1.5 Basic Fluid Properties

Density (ρ = rho) => a ratio of mass and volume of material

ρ = m/V ; m = mass (kg)

V = volume (m3)

ρ = density (kg/m3)

“Density of water will be decreasing when temperature is increasing”

4 ⁰C, ρ = 1,000 kg/m3

Specific Volume (Vs) => volume of substance in one mass unit=> invert of density => mostly use with gas

Vs = V/m = 1/ρ ; Vs = specific volume (m3/kg)

V = volume (m3)

m = mass (kg)

ρ = density (kg/m3)

“The specific volume will be increasing when temperature is increasing”

Specific Weight (γ = gamma) => Force that is caused by gravity reacts with one unit of fluid.

=> Unit N/m3

γ = W/V = mg/V = ρg ; γ = specific weight (N/m3)W = weight (N)V = volume (m3)m = mass (kg)g = gravity (m/s2)ρ = density (kg/m3)

“The specific weight will be decreasing when temperature is increasing”

Specific Gravity (S) => ratio of substantial density and pure water density

=> pure water density = 1000 kg/m3

=> ratio of specific weight of substance and specific weightof water at standard temperature

=> specific weight of water at standard temperature = 9.81×103 N/m3

=> “specific weight of water at 4⁰C is 1”

S = ρ/ρw = γ / γ w; S = Specific gravity (No unit)

Ex. Substance has 2940 kg/m3. Find

A. Specific gravity (S)

B. Specific volume (Vs)

C. Specific weight (γ)

1.6 Viscosity

Viscosity => Property of the resistance of deformation

=> Caused by intermolecular forces- more intermolecular forces => more viscosity- less intermolecular forces => less viscosity

“increasing temperature => increasing molecular force => increasing viscosity”

F

Y

duu

y

dy

U

Viscosity Experiment

- 2 plates place in parallel with difference of “Y”

- Pull top plate with F, and it has velocity (U)- Area of fluid particles touch each plate

is “A”, and it sticks with plate.- The velocity of each fluid layer will reduce

linearly.

According to Newton’s experiment found that“F” depends on “A” and “U”, but it is inversely proportional to “Y”

𝐹 ∝𝐴𝑈

𝑌

Ratio of U/Y is equal to du/dy. Give “μ” is the “Coefficient of viscosity” or “dynamic viscosity”.

𝐹 = μ𝐴𝑑𝑢

𝑑𝑦

(1)

From shear Stress is

τ =𝐹

𝐴(2)

So

𝐹

𝐴= μ

𝑑𝑢

𝑑𝑦

or

τ =𝐹

𝐴= μ

𝑑𝑢

𝑑𝑦

F

Y

duu

y

dy

U

Viscosity Experiment

Viscosity Experiment

Where τ = Shear stress (Pa)F = Action force (N)A = Touch area m2

μ = Dynamic viscosity (Pa•s)du = velocity (m/s)dy = fluid thickness (m)

τ =𝐹

𝐴= μ

𝑑𝑢

𝑑𝑦

“Newton’s Equation of Viscosity”

τ

μ2

μ1

μ3

μ = Slope

du/dy

Oil

Water

Air

“Newtonian Fluid”

Fluid which is not like Newtonian fluid is called “Non-Newtonian fluid”

1. Dilatant Substance => Fluid has more viscosity when flow angular velocity increase. Ex. Quick sand

2. Pseudo Plastic => Fluid has less viscosity when flow angular velocityincrease. Ex. Semen, Milk

3. Plastic => Fluid must have yield stress, τp before flow angularvelocity changes. Ex. Plastic

τ

du/dy

Newtonian fluid

Pseudo plastic

Dilatant substance

Plastic

Solid (μ=∞)

Fluid without viscosity (μ=0)Kinematic Viscosity (ʋ = nu)= ratio of dynamic viscosity and density

ʋ =μ

ρ

EX. Pull the plate with 1 N. The touch area is 0.5 m2. the plate has 0.5 m/sof velocity and 0.5 mm of fluid thickness. Find the dynamic viscosity ?

F=1N

Fluid, dy=0.5mm

du=0.5 m/s

A=0.5m2

1.7 Equation of state for gases

Gas cannot measure its exact value like liquid. So, in the calculation, we use “Perfect gas”

𝑃 =𝑚

𝑉𝑅𝑇

𝑃𝑉 = 𝑚𝑅𝑇

𝑃 = ρ𝑅𝑇

Where P = absolute pressure (Pa)ρ = gas density (kg/m3)R = gas constant (J/kg•K)T = absolute temperature (K)

Ex. Air has specific weight 18.27 N/m3. It has absolute pressure 1,600 mbar.Find its temperature in ⁰C (Gas constant R=287 J/kg•K)