chapter 2 physical properties of hydraulic fluid

Chapter 2: Physical properties of hydraulic fluid 2011

Hydraulic and pneumatic control lecture notes by Siraj K.

Chapter 2: Physical properties of hydraulic fluid

Learning objectives

At the end of this chapter, you should be able to:

1. Explain the primary functions of a hydraulic fluid.

2. Define the term fluid.

3. Distinguish between a liquid and gas.

4. Appreciate the properties of desired of a hydraulic fluid.

5. Define the term specific weight, density, and specific gravity.

6. Understand the terms pressure, head and absolute pressures.

7. Differentiate between gage pressures and absolute pressures.

8. Calculate the force created by a pressure.

9. Understand the kinematic viscosity and absolute viscosity

10. Convert viscosity from one set of units to another set of units.

11. Explain the difference between viscosity and viscosity index.

2.1 Introduction

The single most important material in a hydraulic system is the working fluid itself. Hydraulic

fluid characteristics have a crucial effect on equipment performance and life. It is important to

use a clean, high-quality fluid in order to achieve efficient hydraulic system operation. Most

modern hydraulic fluids are complex compound that have been carefully prepared to meet their

demanding task. In addition to having a base fluid, hydraulic fluids contain special additives to

provide desired characteristics.

A hydraulic fluid has the following four primary functions:

1. Transmit power

2. Lubricate moving parts

3. Seal clearances between mating parts

4. Dissipate heat

In addition a hydraulic fluid must be inexpensive and readily available. To accomplish

properly the four primary functions and be practical from a safety and cost point of view, a

hydraulic fluid should have the following properties:

1. Good lubricity



2. Ideal viscosity

3. Chemical stability

4. Compatibility with system materials

5. High degree of incompressibility

6. Fire resistance

7. Good heat-transfer capability

8. Low density

9. Foam resistance

10. Non toxicity

11. Low volatility

This is a challenging list, and no single hydraulic fluid possesses all of these desirable

characteristics. The fluid power designer must select the fluid that comes the closest to being

ideal overall for a particular application.

Hydraulic fluid must also be changed periodically, the frequency depending not only on the

fluid but also on the operating conditions. Laboratory analysis is the best method for the

determining when a fluid should be changed. Generally speaking, a fluid should be changed

when its viscosity and acidity increase due to fluid breakdown or contamination. Preferably,

the fluid should be changed while the system is at operating temperature. In this way, most of

the impurities are in suspension and will be drained off.

Much hydraulic fluid has been discarded in the past due to the possibility that contamination

existed- it costs more to test the fluid than to replace it. This situation has changed as the

need to conserve hydraulic fluids has developed. Figure 2.1 shows a hydraulic fluid test kit

that provides a quick, easy method to test system contamination. Even small hydraulic

systems may be checked. The test kit may be used on the spot to determine whether fluid

quality permits continued use. Three key quality indicators can be evaluated: viscosity, water

content, and foreign particle contamination level.



Figure 2.1. Hydraulic fluid test kit. (Courtesy of Gulf Oil Corp. , Houston, Texas.)

2.2. Fluids: Liquids and Gases

Liquids

The term fluid refers to both gases and liquids. A liquid is a fluid that, for a given mass, will

have a definite volume independent of the shape of its container. This means that even

though a liquid will assume the shape of the container, it will fill only the part of the

container whose volume equals the volume of the quantity of the liquid. For example, if

water is poured into a vessel and the volume of water is not sufficient to fill the vessel, a free

surface will b e formed, as shown in the figure 2.2(a). A free surface is also formed in the

case of a body of water, such a lake, exposed to the atmosphere [figure 2.2(b)]. Liquids are

considered to be incompressible so that their volume does not change with pressure changes.

This is not exactly true, but the change in volume due to pressure change is small that it is

ignored for most engineering applications.

Figure 2.2 free surface of a liquid



Gases

Gases, on the other hand, are fluids that are readily compressible. In addition, their volume

will vary to fill the vessel containing them. This is illustrated in figure 2.3, where a gas is

allowed to enter an empty vessel. As shown, the gas molecules always fill the entire vessel.

