Fluid Power i

7/31/2019 Fluid Power i

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IIISEM NAGPURINSTITUTEOFTECHNOLOGY,NAGPUR FP I

THE ABOVE QUESTIONS AND ANSWERS HAVE BEEN COMPILED FOR STUDENTS BENEFIT, AND ARE FORINTERNAL PRIVATE CIRCULATIONS ONLY, AND NOT FOR SALE. Page 1


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3RDSEM.MECHANICALENGG.,FLUIDPOWERI,UNITI.

Q. 1. Write a short note on

i) Newtons law of viscosity ii) Pascals law iii) Capillarity

i) Newtons law of viscosity:- It states that the shear stress on a fluid element layer isdirectly proportional to the rate of shear strain. The constant of proportionality is called the

co efficient of viscosity. Mathematically, it is expressed as given by equation.

ii) Pascals law :- It states that the pressure or intensity of pressure at a point in a state fluid

is equal in all directions, this is proved as:

The fluid element is of very small dimensions i.e., dx ,dy

and ds.

Consider an arbitrary fluid element of wedge shape in a

fluid mass at rest as shown in fig. let the width of the

element perpendicular to the plane of paper is unity and

px , py and pz are the pressures or intensity of pressure

acting on the face AB ,AC and BC respectively. Let , then the force acting on theelement are:

1) Pressure forces is normal to the surface2)

Weight to element in the vertical directionThe forces on the faces are:-

Force on the face AB = px area of face AB= pxdy 1

Similarly force on the face AC= pydx 1Force on the face BC = pz ds 1Weight of element = (Mass of element) g

= ( volume p)g = 1 Where

= density of fluid

Resolving the forces in x- direction, we have 1 1 sin90 0 1 1 0But ds cos = AB = dy 1 1 0


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Similarly, resolving the forces in y direction ,we get,

1 1 cos90 2 1 0 2 0

But ds

= dx and also the element is very small and hence weight is negligible.

0py = pzFrom equation, we have

px= py = pz

The above equation shows that, the pressure at any point in x,y, and z directions is

equal since the choice of fluid element was completely arbitrary, which means the

pressure at any point is the same in all directions.

iii) Capillarity:- Capillarity is defined as a phenomenon of rise or fall of a liquid

surface in a small tube relative to the adjacent general level of liquid when the tube isheld vertically in the liquid, the rise of liquid surface is known as capillarity rise, while

the fall of the liquid surface is known as capillarity depression. It is expressed in terms

of cm or mm of liquid. Its value depends upon the specific weight of the liquid,

diameter of the tube and surface tension of the liquid.

Expression for Capillary Rise:- Consider a glass tube of

small diameter d opened at both ends and is inserted in

a liquid, say water. The liquid will rise in the tube above

the level of the liquid.

Let h = height of the liquid in the tube. Under a state of

equilibrium the weight of liquid of height h is balanced

by the force at the surface of the liquid in the tube. But the force at the surface of the

liquid in the tube is due to surface tension.

Let = Surface tension of liquid = Angle of contact between liquid and glass tube.The weight of liquid of height h in the tube = (Area of tube x h)x x g

Where = density of liquidVertical component of the surface tensile force

= ( =

For equilibrium, equating the equations, we get, cos


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cos 4 4

between water and clean glass tube is approximately equal to zeroand hence cos is equal to unity, then rise of water is given by

4

Expression for capillary fall: if the glass tube is dipped in mercury, the level ofmercury in the tube will be lower than the general level of the outside liquid, as shown

in the figure.

let h= height of depression in tube

Then in equilibrium, two forces are acting on the

mercury inside the tube. First one is due to surface

tension acting in the downward direction and is equal to cos Second force is due to hydrostatics force acting upward

and is equal to intensity of pressure at a depth h x area

= p x

Equating the two, we get

cos 4 4

value of for mercury and glass tube is 1280.Q.2.

Write

ashort

note

on

variation

of

viscosity

in

case

of

gas

and

liquid.

Ans:-Variationof viscositywith temperature:- Temperature affects the viscosity.The viscosity of liquids decreases with the increase of temperature while the viscosityof gases increase with the increase of temperature. This is due to reason that theviscous forces in a fluid are to due cohesive forces and molecular momentum transfer.In liquids the cohesive forces predominates the molecular momentum transfer, due toclosely packed molecules and with the increase in temperature, the cohesive forces

decreases with the result of decreasing viscosity. But in case of gases the cohesive forceare small and molecular momentum transfer predominates. With the increase intemperature, molecular momentum transfer increases and hence viscosity increase.The relation between viscosity and temperature for liquids and gases are.i) For liquids, )

Where viscosity of liquidat t0 C, in poise


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viscosity of liquidat 00 C, in poise, are constant for the liquid

For water, 1.7910 , 0.03368 0.000221The equation show that with the increases of temperature, the viscosity decreases.ii) For a gas, where for air, 0.000017, 0.000000056, 0.1189 10The equation show that with the increase of temperature, the viscosity increases.Q.3. Explaintypesoffluids

Ans:- Types of fluid: The fluid may be classified into the following types,1)Ideal fluid2)Real fluid3)Newtonian fluid4)Non Newtonian fluid and5)Ideal plastic fluid

Ideal fluid:-A fluid, which is incompressible and ishaving no viscosity, is known as an ideal fluid. Ideal fluidis only an imaginary fluid as all the fluids, which exist,have some viscosity.Realfluid: - A fluid, which possesses viscosity, is known as real fluid. All the fluids, inactual practice are real fluids .Newtonianfluid:- A real fluid in which the shear stress is directly, proportional to therate of shear strain (or viscosity gradient), is known as a Newtonian fluid.NonNewtonianfluid:-

A real fluid, in which the shear stress is not proportional to therate of shear strain (or viscosity gradient), is known as a NonNewtonian fluid.Idealplasticfluid:- A fluid, in which is shear stress is more than the yield value andshear stress is proportional to the rate of shear strain (or viscosity gradient), is knownas a Ideal plastic fluid.


