CC303 HYDRAULIC 1 (CRTH) 2011

1.0 FLUID CHARACTERISTIC

1.1 FLUID MECHANICS AND HYDRAULICS

Fluid mechanics and hydraulics represent that branch of applied mechanics that deals with the behavior of fluids at rest and in motion. In the development of the principles of fluid mechanics, some fluid properties play principle roles, others only minor roles or no roles at all. In fluid statics, specific weight (or unit weight) is the important property, whereas in fluid flow, density and viscosity are predominant properties. Where appreciable compressibility occurs, principles of thermodynamics must be considered. Vapor pressure becomes important when negative pressures (gage) are involved, and surface tension affects static and flow conditions in small passages.

Fluid mechanics is the science of the mechanics of liquids and gases, and is based on the same fundamental principles that are employed in the mechanics of solids. The mechanics of fluids is a more complicated subject than the mechanics of solids, however, because with solids one deal with separate and tangible elements, while with fluids there are no separate elements to be distinguished.

1.1.1 DEFINITION OF A FLUID

Fluid is substances that are capable of flowing and conform to the shape of containing vessels. When in equilibrium, fluids cannot sustain tangential or shear forces. All fluids have some degree of compressibility and offer little resistance to change of form. Fluids can be classified as liquids or gases.

1.1.2 DISTINCTION BETWEEN A GAS AND A LIQUID

The differences between liquids and gases are:

LIQUIDS GAS1. Compression is too small. 2. Takes the form of container contains. 3. Has a free surface.

1. Easily compressed. 2. Always expanding and filled the container. 3. Do not have a surface free.

A fluid may be either a gas or a liquid. The molecules of a gas are much farther apart than those of a liquid. Hence a gas is very compressible, and when all external pressure is removed, it tends to expand definitely. A gas is therefore in equilibrium only when it is completely enclosed. A liquid is relatively incompressible, and if all pressure, except that of its own vapor pressure, is removed, the cohesion between molecules holds them together, so that the liquid does not expand indefinitely.

A vapor is gas whose temperature and pressure are such that it is very near the liquid phase. Thus steam is considered a vapor because its state is normally not far from that of water. A gas may be defined as a highly superheated vapor; that is, its state is normally very far from that of liquid air. The volume of a gas or vapor is greatly affected by changes in pressure or temperature or both. It is usually necessary, therefore, to take account of changes in volume and temperature in dealing with gases or vapors. Whenever significant temperature or phase changes are involved in dealing with vapors and gases, the subject is largely dependent on heat phenomena (thermodynamics). Thus fluid mechanics and thermodynamics are interrelated.

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CC303 HYDRAULIC 1 (CRTH) 2011

1.1.3 MEASURE OF FLUID

i. DensityThe density (or mass density) of fluid is its mass per unit volume

unit : kg/m3

ii. Specific volumeIs the volume occupied by a unit mass of fluid.

or

unit : m3/kg

iii. WeightMass multiplied by the gravitational acceleration

w=mg unit : N

iv. Specific weightThe specific weight(or weight density) of a liquid defined as the weight per unit volume.

unit : N/m3

v. Specific GravityRatio of the density of the fluid to the density of water at some specified temperatureSpecific weight of the liquid divided by the specific weight of water

or

unit: none

*note: pwater=1000 kg/m3 , γwater=9810 N/m3

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CC303 HYDRAULIC 1 (CRTH) 2011

1.1.4 Relationship between specific weight, specific gravity, density and solve problems

Example 1If 6m3 of oil weighs 47kN, calculate its specific weight γ, density, ρ and specific gravity.

Specific weight

γ = = 7.833kN/m 3

Density

=798 kg/m 3

Specific gravity

= 0.8

Example 2The specific weight of water at ordinary pressure and temperature is 62.4N/m3. The specific gravity of mercury is 13.56. Calculate the density of water and the specific weight and density of mercury.

Weight wwater=mg

= 6.361kg

Density

= 6.361kg/m 3

Specific gravity Specific gravity

Specific weight =133023.6N/m 3 Density =13560kg/m 3

Example 3

Derive formula of acceleration and force into SI based unit and state the dimension and the units of each quantity.

