fluid mech. chapter6

Fluid Mechanics Fluid Mechanics Chapter 6Chapter 6

Energy Equation and its ApplicationsEnergy Equation and its Applications

FOSTEMFOSTEMINTI International UniversityINTI International University

Energy of a Flowing FluidEnergy of a Flowing Fluid For an element of fluid whose weight W = mgFor an element of fluid whose weight W = mg

a) Potential energy = mgza) Potential energy = mgz (or) (or) Potential energy/wt = zPotential energy/wt = z

b) Kinetic energy = ½ mvb) Kinetic energy = ½ mv22 (or) (or) Kinetic energy/wt = vKinetic energy/wt = v22 /2g /2g

c) Pressure energy (Work done by the fluid pressure)c) Pressure energy (Work done by the fluid pressure)

Fig. 6.1: Energy of a flowing fluidFig. 6.1: Energy of a flowing fluid

Energy of a Flowing FluidEnergy of a Flowing Fluid

A flowing fluid can do work due to its pressureA flowing fluid can do work due to its pressurePressure at section AB = p Cross-sectional Area = APressure at section AB = p Cross-sectional Area = AForce exerted on AB = pAForce exerted on AB = pA

After the fluid has flowed along the streamtube, section (AB) will have After the fluid has flowed along the streamtube, section (AB) will have moved to (A’B’)moved to (A’B’)

Volume passing section (AB) = Wt / Volume passing section (AB) = Wt / g g = mg/= mg/g = m/g = m/

Distance AA’ = Volume/Area = m/Distance AA’ = Volume/Area = m/AA

Work done = Force x distance = Work done = Force x distance =

W.D./wt due to pressure = W.D./wt due to pressure = = = Pressure energy/wt = = Pressure energy/wt

)A

m)(pA(

gp

)mg1)(

Am)(pA(

Bernoulli’s equationBernoulli’s equation

For steady flow of a frictionless fluid along a streamline, the total energy per For steady flow of a frictionless fluid along a streamline, the total energy per unit weight remains constant from point to point although its division unit weight remains constant from point to point although its division between the three forms of energy may vary:between the three forms of energy may vary:

Pressure Energy/Wt. + Kinetic Energy/Wt. + Potential Energy/Wt. = Pressure Energy/Wt. + Kinetic Energy/Wt. + Potential Energy/Wt. = Total Energy/Wt. = Constant (for frictionless fluid)Total Energy/Wt. = Constant (for frictionless fluid)

Pressure head + Velocity head + Elevation head = Total headPressure head + Velocity head + Elevation head = Total head

(m)(m)Hzg2

vg

p 2

Steady Flow Energy EquationSteady Flow Energy Equation

In general, energy could have been lost by doing work against friction In general, energy could have been lost by doing work against friction or energy loss in a turbine or energy could have been gained by or energy loss in a turbine or energy could have been gained by having a pump.having a pump.

Bernoulli’s equation can be expanded to include these conditions, Bernoulli’s equation can be expanded to include these conditions, such as,such as,

Total energy/wt at 1 = Total energy/wt at 2 + Friction Loss/wt + Total energy/wt at 1 = Total energy/wt at 2 + Friction Loss/wt + W.D./wt by a turbine – Energy supplied/wt by a pumpW.D./wt by a turbine – Energy supplied/wt by a pump

ptf hhhzgv

gpz

gv

gp

2

222

1

211

22

Flow of water between Two ReservoirsFlow of water between Two ReservoirsEnergy lost due to pipe frictionEnergy lost due to pipe friction

Fig. 6.5: Flow between two reservoirs open to atmosphere

fhzgv

gpz

gv

gp

2

222

1

211

22

Energy subtracted by a TurbineEnergy subtracted by a Turbine

Fig. Energy subtracted by a turbineFig. Energy subtracted by a turbine

tf hhzgv

gpz

gv

gp

2

222

1

211

22

1

2

Energy added by a PumpEnergy added by a Pump

Fig. Energy added by a pumpFig. Energy added by a pump

2 1 Pump

fp hzgv

gphz

gv

gp

2

222

1

211

22

Example 6.1Example 6.1


(a) (a) To determine the velocity of jet at CTo determine the velocity of jet at C:: Applying Bernoulli’s equation to A and C: Applying Bernoulli’s equation to A and C:

Total energy/wt at A = Total energy/wt at C + Loss in inlet pipe – Energy/wt supplied by Total energy/wt at A = Total energy/wt at C + Loss in inlet pipe – Energy/wt supplied by pump + Loss in discharge pipepump + Loss in discharge pipe (I)(I)