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Q.4.WhatdoyouunderstandbydimensionalHomogeneity?Whatareitsapplications?

Ans:-Dimensional Homogeneity means the dimensions of each term in an equation onboth the side are equal. Thus if the dimensions of each term on both the sides of an

equation are the same, the equation is known as dimensionally homogeneous. In otherwords the powers of fundamental dimensions (i.e. L, M, T) on both sides of the equationwill be identical for a dimensionally homogenous equation.Let us consider the equation p=whDimensions of L.H.S. = ML1T2Dimension of R.H.S = ML2T2 x L = ML1T2Dimension of LHS = Dimension of R.H.SEquation p=w h is dimensionally homogenous so it can be used in any system of units.Applicationofdimensionalhomogeneity:-1 It facilitates to determine the dimension of a physical quantity.2 It is helpful in checking whether an equation of any physical phenomenon isdimensionally homogenous or not?3 It assist in the conversion of units from one system to another.4 It produces a step towards dimensional analysis which is fruitfully employed

to plan experiments and to present the result meaningfully.Q.5.Whatarethemethods ofdimensionalanalysis?Ans:If the number of variable involve in a physical phenomenon are known, thenthe relation among the variables can be determined by the following methods1 Rayleighs method2 Buckinghams method

ExplainRayleighsmethodofdimensionalanalysiswithexample?

Ans:- The method is used for determining the expression for a variable whichdepend upon maximum three or four variables only. If the number ofindependent variable exceeds four then it becomes quite difficult to find theexpression for the dependent variable.Let V d be a variable, which depends on V1 ,V2 and V3 variables, then according toRayleighs method Vd is a function of V1 , V2 and V3 and mathematically it isexpressed as.


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, , ,, , Where K is constant and a, b, c, are arbitrary powers. The values of a, b and c areobtained by comparing the powers of the fundamental dimensions on both sides.Thus the expressions is obtained for dependent variable.

For example1) Drag force: f is a function of (i) Diameter, D, (ii)Velocity, Viii) Density, iv) Viscosity, Here, f is a dependent variable while diameter, velocity, density and viscosity areindependent variablesF = K Da Vb c d (1)

Where K is non dimensional factor.

According to Rayleighs method, the powers on both sides are equated Equating the powers of M, L and T on both sides.Power of M. 1 = c + dPower of L. 1 = a + b 3c d,Power of T. 2 = b dAs a such there are four unknowns (a, b, c , d) and number of equation are three, henceit is not possible to find the value of a, b, c and d. But three of them can be expressed interms of fourth variable which is most important. In this example, viscosity is playingan important role and hence a, b, c, are expressed in terms of d, which is the power toviscosity,

C = 1 d, b =2 dA = 1 b + 3 c + d = 1 2 + d +3 (1 d ) + d= 1 2+ d + 3 d + d = 2 dSubstituting these values of a, b and c, in (1) we get

F = k D2dV2d, 1d d= k D2 V2 (Dd Vd d d)= k D2 V2 )d


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Q.6.WhatisthedrawbackofRayleighsmethod?

Ans:- The Rayleighs method of dimensional analysis becomes more laborious if thevariable are more than the number of fundamental dimension (M.L.T.)ExplainBuckinghams

method.

Ans:- The drawback Rayleighs method is overcome by using Buckinghams metod,which states if there are n variable (both independent and dependent) in a physicalphenomenon and the variables contain m fundamental dimensions (M.L.T.)then thevariables are arranged into (n m)dimensionless terms. Each term is called as term.Let: Vd , V1, V2 ,V 3 Vn are the variable involved in a physical problem, let Vd bethe dependent variable and V1 ,V2Vn are the independent variable on which

Vd

depends then Vd

is a function of V1

, V2

Vn

and mathematically it is expressed asVd = (V1 , V2 , Vn) (1)Equation (1) can also be written as1 (Vd , V1 , V2 , Vn) = 0 (2)Equation (ii) is a dimensionally homogenous equation. It contains n variables. If thereare m fundamental dimensions (i.e. M L T) then according to Buckinghams theorem,equation (i) can be written in terms of number of dimension less groups of terms inwhich the number of terms is equal to (n m). Hence equation (ii) becomes. (1, 2 (nm))= 0Each of terms is dimension less and is independent of the system. Division ormultiplication by a constant does not change the character of the term. Each termscontain m+1.Where m is the number of fundamental dimensions and is also called a repeatingvariable. Let in the above case V1,V2 and V3 be the repeating variables if fundamentaldimension in (M.L.T.) = 3Then each terms is written as:


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Each equation is solved by the principle of dimensionless homogeneity and values of a1,b1, c1 etc. are obtained.These values are substituted in the above equationand values of , , are obtained. These values of are substituted in

equation (2) and the final equation for the phenomenon is obtained by expressing anyone of the terms as a function of other as , . . , . .

Q.7.Writeashortnoteonsinglecolumnmanometer.

Ans:-Singlecolumnmanometer:-

Single column manometer is a modified form of a Utube manometer in which a reservoir, having a large cross sectional area (about 100times) as compared to the area of the tube is connected to one of the limbs (say leftlimb) of the manometer as shown in fig. due to large crosssectional area of thereservoir, for any variation in pressure, the change in the liquid level in the reservoirwill be very small which may be neglected and hence the pressure is given by the heightof liquid in the other limb. The other limb may be vertical or inclined. Thus there aretwo types of single column manometer as:1

Vertical single column manometer2 Inclined single column manometerVerticalsinglecolumnmanometer:-Fig. shows the vertical single column manometer, let XX be the datum line in thereservoir. When the manometer is connected to the pipe, due to high pressure at A,the heavy liquid in the reservoir will be pushed downward and will rise in theright limb.

