hydraulics of structures -...
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Hydraulics of structures
K141 HYAE Hydraulics of structures 1
OUTFLOW FROM ORIFICE
K141 HYAE Hydraulics of structures 2
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)
K141 HYAE Hydraulics of structures 3
STEADY FREE OUTFLOW (SFO) OF IDEAL LIQUID
BE surface – outflow:
Torricelli (1608 - 1647) equation for outlet velocity of ideal liquid vi
For large reservoirs with free level:
outlet discharge of ideal liquid Qi:
for small orifice (bottom and wall):
2 20 iv vp
h+ + =g 2g 2g
∆ρ
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 orificesection
overpressure
i iu v≈h… depth of centoid of orifice
hh,0g2
v,0p E
20 =≈=∆
STEADY OUTFLOW FROM ORIFICE
K141 HYAE Hydraulics of structures 4
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
K141 HYAE Hydraulics of structures 5
SFO OF REAL LIQUID FROM ORIFICE AT THE BOTTOM OF TANK
g2v
g2v
gp
gp
g2v
lh2
c2
ca0s2
0c
α⋅ζ+α+ρ
=ρ
+α++
BE 0 - 1
lc ~ 0,5·D
ρ−
ρ+α++⋅
ζ+=
gp
gp
g2v
lhg211
v a0s2
0cc
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=⇒
K141 HYAE Hydraulics of structures 6
- Large orifice hT < (2 - 3)·a ⇒ change of outlet 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 -h
32
Q= 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=⇒
K141 HYAE Hydraulics of structures 7
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 solution dosing,
Q = const.
φ, ε, µv for imperfect and partial contraction > φ, ε, µv for full contraction
empirical formulas
K141 HYAE Hydraulics of structures 8
OUTLET FROM SUBMERGED ORIFICE
For both small and large orifices ofwhatever shape
for small orifice
Note:resolution 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
=
=
K141 HYAE Hydraulics of structures 9
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), extinguishing, 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
K141 HYAE Hydraulics of structures 10
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 t
1y = gt
2
theoretical
20
0
gt21
sinδtvy
cosδtvx
−=
=
δv0cosδ
v 0si
nδ
v0
K141 HYAE Hydraulics of structures 11
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)
K141 HYAE Hydraulics of structures 12
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
K141 HYAE Hydraulics of structures 13
OVERFALL
K141 HYAE Hydraulics of structures 14
TYPES OF SPILLWAYS- sharp-crested, t < 2/3 h, measuring spillways (a)- weir, spillway of practical profile (b,c), streamlined spillway (d)- broad-crested, 2 h < t < 10 h (e)- special - shaft (f), side (g), bed drop (h) ...
(a)
(g)(f) (h)
(c)(b)
(e)
(d)streamlined weir body
h
- overfall: hydraulic phenomenon- spillway: structure, weir
K141 HYAE Hydraulics of structures 15
FLOW OVER WEIRS
• Q… discharge• h… overflow head
(measurements of non-reducedlevel in distance OP = (3 ÷ 4) h)
• v0…approach velocity v0= Q / S0• h0…overfall energetic head
h0 = h + h0d = h + α v02 / 2 g
• b… spillway width, length of crest• b0…active spillway width
(side contraction of overfall jet, b0 < b)• H… gradient head in levels UW and DW• s, sd… spillway height in UW, DW• µp,m… coefficient of discharge (describes hydraulic losses, depends
on type, shape and arrangement of weir and also on further characteristics experiments → TAB.
K141 HYAE Hydraulics of structures 16
• hσ… height of overfall submergence (effect of DW)• σz… coefficient of submergence
- free overfall σz = 1, no effect of DW- submerged overfall σz < 1,
DW → decreases Q → enlarges h
(for submergence of overfall, H < h, i.e. hσ > 0, is not sufficient)
K141 HYAE Hydraulics of structures 17
EQUATION OF OVERFALL
Weisbach equation
Bazin equation
Steady flow, rectangular overflowing jet-width b,
BE for O-P, analogy of outflow from large rectangular orifice in vertical wall.
( ) ( )
( )
1/ 20d 0 p 0 0d
h 1/ 2p 0 0d
0
u= 2g z+h dQ=u b dz= b 2g z+h dz
Q= b 2g z+h dz
ϕ ε µ
µ ∫
free overfall2
300
23
d023
00p
hg2bmQ
hhg2b32
Q
=
−µ=
without side contraction: b0 = b
K141 HYAE Hydraulics of structures 18
⇒ side contraction:
i
n
0 i 01
b= b
b =b-0,1 hξ
∑
∑
n ... number of contractionsξi… pier coefficient (shape, location)
circle-curved intermediate pier ξ = 0,5circle-curved bank pier ξ = 1tapered or streamlined pier with forward nosing ξ ∼ 0
TAB.
