fluid mechanics lecture

FLUID MECHANICSLecture Notes # 1

FLUID MECHANICS- Is a physical science dealing with the action of fluids at rest or in motion, and with applications

and devices using fluids.

TWO MAJOR DIVISIONS OF FLUID MECHANICS:

1. Fluid Statics – deals with fluids at rest2. Fluid Dynamics – deals with fluids in motion

TYPES OF FLUID:

1. Ideal Fluids Assumed to have no viscosity (and hence, no resistance to shear) Incompressible Have uniform velocity when flowing No friction between moving layers of fluid No eddy currents or turbulence

2. Real Fluids Exhibit infinite viscosities Non-uniform velocity distribution when flowing Compressible Experience friction and turbulence in flow Are further divided into Newtonian Fluids and Non-Newtonian Fluids

PROPERTIES OF FLUIDS:

1. MASS DENSITY, ρ (Rho)- The density of a fluid is its mass per unit of volume

Units:English : slugs/ ft3

Metric : gram/cm3

SI : kg/ m3

For an ideal gas, its density can be found from the specific gas constant and ideal gas law:

Engr. Jerome F. Famadico of 62nd Semester S.Y. 2013-2014

ρ = mass of fluid, M

volume, V

ρ = PRT

Equation 1-1

Equation 1-2


Where: P = absolute pressure of gas in Pa

R = gas constant Joule/ Kg-°KFor air:

R = 287 J/ kg - °KR = 1,761 lb-ft/ slug - °R

T = absolute temperature in °Kelvin°K = °C + 273°R = °F + 460

2. SPECIFIC VOLUME, Vs

- Specific volume, Vs, is the volume occupied by a unit mass of fluid.

3. UNIT WEIGHT or SPECIFIC WEIGHT, γ

- Specific weight or unit weight, γ, is the weight of a unit volume of a fluid.

Units:English : lb/ ft3

Metric : dyne/cm3

SI : N/m3 or kN/m3


Vs =1ρ

γ = weight of fluid, W

volume, V

γ = ρ g

Vs =1

mass density

Equation 1-3

Equation 1-4

Equation 1-5


4. SPECIFIC GRAVITY

- Specific gravity, s, is a dimensionless ratio of a fluid’s density to some standard reference density.

- For liquids and solids, the reference density is water at 4°C (39.2°F).

In gases, the standard reference to calculate the specific gravity is the density of air

For water at 4°C:γ = 62.4 lb/ft3 = 9.81 kN/m3ρ = 1.94 slugs/ft3 = 1000 kg/m3s = 1.0

PRACTICE PROBLEMS:

1. If 5.6 m3 of oil weighs 46,800 N, calculate the following:a. Unit Weightb. Densityc. Specific Gravity

2. The volume of a tetrachloride having a mass of 1200 kg is 0.952 m3. Compute the following:a. Mass Densityb. Specific Weightc. Specific Gravity

3. What is the specific weight of air at 480 kPa absolute and 21°C?

4. Find the mass density of helium at a temperature of 4°C and a pressure of 184 kPa gage, if atmospheric pressure is 101.92 kPa. (R = 2079 J/kg - °K)

5. A cylindrical tank 80 cm in diameter and 90 cm high is filled with a liquid. The tank and the liquid weighed 420 kg. The weight of the empty tank is 40 kg. What is the unit weight of the liquid in kN/m3.?

PROPERTIES OF FLUIDS (Continuation..)


s = ρliquid

ρwater

s = ρgas

ρair

Equation 1-6

Equation 1-7


5. VISCOSITY, µ, (MU)

- The property of fluid which determines the amount of its resistance to shearing forces.- A perfect fluid would have no viscosity.

Where:τ = shear stress in lb/ft2 or Paµ = absolute viscosity in lb-sec/ft2 (poises) or Pa-sec.y = distance between the plates in ft or mU = velocity in ft/s or m/s

6. KINEMATIC VISCOSITY, ν, (NU)

- Is the ratio of the dynamic viscosity of the fluid, µ, to its mass density, ρ.

Where:µ = absolute viscosity in Pa-secρ = density in kg/m3

7. SURFACE TENSION, σ (SIGMA)

- The membrane of “skin” that seems to form on the free surface of a fluid is due to the intermolecular cohesive forces, and is known as surface tension.

Pressure inside a Droplet of Liquid:

Where:σ = surface tension in N/md = diameter of the droplet in mP = gage pressure in Pa


µ = τ

dV / dy

s = µρ

Equation 1-8

Equation 1-9

P = 4σd

Equation 1-10


8. CAPILLIARITY (CAPILLIARY ACTION)

- Is a manifestation of surface tension by which the portion of the surface of a liquid coming in contact with a solid is elevated or depressed, depending on the adhesive or cohesive properties of the liquid.

Where:h = capillary rise or depression in mγ = unit weight in N/m3

d = diameter of the tube in mσ = surface tension in Pa

9. BULK MODULUS OF ELASTICITY, EB

- Expresses the compressibility of the fluid- Is the ratio of the change of unit pressure to the corresponding volume change per unit

volume

Where:E = bulk modulusΔP = change in pressureΔV = change in volume

10. COMPRESSIBILITY, β

- Is the fractional change in volume of a fluid per unit change in pressure in a constant temperature process.

Where:β = coefficient of compressibilityEB = bulk modulus of elasticity

PRACTICE PROBLEMS:


h = 4σcosΘ

ydEquation 1-11

EB = ΔP

ΔV/VEquation 1-12

β = 1

EBEquation 1-13


1. A liquid is compressed in a cylinder having a volume of 1 liter at one MN/m2 and a volume of 995 cm3 at 2 MN/m2.

a. Compute the change in volumeb. Compute the change in pressurec. Compute the bulk modulus of elasticity

2. (a) Determine the surface tension in a tube with 0.2 m radius and wetting angle 0° and capillarity rise of 5 mm. (b) Determine the surface tension in lb/ft. (c) If wetting angle Θ = 80°, determine the surface tension.


fluid mechanics lecture

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