Showing posts with label Fluid Statics. Show all posts
Showing posts with label Fluid Statics. Show all posts

Mechanical Properties of Fluids Problems and Solutions Five

We are solving series of problems in Mechanical properties of Fluids.Viscosity is the property of the fluid due to which its relative motion is opposed. When a fluid is moving, it moves like different set of layers, each layer moving in the forward direction opposed the lower layers motion in the forward direction and thus viscous force is generated. It is directly proportional to the area of cross section of the fluid, velocity of the fluid flow and it is inversely proportional to the width of the layer. We can find coefficient of viscosity in this way. When a spherical body is moving in a fluid due to opposing viscous force and upward upthrust and weight in downward direction, some where the resultant force becomes zero and the body acquires a constant velocity called terminal velocity.

Problem

A plate of area 100 square centimeter is placed on the upper surface of castor oil having only 2 mm thickness. Coefficient of thickness is given to us and we need to measure the horizontal force required to move the plate with a certain velocity and the problem is as shown in the diagram below.


Solution

We can solve the problem using the very definition of coefficient of viscosity as shown in the diagram below. We know that the viscous force is directly proportional to the area of cross section,velocity of fluid flow and inversely proportional to the distance between the layers. Solution is as shown in the diagram below.


Problem

A vessel has a height of 40 meter. It has three horizontal tubes each of same diameter and length at different heights from the base as shown in the problem below. We need to find out the length of the single pipe of the same diameter that has to be replaced instead of three pipes so that fluid flow is same.


Solution

The sum of rate of flow through each pipe has to be the fluid flow in the single pipe. We can use Poisellie’s equation and find the length of the new pipe as shown in the diagram below.


Problem

A cylindrical tank has a hole of known area at the bottom. If the water is allowed to flow into the tank from a tube above it with a known rate, we need to find the maximum height of the fluid in the cylinder. Problem is as shown in the diagram below.


Solution

At the maximum possible height of the liquid in the cylinder, the rate of fluid in is equal to the rate of the fluid out. Rate of fluid out can be written in terms of equation of continuity and we also know that the velocity of the fluid coming out of the hole is similar to velocity of the freely falling body. Taking this into consideration, we can solve the problem as shown in the diagram below.



Related Posts

No comments:
Labels: , , ,

Mechanical Properties of Fluids Problems and Solutions Four

We are solving series of problems on fluid dynamics. Viscous force acts on all bodies that have relative motion. If a spherical ball is falling through a fluid, it experience a viscous force and it can be measured using a law called Stroke’s law. According to this rule, viscous force is directly proportional to the square of the radius of the ball. There is weight of the ball acting in the downward direction and upthrust acting always in the upward direction. Viscous force always acts opposite to the relative motion of the body. After some travel, the resultant force acting on the body becomes zero and the body acquires a constant velocity and that velocity is called as terminal velocity. 

Problem

An open U tube contains mercury. When certain amount of water of known height is filled in one tube over mercury, we would like to measure the difference between the liquid levels in both the sides and the problem is as shown in the diagram below.


Solution

We know that at the bottom of the system both the pipes has one common point whose height is same and hence the pressure is same at both the points. We know that the pressure can be expressed as the product of height of the liquid, density of the liquid and acceleration due to gravity at the given place. Thus by equating the pressure at both the points we can solve the problem as shown in the diagram below.


Problem

A sphere of known radius is having a cavity of half the radius of the sphere. It is found that the sphere is just floating in the water with its highest point in touch with the water. We need to find the specific gravity of the material and the problem is as shown in the diagram below.


Solution

We know that when the sphere is just floating, its weight is balanced by the upthrust acting on it. We know that upthrust is the product of volume of the fluid displaced, density of the fluid and acceleration due to gravity. As the body is completely immersed in the fluid, volume of the fluid displaced is equal to the volume of the body itself. We know the values of density of water and acceleration due to gravity.

Weight acting in the downward direction is only due to the mass present in the system. We can write mass as volume and density product of the body. Volume of the content means we shall count only the part of mass present in the system. Thus we can measure the weight and equate it ot the upthrust as shown in the diagram below.


Problem

The upthrust force acting on the wing of aeroplane is given to us. Velocity of the air on its lower surface and area of cross section is also given to us and we need to find the velocity of air on its upper surface ad the problem is as shown in the diagram below.


Solution

We can assume that the wing is horizontal and hence there is no difference in the gravitational energy. As velocity is different kinetic energy per unit mass and hence pressure energy are going to be different. We can apply Bernoullie’s theorem, we can solve the problem as shown in the diagram below.


Problem

Due to a disease main artery expanded in its area of cross section as per the given data in the problem. Velocity of the blood in non expanded area is given to us and blood density is also given to us. We need to find the excess pressure developed due to this and the problem is as shown in the diagram below.
  