Let h = Fall the heavy liquid in reservoirH2 = rise of heavy liquid in right limbH1 = Height of centre of pipe above XXPA = pressure at A, which is to be measuredA = cross sectional area of the reservoira = cross sectional area of the right limb


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S1 = Sp.gr. of liquid in pipeS2 = Sp. gr. Of heavy liquid in reservoir and right limb. = Density of liquid in pipe. = Density of liquid in reservoir.Fall of heavy liquid in reservoir will cause a rise of heavy liquid level in the rightlimb. A x h = a x h2

Now consider the datum line YY as shown in fig. Then pressure in the right limbabove YY Pressure in the left limb above YY + Equating these pressures, we have

+

[ As there area A is very large as compared to a, hence ratio a/A becomes verysmall and can be neglected.

Then, From above equation, it is clear that as a h1 is known and hence by knowing h2 orrise of heavy liquid in the right limb the pressure at A can be calculated.


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Q. 8. A plate 0.025mm distant from a fixed plate,moves at 60 cm/s and

requires a force of 2 N per unit area i.e. 2 N/m2 tomaintain this speed.

Determinethefluidviscositybetweentheplates.

Ans: Distance between plates, dy = 0.025 mm

= 0.025 X 103

mVelocity of upper plate, u =60cm/s =0.6 m/sForce on upper plate 2.0 This is the value of shear stress i.e. Let the fluid viscosity between the plates is

we have,

Where = Change of velocity = 0 0.60 / = Change of distance = 0.25 x 103m = Force per unit area = 2.0 2 0.600.02510 , 2.00.02510

0.60 8.33 10

8.33 10 10 . Poise.


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3RDSEM.MECHANICALENGG.,FLUIDPOWERI,UNITII.

Q.1. Arectangularplanesurface2mwideand3mdeepliesinwaterinsucha

way that its plane makes an angle of 300 , with the free surface of water.

Determinethetotalpressureandpositionofcenterofpressurewhentheupper

edgeis1.5mbelowthefreewatersurface.

Ans:Width of plane surface, b=2m

Depth, d= 3mAngle, =300Distance of upper edge from free water surface = 1.5 m

i)

Total pressure force is given by equation asF =gAhWhere =1000kg/m3A = b x d = 3 x2 = 6 m2 depth of C.G. from free water surface

1.5 1.5 sin 30

1.5 30 1.5 1.5 30=1.5 + 1.5 x = 2.25 mF = 1000 x 9.81 x 6 x 2.25 =132435 N

ii) Centre of pressure (h*)Using equation, we have h + , where I 4.5m

h 4.5sin3062.25 2.25 4.5 1462.25 2.25= 0.0833 + 2.25 = 2.3333m.


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Q.2. Aplatehavinganareaof0.8m2isslidingdowntheinclinedplaneat300to

thehorizontalwithavelocityof0.4m/s, there isacushionof fluid2mmthick

betweentheplaneandtheplate.Findtheviscosityofthefluidiftheweightofthe

planeis300N.

Ans:-Area of plate, A =0.8 m2Angle of plane, = 300Weight of plate, w=300 NVelocity of plate, u = 0.4 m/sThickness of fluid film, t = dy =2 mmt = 2 x 10 3m

Let the viscosity of fluid between plate and inclined plate is , component ofweight, along the plane = w cos 60= 300 cos 60 = 150 NThus the shear force, F, on the bottom surface of the plate = 150 NAnd shear stress, . 1875 N/m2Now, we know,

Where du = change in viscosity= u 0 = 0.4 m/sdy = t = 2 x 103m

1875 = . = 0.9375 Ns/m2

= 9.375 poise.


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Q.3.Fig.showsagatehavingaquadrantshapeofradius2m.Findtheresultant

forceduetowaterpermeterlengthofthegate.Findalsotheangleatwhichthe

totalforcewillact.

Ans:- Radius of gate =2m

Width of gate= 1mHorizontalforceFx = force on the projected area of thecurved surface on vertical plane

= Force on BO = Where A = Area of BO = 2 1 2,

2 1;Fx = 1000 x 9.81 x 2 x 1 = 19620 NThis will act at a depth of 2 m from free surface of liquid

VerticalforceFy

Fy= Weight of water (imagined) supported by AB

= x Area of AOB x 1.0= 1000 x 9.81 2 1.0 30819 This will be act at a distance of = 0.848 m from OBResultant force, F is given by,

F =

= 19620 30819 384944400949810761=36534.4NTheangle made by the resultant with horizontal is given bytan = 1.5708 1.5708 5731


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Q. 4. Determine the total pressure and centre of pressure on an isosceles

triangularplateofbase4mandaltitude4mwhenitisimmersedverticallyinan

oilofsp.gr. 0.9. Thebaseoftheplatecoincideswiththefreesurfaceofoil.

Ans:-

Base of plate, b = 4 mHeight of plate, h = 4 mArea, 8.0 Sp.gr. of oil, S = 0.9

Density of oil, = 900 kg/m3The distance of C.G. from free surface of oil,

13 13 4 1.33 Total pressure(F) is given by, F = = 900 x 9.81 x 8.0 x 1.33N = 9597.6 N.

Centre of pressure (h*) from free surface of oil is given by,

Where = M. I. of triangular section about its C. G.= 7.11 7.118.01.33 1.33 0.6667 1.33 1.99 .


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Q. 5. A rectangular pontoon 10.0 m long, 7m broad and 2.5m deep weighs

686.7kN. Itcarriesonitsupperdeekanemptyboilerof5.0mdiameterweighing

588.6 kN. The centre of gravity of the boiler and the pontoon are at their

respective centre along a vertical line. Find the meta-centric height. Weight

densityofseawateris10.104kN/m3.

