The phenomenon of surface tension is explained by Fig. Consider three moleculesA, 8, C of a liquid in a mass of liquid. The molecule A is attracted in all directions equally by the surrounding molecules of the liquid.
Thus the resultant force acting on the molecule A is zero. But fhe molecule B , which is situated near the free surface, is acted Jpon by upward and downward 'forces which are unbalanced. Thus a net resuitant force on molecule B IS acting in the downward The molecule C, situated on the free surface of liquid, doesexperience a resultant downward force.
All e molecules Fig. Surface Te ns ion on liquid Droplet. Let the droplet is cut into two halves. P bubbJe n air has tWo surfaces in contact with air, one inside and other outside. Thus two surfaces are subjected tuurface 'teDsro ii. Comider a liquid jet of diameter d' and length 'L' as shown in Fig. The surface Tension of water in contact with ai0Jt 2flC is 0.
Tlte pressure inside a droplet of water is to be 0. Calculp: e.. Calculate the pressure within the droplet if surface tension is given as Nlm of water. The rise pf liquid. Consider a glass tube of small diameter 'd' opened at both ends and is inserted in.
The liquid will rise in the tube above the level. Then rise of water is given by 4 cr. Expression for Capillary Fall.
Let h Height oJ depression in tube. Then in equilibrium, two forces are acting on the mercury inside the tube. Take surface tensions cr 0. Given : Dia. The temperature of the liquid is 20C and the values of the surface tension of water and mercury at 20C in contact with air are 0.
The angle of contact for water is zero that for mercury 1. Take density of water at U. The capillaty rise in the glass tube is not to exceed 0. Given : Capillary rise, Surface tension,. Let dia. Consider surface tension of water in contact. Converted to SI Units, A. Given : Capillary rise,. Surface tension, cr Let dia. The diameter of the shaft is 0. Calculate the power lost in oil for a sle. Given : Viscosity,.
Consider a liquid say water which is confined in a closed vessel Let the temperature of liquid is 20"C and pressure is atmospheric. This liquid will vaporise at lOO"C. When vaporization takes place, the molecules escapes from the free surface of the liquid. These vapour molecules get accumulated in the space between the free liquid surface and top of the vessel. These accumulated vapours exert a pressure on the liquid surface. This pressure is known as vapour pressure of the liquid.
Or this is the pressure at which the liquid is converted into vapours. If the press Jre above the liquid surface is reduced by some means, the boiling temperature will also reduce. If the pressure is reduced to such an extent that it becomes equal to or less than the vapour pressure, the boiling of the liquid will start, though the temperature of the liquid is 20C.
Thus a liquid may boil even at ordinary temperature, if the pressure above the liquid surface is reduced so as to be equal or less than the vapour pressure of the liquid at th.! Now consider a flowing liquid in a system. The bubbles of these vapours are carried by the flowing liquid into the region of high pressure where they collapse, giving rise to high impact pressure.
The pressure developed by the collapsing bubbles is so high that the material from the adjoining boundaries gets eroded and cavities are formed on them. This phenomenon is known as cavitation. Hence the cavitation is the phenomenon of formation of vapour bubbles of a flowing liquid in a region where the pressure of the liquid falls below che vapour pressure and sudden collapsing of these vapour bubbles in a region of higher pressure.
When the vapour bubbles collapse, a very high pressure is created. The metallic surfaces, above which the liquid is flowing, is subjected to these high pressures, which cause pitting action on the surface. Thus cavities are formed on the metallic surface and hence the name is cavitation. The weight density or specific weight o a fluid is equal to weight per unit volume. M ath emauc all. Poise and stokes are the units of viscosity and kinematic viscosity respectively. For a perfect gas, the equatio11 of state is.
For-isothermal process. IS gtven. Define the following fluid properties :. State their me surements. Explain the terms : i Dynamic viscosity, and ii Kinematic viscosity. Give their dimensions. State the Newton's law of viscosity and give examples of its application. Delhi University, June 6. Enunciate Newton's law of viscosity. Explain the importance of viscosity in fluid motion. What is the effect of temperature on viscosity of water and that of air? Define Newtonian and Non-Newtonian fluids.
