Fluid discharge

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FLUID DISCHARGE

Keith Vaugh BEng (AERO) MEng

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OBJECTIVESIdentify the unique vocabulary used in the description and analysis of fluid flow with an emphasis on fluid discharge

Describe and discuss how fluid flow discharge devices affects fluid flow.

Derive and apply the governing equations associated with fluid discharge

Determine the power of flow in a channel or a stream and how this can be affected by height, pressure, and/or geometrical properties.

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FLOW THROUGH ORIFICES and

MOUTHPIECES

An orifice Is an opening having a closed perimeter

A mouthpiece Is a short tube of length not more than two to three times its diameter

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THEORY OF SMALL ORIFICES

DISCHARGING

u1

z1

z2 ②

H

u = u2

①

Orifice area A

Applying Bernoulli’s equation to stations ① and ②

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DISCHARGING

u1

z1

z2 ②

H

u = u2

①

Orifice area A

z1+p1

ρg+u12

2g= z

2+p2

ρg+u22

2g


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DISCHARGING

u1

z1

z2 ②

H

u = u2

①

Orifice area A

z1+p1

ρg+u12

2g= z

2+p2

ρg+u22

2g


putting z1 - z2 = H, u1 = 0, u2 = u and p1 = p2

Velocity of jet, u = 2gH( )

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TORRICELLI’s THEOREM

Torricelli’s theorem states that the velocity of the discharging jet is proportional to the square root of the head producing flow. This is support by the preceding derivation;

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Velocity of jet, u = 2gH( )

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Velocity of jet, u = 2gH( )The discharging flow rate can be determined theoretically if A is the cross-sectional area of the orifice

!V = Area × Velocity = A 2gH( )

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Velocity of jet, u = 2gH( )The discharging flow rate can be determined theoretically if A is the cross-sectional area of the orifice

!V = Area × Velocity = A 2gH( )Actual discharge

!Vactual

= CdA 2gH( )

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The velocity of the jet is less than that determined by the velocity of jet equ because there is a loss of energy between stations ① and ② i.e actual velocity, u = Cu √(2gH)

where Cu is a coefficient of velocity which is determined experimentally and is of the order 0.97 - 0.98

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The velocity of the jet is less than that determined by the velocity of jet equ because there is a loss of energy between stations ① and ② i.e actual velocity, u = Cu √(2gH)

where Cu is a coefficient of velocity which is determined experimentally and is of the order 0.97 - 0.98

The paths of the particles of the fluid converge on the orifice and the area of the discharging jet at B is less than the area of the orifice A at C

actual area of jet at B = Cc A

where Cc is the coefficient of contraction - determined experimentally - typically 0.64

Vena contracta

C BpB

uuC

pC

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Actual discharge = Actual area at B × Actual velocity at B

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= Cc× C

uA 2gH( )

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= Cc× C

uA 2gH( )

we see that the relationship between the coefficients is Cd = Cc × Cu

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= Cc× C

uA 2gH( )


To determine the coefficient of discharge measure the actual discharged volume from the orifice in a given time and compare with the theoretical discharge.

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= Cc× C

uA 2gH( )



Coefficient of discharge, Cd= Actual measured discharge

Theoretical discharge

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= Cc× C

uA 2gH( )





Coefficient of contraction, Cc= Area of jet at vena contracta

Area of orifice

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= Cc× C

uA 2gH( )





Coefficient of contraction, Cc= Area of jet at vena contracta

Area of orifice

Coefficient of velocity, Cu= Velocity at vena contracta

Theoretical velocity

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EXAMPLE 1(a) A jet of water discharges horizontally into the atmosphere

from an orifice in the side of large open topped tank. Derive an expression for the actual velocity, u of a jet at the vena contracta if the jet falls a distance y vertically for a horizontal distance x, measured from the vena contracta.

(b) If the head of water above the orifice is H, calculate the coefficient of velocity.