Ans: Dimension of pontoon =10 x 7 x 2.5Weight of pontoon, W1= 686.7 kNDia .of boiler, D = 5.0 mWeight of boiler, W2 = 588.6 kNwfor sea water = 10.104 kN/m3To find the metacentric height, first determine the commoncentre of gravity G and common center of buoyancy B of theboiler and pontoon. Let G1 and G2 are the centre of gravities ofpontoon and boiler respectively. Then,

2.52 1.25 2.5 5.02 2.5 2.5 5.0

The distance of common centre of gravity G from A is given as . . . .. . 2.98

Let h is the depth of immersion. Then,Total weight of pontoon and boiler = Weight of sea water displaced

(686.7 + 588.6) =w

x Volume of the pontoon in water= 10.104 x L x b x Depth of immersion 1275.3 = 10.104 x 10 x7 x h 1275.310710.104 1.803


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The distance of the common centre of buoyancy B from A is 2 1.8032 0.9015

BG = AG AB = 2.98 0.9015 = 2.0785m = 2.078m

Metacentric height is given by Where l= M.I. of the plan of the body at the water level along y y

= 10.07 Volume of the body in water= L x b x h = 10.0 x 7 x 1.857

1 10497121071.857 49121.857 2.198 2.198 2.078 0.12 Metacentric height of both the pontoon and boiler =0.12m.


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3RDSEM.MECHANICALENGG.,FLUIDPOWERI,UNITIII.

Q.1.Stateandexplaintypesoffluidflow.

Thefluidflowisclassifiedas:

(1) Steadyandunsteadyflows(2) Uniformandnonuniformflows(3) Laminarandturbulentflows(4) Compressibleandincompressibleflows(5) Rotationalandirrotationalflows; and(6) One,twoandthreedimensionalflows.

Steadyandunsteadyflows:- Steady flow is defined as that type of flow in which fluidcharacteristics like viscosity, pressure, density, etc. at a point do not change with time.Thus for steady flow, mathematically, we have,, , 0, , , 0, , , 0,

Where (, , ) is a fixed point in fluid field.Unsteady flow is that type of flow, in which the velocity, pressure or density at a pointchanges with respect to time. Thus, mathematically, for unsteady flow:

, ,

0,

, ,

0, .Uniform andNon uniform flow:- Uniform flow is defined as that type of flow inwhich the velocity at any given time does not change with respect to space(i.e. length ofdirection of the flow). Mathematically, for uniform flow:

0Where Change of velocity

Length of flow in the direction S.Non uniform flow is that type of flow in which the velocity at any given time changewith respect to space. Thus, mathematically, for uniform flow: 0


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LaminarandTurbulentflows:Laminar flow is defined as that type of flow in whichthe fluid particles move along well defined paths or stream line and all the stream linesare straight and parallel. Thus the particles move in laminas or layers gliding smoothlyover the adjacent layer. This types of flow is also called as streamline flow or viscousflow.

Turbulent flow is that type of flow in which the fluid particles move in a zigzag way.Due to the movement of fluid particles in a zigzag way, the eddies formation takesplace which are responsible for high energy loss. For a pipe flow, the type of flow isdetermined by a non dimensional number called the Reynoldss numberThere D= Diameter of pipe.V =Mean velocity of flow in pipeAnd v= Kinematic viscosity of fluidIf the Reynoldss number is less than 2000, the flow is called laminar. If the Reynoldssnumber lines between 2000 and 4000, the flow may be laminar or turbulentCompressibleand Incompressible flow: Compressible flow is that type of flow inwhich the density of the fluid changes from point to point or in other words the density() is not constant for the fluid. Thus, mathematically, for Compressible flow: constant

Incompressible flow is that type of flow in which the density is constant for the fluidflow. Liquids are generally incompressible while gases are compressible. ThusMathematically, For incompressible flow: = constantRotational and Irrotational flows :- Rotational flow is that type of flow in which the fluid

particles while flowing along stream lines, also rotate about their own axis, and if the fluid

particles while flowing along stream lines, do not rotate about their own axis, that type of

flow is called irrotational flow.

One, two and three Dimensional flows:- One dimensional flow is that type of flow inwhich the flow parameter such as a velocity is a function of time and one space coordinate

only, say x. For a steady one dimensional flow, the velocity is a function of one space co-

ordinate only. The variation of velocities in other two mutually perpendicular directions is

assumed negligible. Hence mathematically, for two dimensional flow:

u , 0 , Where u, v and w are velocity components in x, y and z directions respectively.


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Two dimensional flow:- Is that type of flow in which the velocity is a function of time and

two rectangular space co-ordinates say x and y. For a steady two dimensional flow the

velocity is function of two space coordinates only. The variation of velocity in the third

direction is negligible. Thus, mathematically for two dimensional flow:

, , , , Three dimensional flow:- Is that type of flow in which the velocity is a function of time andthree mutually perpendicular directions. But for a steady three dimensional flow the fluid

parameters are functions of three space co-ordinates(x, y and z) only. Thus, mathematically,

for three dimensional flow:

,,, ,,, ,,

Q.2. Explain the following

A) Vortex flow:- Vortex flow is defined as the flow of a fluid along a curve path or the flow

of rotating mass of fluid is known as, vortex flow. The vortex flow is of two types namely:

1 Forced vortex flow and2 Free vortex flow

1. Forced vortex flow:- Forced vortex flow is defined as that type of vortex flow, in which

some external torque is required to rotate the fluid mass. The fluid mass in this type of flow

rotates at constant angular velocity,

. The tangential velocity of any fluid particles is given

by:

V = x rWhere, r = Radius of fluid particles from the axis of rotation

Hence angular velocity is given by:


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Examples of forced vortex are:-

1 A vertical cylinder containing liquid which is rotated about its central axis witha constant angular velocity, as shown in fig.

2 Flow of liquid inside the impeller of a centrifugal pump.3 Flow of water through the runner of a turbine.

2. Free vortex flow:- When no external torque is required to rotate the fluid mass, that typeof flow is called free vortex flow. Thus the liquid in case of free vortex is rotating due to the

rotation which is imparted to the fluid previously. Examples of the free vortex flow are.:

1 Flow of the liquid through a hole provided at the bottom of a container.2 Flow of liquid around the circular bend in a pipe.3 A whirlpool in a river.4 Flow of fluid in a centrifugal pump casing.

