Coefficient of Viscosity and Terminal Velocity

Viscosity

The viscous force acting between two adjacent layers of a liquid is directly proportional  to the surface area of the layers in contact and the velocity gradient. The viscous force acts tangential to the liquid and in opposite direction to the direction of flow of liquid.  Hence negative sign is used. It is like a frictional force acting against the relative motion.

It can  be defined as the tangential force per unit area required to maintain unit velocity gradient. (or) It is the ratio between tangential stress and velocity gradient.

Effect of temperature on Viscosity

In the case of liquids, coefficient of viscosity decreases with increase  of temperature as the  cohesive forces decrease with increase of temperature.In the case of gases,coefficient of viscosity increases with increase of temperature because the change in momentum of molecules increases with increase of temperature.

In the case of liquids with increase in temperature, distance between molecules increases . This leads to  the decrease of force of attraction .

In the case of gases, with the increase in temperature, random motion of molecules increases and collisions also increases. This increases viscous nature.

Effect of pressure

In a liquids the value of increases with increase of pressure. For gases, its value of  increases with increase of pressure at low pressure.  But at high pressure, it is independent of pressure.




Raynold’s Number

We can decide the flow of the fluid as is a streamlined flow or not basing on a value called Raynold’s number.

The velocity at which the streamlined flow turns into a non-streamline flow is called as critical velocity. Critical velocity of a fluid is directly proportional to coefficient of viscosity, inversely proportional to diameter of the fluid flow and also inversely proportional to the density of the fluid. We can write the equation by keeping all of these things together. The proportionality can be eliminated with a constant as shown below.




Terminal velocity

A constant velocity is acquired by a body when the resultant force acting on it is zero and it is called as terminal velocity. When a spherical body is moving in a fluid its weight always acts in downward direction and it’s upthrust always acts in upward direction. The medium applies a force against its motion and it is called viscous force. The direction of the viscous forces always against the relative motion of the body in the fluid.

When the body is initially in the state of rest there is no viscous force acting on it. When the body starts coming down it’s velocity increases due to the gravitational force and automatically viscous force also starts increasing against  the gravitational force.

At a certain stage the downward force is balanced by the up ward force therefore the body will acquire a constant velocity and that velocity is called as terminal velocity. By drawing the condition for equilibrium we can derive the equation further terminal velocity as shown below.


Special Case 

When multiple drops are falling with a different terminal velocities and if the drops are combined together to form a big drop we can derive a equation further terminal velocity of the big drop as shown below.

While solving this problem we are going to depend on the simple concept that volume of the big drop is equal to the sum of the volume of all the small drops together. We can also express the terminal velocity in terms of the mass is shown in the below diagram.



Poisellie’s Equation

This equation helps in identifying the volume of the fluid flowing through a given hole per second.It depends on radius of tube with power four,pressure and inversely proportional to  its length and coefficient of viscosity.Keeping only pressure in the numerator, every thing can be shifted to denominator and hence it oppose the flow of fluid.Then it is being called as fluid resistance.



When the pipes are connected in series, the same volume of the fluid flows through them and pressure at different points is going to be different.

When the pipes are connected in parallel, volume of fluid will be shared across them but pressure across two pipes is going to be the same.





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Problems on Bernoulli's theorem and Its Applications

Problem and solution

A rectangular vessel when full of water and it takes 10 minutes to be emptied through a small hole. If the same vessel is only half filled , calculate the time taken to empty the vessel?

Basing on the concept of Bernoulli’s Theorem it is proved that the time taken to empty the tank is the difference between the Square route of heights of the fluid filled.

We can solve one more problem basing on the same concept. 

 Water in a tank flowing through a hole of diameter 2 cm under a constant pressure difference of 10 cm water column. What is the rate of the flow of the water through the hole ?




Problem and solution

A hole is made at the bottom of the tank filled with water. If the total pressure at the bottom of the tank is three  tmes of atmospheric pressure what is the velocity of the efflux?

The velocity of the water with which it comes out through the hole is similar to the velocity of a freely falling body. The pressure due to 10 meter of water  is mathematically equal to one atmospheric pressure. It is proved in the following diagram.




Problem and solution

A in compressible liquid flows in a horizontal tube as shown. Find the velocity of the fluid?

To explain this concept we shall use the equation of continuity. As per this concept the mass of the fluid that enters through the system is equal to the mass of the fluid that exits through the system in one second.

We have one more problem to solve in this attached paper.

An aeroplane of certain mass and certain area of cross-section can experience a certain pressure in the up thrust. 

 As there is no information is given in terms of velocities we have to deal it only in terms of pressure as pressure is defined as the force per unit area.




