outflow from orifice -...

K141 HYAE Outflow from orifice 0

OUTFLOW FROM ORIFICE


TYPES OF OUTFLOW

Outflow

Outflow

steady: z = const, hE = const(H = const, HE = const) Qp = Q

quasi-steady: z ~ const., phenomenon of large reservoirunsteady: z ≠ const (H ≠ const)

Qp ≠ Q, filling and drawdown of tank (reservoir)

free (a) → free outlet jetsubmerged (b) → submerged outlet jetpartly 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 vph+ + =

g 2g 2gΔρ

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,Qipa(=0)

S orificesection

overpressure

i iu v≈

h… depth of cente of orifice

hh,0g2

v,0p E

20 =≈=Δ

STEADY OUTFLOW FROM ORIFICE


Hydraulic lossesoutlet loss ζv ... depends on shape, setup and

size of orifice (structure), Re

CONTRACTION OF OUTLET JET

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

well mouthed orifice

partial contraction

re-entrant streamlined mouthpiece

external mouthpiece ∅D

sharp edged orifice

TAB.

imperfect contraction

2

vvcZ=2g

ζ


SFO OF REAL LIQUID FROM ORIFICE AT THE BOTTOM OF TANK

g2v

g2v

gp

gp

g2vlh

2c

2ca0s

20

cα⋅ζ+

α+

ρ=

ρ+

α++

BE 0 - 1

lc ~ 0,5·D

⎟⎟⎠

⎞⎜⎜⎝

⎛

ρ−

ρ+

α++⋅

ζ+=

gp

gp

g2vlhg2

11v a0s

20

cc

φ ... velocity coefficientα

c c c vQ v S , S S, ... orifice discharge

coefficient= ⋅ = ε ⋅ ε ⋅ ϕ = μ

contraction coefficientφ, μv, ε ... TAB.

Simplification:free level → ps0 = pa →S0 >> S → v0 ~ 0lc << hE → lc ~ 0

0g

pp a0s =ρ−

hg2SQ,hg2v

v

c

⋅⋅μ=

⋅ϕ=⇒


- 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= 2ghϕ1/2

v EQ= 2g h dSS

μ ∫

E2

E1

h1/2

v E Eh

Q= b 2g h dhμ ∫ ( )( )

3/2 3/2v E2 E1

3/2 3/2v 2 1

2Q= b 2g h -h32Q= b 2g h -h3

μ

μ

EdS=bdh S=ba

hh02gv

E

20 =⇒≈

SFO OF REAL LIQUID FROM ORIFICE IN VERTICAL WALL OF TANK

- Small orifice hT > (2 - 3)afor S0 >> S → v0 ~ 0 tv

tc

hg2SQ

,hg2v

⋅⋅μ=

⋅ϕ=⇒


Coefficients for discharge determination

- small sharp-edged orificewith full contraction 0,97 0,63 0,61

- external cylindrical mouthpiece L/D = 2 ÷ 4 0,81 1,00 0,81- streamlined mouth piece jet tube 0,95 1,00 0,95- large orifices at the bottom with significant 0,65 až 0,85

or continuous side contraction

- outlet tube of diameter D and length L with free outflow v

i

1=L1+ +D

μλ ζ∑

ϕ ε μv

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 ofwhatever 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 reservoirH = H0

02gHvu ϕ==

2gHSμQ

2gHSμQ

v

0v

=

=


OUTFLOW JETSFree 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), extinguishing, irrigation jets …

- submerged jets - dosing, mixing, rectifying, …

type: water - air

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

L = sin2 =2h sin2g

δ δ

22 20

0 dv

y = sin =h sin2g

δ δ

20

dv

=h2g

energetic head of jet

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

For δ = X° a δ = 90 -X° → same arcing distanceFor δ = 90° vertical jet → y0max = v0

2/2g = hd

For δ = 0° horizontal jet (horizontal projection)

real liquid, large reservoirp d TL =2 h y

p T TL =2 h yϕ

0

2

x =v t1y = gt2

theoretical

20

0

gt21sinδtvy

cosδtvx

−=

=

δv0cosδ

v 0si

nδ

v0


UNSTEADY OUTFLOW FROM ORIFICE

Differential equation of unsteady flowQp < 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 -Qthe 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 Δt

(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 2gμ∫ ( )0

1 2v

2St = h - h

S 2gμ

0 1 0 1 1

01v v 1

2S h 2S h 2 VT= = =QS 2g S 2ghμ μ

0 vQ = S 2ghμ

outflow from orifice -...

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