1.9 Bulk Modulus and Compressibility

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Bulk modulus:

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It is defined as the ratio of change in pressure to the volumetric strain of the fluid.
It should be noted that the volumetric strain is the ratio of change in volume to its original volume.
 Bulk modulus is represented by B and by some authors as K.

This mathematically bulk modulus can be given as

where dp is the change in pressure.
 dV  is the change in volume.
V is the original volume.

It should be noted that here a negative sign is added to the bulk modulus mathematical relation(formula) because the stain is considered to be compressive strain. If the strain is tensile then the negative should not be considered. 

In SI units its dimensional formula can be given as M¹L⁻¹T⁻².

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Compressibility:

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It is defined as the inverse of the bulk modulus. Thus it can be defined as the ratio of the volumetric strain to the change in pressure. Thus mathematically it can be given as

Compressibility = 1/K or 1/B

In SI units its dimensional formula can be given as M^-1L¹T².

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References:
http://en.wikipedia.org/wiki/Bulk_modulus
http://www.britannica.com/EBchecked/topic/84278/bulk-modulus
http://dictionary.reference.com/browse/bulk+modulus
Img Credits: http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/bulk3.gif

1.8 Types of fluids based on viscosity

There are five types of fluids in fluids mechanics

1. Ideal Fluid
2. Real Fluid
3. Ideal Plastic Fluid
4. Newtonian Fluid
5. Non-Newtonian Fluid

1.Ideal Fluid

A Fluid which has an ideal flow and which is incompressible is called as ideal fluid. Ideal fluids are assumed to have zero viscosity.
It should be noted that in practice there is no ideal fluid. 

2. Real Fluid

A fluid which doesn't have ideal flow and which is compressible is called as real fluid. Real Fluids have viscosity and they face friction while they are flowing. 
All the fluids in real practice are Real Fluids

3. Ideal Plastic Fluid

A fluid in which shear stress is more than the yield value and the shear stress is directly proportional to the shear strain(du/dy) is called as Ideal Plastic Fluid.
 One of the example of the Ideal Plastic Fluid is Bingham Plastic. 

4. Newtonian Fluid

The Fluid which obeys the newtons law of viscosity is called as Newtonian Fluid. It should be remembered that the Newtons law o viscosity can be given as

And thus the fluid which obeys this law is called as Newtonian fluid.

5. Non Newtonian Fluid

A Fluid which does not obey the above equation(Newtons law of viscosity) is called as Non-Newtonian Fluid. Thus Non Newtonian fluids deviate from the above equation.

References:
http://en.wikipedia.org/wiki/Bingham_plastic
http://www.definition-of.com/Ideal+Plastic+fluid 
http://encyclopedia2.thefreedictionary.com/Ideal+Fluid  
http://mechteacher.com/fluid/

1.7 Newtons law of viscosity.

It states that the shear stress of  a fluid element in a fluid layer which is flowing with some velocity is directly proportional to the rate of shear strain. The equation we have already seen in viscosity concept is the one proposed by newton. It is given as

And if the constant of proportionality is introduced then the equation becomes

Where, μ is the constant of proportionality and is called as the coefficient of dynamic viscosity or viscosity.
τ is the shear stress and
du/dy is the rate of shear strain 

Types of fluids based on newtons law

1. Newtonian Fluids
2. Non-Newtonian Fluids

1.Newtonian Fluids

Fluids which behave according to the equation of Newtons law of viscosity are called as Newtonian Fluids. 

2. Non Newtonian Fluids

Fluids which do not satisfy the newtons law of viscosity or the newtons equation on viscosity are called as non-newtonian fluids.

Note:

1. All real fluids are Non Newtonian Fluids.
2. All ideal fluids accept the Newtons law of viscosity and are called as Newtonian fluids.
3. Water, thin motor oil are considered to be partially Newtonain Fluids.

Example Problem:
This is an example problem and remember that the values in the real life will be in the order of exponential. All the values given here are for the sake of beginners.
Consider a fluid flowing in a river. If the shear stress is  equal to 10N/m² distance between the fluid layers is 2m. Fluid is flowing with a velocity of 5m/s. 

1. Find whether the fluid is Newtoinain Fluid or non Newtonian fluid if the coefficient of dynamic viscosity is 40N.s/m².
2. Find the viscosity of the fluid if it were a Newtonian fluid.

Sol: 

1.We know from the equation that   

Therefore we can observe that

 10 ≠ 40x5/2

Therefore the given fluid is not a Newtonian Fluid.

2.If it was a Newtonian Fluid then from the above equation it can be found that 

μ = 4N.s/m².

References:

http://en.wikipedia.org/wiki/Non-Newtonian_fluid

http://en.wikipedia.org/wiki/Newtonian_fluid

http://www.enchantedlearning.com/math/symbols/

 

1.6 Kinematic Viscosity

Kinematic viscosity is defined as the ratio of the dynamic viscosity to the density of the given fluid. It is denoted by the symbol v(nu).

v = Viscosity/Density

v = μ/ρ

Video:
It should be noted that the video here also contains about the what is viscosity. I suggest you to watch about the viscosity, to refresh about the viscosity as kinematic viscosity is dependent on viscosity.
 