Therefore unlike liquids, which have a definite volume for a given mass, the volume of the

given mass of gas will increase to fill the vessel that contains it. Gases are generally

influenced by the pressure to which they are subjected. An increase in pressure causes the

volume of the gas to decrease and vice versa. Figure 2.4 summarizes the key physical

differences between liquids and gases for a given mass. Air is the only gas commonly used

in fluid power systems because it is inexpensive and readily available. Air also has the

following desirable features as a power fluid:

1. It is fire resistant

2. It is not messy

3. It can be exhausted back into the atmosphere

The disadvantages of using air versus using hydraulic oil are:

1. Due to its compressibility, air cannot be used in an application where accurate rate of

positioning or rigid holding is required.

2. Because air is compressible, it tends to be sluggish.

3. Air can be corrosive, since it contains oxygen and water.

4. A lubricant must be added to air to lubricate valves and actuators

5. Air pressures of greater that 250 psi are typically not used due to explosion dangers

involved if components such as air tanks should be rupture. This is because air (due to its

compressibility) can store a large amount of energy as it is compressed in a manner

similar to that of a mechanical spring.



Figure 2.3. A gas always fills its entire vessel.

Parameter Liquid Gas

Volume Has its own volume Volume is determined by container

Shape Takes shape of container but only to its

volume

Expands to completely fill and take the

shape of the container

Compressibility Incompressible for most engineering

applications

Readily compressible

Figure 2.4. Physical differences between liquids and gases

2.3. Specific weight, density and specific gravity

Weight versus mass

All objects, whether solids or fluids are pulled towered the center of the earth by a force of

attraction. This force is called the weight of the object and is proportional to the object’s mass as

defined by

mgWF == 2.1

Where, in the English systems of units (also called U.S. customary units used extensively in the

United States) we have

F = force in units of lb,

W = weight in units of lb,

m= mass of object in unit slugs,

g= proportionality constant called the acceleration of gravity, which equals 32.2ft/s2.

From equation (2.1), W equals 32.2 lb if m is 1 slug. Therefore, 1 slug is the amount of mass that

weights 32.2 lb on the surface of the earth.



Specific weight

Figure 2.5 shows a cubic container full of water as an example of discussing the fluid property

called specific weight. Since the container has the shape of rectangular solid, its volume can be

calculated using equation (2.2).

( ) ( )heightxbaseofareavolume = 2.2

Figure 2.5. Cubic container full of water.

Substituting values yields

( ) ( ) 31111 ftftxfftvolume =∗=

It has been found by measurement that 1ft3 of water weighs 62.4 lb. specific weight is defined as

weigh per unit volume. State mathematically, we have

volume

weightweightspecific =

Or V

W=γ 2.3

Where =γ Greek symbol gamma=specific weight (lb/ft3)

=W weight (lb)

=V volume (ft3)

Knowing that 1 ft3 of water weighs 62.4 lb, we can now calculate the specific weight of

water using equation (2.3):

33

/4.621

4.62ftlb

ft

lb

V

Wwater ===γ

If we want to calculate the specific weight of water in units of lb/ft3, we can perform the units

manipulation:



The conversion factor of 1/1728 is valid since 1ft3 = 1728 in3. This provides a consistent set of

units of lb/in3 on both sides of the equals sign since the units of ft3 cancel out in the numerator

and denominator. The resulting units’ conversion equation is

( ) ( )1728

//

33 ftlb

inlbγγ =

Substituting the known value for the specific weight of water in units of lb/ft3, we have

( ) 33 /0361.01728

4.62/ inlbinlbwater ==γ

Most oils have a specific weight of about 56 lb/ft3, or 0.0325 lb/in3. However, depending on the

type of oil, the specific weight can vary from low of 55 lb/ft3 to a high of 58 lb/ft3.

Specific gravity

The specific (SG) of a given fluid is defined as the specific weight of the fluid divided by the

specific weight of water. Therefore, the specific gravity of water is unity by definition. The

specific gravity of oil can be found using:

( )water

oiloilSG

γγ

= 2.4

( ) 899.0/4.62

/563

3

==ftlb

ftlbSG oil

Note that specific gravity is a dimensionless parameter (has no units).