PPAA and P and Pcc = P = Patmatm vvAA = 0 (large sump) = 0 (large sump) zzA A = 0 = 0 zzC C = z= z3 3 = 32= 32

(II)(II)

From continuity equation, QFrom continuity equation, Q11 = Q = Q22 = Q = Q33

andand

)18(g2u12u5u 22

21

23

g2u1250

g2u5)z

g2u

gp()z

g2v

gp(

22

21

3

23C

A

2AA

3232

221

21 ud

4ud

4ud

4

332

32

1

31 u

41u)

15075(u)

dd(u 33

23

2

2

32 u

169u)

10075(u)

dd(u

Example 6.1Example 6.1 Substituting in equation (II)Substituting in equation (II)

uu3 3 = 8.314 m/s = 8.314 m/s (velocity of the jet from the nozzle at C)(velocity of the jet from the nozzle at C)

)18(g2])169(x12)

41(x51[u 222

3

)18(g2u109.5 23


(b) (b) To find the pressure in the suction pipe at pTo find the pressure in the suction pipe at pBB :: Applying Bernoulli’s equation to A and B:Applying Bernoulli’s equation to A and B:

Total energy/wt at A = Total energy/wt at B + Loss in inlet pipe Total energy/wt at A = Total energy/wt at B + Loss in inlet pipe

zz22 = 2 m, u = 2 m, u1 1 = (1/4)u= (1/4)u3 3 = 8.314//4 = 2.079 m/s= 8.314//4 = 2.079 m/s

ppBB//g = - (2 + 6x2.079g = - (2 + 6x2.07922/1000x9.81) = - (2 + 1.32) = -3.32 m/1000x9.81) = - (2 + 1.32) = -3.32 m

ppBB = - 1000x9.81x3.32 = - 32.569 kNm = - 1000x9.81x3.32 = - 32.569 kNm-2-2 (or)(or)

ppBB = 32.569 kNm = 32.569 kNm-2-2 below atmospheric pressure below atmospheric pressure

g2u5)z

g2u

gp(0

21

2

21B

g2u6z

gp 2

12

B

g2u5)z

g2v

gp()z

g2v

gp(

21

B

2BB

A

2AA

Pitot TubePitot Tube

Fig. 6.10: Pitot tube

Pitot TubePitot Tube

The pitot tube is used to measure the velocity of a stream and consists The pitot tube is used to measure the velocity of a stream and consists of a simple L-shaped tube facing into the oncoming flowof a simple L-shaped tube facing into the oncoming flow

Applying Bernoulli’s equation at A and B Applying Bernoulli’s equation at A and B Total energy/wt. at A = Total energy/wt. at BTotal energy/wt. at A = Total energy/wt. at B

uu22 /2g + p/ρg = u /2g + p/ρg = u0022 /2g + p /2g + p00 /ρg /ρg

uu22 /2g + p/ρg = p /2g + p/ρg = p00 /ρg since u /ρg since u00 = 0 = 0 uu22 /2g = p /2g = p00 /ρg - p/ρg /ρg - p/ρg

uu22 /2g = (h +z) – z /2g = (h +z) – z uu22 /2g = h /2g = h

Velocity at A Velocity at A u = √(2gh)u = √(2gh)

Actual velocity u = C√(2gh)Actual velocity u = C√(2gh) where C is the coefficient of the where C is the coefficient of the instrumentinstrument

Changes of Pressure in a tapering PipeChanges of Pressure in a tapering PipeExample 6.2Example 6.2

A pipe inclined at 45A pipe inclined at 45 to the horizontal (Fig. 6.12) converges over a length l of 2m from a to the horizontal (Fig. 6.12) converges over a length l of 2m from a diameter ddiameter d11 of 200 mm to a diameter of d of 200 mm to a diameter of d2 2 of 100 mm at the upper end. Oil of relative of 100 mm at the upper end. Oil of relative density 0.9 flows through the pipe at a mean velocity vdensity 0.9 flows through the pipe at a mean velocity v11 at the lower end of 2 m/s. at the lower end of 2 m/s.

Find the pressure difference across the 2 m length ignoring any loss of energy, and the Find the pressure difference across the 2 m length ignoring any loss of energy, and the difference in level that would be shown on a mercury manometer connected across this difference in level that would be shown on a mercury manometer connected across this length. The relative density of mercury is 13.6 and the leads to the manometer are filled length. The relative density of mercury is 13.6 and the leads to the manometer are filled with oil.with oil.