It's units in SI and MKS system are given by m²/s. In CGS system its units are given by cm²/s.
It should be noted that cm²/s is also called as stoke and 
stoke = 10^-4 m²/s.

Note:

1. 1Stoke = 100 Centistoke

1.5 Viscosity

Viscosity is the measure of the fluid resistance to flow.

Video:
This video also contains about the kinematic viscosity which we are going to cover in the next post. If you want to watch it you can watch it.


It can be said that the resistance is due to the intermolecular forces between the fluid layers. Thus in the figure the top fluid layer only has force on it by the fluid layer under it. But for the fluid layer under it has forces above it and layer under it.


To understand this properly let us consider an example.

Let us consider a fluid moving in x direction as shown in the figure. 

Let us assume that the lines represents the fluid layers and they are dy distance apart. And also assume that the top layer of the fluid is moving at a velocity of u. Now the viscosity is that the layer 2 pulls back the layer 1 for its movement. Thus the velocity of the second layer will be u-du. Therefore the shear stress on the first layer will be towards the left hand side and on the second layer will be towards the right hand side for this example. And we know that the shear stress is defined as the force acting on the given object per unit area. Therefore shear stress can be given as 

Here the proportionality constant becomes the coefficient of viscosity. and is denoted by μ.

Thus the equation can be represented as 

where,

τ is the shear stress 

μ is the coefficient of dynamic viscosity or viscosity.

and du/dy is the rate of shear strain or velocity gradient.

Thus we can write the equation of viscosity can be given as 

Units of viscosity can be given as

and in CGS system can be given as

Where 1 Newton is equal to 100000 dynes and it should be noted that dyne.s/cm² is also called as poise

in MKS system the units can be given as 

and the relation between the MKS system and the SI system can be given as 

Note:

1. All Real fluids have viscosity.
2. Only in ideal fluids viscosity is assumed to be zero.
3. Viscosity is also called as dynamic viscosity. 
4. The More the viscosity the slower the motion of the liquid.

1.4 Specific Gravity

Specific gravity is defined as the ratio of the density of the given substance to that of the standard substance. It is denoted by the symbol SG. It should be noted that the standard substances are water for solids and liquids whereas dry air is taken as standard for gaseous fluids. Therefore Specific Gravity can be written as

For Solids or liquids:

SG_{solids or fluids} = (Density_{solid or fluid} / Density _water)

For Gaseous Substances:
 
SG_Fluid = (Density_Fluid / Density _air)
   
_{It can be seen that as Specific Gravity is the ratio of two densities there will be no units. 
  Note:}

1. _{Remember that the density of water is 1000 kg/m3, and that of air is 1.28 kg/m3.}
2. _{Thus the specific gravity of mercury becomes 13.6, as density of mercury is 13600 kg/m3.}
3. _{Specific Gravity is an important topic in the concept of buoyancy.}
4. _{Specific gravity is also called as relative density.}

 <sub>Video:

 
Example problem:
Find the specific gravity of gold whose density is measured to be 19,300kg/m3.(Take density of water to be 1000kg/m3).
Solution:
We know that specific gravity is the ratio of the density of the substance to that of the water for solid substances.
Therefore 
SGGold = 19300/1000
=>19.3
Therefore the specific gravity of gold is found to be 19.3(No units).</sub>

Ignore the program and continue to the next topic if you don't know programming in python. This is just for the sake of programmers.

Python Program

def specific_gravity_solid(density):
 """Specific Gravity for solid or liquid
 density = The density of the substance"""
 sg = density/1000
 return sg

def specific_gravity_gas(density):
 """Specific Gravity for gaseous substances
 density = The density of the gas"""
 sg = density/1.28
 return sg

#Find the density of carbon di-oxide

print specific_gravity_gas(1.98) #Prints 1.546875

Output:

1.546875

References:

http://www.britannica.com/EBchecked/topic/558700/specific-gravity
http://dictionary.reference.com/browse/specific+gravity
http://en.wikipedia.org/wiki/Specific_gravity

3. Specific Volume

Specific Volume is defined as amount of volume occupied by unit mass of the fluid. It can also be defined as the ratio of volume and mass. It is denoted by v.

v = v/m

where v is the specific volume.
m is the mass of the body.
v is the volume of the body.

In SI units mass has units of Kg.
In SI units volume has the units of m³ 
Therefore units of specific volume becomes v = v(m³)/m(kg).

It's units are m³/kg in SI units.