Density

In addition to specific weight, we can also talk about the fluid property called density, which is

defined as the mass per unit volume:

V

m=ρ 2.5

Where ρ = Greek symbol rho = mass density (kg/m3),



m= mass (kg),

V= volume (m3).

Since weight is proportional to mass (from the equation W= mg), it follows that specific gravity

can also be defined as the density of the given fluid divided by the density of water. This is

shown as follows:

mgW =

Or

VgV ργ =

Solving for the density we have

g

γρ = 2.6

Where γ has a unit of lb/ft3,

g has a unit of ft/s2,

ρ has a unit of slugs/ft3.

Hence density equals specific weight divided by the acceleration of gravity. This allows us to

obtain the desired result.

waterwaterwater g

gSG

ρρ

ρρ

γγ === 2.7

The density of oil having a specific weight of 56 lb/ft3 can be found from equation (2.6).

34

22

3

2

3

74.1.

74.174.12.32

56

ft

slugs

ft

slb

ft

s

ft

lb

sftft

lb

oil ==∗==ρ

Note that from equation (2.1) mgW = . Thus, we have the following equality of units between

weight and mass: 2/ sftslugslb ∗=

2.4. Force, pressure and head

Force and pressure

Pressure is defined as force per unit area. Hence, pressure is the amount of force acting over a

unit area, as indicated by

A

Fp = 2.8



Where =p pressure,

=F Force,

=A Area.

p will have units of lb/ft2 F and Area have units of lb and ft2, respectively. Similarly, by

changing the units of A from ft2 to in2, the units for p will become lb/in2. Let’s go back to our 1-

ft3 container of figure 2.5. The pressure acting on the bottom of the container can be calculated

using equation (2.8), knowing that the total force acting at the bottom equals the 62.4-lb weight

of the water:

psfftlbft

lbp 4.62/4.62

1

4.622

2===

Note that units of lb/ft2are commonly written as psf. Also, since 1 ft2 = 144 in2, the pressure at

the bottom of the container can be found in units of lb/ft2 as follows using equation (2.8):

psiinlbin

lbp 433.0/433.0

144

4.622

2===

Units of lb/in2 are commonly written as psi.

Head

We can now conclude that, due to its weight, a 1-ft column of water develops at its base

pressure of 0.433 psi. The 1-ft height of water is commonly called a pressure head.

Let’s now refer to figure 2.6, which shows a 10-ft high column of water that has a cross-sectional

area of 1 ft2. Since there are 10 ft3 of water and each cubic foot weighs 62.4 lb, the total weight

of water is 624 lb. the pressure at the base is

psiinlbin

lbp 33.4/33.4

144

6242

2===

Thus each foot of the 10-ft head develops a pressure increase of 0.433 psi from its top to bottom.

What happens to the pressure if the fluid is not water? Figure 2.7 shows at 1-ft3 volume of oil.

Assuming a weight of density of 57 lb/ft3, the pressure at the base is liquid column changes it

weight (and thus the force at its bottom) by a proportional amount. Hence F/A (which equals

pressure) remains content.



psiin

lb

A

Fp 40.0

144

572

===

Hp ∗= γ

Figure 2.6. Pressure developed by a 10-ft column of water. (Courtesy of Sperry Vickers, Sperry

Rand Corp., Troy, Michigan)

Figure 2.7. Pressure developed by 1-and 2-ft columns of oil. (Courtesy of Sperry Vickers, Sperry

Rand Corp., Michigan.)

Substituting the correct units for γ, and into equation (2.9) produces the proper unit for pressure:



( ) ( ) ( ) ( )inHinlbpsipinlbp ∗== 32 // γ

Or

( ) ( ) ( ) ( )ftHftlbpsfpftlbp ∗== 32 // γ

Figure 2.8. The atmosphere as a pressure head. (Courtesy of Sperry Vickers, Sperry Rand Corp.,

Troy, Michigan.)

Atmospheric pressure

What about the pressure developed on the surface of Earth due to force of attraction between the

atmosphere and the Earth? For all practical purposes we live at the bottom of a huge sea of air,

which extends hundreds of miles above us. Equation (2.9) cannot be used to find this pressure

because of the compressibility of air. As a result the density of the air is not constant throughout

the atmosphere.

