Outflow from orifice - Katedra hydrauliky a hydrologie - …hydraulika.fsv.cvut.cz/Hydraulika/Hydraulika/Predmety/...K141 HYAE Outflow from orifice 6 Coefficients for discharge determination

Outflow from orifice

K141 HYAE Outflow from orifice 1

TYPES OF OUTFLOW

Outflow

Outflow

steady: z = const, hE = const

(H = const, HE = const) Qp = Q

quasi-steady: z ~ const., phenomenon of large reservoir

unsteady: z const (H const)

Qp Q, filling and drawdown of tank (reservoir)

free (a) free outlet jet

submerged (b) submerged outlet jet

partly submerged, e.g. outflow from large orifices

at the bottom (slide gate)


STEADY FREE OUTFLOW (SFO) OF IDEAL LIQUID

BE surface – outlet:

Torricelli (1608 - 1647) equation for

outflow velocity of ideal liquid vi

for large reservoirs with free level:

outflow discharge of ideal liquid Qi:

for small orifice (bottom and wall):

2 20 iv vp

h+ + =g 2g 2g

2i

E

i E

vh =

2 g

v = 2gh

iv = 2gh

i i iS S

Q = dQ = u dS

i iS

Q =v dS i iQ =v S=S 2gh

ui,vi,dQi,Qi

pa(=0)

S orifice

section

overpressure

i iu v

h… depth of cente of orifice

hh,0g2

v,0p E

20

STEADY OUTFLOW FROM ORIFICE


Hydraulic losses

outlet loss v ... depends on shape, setup and

size of orifice (structure), Re

CONTRACTION OF OUTLET JET

Strip area Sc < S, Sc = · S, contraction coefficient 1

well mouthed

orifice partial

contraction

re-entrant

streamlined

mouthpiece

external

mouthpiece D

sharp edged

orifice

TAB.

imperfect

contraction

2

v

vc Z= 2g


SFO OF REAL LIQUID FROM ORIFICE AT THE BOTTOM OF TANK

g2

v

g2

v

g

p

g

p

g2

vlh

2c

2ca0s

20

c

BE 0 - 1

lc ~ 0,5·D

g

p

g

p

g2

vlhg2

1

1v a0s

20

cc

... velocity coefficient

contraction coefficient φ, μv, ε ... TAB.

Simplification:

free level → ps0 = pa →

S0 >> S → v0 ~ 0

lc << hE → lc ~ 0

0g

pp a0s

hg2SQ

,hg2v

v

c

Q = vcSc, Sc = εS, εφ = μv … orifice discharge

coefficient


- Large orifice hT < (2 - 3)·a

change of outflow velocity u

with height of orifice

- Open reservoir and large rectangular orifice in vertical wall:

for large tank:

Eu= 2gh1/2

v EQ= 2g h dSS

E2

E1

h1/2

v E Eh

Q= b 2g h dh

3/2 3/2v E2 E1

3/2 3/2v 2 1

2Q= b 2g h -h

3

2Q= b 2g h -h

3

EdS=bdh S=ba

hh02g

vE

2

0

SFO OF REAL LIQUID FROM ORIFICE IN VERTICAL WALL OF TANK

- Small orifice hT > (2 - 3)a

for S0 >> S → v0 ~ 0 tv

tc

hg2SQ

,hg2v

3/2 3/2v E2 E1

3/2 3/2v 2 1

2Q= b 2g h -h

3

2Q= b 2g h -h

3


Coefficients for discharge determination

- small sharp-edged orifice

with full contraction 0,97 0,63 0,61

- external cylindrical mouthpiece L/D = 2 4 0,81 1,00 0,81

- streamlined mouthpiece jet tube 0,95 1,00 0,95

- large orifices at the bottom with significant 0,65 to 0,85

or continuous side contraction

- outlet tube of diameter D and length L

with free outflow

v

i

1=

L1+ +

D

v

Note: special application of outflow through mouthpiece -

- Mariotte vessel - with function of dilution dosing,

Q = const.

φ, ε, μv for imperfect and partial contraction > φ, ε, μv for full contraction

empirical formulas


OUTFLOW FROM SUBMERGED ORIFICE

for both small and large orifices of

whatever shape

for small orifice

Note:

solution for partial submergence: Q = Q1 + Q2

(Q1 outflow from free part of orifice, Q2 outflow from submerged

part of orifice).

for large reservoir

H = H0

02gHvu

2gHSμQ

2gHSμQ

v

0v


OUTFLOW JETS

Free outflow jet

Supported outflow jet Submerged outflow jet

different functions of jet requirements for outlet equipment

and outlet velocity

- free jets – cutting, drilling, hydro-mechanization

(unlinking), firefighting, irrigation jets …

- submerged jets - dosing, mixing, rectifying, …

type: water - air

type: water – air – solid surface type: water - water

jet core with constant velocity

pulsating margin of

boundary layer

(mixing regions)

theoretical trajectory (parabola 2°)

decay of jet, aeration, drops

connected part


hd

Theoretical shape of outflow jet (projection at an angle) arcing distance of jet

maximum height

20

p0 d

vL = sin2 =2h sin2

g

22 20

0 d

vy = sin =h sin

2g

20

d

v=h

2g

energetic head

of jet

For = 45° Lp0max = v02/g = 2hd, y0 = 0,5 hd

For = X°, = 90 -X° same arcing distance

For = 90° vertical jet y0max = v02/2g = hd

For = 0° horizontal jet (horizontal projection)

real liquid, large reservoir p d TL =2 h y

p T TL =2 h y

0

2

x =v t

1y = gt

2

theoretical

2

0

0

gt2

1sinδtvy

cosδtvx

δ v0cosδ

v0sin

δ v0


UNSTEADY OUTFLOW FROM ORIFICE

Differential equation of unsteady flow

Qp < Q0 drawdown, Qp > Q0 filling

0 p 0

p 0 0

Q dt -Q dt =-S dh

Q dt -Q dt =S dh (filling: t1 ↔ t2, h1 ↔ h2)

0 0

0 p p 0

S dh S dhdt =- =

Q -Q Q -Q

the same equation

for drawdown and

filling

1 1

2 2

h h0 0

2 1h h0 p p 0

S dh S dht =t - t = =

Q -Q Q -Q

For Qp const., S0 const., irregular reservoir

numerical solution in intervals t

(drawdown)


Drawdown of prismatic tank (S0 = const.), at Qp= 0

Assumptions:

- outflow from small orifice, mouthpiece, tube

- free level

- S0 >>S → v0 ~ 0

Time of total emptying (h2 = 0):

1

2

h-1/20

hv

St = h dh

S 2g 0

1 2

v

2St = h - h

S 2g

0 1 0 1 1

01v v 1

2S h 2S h 2 VT= = =

QS 2g S 2gh

0 vQ = S 2gh

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Outflow from orifice - Katedra hydrauliky a hydrologie - …hydraulika.fsv.cvut.cz/Hydraulika/Hydraulika/Predmety/...K141 HYAE Outflow from orifice 6 Coefficients for discharge determination