(c) If the orifice has an area of 650 mm2 and the jet falls a distance y of 0.5 m in a horizontal distance x of 1.5 m from the vena contracta, calculate the values of the coefficients of velocity, discharge and contraction, given that the volumetric flow is 0.117 m3 and the head H above the orifice is 1.2 m

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

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x = ut→ u = xt

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x = ut→ u = xt

y = 12gt2 → t = 2y

g

⎛⎝⎜

⎞⎠⎟

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x = ut→ u = xt

Velocity at the vena contracta, u = gx 2

2y

⎛

⎝⎜⎞

⎠⎟

y = 12gt2 → t = 2y

g

⎛⎝⎜

⎞⎠⎟

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x = ut→ u = xt


2y

⎛

⎝⎜⎞

⎠⎟

y = 12gt2 → t = 2y

g

⎛⎝⎜

⎞⎠⎟

Theoretical velocity = 2gH( )

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x = ut→ u = xt


2y

⎛

⎝⎜⎞

⎠⎟

y = 12gt2 → t = 2y

g

⎛⎝⎜

⎞⎠⎟


Coefficient of velocity, Cu= Actual velocity


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x = ut→ u = xt


2y

⎛

⎝⎜⎞

⎠⎟

y = 12gt2 → t = 2y

g

⎛⎝⎜

⎞⎠⎟




= u

2gH( )

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x = ut→ u = xt


2y

⎛

⎝⎜⎞

⎠⎟

y = 12gt2 → t = 2y

g

⎛⎝⎜

⎞⎠⎟




= u

2gH( )= x 2

4yH

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putting x = 1.5m, H = 1.2m and Area, A = 650×10-6m2

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Coefficient of velocity, Cu= x 2

4yH


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4yH


= 1.52

4 × 0.5×1.2( )= 0.968

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4yH


= 1.52

4 × 0.5×1.2( )= 0.968

Coefficient of discharge, Cd=

!V

A 2gH( )

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4yH


= 1.52

4 × 0.5×1.2( )= 0.968


!V

A 2gH( )

=

0.11760

⎛⎝⎜

⎞⎠⎟

650 ×10−6 2 × 9.81×1.2( )= 0.618

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4yH


= 1.52

4 × 0.5×1.2( )= 0.968


!V

A 2gH( )

=

0.11760

⎛⎝⎜

⎞⎠⎟

650 ×10−6 2 × 9.81×1.2( )= 0.618

Coefficient of contraction, Cc=Cd

Cu

= 0.6180.968

= 0.639

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THEORY OF LARGE

ORIFICES

H1

H2

h

δhD

B

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(a) A reservoir discharges through a sluice gate of width B and height D. The top and bottom openings are a depths of H1 and H2 respectively below the free surface. Derive a formula for the theoretical discharge through the opening

(b) If the top of the opening is 0.4 m below the water level and the opening is 0.7 m wide and 1.5 m in height, calculate the theoretical discharge (in meters per second) assuming that the bottom of the opening is above the downstream water level.

(c) What would be the percentage error if the opening were to be treated as a small orifice?

EXAMPLE 2

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Given that the velocity of flow will be greater at the bottom than at the top of the opening, consider a horizontal strip across the opening of height δh at a depth h below the free surface

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Area of strip = Bδh

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Velocity of flow through strip = 2gH

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Discharge through strip, δ !V = Area × Velocity = B 2g( )h 12δh

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For the whole opening, integrating from h = H1 to h = H2




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Discharge !V = B 2g( ) h12 dh

H1

H2∫




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Discharge !V = B 2g( ) h12 dh

H1

H2∫




= 23B 2g( ) H232 −H132( )

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putting B = 0.7 m, H1 = 0.4 m and H2 = 1.9 m

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Theoretical discharge !V = 23× 0.7 × 2 × 9.81( ) 1.9

32 − 0.4

32( )

= 2.067 2.619 − 0.253( )= 4.891m3

s

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32 − 0.4

32( )

= 2.067 2.619 − 0.253( )= 4.891m3

s

For a small orifice, !V = A 2gh

A = BD= 0.7 ×1.5

h = 12H

1+H

2( )= 1

20.4 +1.9( ) =1.15m

!V = 0.7 ×1.5 2 × 9.81×1.15

= 4.988 m3

s

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32 − 0.4

32( )

= 2.067 2.619 − 0.253( )= 4.891m3

s

For a small orifice, !V = A 2gh

A = BD= 0.7 ×1.5

h = 12H

1+H

2( )= 1

20.4 +1.9( ) =1.15m

!V = 0.7 ×1.5 2 × 9.81×1.15

= 4.988 m3

s

error =4.988 − 4.891( )4.891

= 0.0198 =1.98%

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NOTCHES & WEIRS

H

h

δhH

b

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Consider a horizontal strip of width b and height δh at a depth h below the free surface.