The relation between velocity and radius, in free vortex is obtained by putting the value of

external torque equal to zero, or, the time rate of change of angular momentum, i.e. momentof momentum must be zero. Consider a fluid particles of mass m at a radial distance r

from the axis of rotation being a tangential velocity, v. Then:

Angular momentum = mass x velocity = m x v

Moment of momentum = momentum x r = m x v x r

Time rate of change of angular momentum = For free vortex Integrating, we get: mvr = constant or vr =

Q.3. Explain the following:

Eulers Equation of motion:- This is equation of motion in which the forces due to gravity

and pressure are taken into consideration. This is derived by considering the motion of a

fluid element along a stream line as:

Consider a stream line in which flow as taking place in S direction as shown in figure.

Consider a cylindrical element of cross section dA and length dS. The forces acting on the

cylindrical element are:

1 Pressure force pdA in the direction of flow.


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2 Pressure force ( dA opposite to the direction of flow.3 Weight of element

Let is the angle between the direction of flow and the line of action of the weight ofelement.

The resultant force on the fluid element in the direction of s must be equal to the mass of

fluid element x acceleration in the direction s.

pdA - ( dA- pgdAds.cos = p dA ds x as

Where as, is the acceleration in the direction of s,

now , where v is a function of s and t,= Ifthe flow is steady, 0

as = Substituting the value of as in equation and simplifying the equation , we get,

Dividing by , cos

cos 0Butfromfigure,wehave: cos

1

0

0

0Equation is known as Eulers equation of motion


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Q.4.DerivetheBernoullisequationfromEulersEquation.

Bernoullis equation is obtained by integrating the Eulers equation of motion as

If flow is incompressible, is constant and = constantOr

2

This equation is a Bernoullis equation in which,= pressure energy per unit weight of fluid or pressure head ./2 = Kinetic energy per unit weight or kinetic head.Z = potential energy per unit weight or potential head.Assumptions:The following are the assumptions made in the derivation of Bernoullis equation:i) The fluid is ideal, i. e., viscosity is zero ii) The flow is steadyiii) The flow is incompressible iv) The flow is irrotational.

Q.5.Waterisflowingthroughapipeof5cmdiameterunderapressureof29.43

N/cmandwithmeanvelocityof2.0m/s.Findthetotalheadortotalenergyper

unitweightofthewateratacrosssection,whichis5mabovethedatumline.

Ans:Diameter of pipe = 5 cmPressure = p =29.43 N/cm 2 = 29.43 X 104 N/m2Velocity v = 2.0 m/sDatum head, z=5 m


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Total head = pressure head + kinetic head + datum headPressure head .. 30 1000 /Kinetic head, . 0.204

Total head 30 0.204 5 . .


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3RDSEM.MECHANICALENGG.,FLUIDPOWERI,UNITIV.

Q.1.WriteashortnoteonPitottube.Pitottube: It is a device used for measuring the velocity of flow at any point in a pipeor channel. It is based on the principle that, if the velocity of flow at a point becomes

zero, the pressure there is increased due to the conversion of the kinetic energy intopressure energy. In its simplest form, the pitot tube consist of glass tube, bent at rightangles as shown in figure.The lower end, which is bent through 900 is directed in the upstream direction asshown in figure. The liquid rises up in the tube due to the conversion of kinetic energyin to pressure energy.The velocity is the determined by measuring the rise of liquid inthe tube.Consider two points (1) and (2) at the same level in such a way that point (2) is just asthe inlet of the pitot tube and point (1) is far away from the tube.Let, P1 = intensity of pressure at point(1)

v1 = velocity of flow at (1)P2 = pressure at point (2)v2 = velocity at point (2), which is zero.H = depth of tube in the liquid.h = rise of liquid in the tube above the free surface.Applying Bernoullis equations points (1) and (2) ,we get:

2

2

But z1 = z2 ,as point (1) and (2) are on the same line and v2 =0 1 2

Substituting these values ,we get, 2 ,

2 2


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This is theoretical velocity, actual velocity is given by 2

Where efficient of pitot tube velocity at any point v = 2Velocity

of

flow

in

apipe

by

pitot

tube:

For finding the velocity at any point in a pipeby pitot tube. The following arrangements are adopted.1 Pitot tube along with a vertical piezometer tube as shown in fig.2 Pitot tube connected with piezometer tube as shown in fig.

3

Pitot tube and vertical piezometer tube connected with a differential U tubemanometer as shown in figure.4 Pitot static tube, which consist of two circular concentric tubes one inside theother with some space in between as shown in fig. The outlet of these twotubes are connected to the differential manometer where the difference ofpressure head h is measured by knowing the difference of the levels of themanometer liquid say x,Then 1


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Q.2. Explainthephenomenonofdischargeoverarectangularnotchorweir.

Ans:Dischargeoverarectangularnotchorweir:

The expression for discharge over a rectangular notch or weir is the same.

Consider a rectangular notch or weir provided in channel carrying water as shown infig:

Let, H = Head of water over the crestL = length of the notch or weir.For finding the discharge of water flowing over the weir or notch, consider anelementary horizontal strip of water of thickness dh and length L at a depth h from freesurface of water as shown in figure.The area of strip = L x dh

And theoretical velocity of water flowing through strip = 2The discharge dQ, through strip is:dQ = Cd x area of strip x theoretical velocity= 2

Where Cd = Co efficient of discharge,The total discharge Q, for the whole notch or weir is determined by integratingequation (i) between the limits 0 and H.Q = 2 . 2 2 = 2

//

= 2/


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Q.3.Determinetheheightofarectangularweiroflength6mtobebuiltacrossa

rectangularchannel.Themaximumdepthofwateron theupstreamsideof the

weir is 1.8 m and discharge is 2000 liters/s. Take =0.6 and neglect endcontractions.