What do you understand by terms : i Isothermal process, it Adiabatic process, and iii Universal-gas constant. Define compressibility. Prove that compressibility for a perfect gas undergoing isothermal compression is. Define surface tension. Explain the phenomenon of capillarity. Obtain an expression for capillary rise of a liquid. Explain the importance of compressibility in fluid flow. Vishwavidyalaya, Bhopal S Define and explain Newton's law of viscosity.
Delhi University. April ] Wimer I 6. Hyderabad S l7. One litre of crude oil weighs 9. Calculate its specific weight, density and specific gravity. Assume 2 dynamic viscosity as 8 poise. Determine the fluid viscosity between the plates in the poise. The lower plate is fixed while the upper plate having surface area 1. Find the force and power required to maintain this speed, if thl!
The weight of the square is Find the dynamic viscosity of the oil. In a stream of glycerine in motion, at a certain point the velocity gradient is 0. The mass density of fluid is Calculate the shear stress at the point. Determine the specific gravity of a nuid having viscosity 0. Determine the viscosity of a liquid having kinematic viscosity 6 stokes and specific gravity 2.
Jf the velocity distribution of a fluid over a plate is g. An oil of viscosity 5 poise is used for lubrication between a shaft and sleeve. The diameter of shaft is 0. The weight of a gas is given as Determine the gas constant and also tbc density of the gas. A cylinder of 0. The air is com3 pressed to 0. Find i the pressure inside the cylinder assuming isothermal process, ii pressure and temperature assuming adiabatic process. Determine the bulk modulus of elasticity. Determine the bulk modulus of elasticity of a nuid which is compressed in a cylinder from a volume of 0.
The surface tension of water in contact with ai r at 20C is given as 0. The pressure inside a droplet of water is to be 0. Find the surface tension in a soap bubble of30 mm diameter when the inside pressure is 1. Tite surface tension of water in contact with air is given as 0. Calculate the capillary rise in a glass tube of 3.
Take surface tensions for mercury and water as 0. Specific gravity for mercury is givep as I 3. The capillary rise in the glass tube used for measuring water level is not to exceed 0. Converted to SJ units, A. SI Units. A piston mm diameter and mm long works in a cylinder of mm diameter.
If the annular space is filled with a lubricating oil of viscosity 5 cp centi-poise , calculate the speed of descent of the piston in vertical position. The weight of the piston and axial load are 9. Find the capillary rise of water io a tube 0. The surface tension of water is 0. Calculate the specific weight, density and specific gravity of two litres of a liquid which weight 15 N. Delhi University, April [Ans. A mm diameter vertical cylinder rotates concentrically inside another cylinder of diameter l 51 mm.
Both the cylinders are of mm height. The space between the cylinders is filled with a liquid of viscosity I 0 poise. Determine the torque required to rotate the inner cylinder at r. Delhi Uni1ersiry, April [A ns. A shaft of diameter mm is rotating inside a journal bearing of diameter I 22 mm at a speed of r. The space between the shaft and the bearing is filled with a lubricating oil of viscosity 6 poise. Find the power absorbed in oil if the length of bearing is mm. Delhi Uni1ersiry, May [Ans.
A shaft of diameter I 00 mm is rotating inside a journal bearing of diameter mm at a space of r. The space between the shaft aod bearing is filled with a lubricating oil of viscosity 5 poise.
The length of the bearing is mm. Find the power absorbed in the lubricating oil. Assuming that the bulk modulus of elastici ty of water is 2. A square plate of size 1 mx I m and weighing N slides down an inclined plane with a uniform velocity of 1. The inclined plane is laid on a slope of 5 vertical to 12 horizontal and has an oil film of 1 mm thickness. Calculate the dynamic 'liscosity of oil.
Consider a small area dA in large mass of fluid. If the fluid is stationary, then the force exerted by the surrounding fluid on the area dA will always be perpendicular to the surface dA.
Let dF is the force the normal direction. It states that the pressure or intensity of pressure at a point in. Consider an a rbitr ary fluid element of wedge shape in a fluid mass at rest as shown in Fig. Then the forces acting on the element are : I. Pressure forces normal to the surfaces. Weight of element in the vertical direction. Resolving the forces in x-direction. Similarly, resolving the forces in. The pressure at any point in a fluid at rest is obtained by the Hydrostatic Law which states that the rate of increase of pressure in a vertically downward direction must be equal to the specific weight of the fluid at that point.