Fig. 6.12: Pressure change in a tapering pipeFig. 6.12: Pressure change in a tapering pipe

Changes of Pressure in a tapering PipeChanges of Pressure in a tapering PipeExample 6.2Example 6.2

Fig. 6.12: Pressure change in a tapering pipeFig. 6.12: Pressure change in a tapering pipe


(a) To determine pressure difference across the length (p(a) To determine pressure difference across the length (p11 - p - p22):): From continuity equation, QFrom continuity equation, Q11 = Q = Q22 andand vv11 = 2 m/s = 2 m/s

Applying Bernoulli’s equation to section 1 and 2 and ignoring losses: Applying Bernoulli’s equation to section 1 and 2 and ignoring losses:

zz2 2 - z- z1 1 = l sin45˚ = 2x0.707 = 1.414 m = l sin45˚ = 2x0.707 = 1.414 m o o = 0.9x1000 = 900 kg/m= 0.9x1000 = 900 kg/m3 3

pp11 - p - p2 2 = 4.472x900x9.81 = 39484 N/m = 4.472x900x9.81 = 39484 N/m22 = 39.484 kN/m= 39.484 kN/m22

2

22

o

21

21

o

1 zg2

vg

Pzg2

vg

P

2221

21 vd

4vd

4

s/m82x)1.02.0(v)

dd(v 2

12

2

12

m472.4414.181.9x228)zz()

g2vv(

gP

gP 22

12

21

22

o

2

o

1


(b) To determine the manometer reading h:(b) To determine the manometer reading h:

)1(hzzgpp

o

m21

o

21

))](zz(gpp[h

om

o21

o

21

)10x9.010x6.13

10x9.0](414.1472.4[h 33

3

gh)hz(gpgzp m2o21o1

)(ghghgh)zz(gpp omom21o21

mm217m217.0h

Venturi meterVenturi meter

Fig. 6.13: Inclined Venturi meter and U-tubeFig. 6.13: Inclined Venturi meter and U-tube


Assuming that there is no loss of energy, and applying Bernoulli’s equation to Assuming that there is no loss of energy, and applying Bernoulli’s equation to section 1 and 2section 1 and 2

pp11//g + vg + v1122/2g + z/2g + z11 = p = p22//g + vg + v22

22/2g + z/2g + z22

vv2222 – v – v11

22 = 2g[(p = 2g[(p1 1 - p- p22)/)/g + (zg + (z1 1 - z- z22)])]

For continuity of flowFor continuity of flowAA11vv11 = A= A22vv22 oror vv22 = (A = (A11/A/A22)v)v11

Substituting in the energy equation Substituting in the energy equation vv11

22[(A[(A11/A/A22))22 – 1] = 2g[(p – 1] = 2g[(p1 1 - p- p22) /) /g + (zg + (z11 – z – z22)])]

vv1122[(A[(A11

22 - A - A22

22)/A)/A22))22] = 2g[(p] = 2g[(p1 1 - p- p22) /) /g + (zg + (z11 – z – z22)])]

)]zzgpp(g2[

)AA(Av 21

212/12

221

21


Discharge Q = ADischarge Q = A11vv11 = [A = [A11AA22/(A/(A1122 – A – A22

22))1/21/2]](2gH)(2gH)

where H = (pwhere H = (p11-p-p22) /) /g + (zg + (z11 – z – z22) or, if m = area ratio = A) or, if m = area ratio = A11/A/A22

(6.15)(6.15)

Actual discharge, QActual discharge, Qactualactual = C = Cdd Q Qtheoreticaltheoretical

Pressure at level X-X must be the same Pressure at level X-X must be the same PPxx = p = p11 + + g(zg(z1 1 - z) = p- z) = p2 2 + + g(zg(z2 2 – z - h) + – z - h) + manmanghgh

pp11 - p - p2 2 + + g(zg(z1 1 - z- z22) = -) = -gh + gh + manmanghgh

H = (pH = (p11-p-p22) /) /g + (zg + (z11 – z – z22) = h() = h(manman// - 1) - 1)

Discharge (6.16)Discharge (6.16))]1(gh2[]

)1m(A[Q man

21

gH2])1m(

A[Q2

1


)]1(gh2[])1m(

A[Q man2

1

Discharge Discharge (6.16)(6.16)


Fig. Venturi meter with a U tube manometerFig. Venturi meter with a U tube manometer

