. .
2.5
S V mg KOH
 

S.V. = 224 mg. KOH
Ans.: Saponification value of oil = 224 mg. KOH
7) 5 gms of oil was saponified using 50 ml alcoholic KOH [1.4 gms per 50 ml solution].
The mixture required 10 ml 0.5 N HCl while blank titration reading was 40 ml of
same HCl. Find the Saponification value of oil.
Solution:
Given:
Weight of oil = 5 gms
Strength of alcoholic KOH = 1.4 gm per 50 ml
= 28 gms per 1000 ml
= 0.5 N
Back titration reading = 10 ml 0.5 N HCl
To find, Saponification value
56
. . KOHVolume of alcoholic KOH N
S V
Weight of Oil in gms
 

  56
. . KOHBlank Back N
S V
Weight of Oil
 

 40 10 0.5 56
. .
5
S V mgs KOH
 

30 0.5 56
. .
5
S V
 

S.V.= 168 mgs KOH.
Ans.: Saponification value of oil = 168 mgs KOH.
8) 6 gms of an oil was saponified with 50 ml 0.5 N alcoholic KOH. After refusing for 2
hrs., the mixture was titrated by 25 ml of 0.5 N HCl. Find Saponification value of oil.
Solution:
Volume of 0.5 N HCl, before Saponification = 50 ml
Volume of 0.5 N HCl after Saponification = 25 ml
Volume of acid reacted with oil = (50-25) = 25 ml
56
. .
Amount of KOH Consumed Normality of KOH
S V
Weight of Oil ingms
 


25 0.5 56
5
 

= 140 mg of KOH
Ans.: Saponification value = 140 mg of KOH.
9) 16 gm of blended oil was heated with 50 ml KOH. This mixture then required 31.5
ml of 0.5 N HCl. 50 ml KOH required 45 ml 0.5 N HCl. Find % cottonseed oil , if
Saponification value = 192 mg.
Solution:
Given: weight of blended oil = 16 gm
Amount of KOH = 50 ml
Amount of HCl = 31.5 ml
Saponification value = 192 mgs
% cottonseed oil =?
To find: % cottonseed oil in blended
50 ml KOH = 45 ml 0.5 N HCl
1 1 2 2N V N V
1 150 0.5 45N   
1
0.5 45
50
N


= 0.45 N
Normality of KOH = 0.45 N
Weight of blended oil = 16 gm
Weight of cottonseed oil = 16 – x gm
Now, Saponification value
 
 
56
16
Blank Back N
x
  


Blended oil contains castor oil & petroleum oil.
Petroleum oil has Saponification value always zero.
Let us calculate Saponification value of the blended oil.
56
. .
Volumeof KOH Normality of KOH
S V
Weight of Blend
 

13.5 0.45 56
16
 

Saponification value of blend = 21.26 mgs of KOH
% castor oil 100
Saponification value of blend
Saponification value of castor oil
 

21.26
100
192
 
= 11.074 %
Ans.: % of castor oil in blend = 11.074%.
10) 3 gm of liquid lubricant was saponified with potassium hydroxide solution. After
Saponification the mixture was titrated against 0.5 N HCL solution. The burette
reading was found to be 12 ml. if blank titration burette reading was 36 ml,
calculate the Saponification number of the lubricant.
Solution:
Given: Weight of lubricant = 3 gm
Blank titration reading = 36 ml
Back titration reading = 12 ml
To calculate Saponification value,
Sap. Value
. 28
.
.
Vol of KOH consumed
mgs of KOH
Wt of Oil


 36 12 28
.
3
mgs of KOH
 

 
24 28
8 28 .
3
mgs of KOH

  
= 224 mgs of KOH
Ans.: Saponification value = 224 mgs of KOH.
11) 1.55 gram of the oil is saponified with 20 ml
2
N
of alcoholic potassium hydroxide
solution. After refluxing the mixture, it requires 15 ml of
2
N
HCl solution. Find
Saponification value of oil.
Solution:
Given: weight of oil = 1.55 gms
Volume of KOH = 20 ml
Normality of KOH =
2
N
= 0.5 N
Volume of HCl = 15 ml
Normality of HCl =
2
N
= 0.5 N
Saponification value = ?