Let’s refer to figure 2.8 which shows a column of air with a cross-sectional area of 1 in2 and as

high the atmosphere extends above the surface of the earth. This entire column of air weighs

about 14.7 lb and thus produces a pressure of about 14.7 lb/in2 on the surface of the Earth at sea

level. This pressure is called atmospheric pressure and the value of 14.7 lb/ln2 is called standard

atmosphere pressure because atmospheric pressure varies a small amount depending on the

weather conditions which affect the density of the air. Unless otherwise specified, the actual

atmospheric pressure will be assumed to equal the standard atmosphere pressure.



Gauge and absolute pressure

Why then, does a deflated automobile tire read zero pressure of 14.7 psi when using a pressure

gauge? The answer lies in the fact that a pressure gauge reads gauge pressure and not absolute

pressure. Gauge pressures are measured relative to the atmosphere, whereas absolute pressures

are measured relative to a perfect vacuum such as that existing in outer space. To distinguish

between them, gauge pressures are labeled psig, or simply psi, whereas absolute pressures are

labeled psi (abs), or simply psia.

Figure 2.9. Operation of a mercury barometer. (Courtesy of Sperry Vickers, Sperry Rand Corp.,

Troy, Michigan)

This means that atmospheric pressure equals 14.7 psia or 0 psig. Atmospheric pressure is

measured with special devices called barometers. Figure 2.9 shows how a mercury barometer

works. The atmospheric pressure to be measured can support a column of mercury equal to 30.0

in. because this head produces a pressure of 14.7 psi. This can be checked by using equation

(2.9) and noting that the specific weight of the mercury is 0.490 lb/in3:

Hp γ=

( )inHinlbinlb ∗= 32 /490.0/7.14

mercuryofinH 00.30=



Figure 2.10 has a chart showing the difference between gauge and absolute pressures. Let’s

examine two pressure levels: p1 and p2. Relative to a perfect vacuum they are

psiap 7.41 = (a pressure greater less than atmospheric pressure)

psiap 7.241 = (a pressure greater than atmospheric pressure)

Relative to the atmosphere, they are

psigp 101 = suction (or vacuum) psig10−=

psigp 101 =

The use of terms suction or vacuum and the use of minus sign mean that pressure p1 is 10 psi

below atmospheric pressure. Also note that the terms psi and psig are used interchangeably.

Hence, p1 also equals -10 psi and p2 equals 10 psi. as can be seen from figure 2.10, the

following equation can be used in converting gauge pressures to absolute pressures, and vice

versa.

atmgaugeabs ppp += 2.10

Vacuum or suction pressures exist in certain locations of fluid power systems (for example, in

the inlet or suction lines of pumps). Therefore, it is important to understand the meaning of

pressures below atmospheric pressure. One way to generate a suction pressure is to remove some

of the fluid from a closed vessel initially containing fluid at atmospheric pressure.

Figure 2.10. Difference between absolute and gauge pressures



2.5. The SI metric system

Introduction

The SI metric system was standardized in June 1960 when the International Organization for

Standardization approved a metric system called Le Systéme International d’Unités. This system,

which is abbreviated SI, has supplanted the old CGS (centimeter-gram-second) metric system,

and U.S. adoption of the SI metric system is considered to imminent. In the metric system, the

units of measurement are as follows:

� Length is the meter (m).

� Mass is the kilogram (kg)

� Force is the newton (N)

� Time is the second (s)

� Temperature is the degree Celsius (0C)

Length, mass and force comparisons with English system

The relative sizes of length, mass and force units between the metric and English systems are

given as follows:

• One meter equals 39.4 in = 3.28 ft

• One kilogram equals 0.685 slugs

• One newton equals 0.225lb

A newton is the force required to give a mass of 1 kg an acceleration of 1 m/s2. Stated

mathematically, we have

2/111 smkgN ∗=

Since the acceleration of gravity at sea level equals 9.80 m/s2, a mass of 1 kg weighs 9.80 N.

Also, since 1 N = 0.225 lb, mass of 1 kg also weighs 2.20 lb.