)]1(gh2[])1m(

A[Q man2

1

pp11//g + Vg + V1122/2g + z/2g + z11 = p = p22//g + Vg + V22

22/2g + z/2g + z22

AA11VV1 1 = A= A22VV22


Fig. Venturi meter with piezometers

pp1 1 - p - p22 = = ghgh

pp11//g + Vg + V1122/2g + z/2g + z11 = p = p22//g + Vg + V22

22/2g + z/2g + z22

AA11VV1 1 = A= A22VV22

Bourdon GaugeBourdon Gauge

Pipe OrificesPipe Orifices

Fig. 6.14: Pipe orifice meter

where H = (p1-p2) /g + (z1 – z2)

(6.15)(6.15)DischargeDischarge gH2])1m(

A[Q2

1

)]1(gh2[])1m(

A[Q man2

1

(6.16)(6.16)

Theory of Small OrificesTheory of Small Orifices

Fig. 6.15: Flow through a small orificeFig. 6.15: Flow through a small orifice


Applying Bernoulli’s equation to A and B, assuming that there is no loss of Applying Bernoulli’s equation to A and B, assuming that there is no loss of energy,energy,

Total energy/wt at A = Total energy/wt at BTotal energy/wt at A = Total energy/wt at B

ppAA//g + vg + vAA22/2g + z/2g + zAA = p = pBB//g + vg + vBB

22/2g + z/2g + zBB

Putting zPutting zAA – z – zBB = H, v = H, vAA = 0, v = 0, vBB = V and = V and

ppAA = p = pBB = p = patmatm = 0 = 0

Velocity of jet, V = Velocity of jet, V = (2gH)(2gH) (6.17)(6.17)

This is called This is called Torricelli’s theoremTorricelli’s theorem, that the velocity of the issuing jet is , that the velocity of the issuing jet is proportional to the square root of the head producing flow.proportional to the square root of the head producing flow.

Discharge, Q = AV = ADischarge, Q = AV = A(2gH)(2gH) (6.18)(6.18)


In practice, the actual discharge is less than the theoretical discharge, which In practice, the actual discharge is less than the theoretical discharge, which must be modified by introducing a must be modified by introducing a coefficient of discharge Ccoefficient of discharge Cdd

Actual discharge, QActual discharge, Qactualactual = C = Cdd Q Qtheo.theo. = C = Cdd A A(2gH)(2gH) (6.19)(6.19)

There are There are two reasonstwo reasons for the difference between the theoretical and actual for the difference between the theoretical and actual discharges. discharges.

(1) velocity of the jet V is less than (1) velocity of the jet V is less than 2gH because there is a loss of energy 2gH because there is a loss of energy between A and B:between A and B:

Actual velocity at B, VActual velocity at B, Vactualactual = C = Cvv V = C V = Cvv(2gH)(2gH) (6.20)(6.20)

where Cwhere Cvv = coefficient of velocity. = coefficient of velocity.

Theory of Small OrificesTheory of Small Orifices (2) The paths of the fluid particles converge on the orifice, and the area of the (2) The paths of the fluid particles converge on the orifice, and the area of the

issuing jet at B is less than the area of the orifice at C.issuing jet at B is less than the area of the orifice at C.

Fig. 6.16: Contraction of issuing jetFig. 6.16: Contraction of issuing jet

In the plane of the orifice, the pressure at C is greater than atmospheric In the plane of the orifice, the pressure at C is greater than atmospheric pressure. At B, the paths of the particles have become parallel. This section is pressure. At B, the paths of the particles have become parallel. This section is called the called the vena contractavena contracta. .

Actual area of jet at B = CActual area of jet at B = Ccc A A (6.21)(6.21)

where Cwhere Ccc = coefficient of contraction = coefficient of contraction and A = area of the orifice.and A = area of the orifice.


Actual discharge = Actual area at B x Actual velocity at BActual discharge = Actual area at B x Actual velocity at B

Actual Q = CActual Q = Ccc A x C A x Cvv (2gH)(2gH)

QQactualactual = C = Ccc C Cvv A A (2gH)(2gH) (6.22)(6.22)

Also, QAlso, Qactualactual = C = Cdd Q Qtheo.theo. = C = Cdd A A(2gH)(2gH)

Therefore the relation between the coefficients is Therefore the relation between the coefficients is

CCdd = C = Ccc.C.Cv v

To determine the To determine the coefficient of dischargecoefficient of discharge, it is necessary to measure the actual , it is necessary to measure the actual volume discharged from the orifice in a given time and compare this with the volume discharged from the orifice in a given time and compare this with the theoretical discharge given by equation, Q = Atheoretical discharge given by equation, Q = A2gH2gH

eargdischlTheoreticaeargdischActualC,eargdischoftcoefficien d


Similarly, the actual area of the jet at the vena contracta can be measured,Similarly, the actual area of the jet at the vena contracta can be measured,