. 56
. . .KOHVol of KOH consumed N
S V mgs of KOH
Weight of Oil
 

 20 15 0.5 56
1.55
  

5 0.5 56
1.55
 

2.5 56
1.55


= 90.32 mgs of KOH
Ans.: Saponification value = 90.32 mgs of KOH
12) 1.25 gram of an oil was saponified with 50 ml 0.1 N potassium hydroxide solution.
After refluxing the mixture required 7.5 ml 0.1 N hydrochloric acid for
neutralization. Find Saponification value of the oil.
Solution:
Given: Weight of oil = 1.25 gm
Volume of 0.1 N KOH = 1V =50 ml
Volume of 0.1 N HCl = 2V = 7.5 ml
Used for neutralization
Volume of 0.1 N KOH = Excess = 7.5 ml = 2V
56
. . KOHVolume of KOH used in saponificaiton N
S V
Weight of oil
 

 50 7.5 0.1 56
1.25
  

42.5 0.1 56
1.25
 

= 190.4
Ans.: Saponification value = 190.4
13) 2.5 of a blended oil was saponified using express alcoholic KOH solution (0.5 N).
after refluxing for two hours, the mixture was titrated against 0.5 N HCl solution.
The burette reading was found to be 24 ml. the blank titration required 40 ml of the
same HCL solution. Find the Saponification value of the oil. If the oil used for
blending has Saponification value of 191, calculated percentage oil in the blend.
Solution:
Given: Wt. of Blended oil = 2.5 gms.
Normality of KOH (alcoholic) = 0.5 N
Normality of HCl soln = 0.5 B
Volume of HCl son = 24 ml
Blank titration (B.R) = 40 ml

  56
. . KOHBlank Back N
S V
Weight of oil
  

 40 24 0.5 56
2.5
  

16 28
2.5


= 179.2
Saponification value = 179.2
Now for blending if oil with Saponification value 191 is used
% blend
1
100
2
Saponification value of sample
Saponification value of sample
 
179.2
100
191
 
= 93.8 %
14) A vegetable oil was tested for its acid value. 10 gms of oil required 0.;2 ml of KOH.
Find its acid value.  0.02KOHN N
Solution:
Volume of KOH = 0.2 ml
To find: Acid Value,
56
. . KOHVolume of KOH N
AV
Weight of oil in gms
 

0.2 0.02 56
. .
10
AV
 

= 0.0224 Mgs /gm
Ans.: Acid value of oil = 0.0224 mgs KOH
15) An oil blend used was analyzed for its acid value. 5 ml of required 2.5 ml
2
N
KOH.
Find its acid value. State whether the blend can still be useful for lubrication.
(density of oil = 0.89)
Solution:
Given: (Weight = Volume in ml  Density )
Weight of oil = 5 ml  0.89 = 4.45 gms

To find: Acid value,
56
AV
 

2.5 0.01 56
. .
4.45
AV
 

= 0.315 mgs/ gms.
Ans.: Acid value = 0.315 mgs KOH
This value is more than 0.1 the oil cannot be used for lubrication.
16) Find the acid value of vegetable oil whose 5 ml required 2 ml of N/100 KOH during
titration . (Density of oil = 0.92).
Solution:
Given: Volume of oil = 5 ml Volume of KOH = 2 ml
Density of oil = 0.92 Normality of KOH = 0.01 N
To find: Acid value
Volume of oil = 5 ml
To find weight of oil,
Density =
Mass
Volume
or Mass = Density  Volume
= 0.92  5
Weight of oil = 4.60 g
 
56.1
. .
Volume of KOH required Normalityof KOH
AV
Weight of oil gm
 

Here acid value
2 0.01 56
0.243
4.60
mg
 
 
Ans.: Acid Value = 0.243 mg of KOH
17) 9 ml oil is taken from machine and it requires 1.5 ml of 0.04 N KOH. Find acid value
(density of oil = 0.81 g/ml)
Solution:
Given: Volume of oil = 9 ml
Density of oil = 0.81 g/ml
Now
56
AV
Weight of oil
 

Here, wt. (mass) of oil = volume  density
9 0.81 7.29 gm  

1.5 0.04 56
.
7.29
AV
 

= 0.461 mgs of KOH
Ans.: Acid value = 0.461 mgs of KOH
18) 20 ml of a lubricating oil was dissolved in alcohol. Solution was titrated against 0.1
N KOH solution. At the end point burette reading was found to 2.5 ml. calculate the
acid value of the oil. (Density of oil = 0.86 gm/ml).
Solution:
Given: Volume of oil = 20
Density of oil = 0.86 gm/ml
Normality KOH = 0.1
Amount of KOH = 2.5 ml
To find: Acid value
Weight of oil = volume  density
= 20  0.86
Weight of oil = 17.20 gm
56
AV
Weight of oil
 

2.5 0.1 56
.
17.20
AV
 

Acid value = 0.814 mgs of KOH
19) Find acid value of 3 gm of oil which required 0.2 m of 0.025 N KOH to neutralize free
acids presents.
Solution:
Given: Weight of oil = 3 gm
Quantity of KOH = 0.2 ml
Acid Value = ?
To find: Acid Value
56
AV
Weight of oil
 