Z Pressure forces on surfaces BC and AD are equal and opposite. For equilibrium of fluid element, we have pM - p. Equation 2. This is Hydrostatic Law. By integrating the above equation 2. From equation 2. Here Z is called p ressure head. Problem 2. Find the weight lifted by the hydraulic press when the force applied at the plunger is N. Due to Pascal's law, the intensity of pressure will be equally transmitted in all directions.
Hence the pressure intensity at the ram 2 It is used for lifting a 11eight of 30 kN. Find the force required at the plunger. Weight lifted, See Fig. Take density of II"Ctter, p kg!
Height of liquid column. Fluid Mechanics The pressure at any point in a liquid is given by equation 2. From equation l. Density of mercury,. Given : Pressure intensity,. AI a poilllthe height of oil is 40 Find the corresponding heighT of lnlfer m the point. Given : S0 0. Dens1ty of water x g x9. Find the pressure intensity i at the inte,face of the two liquids.
A force oj. Find the load lifted by the large piston when : a the pistons are, at the same le1el. The density of the liquid in the jack is given as OOO kg! Pressure intensity on the large piston Fig. Pressure intensity at section A - A. The pressure on a fluid is measured in two different systems.
In one system, it is measured above the absolute zero or complete vacuum and it is called the absolute pressure and in other system, pressure is measured above the atmospheric pressure and it is called gauge pressure.
Thus : I. Absolute pressu r e is defined as the pressure which is measured with reference to absolute vacuum pressure. Gauge pressure is defined-as the pressure which is measured scanned measuring instrument, in which the atmospheric pressure is taken as datum. Vacuum pressure is defined as the pres- w sure below the atmospheric pressure. The specific gravity of mercury is Given : Depth of liquid, Density of liquid, Atmospheric pressure head,.
Pressure head in terms of mercury. A simple manometer consists of a glass tube having one of its ends connected to a point where pressure is to be measured and other end remains open to atmosphere.
Common types of simple manometers are : 1. Piezometer; 2. U-tube Manometer, and. Single Column Manometer. If at a point A, the height of liquid say water is h in piezometer tube, then pressure at A.
Let B is the point at which pressure is to be measured, whose va1ue is p. The datum line is A-A. As the pressure is the same for the horizontal surface. Hence pressure above the horizontal datum line A-A in the left column and in the right column of U-tube manometer should be same. For measuring vacuum P. The centre of the pipe is 12 em below the level of mercury i11 the right limb.
Find the pressure offluid in the pipe if the difference of mercUI:r level in the tll'o limbs is 20 em. Density of fluid.. The other end of the manometer is open to atmosphere. Find the vacuum pressitre in pipe, if the difference ofmercury level in the two limbs is 40 em and the height of fluid in the left from the cemre of pipe is I 5 em below. Given : Sp. Difference of mercury level. Height of liquid in left limb.
The right limb of tlw manometer contain s mercury and is open to atmosphere. Th e co11taCI betll'een nater and mercury is in the left limb. Determine the pressure of 11ater in the mai11 line, if rhe difference in lel'el of mercury in the limbs of U-tube is 10 em and the. If the pressure o. Sketch the arrangements A. Hence pressure at B should be equal to pressure at C. Hence equating the equations i and ii , we get. The arrangement is shown in Fig. In this case the pressure at A is 98 10 Nlm 2 wh.
Hence mercury in left limb will rise. The rise of mercury in left limb will be equal to the fall of mercury in right limb as the total volume of mercury remains same. Dividing by 9. The reading ofthe manometer given in the figure shows when the vessel is empty.
Find the reading of the manometer when the vessel is completely,filled with water. Vessel is empty. Vessel is full of water.
When vessel is full of water, the Fig. Let the distance through which merc ury goes down in the right limb be. The mercury will rise in the left by a distance of y em. Now the datum line is Z Equating the pr. A red liquid of specific gravity 0. Find the displacement of the surface of separation when the pressure on the surface in Cis g"reater than that in B by an amoimt equal to 1 em head of water.
