Showing posts with label pressure. Show all posts
Showing posts with label pressure. Show all posts

Thermodynamics Problems with Solutions Five

We are solving series of problems based on the concepts of thermodynamics. We also deal about calorimetry in this chapter which deals about conversion of heat energy to other forms of energies and its applications further. We do define terms like specific heat and latent heat to explain this properties and they are the basic terms of heat concepts. When there is a change in the temperature, we need to deal with specific heat concept and when there is a change of state, we need to study it in terms of latent heat and during this process, all the supplied heat energy is used to change the state of the system and hence its temperature remains constant.

Problem

During an adiabatic process,pressure of the gas is proportional to the cube of the temperature and basing on that we need to find the ratio of specific heats of the gas. Problem is as shown in the diagram below.


Solution

We need to take the relation between pressure and temperature and taking that into consideration with the given data, we can get the relation between pressure and temperature in the adiabatic process and the problem can be solved as shown in the diagram below.


Problem

A metal sphere of known radius and specific heat is given to us and it is rotating about its own axis with certain rotations per second. When it is stopped half of its energy is converted into heat and we need to measure the raise in the temperature of the system and the problem is as shown in the diagram below.


Solution

As the body is rotating it has rotational kinetic energy and half of it is converted into heat energy as per the given problem. Taking law of conservation of the energy, we can solve the problem as shown in the diagram below.



Problem

The relation between internal energy,pressure and volume is given to us as shown in the diagram below. We need to find the ratio of specific heats and some constants are also available in the problem.


Solution

We need to differentiate the given equation to get the change in internal energy and hence it can be expressed in terms of specific heat of the gas at constant volume. Problem can be further solved as shown in the diagram below.


Problem

Work done by a system under isothermal conditions has to be determined that change its volume from one to other and satisfy the given equation. Problem is as shown in the diagram below.


Solution

Relation of pressure with other physical quantities is given to us as shown in the diagram below. We need to measure the work done and the pressure is not constant here. So to get the work done we shall integrate the pressure with the change in the volume as shown in the diagram below. By simplifying the equation further, we can get the solution as shown in the diagram below.


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Thermodynamics Problems with Solutions Four

We are solving series of problems based on the concept of thermodynamics. It is a branch of physics that deals with the heat energy conversion into other forms and its applications. Carnot engine is a ideal heat engine that converts the given heat into work under a cyclic process. As the process is cyclic process, internal energy remains same and all the supplied heat is going to be converted into work. But we know that heat is a disordered format of energy and it cannot be completely converted into work. That is the reason why, heat engine cannot have hundred percent efficiency. Heat engine will have three basic parts by name source, working substance and sink. Source do supply heat, working substance do convert the heat into work and the extra generated heat to the sink.

Problem

Pressure and volume graph for two different gases during adiabatic process is given to us as shown in the diagram below. We need to know which graph belongs to which gas.


Solution

From the pressure and volume graph, it can be found that the slope of the graph is directly proportional to the ratio of specific heats of the gas. It is clear from the graph that the slope of the second curve is more and hence it shall be the gas with more ratio of specific heat value. Solution of the problem is as shown in the diagram below.


Problem

A refrigerator is placed in a room of temperature of 300 kelvin and the system temperature is 264 kelvin. We need to measure the how many calories of heat shall be delivered to the room to the room for each kilo kelvin of energy consumed by refrigerator by the system ideally. Problem is as shown in the diagram below.


Solution

We know that proficiency of a refrigerator is the amount of work done when compared with the heat energy supplied to the system. It can also be expressed in terms of temperature as shown in the diagram and is solved as shown in the diagram below.


Problem

In carnot’s engine efficiency is 40 % for a certain temperature of the hot source. To increase the efficiency of the system by 50 %, what shall be the source temperature and the problem is as shown in the diagram below.


Solution

We know that the efficiency of heat engine is the magnitude of the work done when compared with the heat energy supplied and it can be expressed in terms of absolute temperature as shown in the diagram below.

We need to apply it for two cases and solve the problem as shown in the diagram below.


Problem

The relation between temperature and pressure is given to us for a certain gas and we need to find the specific heats ratio of that gas. Problem is as shown in the diagram below.


Solution

This process is adiabatic process and we need to write the given relation between pressure and temperature from the given format to the standard format so that we can solve and find the ratio of specific heats of the given gas as shown in the diagram below.


Problem

When a mono atomic gas expands at constant pressure, the percentage of the heat supplied that increases the temperature of the gas and in doing external work is how much is the problem and it is as shown in the diagram below.


Solution

We know that if volume is kept constant, work done is zero and corresponding specific heat is defined as per it and vice versa. When we write ratio of specific heats, it becomes the ratio of heat supplied to the change in the internal energy and internal energy change the temperature of the system. Thus we can find the percentage value basing on the specific heats ratio value as shown in the diagram below.



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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.



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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.



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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.





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