0.2 0.025 56
.
3
AV
 

= 0.0933 mgs of KOH

20) Find acid value of a vegetable oil whose 10 ml required 4.0 ml of 0.01 N KOH during
titration. (density of the oil = 0.92)
Solution:
Volume of KOH = 4 ml
Density of oil = 0.92
To find: Acid Value
56
. .
Volume of KOH N KOH
AV
Weight of oil in gm
 

Here weight of oil 0.92 10 
9.2 gm
4 0.01 56
. .
9.2
AV
 

= 0.243 mgs of KOH.
Ans.: Acid Value = 0.243 mgs of KOH
21) Find acid value of given oil when 20 ml required 2.8 ml of KOH during titration
(density of oil = 0.86 gm/ml) state whether oil is proper for lubrication or not from
acid value.
Solution:
Normality of KOH =
10
N
Density of oil = 0.86 gm/ml
= Acid value
To find,
And whether oil is proper for lubrication?
56
. .
KOHVolume of KOH N
AV
Weight of oil in gm
 

Weight of oil = (Volume  Density) gms
20 0.86 1.720 gms  
2.8 0.1 56
. . 0.911
1.72
AV mg of KOH
 
 
Ans.: (i) Acid value = 0.911 mgs of KOH
(ii) Since, acid value exceeds 0-1, the oil is not proper for lubrication.

22) Find the acid value of oil sample whose 7 ml required 3.8 ml N/50 KOH during
titration.
(density of oil= 0.88) state whether the oil is suitable for lubrication or not.
Solution:
Given: Weight of oil =  .Vol in ml Density
7 0.88 
= 6.16 gms
Normality of KOH = N/50 = 0.20 N
56
. .
KOHVolume of KOH N
AV
 

3.8 0.1 56
. . 0.690 /
6.16
AV mgs gms
 
 
Acid value = 0.690 mgs KOH
This value is more than 0.1 the oil cannot be used for lubrication.
23) 1.3 g of a gear box oil is taken for acid value determination. It required 0.8 ml of
0.001 N KOH for neutralization. Calculate the acid value & mention whether the oil
is suitable to be used further or not.
Solution:
Given: Weight of oil = 1.3 gms
To find = (a) Acid value of oil
(b) Whether oil is suitable for lubrication?
56
. .
KOHVolume of KOH N
AV
Weight of oil
 

0.8 0.001 56
. . 0.034
1.3
AV mgs of KOH
 
 
Acid value = 0.034 mgs of KOH
Oil is suitable for lubrication since its acid value is less than 0.1 mgs of KOH.
24) Find the acid value of a used lubricating oil sample whose 10 ml required 5 ml of
N/50 of N/50 KOH during titration. (Density of oil = 0.91 g/cc). State whether the
oil is suitable for lubrication or not.
Solution:
Given: Volume of oil = 10 ml , Density of oil = 0.91 gm/cc
Normality of KOH solution = N/50

To find: (a) Acid value of oil (b) whether oil is suitable on lubricant?
Weight (mass) of oil= volume Density
= 10 0.91
= 9.1 gms
56
. .
KOHVolume of KOH N
AV
Weight of oil
 

5 0.02 56
. . 0.615
9.1
AV mgs of KOH
 
 
Ans.:
(a) Acid value of given oil sample is 0.615 mgs of KOH.
(b) Oil is not suitable for lubrication because its acid value is greater than 0.1
mgs of KOH.
 Unsolved Problems:
 Type 1: Problems S.V.:
1) 1.25 gms of an oil was saponified with 50 ml 0.1 N KOH solution. After refluxing,
the mixture required 7.5 ml, 0.1 N HCI for neutralization. Find Saponification
value of the oil. [190.4]
2) 6 gms of oil was saponified with 50 ml. 0.5 N KOH after refluxing for 2 hours, the
mixture was titrated by 25ml of 0.5 N HCI. Find Saponification value of oil. [140]
 Type 2 : Problems A.V. :
1)Find the acid value of given oil whose 20 ml required 2.8 ml of N/10 KOH during
titration (density of oil = 0.86 gm/ml) state whether the oil is proper for lubrication
or not. [ 0.911, Not suitable]
2)Find the acid value of a vegetable oil whose 10 ml required 4 ml of 0.01 N KOH,
density of oil = 0.92 [0.243]
 University Questions:
Dec 2007
1) What are blended oils? How are they superior to vegetable and mineral oils?
(2 Marks)
2) What are lubricants? Explain the mechanism of thin film lubrication in detail.
(4 Marks)
3) Explain any two of the following properties of lubricants :
a. Flash point and fire point temperature
b. Cloud point and pour point temperature
c. Saponification number. (3 Marks)