Problem and solution

A square hole having a certain length is made at the depth y and a circular hole is made at a depth of 4y from the surface of the water tank. If equal amount of the water comes out of the vessel through both the holes, find the radius of the circular hole in terms of the length of the Squire hole?

This problem also can be solved basing on the law of equation of continuity. The concept is simple. The mass of the fluid that enters through one hole per second shall be equal to the mass of the fluid that enters through the other hole also.




Problem and solution

Water is moving with the speed of 5 m/s through a pipe with a cross-sectional area of 4 cm Squire. The water gradually decreased to 10 meters high it as the pipe increases the area to 8 cm Squire. If the pressure at the upper surface is given what is the pressure at the Lower surface?

We can use both equation of continuity and the Bernoulli’s Theorem to solve the problem as shown below.




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Dynamic Lift and Air Foil

Dynamic Lift

Dynamic lift is the uplift experienced by a body when it is placed in a fluid in the state of motion. This can be explained basing on the concept that the total energies of a fluid like pressure energy, potential energy and the kinetic energy always remains constant. To understand this in detail we can deal this in 3 different parts.

In case one let us consider a ball moving with a uniform velocity having only translatory motion in a fluid. This air applies a equal influence both on the upper surface as well as the lower surface of the fluid therefore the ball is not going to experience any effective force.

In case two let us consider the same ball having only spin motion. In this case the body takes the fluid around it in the circular motion as shown. It is similar to the lot of the earth atmosphere revolving around the Sun.Because of the influence of the body some portion of this fluid starts revolving around it. The upper layers of the fluid as well as the lower layers of the fluid has circular motion.They have the same velocity but in the opposite direction. Being the velocities are same, their kinetic energies are also going to be the same and hence the body is not going to experience any resultant force.

In case three let us consider the body having both translatory motion as well as the spin. Because of the translatory motion the fluids upper at layer will get some velocity whereas due to the spin the upper layer will get some another portion of the velocity in the same direction.Thus the effective velocity in the upper portion of the fluid is going to be more which will increase its kinetic energy. As a result its pressure energy decreases at that point. In the similar way there will be more pressure at the bottom and hence the body will experience a uplift. This uplift is called dynamic lift.

 In real life generally all the bodies will have both translatory motion and the spin motion due to the air friction therefore this dynamic lift is also common for all the bodies.





Air foil

Even the wings of the aero plane is designed in such a way that it has to take the advantage of the dynamic lift. Its upper surface is little bit curved whereas the lower surfaces the flat surface. Because of the curved surf race the fluid will cross that part with a higher velocity and hence it will have more kinetic energy. Hence it will have less pressure energy as the total energy of the system shall always remain constant.

Similarly the lower portion of the wing will have more pressure energy which will help the body to take off quite easily.

In a cyclone day if someone is in a hut and close the door of that hut to protect from the cyclone it is going to take the roof of the hut  fly into the air a because of the uplift. 

 It is simply because the strong  air strikes the upper surface of the hut with higher velocity and hence there kinetic energy will be more and pressure energy will be less. As the a flow into the hut is small its kinetic energy is going to be less and the pressure energy is going to be more. This creates a unwanted up thrust on the top of the hut therefore it can fly in the air in a cyclone day.



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Velocity of Efflux and Venturi Meter

Velocity of Efflux

When a vessel is having a small opening above whom there is certain level of water, we can calculate the velocity of the water that is coming out of the opening using this concept. This is nothing but a direct application of  Bernoulli’s theorem.

We can apply that the total energy at the given two points is always constant.One point of consideration is that the surface of the vessel with a larger opening and other point in the consideration is having the small opening. Being the area of the larger par is  bigger the velocity with which it comes down there is negligible.

Similarly being both the parts of the system are open to atmosphere, both of them are having the same pressure which is equal to atmospheric pressure. We can prove mathematically that velocity with which the fluid comes out is equal to the velocity of a freely falling body.



Time taken to empty the tank

Basing on the above concept and using the concept of integration together we can derive the equation for the time taken to empty a vessel as shown below. Here we are using the concept of equation of continuity also.




Venturi Meter

It is a device using which we can calculate the velocity of the fluid. It has a pipe with a larger opening which keeps on decreasing and a particular point the it is having a small opening.

These two points are horizontal to each other and they are having the same potential energy at the two points. We can apply the law of  conservation of energy concept that the two points and by applying the equation of continuity together we can derive the equation for the velocity as shown below.




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Equation of Continuity and Bernoulli's Theorem

Equation of continuity

In a in compressible and non viscous fluid the mass of the fluid that enters at a given point per second is equal to the mass of the point that leaves in the same time.This concept is valid only when the fluid density is constant and it is not experiencing any viscous forces opposite its motion.