Ans: Length of weir = L = 6mDepth of water = H1 = 1.8mDischarge Q = 2000 lit/s = 2m3/s 0.6Let, H is height of water above the crest of weir, and H2 = height of weirThe discharge over the weir is given by the equation as:

Q =

2/

2.0= 0.66.0 29.81/ 10.623/ 2.010.623 H = ..2/3 = 0.328m Height of weir, H2 = H1 H= depth of water on upstream side H

= 1.8 0.328 = 1.472m.Q.4.Writeashortnoteonthefollowing:-

i)Slidingvanetypemeter ii)Rotameter

iii)Turbinemeter

Ans:- i) Sliding vane typemeter:- This type of meter has an accurately machinedbody having a rotor with four evenly spaced slots which form the guides for the vanes.The metered liquid entering the inlet revolves the rotor and the vanes around a camcausing the vanes to move radically. The vane nearest to the inlet port begins to moveoutwards and becomes fully extended at point A. The vane ahead at point B is alreadyfully extended and thus a measuring chamber of known volume is formed between the


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two vanes. A continuous series of chambers at the rate of four per revaluation areformed which delivers the flow at outlet.

This type of meter has low pressure drop and can give an accuracy of the order of 0.2 % of one measured values .ii)Rotameter:- A rotameter is made up of a vertical tube with tapered bore In which afloat assumes a vertical position corresponding to each flow rate through the tube.For a given flow rate, the float remains stationary since the vertical forces of differentialpressure, gravity, viscosity and buoyancy are balanced. This balance is self maintainingsince the meter flow area i.e. the annular area between the float and tube, variescontinuously with vertical displacement. Thus the device may be considered as anorifice of adjustable area. The downward force i.e. gravity force minus buoyancy forceis constant and so the upward force (i.e. pressure drop multiplied by float crosssectionarea) must be constant also. For a fixed flow area varies the square of flow rate, andso to keep constant for different flow rates, the area must very. The tapered tubeprovides this variable area. The float position is the output of the meter and can bemade essentially linear with the flow rate by making the tube area vary linearly withthe vertical distance.Let Q = volume flow rate, m3/s (actual)

Cd = coefficient of dischargeAt = Area of tube, m2Af = area of float, m2g = acceleration due to gravity, m/s2Vf= volume of float


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wf=specific weight of float, N/m3wff= Specific weight of flowing fluid, N/m3ThenQ=

2

The floats of rotameters may be made up of various materials to obtain the desiredensity difference i.e. (wf wff) for metering a particular liquid or gas. Some floatshapes, such as spheres, require no guiding in the tube, others are kept central by guidewires or by internal ribs in the tube. Floats shaped to induce turbulence can giveviscosity insensitivity over a 100 : 1 range. The tubes often are made of high strengthglass to allow direct observation of the float position. Where greater strength isrequired, metal tubes can be used and the float position detected magnetically throughthe metal wall. If a pneumatic or an electrical signal related to the flow rate is desired,the float motion can be measured with a suitable displacement transducer.iii)Turbinemeters:- If a turbine wheel is placed in a pipe containing a flowing fluid,its rotary speed depends on the flow rate of the fluid. By reducing bearing friction andkeeping other losses to a minimum, one can design a turbine whose speed varieslinearly with flow rate. Thus a speed measurement allows a flow rate measurement.The speed can be measured simply and with great accuracy by counting the rate at

which turbines blades pass a given point, using a magnetic proximity pickup to producevoltage pulses. By feeding these pulses to an electronic pulse rate meter, one canmeasure flow rate, by accumulating the total number of pulses during a time interval,the total flow is obtained. This measurement can be made very accurately because oftheir digital nature.If an analog voltage signal is desired, the pulses can be fed to a frequency to voltageconverter.


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Dimensional analysis of the turbine flow meter shows that the following relation holdsgood if bearing friction and shaft power output is neglected.

Where ,Q= volume flow rate, m3/sn= rotor angular velocity r/sD= meter bore diameter, mv= kinematic viscosity m2/sIn actual practice, the effect of viscosity is limited mainly to low flow rates, with high

flow rates being in the turbulent region where viscosity effects are secondary. Fornegligible viscosity effects, a simplified analysis based on strictly kinematic relationshipgives the following result. 1 1 Where L=rotor lead m Dh/D

Db= rotor blade tip diameter,mDh= rotor hub diameter, mt= rotor blade thickness, mQ.5. A venturimeter with inlet and throat diameter 200 cm. and 10 cm. isinsertedinaverticalpipecarryingofsp.gravity0.8,theflowofoilisinupward

direction.Thedifferenceoflevelsbetweenthethroatandinletsectionis50cm.

Theoilmercurydifferentialmanometer gives a readingof30 cm. ofmercury.

Findthedischargeofoil.Neglectlosses.

Ans:-Diameter at inlet, d1 = 20 cm 4 20 314.16 Diameter at throat d2 = 10cm 4 4 10 78.54


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Sp. gravity of oil, S0 =0.8Sp. gravity of mercury, Sg =13.6Differential manometer reading, x=30 cm.

1

30 13.60.8 1 30171 30 16 480 .Cd=1.0The discharge, Q= Cd 2

= . .

.

.

2980480=78725.75 cm3/s =78.725lit/s.


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3RDSEM.MECHANICALENGG.,FLUIDPOWERI,UNITV.

Q.1. A shaftof100mmdiameter rotatesat60 rpm ina200mm longbearing.

Takethatthetwosurfacesareuniformlyseparatedbyadistanceof0.5mmand

takinglinearvelocitydistributionsinthelubricatingoilhavingdynamicviscosity

of4centipoises,findthepowerabsorbedinthebearing.Ans:

- Dia. Of a shaft D = 100mm = 0.1mLength of bearing, L = 200mm = 0.2 mt = 0.5 mm = 0.5 x103m

4 0.04 0.0410 N =60 r. p. m.Using the equation , . ... = 4.961x10-2W.