Solution

We can apply equation of continuity and velocity of the other part of artery as shown in the diagram below. Now knowing the velocity, we can apply conservation of energy concept and solve the problem as shown in the diagram below.


Problem

The coefficients of viscosity of two liquids is given to us as 2:3 densities of the fluids is given to us as 4:5. In the equal time we need to know the rate of fluid flow in the tubes and the problem is as shown in the diagram below.


Solution

We need to use a concept called Poisellie’s equation. As per it we can write the equation for the rate of flow as shown in the diagram below and solve the problem. Length and diameter of tubes is given as same and the problem is solved as shown in the diagram below.



Related Posts

No comments:
Labels: , , , , ,

Mechanical Properties of Fluids Problems and Solutions Three

We are dealing with the series of problems based on fluid statics and fluid dynamics. When a fluid is in the state of motion, it has potential energy,kinetic energy and also pressure energy. We know that the total energy of the system is constant. As per Bernoulli's theorem, the sum of potential energy,kinetic energy and pressure energy per unit mass always remains constant. We can also consider the viscous force acting opposite to the motion. It is the force that always opposes the relative motion. It depends on the area of cross section of the fluid, velocity of the fluid flow and the distance between the two layers of the fluid in the motion.

Problem

A vessel is kept on a table and filled with water. From the bottom of the vessel, a orifice is made and its location is given in the problem as shown in the diagram below. We need to measure the horizontal distance at which it is going to strike the floor.


Solution

We know that the velocity of the fluid coming out of orifice is similar to a velocity of a freely falling body and it depends on the height of the fluid above the opening. Once it is out, it is under the influence of the gravity and its horizontal distance distance can be measured with the equations of motion.There is no gravity acting on it so its velocity is uniform along horizontal direction. Further problem can be solved as shown in the diagram below.


Problem

Velocity of air flow on the upper surface of the wing of airplane is 40 meter per second and on the lower surface is 30 meter per second and its area of cross section is given to us and its mass is also given to us. We need to measure the force experienced by the wing and the problem is as shown in the diagram below.


Solution

As the wing is airplane is horizontal, there is no change in potential energy and we can remove that components of Bernoulli's theorem. Applying kinetic energy data, we can get the difference in the pressure at both the cases. We know that pressure is the force per unit area and hence we can measure the force acting on the system as shown in the diagram below.


Problem

Specific gravity of two liquids combined together when they have same mass and volume is given to us and we need to measure the density of each liquid. The problem is as shown in the diagram below.


Solution

We know that the density is the ratio of mass to the volume. We can find the effective density of the system when they have equal volume and equal mass can be found as shown in the diagram below. We have all ready derived equations for both of them and it can be done as shown below.



Problem

A sphere has known density and it is falling through a fluid of known density and we need to measure the acceleration of the body and the problem is as shown in the diagram below.


Solution

When a body is moving in a fluid, its weight acts in the downward direction and the upthrust acts in the upward direction. We need not consider viscous force as there is no data about coefficient of viscosity. We can write equation for the resultant force as the difference between weight and upthrust. We can use Newton’s second law and find as shown in the diagram below.


Problem

When a polar bear jumps on to ice block it just sinks and we need to measure the weights of that ice block and the specific gravity of ice and sea water is given to us. Problem is as shown in the diagram below.


Solution

We know that weight of the liquid displaced is the sum of weights of ice block and polar bear. We can equate the data and solve the problem as shown in the diagram below.



Related Posts

No comments:
Labels: , , , ,

Mechanical Properties of Fluids Problems and Solutions Two

Fluids are some one which can flow from one place to other and both liquids and gases falls under this category. Fluids in the state of rest and their properties were studied in fluid statics and as we deal more about water, it is also called as hydro statics. Fluids in the state of motion are studied in fluid dynamics. Apart from pressure, here we need to talk about their energies also. They have potential energy,kinetic energy and pressure energy and satisfy law of conservation of energy. That is explained in terms of Bernoullie’s theorem. If the density of the fluid is constant they are called in compressible. We can ignore their viscous property and they are then called non viscous fluids.

Problem

A horizontal pipe of non uniform area of cross section has water flows through it at a point with a speed of 2 meter per second where the pressures is 40 kilo pascal. We need to know the pressure at a point where the velocity is 3 meter per second.


Solution

As the tube is horizontal, there is no potential difference between the two points. We need to apply Bernoullie’s theorem according to which the total energy of the system that is the sum of potential energy,kinetic energy and pressure energy is the constant. If one energy increases, other energy decreases but their sum remains constant.

We can apply this with the given data barring potential energy part as it is same on both the sides and we can solve the problem as shown in the diagram below.


Problem

Two water pipes of diameters 4 and 8 centimeter are connected in series to a main pipe we need to find the ratio of velocities of the water in these two pipes and the problem is as shown in the diagram below.