It can be mathematically proved that as per this concept the area of the cross-section of the fluid flow is inversely proportional to its velocity. If the area of cross-section is more velocity is less and vice versa.

It can be quite easily observed in daily life also. If the opening of a water tap is completely opened, water comes out with a certain velocity. If half of its opening is closed with the finger we can quite easily noticed that water is coming with better velocity.



Bernoulli’s Theorem

In a in compressible, non viscous, a rotational and streamlined fluid flow of the some of potential energy, kinetic energy and pressure energy per unit mass is always constant.This fundamental concept is valid only when the fluid is

1.Having a constant density,
2. No opposition appositive to its  motion,
3.The fluid particles are having only translatory motion and no rotatory motion,
4. All particles of the fluid are having the same velocity is when they are passing a particular point.

This is nothing but law of conservation of energy. Being a fluid it has not only potential and kinetic energy, it is also having pressure energy. The total of all the energies is always constant. If any one energy increases is obvious that the other energy decreases so that the total energy always remains constant. It is similar to law of conservation of energy that the energy is neither created, nor destroyed it just converts from one form to another form.

We can prove this theory basing on the concept of work energy theorem. It simply states that the work done is equal to changing its energy.We have further derived it in the following diagram that changing kinetic energy is equal to sum of difference in potential energies and pressure energies.




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Up thrust or the Buoyant Force and Problems with Solutions

Whenever a body is placed in a fluid, the fluid applies resultant up ward directional force on the body and it is called up thrust. It’s numerical value which is the product of volume of the fluid displaced, density of the fluid and acceleration due to gravity at the given place.Volume of the fluid displaced it depends on portion of the body that is there in the fluid. When the body is completely immersed in the fluid volume of the fluid displaced is equal to the volume of the body itself. When the body is partially immersed in the fluid , volume of the fluid displaced equal to the volume of the body that is there inside the fluid.

According to Archimedes principle ,up thrust is  equal to the weight of the fluid displaced by a static fluid.

Basing on this concept we can find out the relative density of a solid as well as a fluid. Relative density of the fluid is equal to the ratio of the weight of the body in air to the loss of the weight of the body in water.

Relative density of the fluid can be determined as the ratio of loss of the weight of the body in a liquid of the loss of the weight of the body in water.

Up thrust depends on the density of the fluid. As density of the sea water is more than that of ordinary water, upthrust in the case of the sea is more than that of river. Upthrust also depends on acceleration due to gravity. If the arrangement is placed in a lift and the lift is moving in  upward direction than in the place of the acceleration ,we have two consider both acceleration due to gravity as well as the acceleration of the lift.

Under different circumstances we can choose to the equation summed up thrust as shown below.





Determining the volume of the cavity

Let us consider a body having some cavity inside. We can calculate the value of upthrust when it is immersed in water. In the case of the volume of the fluid with can write the volume of the body and in the case of the density of the fluid we can write the density of water. Therefore we can get the volume of the body is the ratio of upthrust to the density of the water and gravity as shown.

Upthrust is nothing but the difference in the weight of the body near to the weight of the body in water. By substituting this terms we can get the volume of the cavity as shown below.




Problems on up thrust

Whenever a body is in a fluid, the fluid applies upthrust, force in the upward direction. The direction of top thrust is always the same. Depending on the volume of the body inside the fluid up thrust will change.

Problem one

A body is floating in water such that 6/10 parts of its volume is under water what is the density of the body?

In solving this problem we have to consider that weight of the body is equal to upthrust as the body is an equilibrium state. In the place of the volume of the fluid displaced, we have  to write only six by 10th of the volume of the body cause only that much of the bodies inside the water.

Problem two

Two solids A and B float on water. It is observed that A floats with half of its volume inside water  and B floats with 2/3  its volume immerse it find the ratio of their densities ?

Even in solving this problem also we are having the same approach. The simple concept is when the body is an equilibrium state the weight acting in the downward direction is equal to the up thrust acting in the upward direction.

Problem three

A cubicle block of wood with each side a 10 cm long floats are at the interface of water and oil. The lower surface is 2 cm below the interface of the liquid. The height of oil and water columns each are 10 cm. Density of the oily is 0.8 g per cc then what is the mass of  block ?

In this problem the weight of the body is equal to the upthrust provided by both water and oil. In the case of the water volume of the fluid displaced is equal to 2 by 10th of the volume of the total body whereas in the case of the oil volume of the fluid displaced is equal to 8 by 10th of the total volume of the body. The problem can be solved as shown below.



It is the study of behavior of liquids in motion.

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