Q.2. Asleeve, inwhichashaftofdiameter75mm, isrunningat1200r.p.m., is

havingaaxialclearanceof0.1mm.Calculatethetorqueresistanceifthelengthof

sleeveis100mmandthespaceisfittedwithoilofdynamicviscosity0.96poise.

Ans:Dia. Of shaft = D=75 mm =0.075 mN=1200 r. p. m.t=0.1mm =0.1 x 103mL=100mm =0.1m

0.96 .

Tangential velocity of shaft, . 4.712/Shear stress . .. 4523.5 /Shear force, F=

=4523.5 0.075 0.1 106.575


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Torque resistance, = = 106.575 . =3.996Nm.Ans.

Q.3. The external and internal diameters of a collar bearing are 200mm and

150mmrespectively.

Between

the

collar

surface

and

the

bearing,

an

oil

film

of

thickness0.25mmandofviscosity0.9poise,ismaintained.Findthetorqueand

thepower lost inovercomingtheviscousresistanceoftheoilwhentheshaft is

runningat250r.p.m.

Ans:- D2 = 200mm = 0.2m R2 = . 0.1

D1=150 mm= 0.15m R1 = . 0.075t=0.25mm=0.00025m

0.9 . Torque required is given by equation

60 0.910 2500. 1 0.075600.00025 =14804.4[0.0001 0.00003164]=1.0114Nm Power lost in viscous resistance

= . =26.48 W.Q.4.Waterisflowingthrougha200mmdiameterpipewithcoefficientoffriction

f=0.04.Theshearstressatapoint40mm fromthepipeaxis is0.00981N/cm2.

Calculatethe

shear

stress

at

the

pipe

wall.

Ans: Dia. Of pipe, D=200mm =0.20mCoefficient of friction, f= 0.04Shear stress at r = 40mm, 0.00981 N/cm2Let the shear stress at pipe wall =


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First find weather the flow is viscous or not. The flow will be viscous if Reynoldsnumber R1 is less than 2000.Using equation, we get or 0.04 . 400

This means flow is viscous. The shear stress in case of viscous flow though a pipe isgiven by a equation as: 2

But is constant across a section. Across a section, there is no variation of x andthere is variation of p

At the pipe wall, radius =100 mm and shear stress is . . =0.0245N/cm2.

Q.5.Explainboundarylayerconceptandwallshear.

Ans:- When a real fluid flows over a solid wall, the fluid particles closed to theboundary get adhered to the boundary and as a result of this condition of no slipoccurs. In other words the velocity of fluid close to the boundary will be the same asthat of the boundary, i.e. if the boundary is stationary the velocity of fluid at theboundary will be zero. As we move further away from the boundary, the velocity will behigher and as a result of this variation of velocity, the velocity gradient will exist.Thus the velocity of fluid increase from zero velocity on the stationary boundary to freestream velocity (v) of the fluid in the direction normal to the boundary. This variationof velocity from zero to free stream velocity in the direction normal to the boundarytakes place in a narrow region in the vicinity of solid boundary. This narrow region ofthe fluid is called boundary layer.Hence according to the concept of boundary layer, the flow of fluid in the neighborhoodof the solid boundary may be divided into two regions.


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Region:-01: A very thin layer of the fluid called the boundary layer, in the immediateneighborhood of the solid boundary, where the variation of velocity from zero at thesolid boundary to the free stream velocity in the direction normal to the boundarytakes place.In this region, the velocity gradient exist and hence the fluid exerts a shear stress onthe wall (wall shear) in the direction of the motion. The value of shear stress is givenby: Region:-02: The remaining fluid, which is outside the boundarylayer . The velocityoutside the boundary layer is constant and equal to free stream velocity. As there is novariation of velocity in this region, the velocity gradient becomes zero, as a result ofthis the shear stress is also zero.Q.6.Explaintheforceexertedbyaflowingfluidonastationarybody.Ans: As shown in figure below consider a body held stationary in a real fluid, which isflowing at a uniform velocity U.The fluid exerts a force on the stationary body. The total force(FR) exerted by the fluidon the body is perpendicular to the surface of the body. Thus the total force is inclinedto the direction of motion. The total force a can be resolved in two components, one inthe direction of motion i.e. drag force and other perpendicular to the direction ofmotion i.e. lift forceDrag:- The component of the total force (FR) in the direction of motion is called drag.This component of force is denoted by FD. Thus drag is the force exerted by the fluid inthe direction of motion.Lift:- The component of the total force (FR) in the direction perpendicular to thedirection of motion is known as lift. This is denoted by FL. Thus lift is the force exerted

by the fluid in the direction perpendicular to the direction of motion. Lift force occursonly when the axis of the body is inclined to the direction of fluid flow. If the axis of thebody is held parallel to the direction of fluid flow, lift force is zero. In this case only dragforce acts.


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3RDSEM.MECHANICALENGG.,FLUIDPOWERI,UNITVI.

Q.1.WhatdoyouunderstandbyHydraulicgradientlinesandTotalenergylines?

Ans: Hydraulic gradient lines and Total energy lines are very useful in the analysis offluids through pipes.

1)Hydraulicgradientlines:-

Hydraulic gradient lines is defined as the line whichrepresents the sum of pressure head (P/W) and datum head (z) of the flowingfluid in a pipe with respect to some reference line or it is the line which isobtained by joining the top of all vertical ordinates, showing the pressure head(p/w) of a flowing fluid in a pipe from the centre of the pipe. It is briefly writtenas H.G.L.2)Totalenergy lines:- Total energy lines is defined as the line which representsthe sum of pressure head and datum head and kinetic head of a flowing fluid in

the pipe with respect to some reference line. It may also be defined as the linewhich is obtained as the result of joining of all the tops of vertical ordinatesshowing the sum of pressure head and kinetic head from the centre of the pipe. Itis briefly written on T.E.L.Q.2.Writeashortnoteonflowthroughsyphon.

