Outflow from orifice
TYPES OF OUTFLOW
Outflow
Outflow
K141 HYAE
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)
Outflow from orifice
1
STEADY OUTFLOW FROM ORIFICE
STEADY FREE OUTFLOW (SFO) OF IDEAL LIQUID
BE surface – outlet:
p v 02 v i2
h+
+ =
g 2g 2g
overpressure
S orifice
section
ui,vi,dQi,Qi
pa(=0)
v i2
hE =
2g
v i = 2 ghE
Torricelli (1608 - 1647) equation for
outflow velocity of ideal liquid vi
for large reservoirs with free level:
2
v0
vi = 2 gh
p  0,
 0, hE  h
2g
outflow discharge of ideal liquid Qi:
Qi =  dQi =  ui dS
S
S
for small orifice (bottom and wall): ui  vi
Qi = v i  dS
Qi = vi S=S 2 gh
h… depth of cente of orifice
S
K141 HYAE
Outflow from orifice
2
CONTRACTION OF OUTLET JET
Strip area Sc < S, Sc =  · S, contraction coefficient   1
sharp edged
orifice
well mouthed
orifice
re-entrant
streamlined
mouthpiece
Hydraulic losses
2
v
outlet loss
Z=  v c
2g
K141 HYAE
partial
contraction
TAB.
imperfect
contraction
external
mouthpiece D
v ... depends on shape, setup and
size of orifice (structure), Re
Outflow from orifice
3
SFO OF REAL LIQUID FROM ORIFICE AT THE BOTTOM OF TANK
BE 0 - 1
v 0 2 p s 0 p a v c 2
v c 2
h  lc 




2g
g g
2g
2g
lc ~ 0,5·D
2


1

v
p
p
0
s
0
a
vc 
 2 g  h  lc 

 


1 
2
g

g

g



 ... velocity coefficient
Q = vcSc, Sc = εS, εφ = μv … orifice discharge
coefficient
contraction coefficient
φ, μv, ε ... TAB.
Simplification:
free level → ps0 = pa →
S0 >> S → v0 ~ 0
lc << hE → lc ~ 0
K141 HYAE
p s0  pa
0
g
 v c    2 g h,
Q  v  S  2 gh
Outflow from orifice
4
SFO OF REAL LIQUID FROM ORIFICE IN VERTICAL WALL OF TANK
- Large orifice hT < (2 - 3)·a
 change of outflow velocity u
with height of orifice
u=  2 ghE
Q =  v 2 g  h1/2
E dS
S
- Open reservoir and large rectangular orifice in vertical wall:
dS=b dhE S=ba
2
hE2
3/2
3/2
Q
=
 v b 2 g hE2
- hE1
1/2
Q =  v b 2 g  hE dhE
3
hE1
2
v 02
3/2
Q
=
 v b 2 g h3/2
h
 0  hE  h
for large tank:
2
1
3
2g


- Small orifice hT > (2 - 3)a
for S0 >> S → v0 ~ 0
K141 HYAE

v c    2 g ht ,
Q   v  S  2 g ht
Outflow from orifice
5


Coefficients for discharge determination

- small sharp-edged orifice
with full contraction
- external cylindrical mouthpiece L/D = 2  4
- streamlined mouthpiece jet tube
- large orifices at the bottom with significant
or continuous side contraction
- outlet tube of diameter D and length L
with free outflow

v
0,97
0,81
0,95
0,63
1,00
1,00
0,61
0,81
0,95
0,65 to 0,85
v =
1
1+ 
L
+  i
D
φ, ε, μv for imperfect and partial contraction > φ, ε, μv for full contraction
empirical formulas
Note:
special application of outflow through mouthpiece - Mariotte vessel - with function of dilution dosing,
Q = const.
K141 HYAE
Outflow from orifice
6
OUTFLOW FROM SUBMERGED ORIFICE
for both small and large orifices of
whatever shape
u  v   2gH0
for small orifice
Q  μv S 2gH0
for large reservoir
H = H0
Q  μv S 2gH
Note:
solution for partial submergence: Q = Q1 + Q2
(Q1 outflow from free part of orifice, Q2 outflow from submerged
part of orifice).
K141 HYAE
Outflow from orifice
7
OUTFLOW JETS
Free outflow jet
theoretical trajectory (parabola 2°)
connected part
type: water - air
decay of jet, aeration, drops
Supported outflow jet
Submerged outflow jet
pulsating margin of
boundary layer
(mixing regions)
type: water – air – solid surface
jet core with constant velocity
type: water - water
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, …
K141 HYAE
Outflow from orifice
8
Theoretical shape of outflow jet (projection at an angle)
arcing distance of jet
x  v 0 t cosδ
v0sinδ
1
y  v 0 t sinδ  gt 2
2
hd
x = v0 t
1 2
y= gt
2
K141 HYAE
v0
δ
v0cosδ
v 02
Lp0 = sin 2 =2hd sin 2
g
maximum height
v 02
y0 =
sin2  =hd sin2 
2g
v 02
=hd energetic head
of jet
2g
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)
theoretical Lp =2 hd y T
real liquid, large reservoir
Lp =2  hT y T
Outflow from orifice
9
UNSTEADY OUTFLOW FROM ORIFICE
Qp < Q0 drawdown,
Qp > Q0 filling
Differential equation of unsteady flow
Q0 dt - Qp dt =- S0 dh (drawdown)
Qp dt - Q0 dt =S0 dh (filling: t1 ↔ t2, h1 ↔ h2)
S0 dh
S0 dh
dt ==
Q0 - Qp Qp - Q0
the same equation
for drawdown and
filling
h1
S0 dh h1 S0 dh
t = t 2 - t1 = 
=
h2 Q0 - Qp h2 Qp - Q0
For Qp  const., S0  const., irregular reservoir 
 numerical solution in intervals t
K141 HYAE
Outflow from orifice
10
Drawdown of prismatic tank (S0 = const.), at Qp= 0
Assumptions:
- outflow from small orifice, mouthpiece, tube Q0 = v S 2 gh
- free level
- S0 >>S → v0 ~ 0
t=
h1
S0
v S 2 g
-1/2
 h dh
t=
h2

2g
2S0
v S
h1 - h2

Time of total emptying (h2 = 0):
2S0 h1
2S0 h1
2 V1
T=
=
=
 v S 2 g  v S 2 gh1 Q01
K141 HYAE
Outflow from orifice
11