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Area of strip = bδh

Velocity of flow through strip = 2gh

Discharge through strip, δ !V = Area × Velocity = bδh 2gh( )

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Integrating from h = 0 at the free surface to h = H at the bottom of the notch




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Total theoretical discharge !V = 2g( ) bh12 dh

0

H

∫




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0

H

∫




b must be expressed in terms of h before integrating

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0

H

∫




b must be expressed in terms of h before integrating

!V = B 2( ) g h12

0

H

∫ = 23B 2g( )H 32

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For a rectangular notch, put b = constant = B

b = constant

B

H

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!V = B 2g( ) h12 dh

0

H

∫= 23B 2gH

32

b = constant

B

H

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!V = B 2g( ) h12 dh

0

H

∫= 23B 2gH

32

b = constant

B

H

For a vee notch with an included angle θ, put b = 2(H - h)tan(θ⁄2)b = 2(H - h)tan(θ⁄2)

Hh

θ

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!V = B 2g( ) h12 dh

0

H

∫= 23B 2gH

32

b = constant

B

H

For a vee notch with an included angle θ, put b = 2(H - h)tan(θ⁄2)

!V = 2 2g( ) tan θ2

⎛⎝⎜

⎞⎠⎟

H − h( )h 12 dh0

H

∫

= 2 2g( ) tan θ2

⎛⎝⎜

⎞⎠⎟23Hh

32 − 25h52

⎡

⎣⎢

⎤

⎦⎥0

h

= 815

2g( ) tan θ2

⎛⎝⎜

⎞⎠⎟H52

b = 2(H - h)tan(θ⁄2)

Hh

θ

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In the foregoing analysis it has been assumed that• the velocity of the liquid approaching the notch is very small so that its

kinetic energy can be neglected• the velocity through any horizontal element across the notch will

depend only on the depth below the free surface

These assumptions are appropriate for flow over a notch or a weir in the side of a large reservoir

If the notch or weir is located at the end of a narrow channel, the velocity of approach will be substantial and the head h producing flow will be increased by the kinetic energy;

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In the foregoing analysis it has been assumed that• the velocity of the liquid approaching the notch is very small so that its

kinetic energy can be neglected• the velocity through any horizontal element across the notch will

depend only on the depth below the free surface

These assumptions are appropriate for flow over a notch or a weir in the side of a large reservoir

If the notch or weir is located at the end of a narrow channel, the velocity of approach will be substantial and the head h producing flow will be increased by the kinetic energy;

x = h + αu 2

2g

where ū is the mean velocity and α is the kinetic energy correction factor to allow for the non-uniform velocity over the cross section of the channel

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Therefore

δ !V = bδh 2gx( )= b 2g( )x 12dx

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Therefore

δ !V = bδh 2gx( )= b 2g( )x 12dx

at the free surface, h = 0 and x = αū2/2g, while at the sill , h = H and x = H + αū2/2g. Integrating between these two limits

!V = 2g( ) bx12

αu 2

2g

H+αu2

2g∫ dx

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Therefore

δ !V = bδh 2gx( )= b 2g( )x 12dx

at the free surface, h = 0 and x = αū2/2g, while at the sill , h = H and x = H + αū2/2g. Integrating between these two limits

!V = 2g( ) bx12

αu 2

2g

H+αu2

2g∫ dx

For a rectangular notch, putting b = B = constant

!V = 23B 2g( )H 32 1+ αu 2

2gH

⎛

⎝⎜⎞

⎠⎟

32

− αu 2

2gH

⎛

⎝⎜⎞

⎠⎟

32⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

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Pressure, p, velocity, u, and elevation, z, can cause a stream of fluid to do work. The total energy per unit weight H of a fluid is given by

H = pρg

+ u2

2g+ z

THE POWER OF A STREAM OF

FLUID

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Pressure, p, velocity, u, and elevation, z, can cause a stream of fluid to do work. The total energy per unit weight H of a fluid is given by

H = pρg

+ u2

2g+ z

If the weight per unit time of fluid is known, the power of the stream can be calculated;

THE POWER OF A STREAM OF

FLUID

Power = Energy per unit time = WeightUnit time

× EnergyUnit weight

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Weight per unit time = ρg !V

Power = ρg !VH

=ρg !V pρg

+ u2

2g+ z

⎛

⎝⎜⎞

⎠⎟

= p !V+12ρu2 !V+ρg !Vz

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In a hydroelectric power plant, 100 m3/s of water flows from an elevation of 12 m to a turbine, where electric power is generated. The total irreversible heat loss is in the piping system from point 1 to point 2 (excluding the turbine unit) is determined to be 35 m. If the overall efficiency of the turbine-generator is 80%, estimate the electric power output.