Ans:- Syphon is nothing but a long bent pipe specifically used to transfer liquid froma reservoir at a higher elevation to another reservoir at a lower level when the tworeservoirs are separated by a hill or high level ground as shown in fig. below.

The point c is at the highest of the siphon is called the high summit. As the point c isabove the free surface of the water in the tank A, the pressure at c will be less thanatmospheric pressure. Theoretically the pressure at c may be reduced to 10.3m ofwater but in actual practice this pressure is only 7.6m of water or 10.3 7.6 = 2.7mof water absolute. If the pressure at c becomes less than 2.7m of water absolute, thedissolved air and other gases would come out from water and collect at the summit.The flow of water will be obstructed.


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Siphonisusedinthefollowingcases-

1)When it is required to carry water from one reservoir to another reservoirseparated by a hill or ridge.2)Whenever it is required to take out the liquid from a tank which is not having anyoutlet.3)In order to empty a channel when it is not provided with any outlet sluice.

Q.3.Writeashortnoteonwaterhammerinpipes.

As shown in fig. consider a long pipe AB, whose one end is connected to a tankcontaining water at a height H from the center of the pipe, while at the other end of thepipe a valve is provided in order to regulate the flow of water. When the wall iscompletely open, the water is flowing with a velocity, V in the pipe. But if now the valveis suddenly closed, the momentum of the flowing water will be destroyed andconsequently a wave of high pressure will be set up. These waves of high pressure willbe transmitted along the pipe with a velocity equal to the velocity of sound wave andmay create noise called knocking. Also these waves of high pressure have the effect ofhammering action on the wall of the pipe and hence it is also known as a waterhammer.The magnitude of pressure rise as a result of water hammer depends upon thefollowing factors.

1)The velocity of flow of water in pipe.2)The length of pipe.3)Time taken to close the valve.4)Elastic properties of the material of the pipe.

Q.4.Tworeservoirsareconnectedbyapipelineofdiameter600mm,andlength

400mm.Thedifferenceofwaterlevelinthereservoiris20m.atthedistanceof

1000mfromthereservoir,asmallpipeisconnectedtothepipeline.Thewater

canbetakenfromthesmallpipe.findthedischargetothelowerreservoir,if

i)

no

water

is

taken

from

the

small

pipe,

and

ii)100lit/sofwateristakenfromsmallpipe.

Takef=0.005andneglectminorlosses.

Soln. dia. of pipe d = 600mm = 0.60mLength of pipe L = 400 m


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Difference of water level, H = 20m, f = 0.005

i) Nowateristakenfromthesmallpipe.The head loss due to friction in pipe AB = 20 4 0.005 4000 0.6 2 9.81

. . . = 2.94 = 1.715 m/s

Discharge, Q = area x V = 0.6 x 1.715 = 0.485 (ans.)ii) 100lit/sofwateristakenfromsmallpipe.

Let, Q1 = Discharge through pipe ACQ2 = Discharge through pipe CB

Then for parallel pipes, Q1 = Q2 + 100 lits / s = Q2 + 0.1

Q2 = (Q1 0.1) Length of pipe AC, L1 = 1000mLength of pipe CB, L2 = 4000 1000 = 3000 mApplying Bernoullis equation to points E and F and taking flow through ABC, wehave

20 =

+

Where V1 = velocity through pipe AC = .d1 = dia. Of pipe AC = 0.6Velocity though pipe CB . .


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d2 = Dia. of pipe. CB=0.6Substituting these values in equation (ii), we get20 . . . .

. . . .

20=21.25 63.75 Q2=Q10.1 or Q1=Q2+0.1Substituting the value of Q1 in equation, we get :

20=21.25(Q2 + 0.1)2 +63.75 Q22=21.55[Q22 +0.01+0.2 Q2] +63.75 Q22=21.55 Q22 +0.2125 +4.250 Q2 +63.75 Q22= 85 Q22 +4.25 Q2 +21.2585 Q22 +4.25 Q219.7875 =0This is a quadratic equation in Q2

4.25 4.25 48519.7875285

4.25 18.06256727.75170 44.2582.13170 .. =0.458m3sDischarge to lower reservoir = =0.458m3/s .

Q.5. EXPLAIN POWERTRANSMISSIONTHROUGHPIPES.

Power is transmitted through pipes by flowing water or other liquids flowing throughthem. The power transmitted depends upon (i) the weight of liquid flowing through thepipe and (ii) the total head available at the end of the pipe. Consider a pipe ABconnected to a tank as shown in fig. the power available at the end B of the pipe andthe condition for maximum transmission of power shall be obtain as mentioned below.Let, L=length of the pipe ,

d=diameter of the pipe.


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H=total head available at the inlet of pipe.V=velocity of flow in pipe,hf=loss of head due to friction, and f =coefficient of friction.

The head available at the outlet of the pipe, if minor losses are neglected= total head at inlet loss of head due to friction Weight of water flowing through pipe per sec.W = volume of water per sec = area velocity

4

The power transmitted at the outlet of the pipe.= weight of water per sec x head of outlet. watts

Power transmitted at outlet of the pipe.

1000

4

4

2 Efficiency of power transmission

=

Condition for maximum transmission of power:- The condition of powertransmission of power is obtained by differentiating equation with respect to V andequating the same to zero.

Thus 0


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1000 4 4

2 0 =0

3 4 2 0 3 0 3

The above equation is the condition for maximum transmission of power. It states thatpower transmitted through a pipe is maximum when the loss of head due to friction isone third of the total head at inlet.MaximumEfficiencyoftransmissionofpower: Efficiency of power transmission

through pipe is given by equation as For maximum power transmission through pipe the condition is given by equation as 3Substituting the value of hfin efficiency, we get maximum /3 1 13 23 66.7%

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