Then Y-Y becomes the final separation level. Now fall in surface level of C multiplied by crosssectional -area of bulb C must be equal to the fall in separation level multiplied by cross-sectional area of limb. Single column manomter is a modified form of. The other limb may be vertical or inclined. Thus there are two rypes of single column manometer as : I. Vertical Single Column Manometer.
Inclined Single Column Manometer. Let X-X be the datum line in the reservoir J scanned by Fahid md in the right limb of the manometer, when it is not connected to the pipe. Then pressure in the right limb above Y-Y. In clined Single Column Manom eter.
This manometer is more sensitive. Substituting the value of h2 , we get. Find the pressure. Differential manometers are the deVices used for measuring the difference of pressures between two points in. A differential mano. Most commonly types of differential manometer'S are : I.
U--tube differential manometer and 2- Imerted U-tube differential manometer. Let the two points A and B are at different level and also contains liquids of different sp. These points are connected to the U-tube differential manometer. Taking datum line at X-X.
A differintial manometer connected at the MO points A and B shows a difference in mercury level as 15 em. Find thescanned differenceby of Fahid pressure at the two points. The pipe A conrain. The pressures at A' and Bare I kgf! At 8 air pressure is 9. Given : Air pressure at or.
The two ends of the tube are connected to the points whose difference of pressure is to be measured. It is used for measuring difference. Let the pressure at A is more than the pressure at B. Taking X-X as datum line. The pressure head in the pipe A is 2m of water, find the pressure in the pipe B for the manometer readings as shown in Fig.
Given : Pressure head at. In Fig. The fluid in manometer is oil of sp. For the manometer readings shown in the figure, find the pressure difference between A and B. OIL OF. Pipes A and B are located at the same level and assume the pressures at A and B to be equal. The vertical distance between the axes of these pipes is 30 em. When an oil of specific gravity 0. Determine the difference of pressure between the pipes.
The points C and D lie on the same horizontal line. Hence pressure at C should be equal to pressure at D. But pressure at C And pressure at D. For compressible fluids. Such problems arc encountered in aeronautics. For gases the equation of state is.
Case I. If temperature T is constant which is true for isothermal process, equation 2. If the datum line is taken at becomes the pressure at datum line. If temperature T is not constant but the process follows adiabatic law then the relation between pressure and density is given by..!!
For the adiabatic process, the temperature at any height in air is calculated as : , Equation of-state at ground level and at a height Z from ground level is written as. It is defined as the rate at which the temperature changes with elevation. To obtain an expression for the temperature lapse-rate, the temperature given by equation 2. In atmosphere, the value of k varies with height and hence the value of temperature lapse-rate also varies.
From the sea-level up to an elevation of about II m or II km. SOC and hence in this range lapse-rate is zero. Temperature rises again. The densiry of air is gile11 as 1. Pressure at any height Z by isothermal law is given by equation 2. Nlcrn 7. The density p for mercury. We know that as the elevation above the sea-level increases, the atmospheric pressure decreases. Pro blem 2. Neglect variatio11 of g with altitude.
S olution. The temperature lapse rate is given as 0. Take density of air at sea-level equal to 1. RT Calculate the pressure around the aeroplane, given the lapserate in the atmosphere as 0.
Neglect variation of g with altitude. Absolute pressure is the pressure in which absolute vacuum pressure is taken as datum while gauge pressure is the pressure in which the atmospheric pressure is taken as datum, Pabs.
Manometer is a device used for measuring pressure at a point in a fluid. Manometers are classified as a Simple manometers and b Differentiat manometers.
Simple manometers are used for measuring pressure at a point while differential man? The pressure at a point in static compressible fluid is obtained by combining two equations, i.
The rate at which the temperature changes with elevation is known as Temperature Lapse-Rate. It is given by. Define pressure. Obtain an expression for the pressure intensity al a point in a fluid. State and prove the Pascal's law. What do you understand by Hydrostatic Law? Differentiate between : i Absolute and gauge pressure, ii Simple manometer and differential manometer, and iii Piezometer and pressure gauges.
What do you mean by vac uum pressure? What is a manometer? What do you mean by single column manometers? How are they used for the measurement of pressure?
What is the difference between U-tube differential manometers and invened U-tube differential manometers? Where are they used? Jical pressure gauges? Derive an expression for the pressure at a height Z from sea-level for a static air when the compression of the air is ;tSSumed isothermal. The pressure and temperature at sea-levels are p0 and T0 respectively.