In the same way, if the actual velocity of the jet at the vena contracta can be In the same way, if the actual velocity of the jet at the vena contracta can be found,found,

If the orifice is not in the bottom of the tank, one method of measuring the actual If the orifice is not in the bottom of the tank, one method of measuring the actual velocity is to measure the velocity is to measure the velocity profilevelocity profile. See Example 6.4. See Example 6.4

orificeofAreacontractavenaatjetofAreaC,ncontractiooftcoefficien c

velocitylTheoreticacontractavenaatVelocityC,velocityoftcoefficien v

Theory of Small Orifices Theory of Small Orifices Example 6.4Example 6.4

A jet of water discharges horizontally into the atmosphere from an orifice in the vertical side A jet of water discharges horizontally into the atmosphere from an orifice in the vertical side of a large open-topped tank (Fig. 6.17). Derive an expression for the actual velocity v of a jet of a large open-topped tank (Fig. 6.17). Derive an expression for the actual velocity v of a jet at the vena contracta if the jet falls a distance y vertically in a horizontal distance x, measured at the vena contracta if the jet falls a distance y vertically in a horizontal distance x, measured from the vena contracta. If the head of water above the orifice is H, determine the coefficient from the vena contracta. If the head of water above the orifice is H, determine the coefficient of velocity.of velocity.

Express velocity profile of a jet ( in x and y ordinates)Express velocity profile of a jet ( in x and y ordinates) Use equations of motion: x = vt and y = ½ gtUse equations of motion: x = vt and y = ½ gt22

Fig. 6.17: Determination of the coefficient of velocityFig. 6.17: Determination of the coefficient of velocity

Theory of Small OrificesTheory of Small OrificesExample 6.4Example 6.4

Let t be the time taken for a fluid particle to travel from the vena contracta A to Let t be the time taken for a fluid particle to travel from the vena contracta A to the point B. Thenthe point B. Then

x = vt x = vt and and y = ½ gty = ½ gt22

oror v = x/t v = x/t and and t = (2y/g)t = (2y/g)0.50.5

Eliminating t,Eliminating t, v = x/(2y/g)v = x/(2y/g)0.50.5 = = (gx(gx22/2y)/2y)

Actual velocity of the jet (at the vena contracta), vActual velocity of the jet (at the vena contracta), vactualactual = = (gx(gx22/2y)./2y). Theoretical velocity of the jet, vTheoretical velocity of the jet, vtheo.theo. = = (2gH)(2gH) (6.17)(6.17)

yH4x

gH2y2

gx

velocitylTheoreticavelocityActualC,velocityoftCoefficien

2

2

v

yH4xC,velocityoftCoefficien

2v

Theory of Small OrificesTheory of Small OrificesExample 6.4Example 6.4

If the orifice has an area of 650 mmIf the orifice has an area of 650 mm2 2 and the jet falls a distance y of 0.5 m in a horizontal and the jet falls a distance y of 0.5 m in a horizontal distance x of 1.5 m from the vena contrata, calculate the values of the coefficients of distance x of 1.5 m from the vena contrata, calculate the values of the coefficients of velocity , discharge and contraction, given that the volume rate of flow is 0.117 mvelocity , discharge and contraction, given that the volume rate of flow is 0.117 m33/min /min and the head H above the orifice is 1.2m.and the head H above the orifice is 1.2m.

Putting x = 1.5 m, y = 0.5 m, H = 1.2 m, A = 650x10Putting x = 1.5 m, y = 0.5 m, H = 1.2 m, A = 650x10-6-6 m m22, Q = 0.117 m, Q = 0.117 m33 /min, /min,Coefficient of velocity, CCoefficient of velocity, Cvv = = √(x√(x22/4yH) = √(1.5/4yH) = √(1.522/4x0.5x1.2) = /4x0.5x1.2) = 0.9680.968

Coefficient of discharge, CCoefficient of discharge, Cdd = Q = Qactualactual/A/A(2gH) (2gH)

= (0.117/60)/[650x10= (0.117/60)/[650x10-3-3 √(2x9.81x1.2)]√(2x9.81x1.2)] = 0.618 = 0.618