EXAMPLE 2

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Assumptions 1. The flow is steady and incompressible2. Water levels at the reservoir and the

discharge site remain constant

Properties We take the density of water to be 1000 kg/m3

(Çengel, et al 2008)

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The mass flow rate of water through the turbine is


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!m = ρ !V = 1000kgm3

⎛⎝⎜

⎞⎠⎟100m

3

s( ) =105 kg sThe mass flow rate of water through the turbine is


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!m = ρ !V = 1000kgm3

⎛⎝⎜

⎞⎠⎟100m

3


We take point ➁ as the reference level, and thus z2 = 0. Therefore the energy equation is


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!m = ρ !V = 1000kgm3

⎛⎝⎜

⎞⎠⎟100m

3


We take point ➁ as the reference level, and thus z2 = 0. Therefore the energy equation is

P1

ρg+α

1

V12

2g+ z

1+ h

pump,u=P2

ρg+α

2

V22

2g+ z

2+ h

turbine ,e+ h

L


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Also, both points ➀ and ➁ are open to the atmosphere (P1 = P2 = Patm) and the flow velocities are negligible at both points (V1 = V2 = 0). Then the energy equation for steady, incompressible flow reduces to

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hturbine ,e

= z1− h

L

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hturbine ,e

= z1− h

L

Substituting, the extracted turbine head and the corresponding turbine power are

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hturbine ,e

= z1− h

L


hturbine ,e

= z1− h

L=120 − 35 = 85m

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hturbine ,e

= z1− h

L


hturbine ,e

= z1− h

L=120 − 35 = 85m

!Wturbine ,e

= !mghturbine ,e

= 105 kgs

⎛⎝⎜

⎞⎠⎟9.81m

s2( ) 85m( )1kJkg

1000m2

s2

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟= 83,400kW

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hturbine ,e

= z1− h

L


hturbine ,e

= z1− h

L=120 − 35 = 85m

!Wturbine ,e

= !mghturbine ,e

= 105 kgs

⎛⎝⎜

⎞⎠⎟9.81m

s2( ) 85m( )1kJkg

1000m2

s2

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟= 83,400kW

Therefore, a perfect turbine-generator would generate 83,400 kW of electricity from this resource. The electric power generated by the actual unit is

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hturbine ,e

= z1− h

L


hturbine ,e

= z1− h

L=120 − 35 = 85m

!Wturbine ,e

= !mghturbine ,e

= 105 kgs

⎛⎝⎜

⎞⎠⎟9.81m

s2( ) 85m( )1kJkg

1000m2

s2

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟= 83,400kW


!Welectric

= ηturbine−gen

!Wturbine ,e

= 0.80( ) 83.4MW( ) = 66.7MW

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hturbine ,e

= z1− h

L


hturbine ,e

= z1− h

L=120 − 35 = 85m

!Wturbine ,e

= !mghturbine ,e

= 105 kgs

⎛⎝⎜

⎞⎠⎟9.81m

s2( ) 85m( )1kJkg

1000m2

s2

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟= 83,400kW


!Welectric

= ηturbine−gen

!Wturbine ,e

= 0.80( ) 83.4MW( ) = 66.7MWNote that the power generation would increase by almost 1 MW for each percentage point improvement in the efficiency of the turbine-generator unit.

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Flow through orifices and mouthpiecesTheory of small orifice discharge

!Torricelli’s theorem

Theory of large orificesNotches and weirsThe power of a stream of fluid

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Andrews, J., Jelley, N., (2007) Energy science: principles, technologies and impacts, Oxford University PressBacon, D., Stephens, R. (1990) Mechanical Technology, second edition, Butterworth HeinemannBoyle, G. (2004) Renewable Energy: Power for a sustainable future, second edition, Oxford University PressÇengel, Y., Turner, R., Cimbala, J. (2008) Fundamentals of thermal fluid sciences, Third edition, McGraw HillDouglas, J.F., Gasoriek, J.M., Swaffield, J., Jack, L. (2011), Fluid Mechanics, sisth edition, Prentice HallTurns, S. (2006) Thermal fluid sciences: An integrated approach, Cambridge University PressYoung, D., Munson, B., Okiishi, T., Huebsch, W., 2011Introduction to Fluid Mechanics, Fifth edition, John Wiley & Sons, Inc.

Some illustrations taken from Fundamentals of thermal fluid sciences

Fluid discharge

Engineering

Transcript of Fluid discharge