Pressure comparisons

The SI metric system uses units of Pascals (Pa) to represent pressure. A pressure of 1 Pa is equal

to a force of 1 N applied over the area of 1 m2 and thus is a very small unit of pressure.

2/11 mNPa =



The conversion between pascals and psi is as follows:

psiPa 000145.01 =

Atmospheric pressure in units of pascals is found as follows by converting 14.7 psi in to its

equivalent pressure in pascals:

( ) ( ) ( )absPapsi

PaabspsiPapatm 000,101

000145.0

17.14 =∗=

Thus, atmospheric pressure equals 101,000 Pa (abs) as well as 14.7 psia. Since the pascal is very

small unit, the bar is commonly used:

psiPamNbar 5.1410/101 525 ===

Thus, atmospheric pressure equals 14.7/14.5 bars (abs), or 1.01 bars (abs).

Temperature comparison

The temperature (T) in the metric system is measured in units of degree Celsius (0C), whereas

temperature in the English system is measured in units of degrees Fahrenheit (0F). figure 2.11

shows a graphical representation of these two temperature scales using mercury thermometer

reading a room temperature of 680F (200C).

Figure 2.11: Comparison of the Fahrenheit and Celsius temperature scales



Relative to figure 2.11 the following should be noted: The Fahrenheit temperature scale is

determined by dividing the temperature range between the freezing point of water (set at 320F)

and the boiling point of water (set at 2120F) at atmospheric pressure into 180 equal increments.

The Celsius temperature scale is determined by dividing the temperature range between the

freezing point of water (set at 00) and the boiling point of water (set at 100 0C) at atmospheric

pressure into 100 equal increments. The mathematical relationship between the Fahrenheit and

Celsius is

( ) ( ) 328.1 00 += CTFT 2.11

Thus to find the equivalent Celsius temperature corresponding to room temperature (680F), we

have:

( ) ( )C

FTCT 0

00 20

8.1

3268

8.1

32 =−=−=

Absolute temperature (Rankine units in English system and Kelvin units in metric system).

SI system prefixes

Figure 2.12 provides the prefixes used in the metric system to represent powers of 10. Thus for

example:

PaPakPa 1000101 3 ==

This means that atmospheric pressure equals 101kPa (abs) as well as 1010 millibars (abs).

Figure 2.12. Prefixes used in metric system to represent powers of 10.



2.6. Bulk modulus

The highly favorable power-to-weight ratio and the stiffness of hydraulic systems make them the

frequent choice for most high-power applications. The stiffness of a hydraulic system is directly

related to the incompressibility of the oil. Bulk modulus is a measure of this incompressibility.

The higher the bulk modulus, the less compressible of stiffer the fluid.

Mathematically the bulk modulus is defined by equation (2.12), where the minus sign indicates

that as the pressure increases on a given amount of oil, the oil’s volume decreases; and vice

versa:

VV

p

/∆∆−=β 2.12

Where =β bulk modulus (psi, kPa)

=∆p Change in pressure (psi, kPa)

=∆V Change in volume (in3, m3)

=V Original volume (in3, m3)

The bulk modulus of an oil changes somewhat with changes in pressure and temperature.

However, for pressure and temperature variations that occur in most fluid power systems , this

factor can be neglected. A typical value for oil is 250,000 psi (1.72 x 106 kPa).

2.7. Viscosity

Introduction

Viscosity is probably the single most important property of a hydraulic fluid. It is a measure of a

fluid’s resistance to fluid. When the viscosity is low, the fluid flows easily and is thin in

appearance. A fluid that flows with difficulty has a high viscosity and is thick in appearance. In

reality, the ideal viscosity for a given hydraulic system is a compromise. Too high a viscosity

results in:

1. High resistance to flow, which causes sluggish operation

2. Increased power consumption due to frictional losses.

3. Increased pressure drop through valves and lines.



4. High temperatures caused by friction.

On the other hand, if the viscosity is too low, the result is

1. Increased oil leakage past seals.

2. Excessive wear due to break down of the oil film between mating moving parts. These

moving parts may be internal components of pump (such as pistons reciprocating in

cylinder bores of a piston pump) or a sliding spool inside the body of a valves as shown

in figure 2.13.