Coefficient of contraction, CCoefficient of contraction, Ccc = C = Cdd/C/Cvv = 0.618/0.968 = 0.639 = 0.618/0.968 = 0.639

yH4xC,velocityoftCoefficien

2v


Fig. 6.16: Contraction of issuing jetFig. 6.16: Contraction of issuing jet

Continuity equationContinuity equation >> Applying Bernoulli’s eqn. at C and B: Applying Bernoulli’s eqn. at C and B:Q = AQ = ACC V VC C = A= ABB V VB B

ppBB//g = 0g = 0

VVBB = = (2gh)(2gh)

VVCC = C = CCC V VBB = C = CCC(2gh)(2gh)

CC

B

B

C CAA

vv

B

2BB

C

2CC z

g2v

gpz

g2v

gp

g2v

g2v

gp 2

B2CC

Theory of Large OrificesTheory of Large Orifices

If the vertical height of an orifice is large, the head producing flow is substantially If the vertical height of an orifice is large, the head producing flow is substantially less at the top of the opening than at the bottom.less at the top of the opening than at the bottom.

The discharge calculated from the formula for a small orifice , using the head h The discharge calculated from the formula for a small orifice , using the head h measured to the centre of the orifice, will not be the true value, since the velocity measured to the centre of the orifice, will not be the true value, since the velocity will vary very substantially from the top to bottom of opening.will vary very substantially from the top to bottom of opening.

Fig. 6.18: Flow through a large orificeFig. 6.18: Flow through a large orifice

Theory of Large OrificesTheory of Large OrificesExample 6.5Example 6.5

A reservoir discharges through a rectangular sluice gate of width B and height D A reservoir discharges through a rectangular sluice gate of width B and height D (Fig. 6.18). The top and bottom of the opening are at depths H(Fig. 6.18). The top and bottom of the opening are at depths H11 and H and H2 2 below the below the free surface. Derive a formula for the theoretical discharge through the opening.free surface. Derive a formula for the theoretical discharge through the opening.

Fig. 6.18: Flow through a large orificeFig. 6.18: Flow through a large orifice

Theory of Large OrificesTheory of Large Orifices

Consider a horizontal strip across the opening of height Consider a horizontal strip across the opening of height h at a depth h below h at a depth h below the free surface:the free surface:

Area of strip, Area of strip, A= BA= Bhh Velocity of flow through strip, v = Velocity of flow through strip, v = (2gh)(2gh)

Discharge through strip, Discharge through strip, Q = Area x velocity = BQ = Area x velocity = B(2g)h(2g)h1/21/2hh For the whole orifice opening, integrating from h = HFor the whole orifice opening, integrating from h = H11 to h = H to h = H22,,

dhhg2BQ,eargDisch2

1

H

H

21

]HH[)g2(B32Q,eargDisch 23

123

2

Example 6.5: Large OrificeExample 6.5: Large Orifice Putting B = 0.7 m, D = 1.5 m, HPutting B = 0.7 m, D = 1.5 m, H11 = 0.4 m, H = 0.4 m, H22 = 0.4 + 1.5 =1.9 m = 0.4 + 1.5 =1.9 m

Theoretical dischargeTheoretical discharge

Q= (2/3)x0.7x(2x9.81)Q= (2/3)x0.7x(2x9.81)1/2 1/2 (1.9(1.93/2 3/2 – 0.4– 0.43/23/2) ) = 4.891 m= 4.891 m33/s/s

For a small orificeFor a small orifice, Q = A, Q = A(2gh) (2gh) where A = area of orifice and h = head above centreline of orifice.where A = area of orifice and h = head above centreline of orifice.

A =BD = 0.7x1.5 = 1.05 mA =BD = 0.7x1.5 = 1.05 m22 h = ½(Hh = ½(H11+ H+ H22) = ½(0.4 + 1.9) = 1.15 m) = ½(0.4 + 1.9) = 1.15 m

Q = 1.05x(2x9.81x1.15)Q = 1.05x(2x9.81x1.15) ½ ½ = 4.988 m = 4.988 m33/s/s

% error = (4.988 – 4.891)/4.891 = 0.0198 = 1.98%% error = (4.988 – 4.891)/4.891 = 0.0198 = 1.98%

]HH[)g2(B32Q 23

123

2

Flow under varying head Flow under varying head Time required to empty a ReservoirTime required to empty a Reservoir

Diameter D, Surface Area ADiameter D, Surface Area A h h

h h – h h – hh

orifice diameter d, area aorifice diameter d, area a Q Q

Time required to empty a ReservoirTime required to empty a Reservoir

At time t, let the head is hAt time t, let the head is hAt time (t+ At time (t+ t), let the level fallen t), let the level fallen hh

