1.0 – Introduction to Fluid Mechanics and Fluid Properties, Slides of Fluid Mechanics

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ENGR. BON RYAN ANIBAN

Properties of Fluid

• Mass Density
• Specific Volume
• Unit Weight of Specific weight
• Specific Gravity
• Viscosity
• Surface Tension
• Capillarity
• Compressibility
• Pressure Disturbances
• Property Changes in Ideal Gas

Principles of Hydrostatic

• Unit Pressure
• Pascal’s Law
• Absolute and Gauge Pressures
• Variations and Pressure
• Pressure Head
• Manometers

Total Hydrostatic Force on Surfaces

• Total Hydrostatic Force on Plane and Curved Surfaces
• Dams
• Buoyancy
• Statical stability of Floating Bodies
• Thin-walled Pressure Vessels

Relative Equilibrium of Liquid

• Rectilinear Translation
• Rotation

ENGR. BON RYAN ANIBAN

MASS DENSITY ( ρ ) – mass per unit volume

mass

Volume

kg

m^3

g

cm^3

slugs

ft^3

IDEAL GAS DENSITY ( ρ )

P = absolute pressure of gas in Pa

recall:

absolute pressure = gauge pressure +

atmospheric pressure

R = gas constant

For Air :

R = 287 Joule/kg-K (SI)

% R = 1716 lb-ft/slug-R (English)

T = absolute temperature

K = °C + 273

R = °F + 460

Densities of Common Fluid Fluid 𝜌 in kg/m^3 Air (STP) 1. Alcohol 790 Ammonia 602 Gasoline 720 Glycerin 1260 Mercury 13600 Water (at 4°C) 1000

SPECIFIC WEIGHT/UNIT WEIGHT/WEIGHT

DENSITY( γ ) – weight per unit volume

weight

Volume

m × g

N

m^3

lb

ft^3

Gravitational acceleration g = 9.81 m/s^2 = 32.2 ft/s^2

Length

1 ft = 0.3048 m 1 mi = 5280 ft. = 1609.344 m 1 nautical mile = 6076 ft = 1852 m 1 yd = 3 ft = 0.9144 m

Mass

1 slug = 14.594 kg 1 tonne = 1000 kg

Velocity

1 ft/s= 0.3048 m/s 1 mi/h = 1.46666 ft/s = 0.44704 m/s

Mass Flow

1 slug/s = 14.594 m/s 1 lbm/s = 0.4536 kg/s

Volume 1 ft^3 = 0.028317 m^3 1 L = 0.001 m^3 = 0.0353515 ft^3

Area 1 ft^2 = 0.028317 m^3 1 mi^2 = 2.78784 x 10^7 ft^2 = 2.59 x 10^6 ft^3

Acceleration 1 ft/s^2 = 0.3048 m/s^2

A glycerin has a mass of 1200 kg and a volume of 0.952 cu.m. (a) Find its weight , (b) unit weight, (c) mass density, (d) and specific gravty.

Solution

a 𝑊 = 𝑚g = 1200 kg = 11772 N = 11.772 kN

b 𝛾 =

11. 772 kN
12. 952 m^3 = 12.366 kN/m^3

c 𝜌 =

1200 kg

1. 952 m^3

= 1260.504 kg/m^3

d 𝑠 =

1260. 504 kg/m^3 1000 kg/m^3

(9.81 m/s^2 )

Specific gravity of a certain oil is 0. (a) Find its weight in kN/m^3 , lb/ft^3 (b) Find its density in kg/m^3 , slug/ft^3

Solution

a 𝑠 =

9.81 kN/m^3 𝛾𝑜𝑖𝑙 = 8.044 kN/m^3

62.4 lb/ft^3

b 𝛾𝑜𝑖𝑙 = 𝜌𝑜𝑖𝑙 g

8.044 × 10^3 N/m^3 = 𝜌𝑜𝑖𝑙(9.81 m/s^2 )

𝛾𝑙𝑖𝑞𝑢𝑖𝑑 = 51.168 lb/ft^3

𝜌𝑙𝑖𝑞𝑢𝑖𝑑 = 819.980 kg/m^3

51.168 lb/ft^3 = 𝜌𝑜𝑖𝑙(32.2 ft/s^2 ) 𝜌𝑙𝑖𝑞𝑢𝑖𝑑 = 1.589 slug/ft^3

𝛾𝑔𝑎𝑠 = 12.498 N/m^3

If a specific volume of a gas is 0.7848 m^3 /kg. What is its unit weight?