Dec 2008
4) Explain the following properties of lubricant with significance. (8 Marks)
a. Viscosity and viscosity index
b. Flash point and fire
c. Saponification value
5) Write note on solid lubricant. (5 Marks)
6) Write note on Extreme pressure Lubrication. (3 Marks)
May 2008
7) What are lubricants? List different functions of lubricants. (3 Marks)
8) Explain any two of the following properties of lubricants
a. Oiliness
b. Acid value
c. Cloud point and pour point temperature (6 Marks)
9) Write short note on Solid lubricant. (5 Marks)
May 2009
10) What is Grease? Under which situation it is used as a lubricant. (3 Marks)
11) What are Lubricants? Explain the Extreme Pressure Lubrication. (4 Marks)
12) Write the definition and significance of the following terms (6 Marks)
a. Flash Point and Fire Point b. S. V. c. Viscosity Index.
Dec 2009
13) Define the terms lubricants and lubrication-Mention the various types of mechanism
involved in the lubrication of Machines. Discuss boundary - film lubrication in
detail. (8 Marks)
May 2010
14) What is the principle of lubrication? Explain mechanism cation. (5 Marks)
15) Give in brief the functions of various additives employed for the improvement of
lubricants. (5 Marks)
16) Write short note on : Liquid Lubricants (5 Marks)
Dec 2010
17) Explain Flash point and Fire point with its significance. (3 Marks)
18) Write a note on Blended oils. (4 Marks)
19) Explain the following types of lubrication: (5 Marks)
a. Boundary lubrication
b. Extreme pressure lubrication.
May 2011
20) Define lubrication and explain the mechanism of hydrodynamic lubrication.
(5 Marks)

21) Write a note on Solid Lubricants. (5 Marks)
22) Write a short note on Blended oils. (4 Marks)
Dec. 2011
23) List any five characteristics of a good lubricant with justification. (5 Marks)
May 2012
24) In what situations are solid lubricants used? Explain structure, properties and uses
of any one solid lubricant. (5 Marks)
25) Explain Boundary lubrication. (5 Marks)
 University Questions (Problems):
Dec 2007
1) 3 gm of liquid lubricant was saponified with potassium hydroxide solution. After
Saponification the mixture was titrated against 0.5 N HCL solution. The burette
reading was found to be 12 ml. If blank titration burette reading was 36 ml,
calculate the Saponification number of the lubricant. (4 Marks)
May 2008
2) 1.55 gram of oil is saponified with 20 ml of
2
N
alcoholic potassium hydroxide
solution. After refluxing the mixture, it requires 15 ml of
2
N
HCI solution. Find
Saponification value of oil. (Pl) (3 Marks)
Dec 2008
3) Find acid value of vegetable oil whose 10 ml required 4.0 ml of 0.01 N KOH during
titration. (Density of the oil=0.92) (3 Marks)
May 2009
4) 9 ml oil is taken from machine and it requires 1.5 ml of 0.04 N KOH. Find acid value
(density of oil = 0.81 g/ml). (4 Marks)
Dec 2009
5) 1.25 gram of the oil was saponified with 50 ml 0.1 N potassium hydroxide solution.
After refluxing the mixture required 7.5 ml 0.1 N hydrochloric acid for
neutralization. Find Saponification value of the oil. (3 Marks)
May 2010
6) Find acid value of given oil whose 20 ml required 2.8 ml of
2
N
KOH during titration
(Density of oil = 0.86 gm/ml) state whether oil is proper for lubrication or not from
acid value. (3 Marks)

Dec 2010
7) Find the acid value of a used oil sample whose 7 ml required 3.8 ml of N/50 KOH
during titration. (Density of oil= 0.88) state whether the oil is suitable for lubrication
or not. (3 Marks)
May 2011
8) Find the acid value of a used oil sample whose 7ml required 3.8 ml N/50 KOH
during titration. (Density of oil = 0.88). State whether the oil is suitable for
lubrication or not. (3 Marks)
Dec 2011
9) 1.3 g of a gear box oil is taken for acid value determination. It 0.8 ml of 0.001 N KOH
for neutralization. Calculate the acid value determination whether the oil is suitable
to be used further or not. (5 Marks)
May 2012
10) Find the acid value of a used lubricating oil sample whose 10 mi required 5 ml of
N/50 KOH during titration. (Density of oil = 0.91 g/cc). State whether the oil is
suitable for lubrication or not. (3 Marks)
11) 2.5 g of a blended oil was saponified using express alcoholic KOH solution (0.5 N).
After refluxing for two hours, the mixture was titrated against 0.5 N HCL solution.
The burette reading was found to be 24 ml. The blank titration required 40 ml of the
same HCL solution. Find the Saponification value of the oil. If the oil used for
blending has Saponification value of 191, calculate percentage oil in the blend.
(4 Marks)

3. lubricants

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3. lubricants