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CC303 HYDRAULIC 1 (CRTH) 2011

Using information we know that :

Dimension of

unit of Accelaration = m/s 2

=

Dimension of

unit of Force = kgm/s 2

EXERCISE

Derive the above formula into SI based unit and state the dimension and the units of each quantity.i. Pressureii. Energyiii. Poweriv. Flowratev. Dynamic viscosity

TUTORIAL

1. If 2.5m3 of certain oil has a mass of 20kg, find its mass density.(Ans ρ=8kg/m3)

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CC303 HYDRAULIC 1 (CRTH) 2011

2. In an experiment, the weight of 2.5m3 of a certain liquid was found to be 18.75 kN. Find the specific weight of the liquid. Also find its density.

(Ans ws=7.5kN/m3, ρ=764.526kg/m3)

3. Find the specific gravity of oil whose specific weight is 7.85kN/m3. (Ans S.G=0.8)

4. A vessel of 4m3 volume contains oil, which weighs 30.2kN. Determine the specific gravity of the oil.

(Ans S.G=0.77)

5. A container weighs 5000N when filled with 60m3 of a certain liquid. The empty weight of the container is 500N. What is the density of the liquid?

(Ans ρ=7.645 kg/m3)

6. Determine the mass density of oil, if 3000kg of the oil occupies a volume of 4m3. (Ans ρ=750 kg/m3)

7. A container of volume 3.0m3 has 25.5kN of oil. Find the specific weight and mass density of the oil.

(Ans ws=8500N/m3,ρ=866.463kg/m3)

8. A drum of 1m3 volume contains 8.5kN oil when full. Calculate its specific weight and specific gravity.

(Ans ws=8500N/m3,S.G=0.866)

C3009/JKA/ASSIGNMENT 1

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CC303 HYDRAULIC 1 (CRTH) 2011

1. If 1m3 of oil has a mass of 860 kg, calculate its specific weight, density and specific gravity.2. The specific weight of a certain liquid is 90.30N/m3. Determine its density and specific gravity.3. The density of a jet fuel is 760 kg/m3. Determine the specific weight and specific gravity of this fuel.4. If 1.0m3 of oil has a mass of 890kg, calculate its specific weight, density and specific gravity.5. A container has a 6m3 volume capacity and weights 1500N when empty and 56000N when filled with

a liquid. What is the density of the liquid?6. A cylindrical tank with a diameter of 80cm is partially filled with SAE 10W oil (ρ=870kg/m 3). If the

mass of the oil in the tank is 340kg, determine the height of the oil in the tank.7. An overhead tank has length, width and height of 2m, 2.5m, and 1.5m respectively. If the water is

filled up to three-quarters of its height, determine the mass and weight of the water inside the tank.8. If a similar tank, as in problem 7 is filled to the same level with oil with a mass of 4000kg, determine

the density and specific gravity of the oil.9. If 6.2m3 of oil weights 52980N, calculate its density and specific gravity.10. A certain liquid, occupying a volume of 1.6m3, weighs 12.8kN. What are the specific weight, density

and specific gravity of the liquid?

1.2 The Physical Nature and Significance of Viscosity in Fluid FlowsViscosity is a measure of the resistance of a fluid which is being deformed by either shear stress or

extensional stress. In general terms it is the resistance of a liquid to flow, or its "thickness". Viscosity

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JABATAN KEJURUTERAAN AWAMPOLITEKNIK TUANKU SULTANAH BAHIYAH

C3009 – HYDRAULICS 1

CC303 HYDRAULIC 1 (CRTH) 2011

describes a fluid's internal resistance to flow and may be thought of as a measure of fluid friction. Thus, water is "thin", having a lower viscosity, while vegetable oil is "thick" having a higher viscosity. All real fluids (except super fluids) have some resistance to stress, but a fluid which has no resistance to shear stress is known as an ideal fluid or in viscid fluid. For example high viscosity magma will create a tall volcano, because it cannot spread fast enough, low viscosity lava will create a shield volcano, which is large and wide.[1] The study of viscosity is known as theology.

where

τ is the shear stress exerted by the fluid [Pa] μ is the fluid viscosity - a constant of proportionality [Pa·s]

is the velocity gradient perpendicular to the direction of shear [s−1]

The viscosity of a fluid is an important property in the analysis of liquid behavior and fluid motion near solid boundaries.

