Lecture -2 ( properties of fluids (part 1) - density, specific weight, specific gravity )

Properties are certain measurable characteristics that can be quantified. with the help of properties we can easily identify the fluids.
Properties of fluids are as follows-

Density or mass density (ρ) 

Density is defined as the ratio of mass per unit volume. The international System of units ( SI units ) is Kilogram per metre cube (kg/m^3).
Density basically represent the number of molecules in a given volume of fluid. As we increase the number of molecules in a given volume its mass will increase and fluid become heavy.
               Density (ρ) = mass/volume 
                                 ρ = m/V
Density of water = 1000 kg/m^3
Density of Mercury = 13600 kg/m^3
Density of air = 1.2 kg/m^3
Density of petrol = 750 kg/m^3
Density of diesel = 850 kg/m^3
Density of solids is always greater than density of liquids and density of liquids is always greater than density of gases.
As pressure of fluid increases it increases the density because with increase in pressure more and more number of molecules will start to accumulate in a given volume of fluid which increases its density.
As temperature of fluid increase its density will decrease because with increase in temperature the randomness of molecules increases they start moving here & there due to which fluid become light therefore its density decreases. 

Specific weight or weight density(w) 

It is defined as the ratio of weight (W) per unit volume (V) and its SI unit is Newton per metre cube (N/m^3).
Basically specific weight represent the force exerted by fluid due to gravity(g) in a given volume. 
 w = W/V ;  W=mg
w = mg/V = ρg  
Specific weight of water = ρwater * g = 1000*9.81 = 9810 N/m^3
Specific weight of Mercury = 13600*9.81 N/m^3
Weight (W) = specific weight (w) * volume
                ; W = w * V
                ; W = ρgV
From the above we can conclude that 'w' is dependent upon 'g' whereas density is an absolute quantity which will notwill not varry with location.

Specific gravity (S) 

specific gravity is defined as the ratio of density of fluid to the density of standard fluid. If fluid is liquid then standard fluid is taken as water and if fluid is gas then  standard fluid is taken as air.
Specific gravity of liquid represent which liquids are heavier then Water and which are lighter then water. Specific gravity of gas represent which gas is heavier then air and which gas is lighter then air.
   S = ρfluid/ρstandard fluid  
Specific gravity is dimensionless quantity.
Example- 
Problem - Find specific gravity for petrol.
Solution - ρpetrol = 750 kg/m^3
                    ρstandard fluid = ρwater = 1000 kg/m^3
                   S = 750/1000 = .75 
If S > 1 than fluid is heavier than water
If S < 1 than fluid is lighter than water
If S = 1 than fluid is water

(Remember all specific gravity are relative density but all relative densities are not specific gravity)
ρsolid/ρliquid , ρliquid/ρgas , ρgas/ρsolid are relative density 

Comments

Post a Comment

Popular posts from this blog

Lecture - 11 (Transmission angle in four bar mechanism)

Image
The angle between coupler link and output link in four bar mechanism is known as transmission angle. It is represented as ' μ'.   In the given figure Let us consider angle between link AO and link OC (fixed link) is represented by ' θ'.   Let us consider length of link OC is 'a', length of link OA is 'b', length of AB is 'c' and length of BC is 'd'.  According to law of cosine we get -  Now  μ will be different for different values of  θ. μ will be maximum for  θ = 180 degrees  & minimum for  θ = 0 degrees.   Problem 1-  Calculate maximum and minimum value of transmission angle for the following mechanism. Length of link OC is 5cm, length of link AO is 1cm, length of link AB is 4cm and length of link BC is 3cm. Solution 1 -   To calculate minimum value of transmission angle observe the mechanism when  θ = 0 degrees   To calculate maximum value of transmission angle observe the mechanism when  θ = 180 degr
Read more

Lecture - 8 ( problems on degrees of freedom)

Image
PROBLEM 1  - For the mechanism shown in figure, calculate it's degrees of freedom (F) SOLUTION 1  -  number of links (L) = 8 Number of binary joints (J) = 10 Number of higher pair (H) = 0 Number of those motions which are not the part of mechanism (N) = 0 KUTZBACH equation: F=3(L-1) - 2(J) - H - N F = 3(8-1) - 2(10) - 0 - 0 = 1 F = 1  PROBLEM 2 -  Calculate degrees of freedom for given mechanism SOLUTION 2 -   L = 4 ; J = 3 ; H = 1 ; N = 1 F = 3(L-1) - 2J - H - N = 3(4-1) -2(3) -1 -1 = 1 F = 1    PROBLEM 3 -  Calculate degrees of freedom for given mechanism.  SOLUTION 3 -   L = 5 ; H = 1 ; J = 5 ; N = 0 F = 3(5-1) -2(5) - 1 - 0 = 1  F = 1 PROBLEM 4 -  Calculate degrees of freedom for given mechanism SOLUTION 4 -    Here one link is spring. Spring is the link of flexible length. Spring can be replaced by 2 links connected with turning pair. L = 5 ; J = 5 ; H = 1 ; N = 0 F = 3(5-1) -2(5) - 1 = 1  F = 1
Read more

Lecture - 15 (Mechanical advantage, mechanism efficiency, toggle mechanism)

Image
Mechanical advantage   It is defined as the ratio of the force produced by a machine to the force applied to it, used in assessing the performance of machine.  The combination of pulley system is shown in the below figure-  Here for lifting download F force is required but the effort given to the system is 8 times less, which is the good example of mechanical advantage. For the above system mechanical advantage is 8 Mechanical advantage = load / effort  Mechanical advantage for linear motion is -  Mechanical advantage for angular motion is -  Efficiency of system or mechanism Efficiency of the system is defined as the ratio of output power to the input power. PROBLEM - Find out mechanical advantage of the given mechanism having 100% efficiency. Solution -  In the sense of angular velocity It means that above mechanism is toogle mechanism.  Toggle mechanism   Those mechanism whose mechanical advantage is so high that it approaches
Read more