Figure 2.13. Fluid film lubricates and seals moving parts. (Courtesy of Sperry Vickers, Sperry

Rand Corp., Troy, Michigan.)

Absolute viscosity

A concept of viscosity can be understood by examining two parallel plates separated by an oil

film of thickness y, as illustrated in figure 2.14. The lower plate is stationary, whereas the upper

plate moves with a velocity v as it being pushed by a force F as shown. Because of viscosity, the

oil adheres to both surfaces. Thus, the velocity of the layer of fluid in contact with the lower

plate is zero, and the velocity of the layer in contact with the top place is v. The consequence in a

linearly varying velocity profile whose slop is v/y. the absolute viscosity of the oil can be

represented mathematically as follows:



profilevelocityofslope

oilinstressshear

yv

AF

yv===

/

/

/

τµ 2.13

Where =τ Greek symbol =tau the shear stress in the fluid in units of force per unit area (lb/ft2,

N/m2); the shear stress (which is produced by the force F) causes the sliding of adjacent layers of

oil;

=v Velocity of the moving parts (ft/s, m/s);

=y Oil film thickness (ft, m);

=µ Greek symbol =mu the absolute viscosity of the oil,

=F Force applied to the moving upper plate (lb, N)

=A Area of the moving plate surface in contact with the oil (ft2, m2).

Figure 2.14. Fluid velocity profile between parallel plates due to viscosity.

Checking units of µ in the English system using equation (2.13), we have

( )2

2

/.//

/ftslb

ftsft

ftlb ==µ

Similarly µ has units of N.s/m2 in the SI metric system. If the moving plate has a unit surface

area in contact with the oil, and the upper plate velocity and oil film thickness’ are given units

values, equation (2.13) becomes

FF

yv

AF ===1/1

1/

/

/µ



We can, therefore, define the absolute viscosity of a fluid as the force required to move a flat

plate (of unit area at unit distance from a fixed plate) with a unit velocity when the space

between the plates is filled with the fluid. Thus, using a fluid of higher viscosity requires a larger

force, and vice versa. This shows that viscosity is a measure of a fluid’s resistance to flow.

Viscosity is often expressed in the CGS (centimeter-gram-second) metric system. In the CGS

metric system, the units per equation (2.13) are

( )2

2

/.//

/cmsdyn

cmscm

cmdyn ==µ

Where a dyne is the force that will accelerate a 1-g mass at a rate of 1 cm/s2. The conversions

between dynes and Newton’s as follows:

dynN 5101 =

A viscosity of 1 dyn.s/cm2 is called a poise. The poise is a large unit of viscosity. A more

convenient unit is the centipoises, abbreviated cP.

Kinematic viscosity

Calculations in hydraulic systems often involve the use of kinematic viscosity rather than

absolute viscosity. Kinematic viscosity equals absolute viscosity divided by density:

ρµ=v 2.14

Where υ = Greek symbol nu = kinematic viscosity.

Units for the kinematic viscosity are given as follows: English: ft2/s, SI metric: m2/s and CGS

metric: cm2/s. A viscosity of 1 cm2/s is called a stoke. Because a stoke is a large unit, viscosity is

typically reported in centistokes (cS).

Saybolt Viscometer

The viscosity of a fluid is usually measured by a Saybolt viscometer, which is shown

schematically in figure 2.15. Basically, this device consists of an inner chamber containing the

sample of oil to tested. A separate outer compartment, which completely surrounds the inner

chamber, contains a quantity of oil whose temperature is controlled by an electrical thermostat



and heater. A standard orifice is located at the bottom of the center oil chamber. When the oil

sample is at the desired temperature, the time it takes to fill a 60-cm3 container through the

metering orifice is then recorded. The time (t), measured in seconds, is the viscosity of the oil in

the official units called Saybolt Universal Seconds (SUS). Since a thick liquid flows slowly, its

SUS viscosity value will be higher that for a thin liquid.

A relationship exists between the viscosity in SUS and cS. This relationship is provided by the

following empirical equations:

SUStt

tcS 100,195

226.0)( ≤−=υ 2.15

SUStt

tcS 100,135

220.0)( >−=υ 2.16

Where the symbol υ represents the viscosity in cS and t is the viscosity is measured in SUS, or

simply seconds.