The viscosity is the fluid resistance to shear or flow and is a measure of the adhesive/cohesive or frictional fluid property. The resistance is caused by intermolecular friction exerted when layers of fluids attempt to slide by one another.

Viscosity is a measure of a fluid's resistance to flow

The knowledge of viscosity is needed for proper design of required temperatures for storage, pumping or injection of fluids.

There are two related measures of fluid viscosity - known as dynamic (or absolute) and kinematic viscosity.

1.2.1 Dynamic viscosity (µ)

Dynamic viscosity and absolute viscosity are synonymous. The IUPAC symbol for dynamic viscosity is the Greek letter eta (η), but it is also commonly referred to using the Greek symbol mu (μ). The SI physical unit of dynamic viscosity is the pascal-second (Pa·s), which is identical to kg·m−1·s−1. If a fluid

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CC303 HYDRAULIC 1 (CRTH) 2011

with a viscosity of one Pa·s is placed between two plates, and one plate is pushed sideways with a shear stress of one Pascal, it moves a distance equal to the thickness of the layer between the plates in one second.

Dynamic viscosity is the tangential force per unit area required to move one horizontal plane with respect to the other at unit velocity when maintained a unit distance apart by the fluid. The shearing stress between the layers of non turbulent fluid moving in straight parallel lines can be defined for a Newtonian fluid as:

Figure 1 : Newtonian Fluid

The dynamic or absolute viscosity can be expressed like

τ = μ dc/dy     (1)

where

τ = shearing stress

μ = dynamic viscosity

Equation (1) is known as the Newtons Law of Friction.

In the SI system the dynamic viscosity units are N s/m2, Pa s or kg/m s where

1 Pa s = 1 N s/m2 = 1 kg/m s

1.2.1 Kinematic viscosity (ν)

Kinematic viscosity is the ratio of absolute or dynamic viscosity to density - a quantity in which no force is involved. Kinematic viscosity can be obtained by dividing the absolute viscosity of a fluid with its mass density as

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CC303 HYDRAULIC 1 (CRTH) 2011

ν = μ / ρ    (2)

where

ν = kinematic viscosity

μ = absolute or dynamic viscosity

ρ = density

In the SI-system the theoretical unit is m2/s

1.2.3 Classification of Fluid

The fluids can be classified into the following four types depending upon the presence of viscosity.

i. Ideal Fluids

All real fluids (except super fluids) have some resistance to stress, but a fluid which has no resistance to shear stress is known as an ideal fluid or in viscid fluid. An ideal fluid is one that is incompressible and has no viscosity. Ideal fluid does not actually exist, but sometimes it is useful to consider what would happen to an ideal fluid in a particular fluid flow problem in order to simplify the problem.

ii. Real Fluids

A fluid having no viscosity is known as a real fluid. In actual practice, all the fluids met with in engineering-science, are real fluids.

iii. Newtonian Fluids

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Figure 2 : Rheological diagram

CC303 HYDRAULIC 1 (CRTH) 2011

Fluids for which the shearing stress is linearly related to the rate of shearing strain are designated as Newtonian Fluids. Newtonian fluid had stress at each point is linearly proportional to its strain rate at that point. The concept was first deduced by Isaac Newton and is directly analogous to Hooke's law for a solid. All gases are Newtonian, as are most common liquids such as water, hydrocarbons, and oils.

iv. Non-Newtonian Fluids

A non-Newtonian fluid is a fluid whose flow properties cannot be described by a single constant viscosity. In a non-Newtonian fluid, the relation between the shear stress and the strain rate is nonlinear, and can even be time-dependent. Therefore a constant coefficient of viscosity can not be defined. A ratio between shear stress and rate of strain (or shear-dependent viscosity) can be defined, this concept being more useful for fluids without time-dependent behavior.