⇒ active width b0< clear width b
effect of intermediate piers and wing walls→separation of flow, wakes
K141 HYAE Hydraulics of structures 19
SUBMERGENCE OF OVERFALL
small gradient H-not solved
3 32 2
z p 0 0 0d
32
z 0 0
2Q= b 2g h h
3
Q= mb 2g h
σ µ −
σ
= K,
hs
,hh
fσ0
d
0
σz
TAB., graphs
data in literature are not distinct and reliable for all cases =>submergence of overfallusually worsens accurency of calculations!
K141 HYAE Hydraulics of structures 20
MEASURING WEIRS
Bazin spillway – rectangle without side contraction
32Q=mb 2g h
20,003 h
m= 0,405+ 1+0,55h h+s
Typical shape of Bazin free owerfall jet
air inlet into space bellow jet!
lower envelope streamlined spillway
0,2 < b [m]< 2,0 0,2 < s [m]< 2,00,1 < h [m]< 1,24
• free overfall, sharp crest• standard spillway geometry,
approach rates• analytical formulae for discharge• empirical formulae for discharge
coefficient• discharge measurement:
registration of level → h → m → Q• Q = Q(h) rating curve of spillway
K141 HYAE Hydraulics of structures 21
BRIDGES AND CULVERTS
K141 HYAE Hydraulics of structures 22
HYDRAULIC CALCULATION OF BRIDGE WITH ONE OPENING
yσ > κE → yσ = yd κ - TAB.
(condition for submerged entrance)
( )2gv
ζαy2gvζ
2gαv
yE2σ
σ
2σ
2σ
σ ++=++=
from BE for pf. 1 - 2:
2σ
2
2
σ S2gQ
yEϕ
+=
backwater by bridge:
σ
20 yy∆H;
2gαv
Ey −=−=
Subcritical flow → flow in opening usually affected by DW
K141 HYAE Hydraulics of structures 23
FLOW IN CULVERTS
Flow in culvert is effected by:- geometry, arrangement and hydraulic conditions of entrance- geometry, roughness, longitudinal slope of culvert- downwater position, rates bellow outlet from culvert.
Sizing and flow resolution → simple schemesMore complicated cases → hydraulic jump, inlet vortex, ... →
→ instability of flow
culv
ert
with free level
with pressure flow
with free–surface entrance
with submergedentrance
in whole culvert in part of culvert
affected by DWnon-affected by DW
Q > QD, QD ... capacitive discharge
affected by DWnon-affected by DW
K141 HYAE Hydraulics of structures 24
solution: inlet, flow in culvert, outlet
1) free level and free-surface inlet flow
(a) non-affected by DWyc < 1,25 yk Ø
yc < 1,1 yk □
kc yyD,βy ⋅κ=⋅<
(b) affected by DWyc > 1,25 yk Ø
yc > 1,1 yk □
HYDRAULIC CALCULATIONS OF CULVERTS
( )2
22
2222
Sg2
Q
ϕ+=++=++= 2
22
222 y
2gv
ζαy2gvζ
2gαv
yE
orig
20 yy∆H,
2gαv
Ey −=−=y2 = yc (a), y2 = yσ (b)
according to type of entrance0,9, β (1,2 1,4)Dκ ≈ = −
K141 HYAE Hydraulics of structures 25
2) free level and submerged entrance
(a) non-affected by DWyc < 1,25 yk Ø
yc < 1,1 yk □
(b) affected by DWyc > 1,25 yk Ø
yc > 1,1 yk □
( )2
22
2222
Sg2
Q
ϕ+=++=++= 2
22
222 y
2gv
ζαy2gvζ
2gαv
yE
orig
20 yy∆H,
2gαv
Ey −=−=y2 = yc (a), y2 = yσ (b)
y > β·D, yc = 0,62·D for Ø
yc = 0,60·a for □
β (1,2 1,4)D= −according to type of entrance
K141 HYAE Hydraulics of structures 26
Coefficients for hydraulic calculation of culverts with free level
1,10,870,75without portal, splayed
1,20,900,85common, frontal portalRectangle: width b and height a
1,40,950,95conical, skew wings
1,20,960,85common, frontal portal
Pipe culvert with inner diameter D
βκϕEntranceCulvert
K141 HYAE Hydraulics of structures 27
3) pressure flow in culvert
(a) outlet non-affected by DW
solution:drawdown + short pipeline
Bernoulli equation- short pipeline (at the end of
culvert – „free outlet“)
(b) outlet affected by DW
( ) ( ) min
2
0E Dii ∆−∆++⋅++⋅−=2gv
ζ1LE
( )g
vvv ddmin
−=∆>∆