Solution

We know that as the two pipes are connected in series the rate of flow of water is same in both of them and hence the equation of continuity is very well valid concept here. According to it the volume of the water passes through a given system is always constant. It is nothing but the product of area of cross section of the pipe and the velocity of the pipe is constant. Area can be written as circular and it is proportional to the square of radius or diameter. Problem can be further solved as shown in the diagram below.


Problem

Two equal  drops of water are falling through with a velocity 10 meter per second and that is constant. If these two drops are combined together and a single drop is formed, we need to find the velocity of that combined drop and the problem is as shown in the diagram below.


Solution

The constant velocity acquired by a spherical drop while passing through a medium is called terminal velocity and at that instant the resultant force acting on the system is zero. We have proved earlier in this chapter concepts that the terminal velocity is directly proportional to the square of the radius of the drop.

When two drops are combined as both of them are of same density, its total volume is the sum of the volumes of two drops. Taking that into consideration, we can find the relation between small and large radius and the problem can be solved as shown in the diagram below.


Problem

A large tank is filled up to certain height and we need to know the ratio of time taken by a small hole placed at the bottom to empty first half of the height of the water when compared with the second height of the water. The problem is as shown in the diagram below.


Solution

We can prove and we have proved while handling this chapter that time taken to empty the tank is directly proportional to the difference of height of fluid from where to where they are emptied in terms of heights of fluid under square root. In the first case, it is emptied from full to half and in the second case, it is emptied from half to zero. By writing that data, we can solve the problem as shown in the diagram below.



Related Posts

No comments:
Labels: , , , , ,

Mechanical Properties of Fluids Problems and Solutions One

Fluid statics is a branch of physics that deals with the fluids in the state of rest and its properties like density,pressure and upthrust. Density is the property of a body which measures how its mass is distributed over its volume. If density is more, its more mass is concentrated in a small volume and vice-versa. Pressure is defined as the force experienced by a body per unit surface area. In this case, force is acting on the body in all directions and hence it changes the volume of the body. Every body in a fluid experience a resultant upward force and it is called upthrust or buoyant force.It depends on the volume of the fluid displaced by the body when it is immersed in the fluid, density of the fluid and acceleration due to gravity. If the body is completely immersed in the fluid, volume of the fluid displaced is equal to the volume of the body itself.

Problem

A body is floating in the water in such a way that six parts out of ten parts of its volume is inside the water. We need to measure the density of the water and the problem is as shown in the diagram below.




Solution

When a body is under water, we know that it experience upthrust. As the body is in the equilibrium state, its weight is balanced by the upthrust and they can be equated mathematically. In the place of weight, we can write the product of volume of the body with its density. Volume of the fluid displaced is six parts of volume of the body as only that much is immersed in the fluid. By applying this data, we can solve the problem as shown in the diagram below.



Problem

A brass sphere weights 100 gram weight in air. It is suspended by a thread in a liquid of known specific density. If the specific density of the brass is also known, we need to find the tension in the thread and the problem is as shown in the diagram below.



Solution

Weight is given in gram weight. We need to convert not only gram into kilogram, we also need to multiply it with acceleration due to gravity so that we are converting it into force that is measured in newton. Tension and upthrust acts in upward direction and weight of the body in the downward direction. As the system is in equilibrium position, we can equate this upward and downward force and solve the problem as shown in the diagram below.


Problem

Ninety grams of sulfuric acid is mixed with ninety grams of water and both the densities are known to us. If the specific gravity of the mixture is given to us, we need to find the loss of volume of the system due to mixing and the problem is as shown in the diagram 
below.



Solution

We need to write the total volume initially is the sum of volumes of the first and second ones and it is to found using the definition as the ratio of mass to the density of the each one as shown in the diagram below.



This much volume is supposed to be there but we can also find out the actual volume available using the data like total mass of the system and the effective density of the mixture and as shown below it is different from the supposed to be the volume measured in the first case. Hence the difference between is the loss of the volume. It is further solved as shown in the diagram below.



Problem

Weight of a solid in air, in water is given to us in the problem as shown in the diagram below and we need to measure the weight of that body when it is completely immersed in a liquid of known specific gravity.



Solution

We can find the loss of the weight of the system as the difference between weight of the body in air to the water. Loss of that weight multiplied with the density of the new system, we can find the loss of the weight of the body in that fluid. Actual mass of the body in the fluid is the difference between actual mass of the body in air to the loss of the weight in the fluid. The solution is as shown in the diagram below.


Problem

A liquid is placed in a cylindrical jar and it is rotating about its axis. If radius and angular velocity of the cylinder is know to us, we need to know the difference between the center of the liquid and rise of the liquid at the sides and the problem is as shown in the diagram below.


Solution

The kinetic energy of the fluid is converted into penitential energy and hence as per the law of conservation of energy, we can equate them. We need to replace linear velocity in terms of angular velocity and radius of the system and the problem can be solved as shown in the diagram below.





Related Posts



No comments:
Labels: , , , , , ,
Subscribe to: Comments (Atom)