1.2.4 Solve Problems Related to Fluid PropertiesExample 1

A liquid has a dynamic viscosity of 0.005 kg/ms and a density of 850 kg/m3. Calculate the kinematic viscosity.

= 5.882x10 -6 m 2 /s

Example 2

Find the kinematic viscosity of an oil , whose specific gravity is 0.95 and dynamic viscosity 0.0011Ns/m2.

ρ = 0.95*1000 = 950kg/m3

kinematic of viscosity = 0.0116m 2 /

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CC303 HYDRAULIC 1 (CRTH) 2011

TUTORIAL

1. Calculate the dynamic viscosity if kinematic viscosity 5.5x10-4 m2/s and specific gravity of this fluid is 1.8. (Ans μ= 0.99kg/ms)

2. A liquid has a density 576 kg/m3. Find its dynamic viscosity if its kinematic viscosity is 6x10-6 m2/s. (Ans μ=3.456x10-3 kg/ms)

3. Calculate specific gravity of a liquid if its dynamic viscosity=0.005 Ns/m2 and kinematic viscosity = 3.2mm2/s. (Ans S.G= 1.563)

4. a. A vessel of 7.3m3 cointains an oil, which mass 6500 kg. Determine the specific gravity, density and specific weight of the oil. (Ans ρ= 890.4kg/m3, S.G=0.89,ws=8734.93 N/m3)

b. An oil as question 4a has dynamic viscosity 1x10-3 N/m2. Find its kinematic viscosity.

(Ans v=1.12x10-6 m2/s)

5. A 1600kg of oil are pour into a cylinder until 1000ml. If the diameter of the cylinder is 50mm and the kinematic viscosity of the oil is 6x10-6 m2/s. Calculate :

i. specific weight of the oil

ii. the height of the oil inside the cylinder

iii. oil dynamic viscosity (Ans ws=1.57x107N/m3,t=0.51m,μ 9.6kg/ms)

6. Discuss the shear characteristics of the fluids for which the curves have been drawn in figure below.

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CC303 HYDRAULIC 1 (CRTH) 2011

Solution :

a. The Newtonian fluids behave according to the law τ= μ (dV/dy), or the stress is proportional to the velocity gradient or rate of shearing strain. Thus for these fluids the plot of shear stress against velocity gradient is a straight line passing through the origin. The slope of the line determines the viscosity

b. For the ideal fluid, the resistance to shearing deformation is zero, and hence the plot coincides with the x axis. Although no ideal fluids exist, in certain analyses the assumption of an ideal fluid is useful and justified.

c. For the ideal or elastic solid, no deformation will occur under any loading condition, and the plot coincides with the y axis. Real solids have some deformation, and within the proportional limit (Hooke’s Law) the plot is a straight line that is almost vertical.

d. Non-Newtonian fluids deform in such a way that shear stress is not proportional to rate of shearing deformation, except perhaps at very low shear stresses. The deformation of these fluids might be classified as plastic

e. The ideal plastic material could sustain a certain amount of shear stress without deformation, and thereafter it would deform in proportion to the shear stress.

ANSWER TOTURIAL VISCOSITY

5. Given :

m= 1600 kg V=1000 ml

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CC303 HYDRAULIC 1 (CRTH) 2011

= 1 x 10-3 m3

d = 50 mm = 0.05 m ν oil = 6 x 10-6 m2/s

Solution :

i. Weight, w = mass x gravity acceleration = 1600 x 9.61 = 15696 N

Specific weight, ws =

=

= 1.57 x 10 7 N/m 3

ii. The height of the oil inside the cylinder

= 1 x 10-3

=1 x 10-3

t =

= 0.51 m

iii. density, ρ =

=

= 1.6 x 10 6 kg/m 3

Dynamic viscosity,µ = kinematic viscosity x ρ fluid

= 6x10-6 x1.6x106

= 9.6 kg/ms

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