It can be found that it is the reciprocal of density(ρ).
Therefore, (Specific volume)v = 1/(density)ρ 

Note:

1.   Specific volume of water = 0.001m³/kg
2. Specific volume of air = 0.78125 m³/kg

Video:

Example Problem:
 Find the specific volume of mercury whose density is found to be 13600kg/m^{3
Sol: 
We know that specific volume is equal to the inverse of density. Therefore,
specific volume is equal to 1/13600}m³/kg
=>v_mercury = 0.000073529m³/kg
 ^{Python Program:
Input:}

def specific_volume(mass,volume):
 return (volume/mass)

def specificVolume(density):
 return (1/density)

 

2. Specific Weight

Specific weight is defined as the weight per unit volume. It can also be defined as the product of Density and gravitational force. It is denoted by the symbol γ(Pronounced as Gamma). Mathematically it can be shown as:

γ=ρ.g 

Its units are kg/s².m²

It's units can also be expressed as N/m³

Note:

1. Specific weight of the substance changes from place to place as there is a change in gravity from one place to other.
2. It is also called as weight density. 
3.  It can also be expressed as γ = w/v, Where w-weight of the body(m.g),v-volume of the body.
4. If we consider the ideal case of gravity as 9.81m². Then γ_air = 12.5568, γ_water = 9810, γ_mercury = 133416. All in N/m³

 Example Problem

Find the specific mass of the substance whose density is 0.085kg/m².s² .(Consider the gravity at the place is 9.81 m²).

Solution:
Specific weight (γ) = ρ.g 
=> 0.085x9.81
=>0.83385N/m³.
Therefore the specific mass of the body is found 0.83385N/m³.


Python Program:

 #!/usr/bin/python
 # -*- coding: utf-8 -*-
def specific_mass(density,gravity=9.81):
 specific_mass = density*gravity
 print u"The specific mass(γ) of the substance is found to be {}".format(specific_mass)

def weight_density(mass,volume,gravity=9.81):
 weight_density = mass*gravity/volume
 print u"The Weight Density/Specific mass(γ) of the substance is found to be {}".format(weight_density)

def take_user_input():
 print "Choose with what you want to calculate specific_mass/weight_density."
 print "Type 1 for: denstiy and gravity input."
 print "Type 2 for: mass, volume and gravity input."
 print "However entering the gravity value is optional."
 choice = raw_input("> ")
 print "Do you want to enter the gravity value? 'y'-Yes, 'n'- No"
 yes_no = raw_input("> ")
 if yes_no == 'Y' or 'y':
  gravity = raw_input("Enter the value of gravity > ")
 elif yes_no == 'N' or 'n':
  gravity = 9.81
 else:
  print "Wrong input please choose again"
  take_user_input()

 if choice == "1":
  density = raw_input("Enter the value of density > ")
  specific_mass(float(density),float(gravity))
 elif choice == "2":
  mass = raw_input("Enter the mass value > ")
  volume = raw_input("Enter the volume > ")
  weight_density(float(mass),float(volume),float(gravity))
 else:
  print "Your choice was not in the given options. Please choose again."
  take_user_input()

take_user_input()

 

Output:

Choose with what you want to calculate specific_mass/weight_density.
Type 1 for: denstiy and gravity input.
Type 2 for: mass, volume and gravity input.
However entering the gravity value is optional.
> 1
Do you want to enter the gravity value? 'y'-Yes, 'n'- No
> y
Enter the value of gravity > 9.63
Enter the value of density > 98653
The specific mass(γ) of the substance is found to be 950028.39

 

1.Density

Density: It is defined as the ratio of mass per unit volume of the substance. It is denoted by the symbol ρ can be pronounced as rho. Mathematically it can be written as 

ρ = (m/v)

Where,

ρ - Density of the substance.

m - Mass of the substance.

V - Volume of the substance. 

Units:
We will derive the units in S.I. system. 
 In SI system mass is in Kg.
In SI system Volume(v) is in m³.

Therefore,

Density(ρ) = mass(kg)/Volume(m³)

Units of ρ are kg/m³.

Note:

1. Density is also called as mass density or volume mass density.
2. Density of pure water is 1000kg/m³.
3. Density of air is 1.28 kg/m³.
4. Density of mercury is 13600kg/m³.  

Video:

Example Problem:

Find the density of the substance whose mass is measured to be 5kg and a volume of 3.91 m³.

Solution: We know that density is the ratio of mass and volume. Therefore,density(ρ) =mass(5kg)/volume(3.91m³).
 = 5/3.91(kg/m³)
 = 1.28(kg/m³) = Density of air.

The density of the fluid is found to be 1.28(kg/m³)

Python Program:

#!/usr/bin/python
# -*- coding: utf-8 -*-
def density(mass,volume):
 density = float(mass)/float(volume)
 print u"Density(ρ) of the substance is found to be {}".format(density)

mass = raw_input("Enter the value of mass > ")
volume = raw_input("Enter the value of volume > ")

density(mass,volume)

 Output:

Enter the value of mass > 46546
Enter the value of volume > 6546
Density(ρ) of the substance is found to be 7.11060189429