What do you understand by the term, 'Temperature Lapse-Rate'? Obtain an expression for the temperature Lapse-Rate. What is hydrostatic pressure distribution? Give one example where pressure distribution is non A. Winter hydrostatic. A hydraulic press has ram of 30 em diameter and a plunger of 5 em diameter.
Itby is used for lifting a PDF created by[Ans. Calculate the pressure due to a column of 0. The pressure intensity at a point in a fluid is given 4. Find the corresponding height uf fluid when it is: a water, and b an oil of sp.
At a point the height of oil is 20 m. Find the corresponding height of water at that point. Find the pressure intensity : 1 at the interface of the two liquids, and ii at the bottom of the tank. A force of 60 N is applied on the small piston. Find the load lifted by the large piston, when : a the pistons are at the same level, and b small piston is 20 em above the large piston.
The densiiy of the liquid in the jack is given as Detennine the gauge and absolute pressure at a point whlch is 2. Take atmospheric pressure as Its right.
The centre of the pipe is 9 em below the level of mercury. The pressure in the pipe is vacuum. The other end of the manometer is open to the atmosphere.
Find the vacuum, pressure in pipe, if the difference of mercury level In the two limbs is 20 em [Ans. A single column vertical manometer i. The area of the reservoir is 80 times the area of the manometer tube. The resenoir contains mercury of sp. The level of mercury in the reservoir is at a height of 30 em below the centre of the pipe and difference of mercury levels in the reservoir and right limb is 50 em.
Find the pressure in the pipe. A differential manometer connected at the two points A and B of the pipe shows a difference in mercury level as 20 em. Find the difference of pressure at the two points. Pipe A contains carbon tetrachloride having a specific gravity 1.
The pipe A lies 2. Find the difference of pressure A. December, [An s. A differential manometer is connected at the two points A and B as shown in Fig. At B air pressure 2 [Ans. An inverted differential m;mometer containing an oil of sp. If the manometer reading is 40 em, find the difference of pressures. In above Fig. The fluid in manometer is oil. If the atmosphaic pressure at sea-level is I0. The density of air is given as 1. Calculate the pressure at a height of m above sea-level if the atmospheric pressure is I 0 1.
Take the density of air at the sea-level as equal to 1. Calculate the pressure and density of air at a height of m above sea-level where pressure and temperature of the air are The temperature lapse-rate is given as [Ans.
An aeroplane is flying at an altitude of m. Calculate the pressure around the aeroplane, given the lapse-rate in the atmosphere as 0. Take pressure and temperature at ground lc'el as The density of air at ground level is given as 1. The atmosphere pressure at the sea-level is Assume density of ai r at sealevel as 1. Neglect variation of 'g' with altitude. An oil of sp. I 14m, ii The atmospheric pressure at the sea-level is Calcul ate the pressure m above sea-level, assuming : i isothermal variation of pressure and density, and i1 adiabatic variation of pressure and density.
Assume density of air at sea-level as 1. Derive the formula that you may use. What are the gauge pressure and absolute pressure at a point 4 m below the free surface of a liquid of specific gravity I. May J.. A tank contains a liquid of specific gravity O.
The atmospheric pressure head is equivalent to mm of mercury. Delhi Unin! June IAns. This chapter deals with the fluids i. The velo9ity gradient, which is equal to the. This force always acts normal to the surface. Cen tre of pressu re is defined as the point of application of the total pressure on the surface. There are four cases of submerged surfaces on which the total pressure force and centre of pressure is to he determined. The submerged surfaces may bl! Vertical plane surface, 2. Horizontal plane surface, 3.
Inclined plane surface, and 4. Curved surface. Consider a plane vertical surface of arbitrary shape immersed in a liquid as shown in Fig. The total pressure on the surface may be determined by dividing the entire surface into a number of small parallel strips. The force on small strip is then calculated and the total pressure force on the whole area is calculated by integrating the force ori small strip.
Consider a strip of thickness dh and width b at a depth of h from free surface of liquid as shown in Fig. Centre of pressure is calculated by using the "Principle of Moments" , which states that the moment of the resultant force about an axis is equal to the sum of moments of the components about the same axis.