Kinematic viscosity is defined as absolute viscosity divided by density since, in the CGS metric

system, density equals specific gravity (because 3/12

cmgOH =ρ and thus has a value of one),

the following equation can be used to find the kinematic viscosity in cS if the absolute viscosity

in cP is known, and vice versa.

( )SG

cPcS

µυ =)( 2.17

As previously noted, in the SI metric system, N.s/m2 and m2/s are the units used for the absolute

and kinematic viscosity, respectively. Although these are good units for calculations purposes, it

is common practice in the fluid power industry to use viscosity expressed in units of SUS or cS.



Figure 2.15. Saybolt viscometer. (Courtesy of USX Corp., Pittsburgh, Pennsylvania.)

Capillary tube viscometer

A quick method for determining the kinematic viscosity of fluids in cS and absolute viscosity in

cP is shown in figure 2.16. This test measures the time it takes for a given amount of fluid flow

through a capillary tube under the force of gravity. The time in seconds is then multiplied by the

calibration constant for viscometer to obtain the kinematic viscosity of the sample fluid in

centistokes. The absolute viscosity in centipoises is then calculated using equation (2.17).



Figure 2.16. Capillary tube viscometer. (Courtesy of USX Corp., Pittsburgh, Pennysylvania.)

2.8 Viscosity index

Oil becomes thicker as the temperature decreases and thins when heated; hence, the viscosity of

given oil must be expressed at specified temperature. For most hydraulic applications, the

viscosity normally equals about 150 SUS at 100 0F. it is a general rule of thumb that the viscosity



should never fall below 50 SUS or rise above 4000 SUS regardless of the temperature. Where

extreme temperature changes are encountered, the fluid should have a high viscosity index.

Viscosity index (VI) is a relative measure of oil’s viscosity change with respect to temperature

change. Oil having low VI is one that exhibits a large change in viscosity with temperature

change. High-VI oil is one that has a relatively stable viscosity, which does not change

appreciably with temperature change. The original VI scale ranged from 0 to 100, representing

the poorest to best VI characteristics known at that time. Today, with improved refining

techniques and chemical additives, oils exist with VI values well above 100. High-VI oil is good

all-weather-type oil for use with outdoor machines operating in extreme temperature swings.

This is where viscosity index is especially significant. For a hydraulic system where the oil

temperature does not change appreciably, the viscosity index of the fluid is not critical. The VI of

any hydraulic oil can be found by using

100∗−−=

HL

ULVI 2.18

Where =L Viscosity in SUS of 0-VI at 100 0F,

=U Viscosity in SUS of unknown-VI oil at 100 0F,

=H Viscosity in SUS 100-VI oil at 100 0F

The VI of unknown-VI oil is determined from tests. A reference oil of 0 VI and a reference oil of

100 VI are selected, each of which has a uniquely the same viscosity at 210 0F as the unknown-

VI oil. The viscosities of three oils are then measured at 100 0F to give values for L, U and H.

Figure 2.17 provides a pictorial representation of the H, L and U terms in equation (2.18). Note

that the change in viscosity of a hydraulic oil as a function of temperature is represented by a

straight lined when using American Society for Testing and materials (ASTM) standard viscosity

temperature charts. Figure 2.18 shows the results of a viscosity index test where oil A is the 0-VI

have VI values of 50 and 140, respectively. For example, the VI for oil B is calculated as

following using the viscosity values at 1000F for oils A (L value), B (U value) and C (H value)

from figure 2.18:

( ) 5010010002400

17002400100 =∗

−−=∗

−−=

HL

ULBoilVI



Figure 2.17. Typical curves for a viscosity index test

Another characteristics relating to viscosity is called the pour point, which is the lowest

temperature wt which a fluid will flow. It is a very important parameter to specify for hydraulic

systems that will be exposed to extremely low temperatures. As a rule of thumb, the pour point

should be at least 200F below the lowest temperature to be experienced by the hydraulic system.



Figure 2.18. Viscosity index. (Courtesy of USX Corp., Pittsburgh, Pennsylvania.)

chapter 2 physical properties of hydraulic fluid

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