Change in volume of tank = Amount discharge through the orificeChange in volume of tank = Amount discharge through the orifice

A(h – A(h – h) – Ah = Q {(t + h) – Ah = Q {(t + t) – t} t) – t} - A- Ah = Qh = Qtt

Discharge from an orifice, Q = CDischarge from an orifice, Q = Cd d aa(2gh)(2gh)

dhQAdt

dhhg2aC

Adhgh2aC

Adt 2/1

dd

h - dh

dh

Q

h

Area A

area a

Time required to empty a ReservoirTime required to empty a Reservoir

If HIf H11 and H and H22 are initial and final values of h, then the time required to empty are initial and final values of h, then the time required to empty the reservoir is,the reservoir is,

where T = time required to empty a reservoir (sec)where T = time required to empty a reservoir (sec)HH11 = initial head and H = initial head and H22 = final head in m = final head in m

A = surface area of reservoir (mA = surface area of reservoir (m22) ) a = orifice area (ma = orifice area (m22))CCd d = coefficient of discharge = coefficient of discharge

)HH(g2aC

A2T 2/12

2/11

d

dhh)g2aC

A(dt 2/1

d

H

H

2

1

Theory of Notches and WeirsTheory of Notches and Weirs

dhbhg2QH

0

2/1

Fig 6.19 Discharge from a notchFig 6.19 Discharge from a notch

Area of strip, Area of strip, A = bA = bhh Velocity through strip, v = Velocity through strip, v = (2gh)(2gh) Discharge through strip, Discharge through strip, Q = Area x Velocity = bQ = Area x Velocity = bh h (2gh)(2gh) Integrating from h = 0 at the free surface to h = H at the bottom of the notch,Integrating from h = 0 at the free surface to h = H at the bottom of the notch,

the total theoretical discharge,the total theoretical discharge,

Rectangular Notch (or) WeirRectangular Notch (or) Weir

Theoretical discharge,Theoretical discharge,

For aFor a rectangular notchrectangular notch (Fig. a), put b = constant = B gives,(Fig. a), put b = constant = B gives,

Theoretical discharge (6.25)Theoretical discharge (6.25)

dhhg2BQH

0

2/1

dhbhg2QH

0

2/1

2/3H)g2(B32Q

V - notch (or) Triangular WeirV - notch (or) Triangular Weir

Theoretical discharge,Theoretical discharge,

For aFor a V-notch with an included angle V-notch with an included angle (Fig. b), put b = 2(H-h) tan(Fig. b), put b = 2(H-h) tan/2 in /2 in the equation,the equation,

Theoretical discharge (6.26)Theoretical discharge (6.26)

dhh)hH()2

(tang22QH

0

2/1

2/5H)2

(tang2158Q

dhbhg2QH

0

2/1

dh)hHh()2

(tang22QH

0

2/32/1

Velocity of approach (to the weir)Velocity of approach (to the weir)in the rectangular channelin the rectangular channel

Consider flow over a weir at the end of a long rectangular channelConsider flow over a weir at the end of a long rectangular channel Velocity of approach to the weir will be substantial Velocity of approach to the weir will be substantial Total head x = h + Total head x = h + vv22 /2g where /2g where = K.E. correction factor = 1.1 = K.E. correction factor = 1.1

energy lineenergy line

free surface vfree surface v22/2g /2g

v v weir weir

channel bed channel bedchannel bed channel bed

Hh x

Considering Velocity of approach (to the weir)Considering Velocity of approach (to the weir)

Fig 6.19 Discharge from a notchFig 6.19 Discharge from a notch

Total head x = h + Total head x = h + vv22/2g /2g x = x = hh

Mean Velocity Mean Velocity v = Q/A where A = cross-sectional area of the channelv = Q/A where A = cross-sectional area of the channel

Discharge through strip, Discharge through strip, Q = Q = A. VA. V = b = b h h 2gx2gx = b = b x x 2g x2g x1/2 1/2 Q = b Q = b 2g x2g x1/2 1/2 x (6.28)x (6.28)

hh

v2/2g

Energy line

x

b

Considering Velocity of approach (to the weir)Considering Velocity of approach (to the weir)

LimitsLimits: : At the free surface, h = 0 and x =At the free surface, h = 0 and x = vv22 /2g /2gAt the sillAt the sill level, h = H and x = H + level, h = H and x = H + vv22 /2g /2g

We haveWe have Q = b Q = b 2g x2g x1/2 1/2 x (6.28)x (6.28)