Substituting lc in equation 3. In equation 3. Hence from equation 3. Table 3. Problem 3. It lies in vertical plane in water. ToW pressure is given by equation 3. Find the position of centre ofpressure also. The forceFis acting at a distance of 4. Moment of this force about horizontal diameter X-X. Hence a torque of Nm must be applied on the disc in the clockwise direction. Proble m 3. The pressure at the centre of the pipe is The depth of C. Or centre of pressure is below the centre of the pipe by a distance of 0.
The distance of C. The opening is 2 m wide and 1. On the upstream of the gate, the liquid of sp. Fi,id also the force acting hori:. Assume A. May, that the gate is hinged at the bottom. F2 X 9. Let F is the force required on the top of the gate to open it as shown in Fig. Taking the moments ofF, F 1 and F2 about the hinge,. Find the total pressure and centre of pressure on the caisson if the water on the outside is just level with the top and dock is empty.
Depth of C. L of triangles about the C. A trapezoidal channel 2 m wide at the bollom and 1 m deep has side slopes 1 : 1. Determine : i the total pressure, a,nd ii the centre of pressure on the vertical gate closing the channel when it is full of water. Given : Width at bottom Depth, Side slopes :. Area of trapezoidal ABCD. The diago nals of the aperture are 2 m long and the tank contains a liquid of specific gravity 1.
Calculate the thrust exerted on the plate by the liquid and position of its centre of pressure. An immiscible liquid of sp. Calculate : ' i total pressure on one side of the tank, ii the position of centre of pressure for one side of the tank, which is 2 wide. It contains water for the lower 0.
The upper remaining parr is filled with oil of specific gravity. Calculate for one vertical side of the tank: a total pressure, and A. Given : Cubical tank of sides 1. This force will be acting at the C. Consider a plane horizontal surface itnmersed in a static fluid. Fahid scanned by. Find: Fig. Width of tank is 2 in. This is known as Hydrostatic paradox.
Consider a plane surface of arbitrary shape immersed in a liquid in such a way that the plane of the surface makes an angle with the free surface of the liquid as shown in Fig. Then is the axis perpendicular to the plane of the surface. Consider a small strip of area at a depth 'h' from free surface and at a distance y from the axis as shown in Fig. Pressure force, dF, on the strip, dF p x Area of strip pgh x dA.
Equating the two values given by equations 3. Determine the total pressure and position of centre of pressure wizen the upper edge is 1. Pro blem 3. Determine the total pressure force and position of centre of pressure, when the upper edge is 2 m below the free surface.
Delhi University, Dec. Determine the total pressure on one. Given: [Refer to Fig. Area of the given plate. Its greatest and least depths are below the surfaces being 2 metre and I metre respectively. Find: i the total pressure on front face of the plate, and ii the position of centre of pressure. Given : E Dia. The distAnce of centre of gravity from the free surface of the-water is given by. Then using equation 3.
To keep the gate in a stable position, a counter weight of kgf is attached at the upper end of the gate as shown in figure. Find the depth of water at. Neglect the weight of the gate andfriction at the hinge and pulley.
First fino tQ. The end A is hinged. Determine the force normal to the. B to open it. Civen :. Find the height h ofthe water so that the gate tips about the hinge. Take the width of the gate as unity. Solut ion. The gate will start tipping about hinge B if the resultant pressure force acts at B.
If the resultant pressure force passes through a point which is lying from B to C anywhere on the gate, the gate will tip over the hinge.
Hence limiting case is when the resultant force passes through B. But the resultant force passes through the centre of pressure. Hence for the given position point B becomes the centre of pressure. A rectangular sluice gate AB, 2 m wide and 3 m long is hinged at A as shown in Fig. It is kept closed by a weight fixed to the gate. The total weight of the gate and weight fixed to the gate is N. The centre of gravity of the weight and gate is at G.
The depth. Substituting this value in ii ,' we get. The base of the plate is parallel to water surface and at a depth of 2. Let dAis the area of a small strip at a depth of h from water surface. Hence integration of equation 3. The problem can. The total force in the x andy directions, i. Then total force on the cum! Resolving the force dF gi,en by equation 3. J hdA sin e rep.
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