Solution

𝑉𝑠 =

1. 7848 m^3 /kg =

𝜌= 1.274 kg/m^3 𝛾𝑔𝑎𝑠= 𝜌𝑔𝑎𝑠 g = 1. 274 kg/m^3 ( 9. 81 m/s^2 )

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

Solution

𝜌 =

184 + 101. 92 103 Pa

( 2079 )(J/kg − K)(4° + 273 )(K)

𝜌 = 0.496 kg/m^3

A

𝐕𝐈𝐒𝐂𝐎𝐒𝐈𝐓𝐘 ( μ ) mu

The property of fluid which determines the amount of its resistance to shearing forces.

y fluid

viscous

Area=A F

U

F ∝ U

y

= 𝜏 (shearing stress)

Where the k is called the

dynamic absolute viscosity

denoted as μ

Where: 𝜏 = shear stress in lb/ft^2 or Pa μ = absolute viscosity in lb sec/ft^2 (poises) or Pa-sec y = distance between the plates in ft or m U = velocity in ft/s or m/s

N

m^2

Two large plane surfaces are 25 mm apart and the space between them is filled with a liquid of viscosity of μ = 0.958 Pa-s. Assuming the velocity gradient to be a straight line, what force is required to pull a very thin plane of 0.37 m2 area at a constant speed of 0.3 m/s if the plate is 8.4 from one of the surfaces Solution

𝐹 = 𝐹1 + 𝐹 𝜇 =

𝑈/𝑦 𝜇^ =^

1. 958 Pa ∙ s =

1. 37 m^2
2. 3 m/s
3. 4 × 10 −^3 m

𝐹 1 = 0. 958 Pa ∙ s

1. 3 m/s
2. 4 × 10 −^3 m (^0.^37 m

N

m^2 𝐹 1 = 12. 659 N

25 mm

8. 4 mm

μ = 0.958 Pa-s

16. 6 mm

F

A = 0.37 m^2

F 1

F 2

1. 958 Pa ∙ s =

1. 37 m^2
2. 3 m/s
3. 6 × 10 −^3 m 𝐹 2 = 6. 406 N 𝐹 = 𝐹 1 + 𝐹 2 𝐹 = 12.659 N + 6.406 N 𝐹 = 19.065 N

A cylinder of 125 mm radius rotates concentrically inside a fixed cylinder of 130 mm radius. Both cylinders are 300 mm long. Determine the viscosity of the liquid which fills the space between the cylinders if a torque of 0.88 N-m is required to maintain an angular velocity of 2π radians/sec. Assume the velocity gradient to be straight line.

Solution

Fixed cylinder

Rotating cylinder

0.3 m

0.125 m

0.13 m

Fixed

F

F

U = 0.

0.005 m

T

= 0. 125 ( 2 π) 𝑈 = 0. 785 m/s

Torque = Force(radius)

1. 88 N ∙ m = 𝐹( 0. 125 m) 𝐹 = 7. 04 N

𝜏 =

7. 04 N

2 π 𝑟rotating cylinder (𝐿)

=

7. 04 N

2 π 0 .125m ( 0. 3 m) 𝜏 = 29. 879 N/m^2 𝜏 = 29. 879 Pa

29. 879 Pa
30. 784 m s / 0 .005m

𝜇 = 0. 191 Pa ∙ s

What is the value of the surface tension of a small drop of water 0.3 mm in diameter which is in contact with air if the pressure within the droplet is 561 Pa?

Solution

σ =

561 Pa( 0. 15 × 10 −^3 m)

𝜎 = 0. 042 N/m

p𝑟

N/m

𝐂𝐀𝐏𝐏𝐈𝐋𝐀𝐑𝐈𝐓𝐘/ CAPILLARY ACTION

Is the name given to the behavior of the liquid in a thin-bore tube. The rise or fall of a fluid in a capillary tube is caused by surface tension and depends on the relative magnitudes of the cohesion of the liquid and the adhesion of the liqud to the walls of the containing vessel.

ADHESION

The attraction force between different molecules.

COHESION

The attraction force between molecules of the same substance

H 20 Hg

h

rise

h

fall

Adhesion > Cohesion Cohesion > Adhesion

Densities of Common Fluid Fluid 𝜌 in kg/m^3

Mercury 13600

Water (at 4 °C)

1000

Capillary Tube