Integrating (6.28) between these limitsIntegrating (6.28) between these limits

dxbxg2Qg2/vH

g2/v

2/1

2

2

Velocity of approach (to the weir)Velocity of approach (to the weir)in the rectangular channelin the rectangular channel

Considering velocity of approach to the weir, total discharge is,Considering velocity of approach to the weir, total discharge is,

where x = h + where x = h + vv22/2g/2g

For a rectangular notch, b = B = constant, and the discharge is,For a rectangular notch, b = B = constant, and the discharge is,

(6.29)(6.29)

dxbxg2Qg2/vH

g2/v

2/1

2

2

])gH2v()

gH2v1[(H)g2(B

32Q 2/3

22/3

22/3

Example 6.7: Velocity of ApproachExample 6.7: Velocity of Approach

vv22/2g /2g energy lineenergy line

vv

H = 0.25 mh

0.2 m

weir

1.2 m

x

0.9 m

Francis formula for rectangular weirFrancis formula for rectangular weir

If L is the length of the weir and H the head over sill, the Francis formula for If L is the length of the weir and H the head over sill, the Francis formula for the discharge is,the discharge is,

Q = 1.84(L – 0.1nH)HQ = 1.84(L – 0.1nH)H3/23/2

where n = number of end contractions.where n = number of end contractions.

Power of a stream of fluidPower of a stream of fluid

Total Energy/weight (or head) H of the fluid isTotal Energy/weight (or head) H of the fluid is

Total HeadTotal Head H = p/H = p/g + vg + v22/2g + z/2g + z If the weight per unit time of fluid flowing is known, the power of the stream,If the weight per unit time of fluid flowing is known, the power of the stream,

Power = Energy/time = (Weight/time)(Energy/weight)Power = Energy/time = (Weight/time)(Energy/weight)

If Q is the volume rate of flow, Weight /time = If Q is the volume rate of flow, Weight /time = gQgQ

Power P = Power P = gQH = gQH = gQ (p/gQ (p/g + vg + v22/2g + z) (6.30)/2g + z) (6.30)


z = 240 m Q = 0.13 mz = 240 m Q = 0.13 m33/s Pelton wheel/s Pelton wheel

Datum levelDatum level v = 66m/s bucketv = 66m/s bucket


Power P = Power P = gQH = gQH = gQ (p/gQ (p/g + vg + v22/2g + z) (6.30)/2g + z) (6.30)

a)a) The jet issuing from the nozzle will be at atmospheric pressure and at the The jet issuing from the nozzle will be at atmospheric pressure and at the datum level, so that in equation (6.30), p = 0 and z = 0.datum level, so that in equation (6.30), p = 0 and z = 0.

Power of jet = (Power of jet = (gQ)(vgQ)(v22/2g) = ½/2g) = ½QvQv22

= ½x10= ½x103 3 x0.13x 66x0.13x 662 2 = 283140 W = 283140 W = 283.14 kW= 283.14 kW

b)b) At the reservoir, the pressure is atmospheric and the velocity of the free At the reservoir, the pressure is atmospheric and the velocity of the free surface is zero, so that in equation (6.30), p = 0 and v = 0.surface is zero, so that in equation (6.30), p = 0 and v = 0.

Power supplied from reservoir = Power supplied from reservoir = gQz gQz =10=103 3 x9.81x0.13x 240 = 306072 x9.81x0.13x 240 = 306072

W W = 306.07 kW= 306.07 kW


c)c) If HIf H11 = total head at the reservoir, H = total head at the reservoir, H2 2 = total head at the jet and = total head at the jet and

hhf f = head lost in the pipeline, then= head lost in the pipeline, then

Power supplied from reservoir = Power supplied from reservoir = gQHgQH1 1 = 306.07 kW= 306.07 kW

Power of issuing jet = Power of issuing jet = gQHgQH22 = 283.14 kW = 283.14 kW

Power lost in pipeline = Power lost in pipeline = gQhgQhf f = 306.07 - 283.14 = 22.93 kW= 306.07 - 283.14 = 22.93 kW

Head lost, hHead lost, hf f = Power lost / = Power lost / gQ gQ

= 22.93x10= 22.93x1033/(10/(1033x9.81x0.13) = 17.98 mx9.81x0.13) = 17.98 m

d)d) Efficiency of transmission = Power of jet / Power supplied by reservoirEfficiency of transmission = Power of jet / Power supplied by reservoir = 283.14/306.07 = 0.925 = 92.5%= 283.14/306.07 = 0.925 = 92.5%

The EndThe End

fluid mech. chapter6

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