‫بسم هللا الرحمن الرحيم‬
‫قياس التصريف في القنوات المفتوحة‬
Discharge Measurement-Open Chanal-3rd Class
Dr. Sataa A. Al-Bayati (09-10)
Why we convert units, & when?
e.g. s → min → hr
m3/s → L/s
Dictionary:
Orifice = ‫فتحة‬, weir = ‫سد غاطس‬, notch = ‫ثلمة‬
1. Orifice in Open Tanks (free orifice):
Fig. (1) Free orifice
____
Q = Cd A √2gH
--- (1)
Where:
Cd = discharge coefficient = 0.6
Range (0.6 & 0.8)
A = cross-sectional area of orifice
H = difference in elevation from water surface to centroid
of orifice shape (circle, square, etc).
Example (1):
15cm circular orifice in a canal. If the orifice center is lower than water
surface by 1m, find the flow rate.
1
Solution:
____
Q = Cd A √2gH
__________
2
Q = 0.6 × (π/4) (0.15) × √2 × 9.81 ×1
= 0.047m3/s
= 47L/s.
*********
2. Weirs)‫ (السذود الغاطسة‬:
Used in water &wastewater treatment plant.
a. Sharp-crested weir)‫(سذ غاطس حاد الحافة‬
Fig. (2) Sharp- crested weir
__
Q = Cd √2g L. H3/2
__
Q = Cd √2g (L – 0.2H) H3/2
--(2a)- suppressed
--(2b)- contracted(rectangular notch)
for H/P ≤ 10
Cd = 0.4 + 0.05(H/P)
or Cd = 0.745
Where: L = weir length = open channel width
H = height of water surface above weir top.
2
Example (2):
Water flows in a rectangular canal at velocity = 3fps & depth of 1 ft.
Determine height of a sharp-crested suppressed weir that must be installed to
raise water depth upstream of the weir to 4 ft. Take Cd =0.745.
Solution:
__
Q = AV = Cd √2g L. H3/2 ____
= (L × 1)(3) = 3L = 0.745√64.4 L H3/2
H3/2 = 0.502
H = 0.63ft
P = 4 – 0.63 = 3.37ft.
****************
Example (3):
Laboratory measurement made on a contracted sharp crested weir shows a
discharge of 0.25m3/s under a head of H=0.2m. Determine the discharge
coefficient in the given units. Contracted length = 1.56m.
Solution:
__
Q = Cd √2g (L – 0.2H) H3/2
____
0.25 = Cd √19.62 (1.56 – 0.2×0.2) (0.2)3/2
Cd = 0.415.
**********
Application: Time of emptying Tank over a Rectangular Notch:
Fig. (3) Water elevations in rectangular notch
1
t  3A
H2

1
H1
----(3)
Cd  L  2 g
Where:
t = time required to lower water level from H1 to H2, sec
A = surface area of the tank,
3
L = notch width = contracted length,
Cd = discharge coefficient,
H1 = water level at t1 = 0,
H2 = water level at t2 = t.
Example (4):
There is a 100×100m2 tank with a 2m long rectangular notch. Find the
required time to lower water level from 2m to 1m elevation, if Cd = 0.6.
Solution:
1
t  3A
H2

1
H1
Cd  L  2 g
1
t  3100  100

1
1
2
0.6  2  19.62
= 1653sec
= 27.6min.
***********
b. V- Notch-Crested Weir(Triangular Weir)
Used for low flow.
Fig. (4) Triangular Weir
8
 
Q  C d 2 g tan  H 2
15
2
5
----(4)
Where: Cd =0.57-0.6
θ = notch angle, range (22.5 - 90o)
H = head of water over invert weir, ft, range (0.2 - 2ft)
4
Example (5):
V- notch with θ = 90o and water level above invert weir by 0.22m and
Cd =0.6. Calculate the flow rate.
Solution:
__
Q = (8/15) Cd √2g tan(θ/2) H5/2
_____
Q = (8/15) (0.6) √19.62 × (1) × (0.22)5/2
= 0.032m3/s
= 32L/s.
*********
Applications: Time of Emptying a Tank through a Triangular Notch
Fig. (5) Water elevations in V-notch
1
1


 3/ 2 
3/ 2

H2
H1
2

t  A
3 8
K 2 g tan / 2
 15



--- (5)
Example (6):
25×15m2 tank with a 90o triangular notch. Find the required time to lower
water level from 1.5m to 0.5m elevation. Use Cd = 0.62.
Solution:
1
1



3/ 2
3/ 2


2
0.5
1.5

t  25  15
3
 8  0.62  19.62 tan 45 o 
 15



= 390sec = 6.5min.
***************
5
c. Trapezoidal Notch
Fig. (6) Trapezoidal Notch
__
__
3/2
Q = 2/3 × Cd × √2g L H + 8/15 × Cd × √2g tan(θ/2) H5/2
-- (6)
Example (7):
A trapezoidal notch with θ = 60o and base width = 30cm, water level above
invert weir by 16cm and Cd = 0.62. Calculate the discharge.
Solution:
__
__
3/2
Q = 2/3 × Cd × √2g L H + 8/15 × Cd × √2g tan(θ/2) H5/2
____
____
3/2
Q = 2/3 × (0.62)* √19.62 (0.3) (0.16)
+ 8/15 * (0.62)* √19.62 (1/√3) (0.16)5/2
= 0.035 + 0.009
= 0.044m3/s
= 44L/s.
**********
d. Cipolletti Sharp Crested Weir
It is a special case of trapezoidal notch. Used in irrigation canals.
28o
Fig. (7) Cipolletti Sharp Crested Weir
6
_
Q = 3.367 L H3/2
(EI)
--- (7)
Example (8):
A Cipolletti weir with base width = 1ft. Water level above invert weir by
0.5ft. Find the flow rate.
Solution:
Q = 3.367 L H3/2
Q = 3.367 (1) (0.5)3/2
= 1.19ft3/s.
************
e. Broad-Crested Weir)‫(سذ غاطس عريض‬
Fig. (8) Broad-Crested Weir
__
Q = 0.385Cd L √2g H3/2
--- (8)
Where:
Cd = discharge coefficient with range (0.85 – 1.05),
L = weir length,
H = height of water surface above weir top.
b = weir width.
If b ≥ 2H → broad crested weir
7
f. Submerged Weir)‫(سذ غاطس مغمور‬
Qfree
Qsubmerged
Fig. (9) Submerged Weir
Free discharge:
__
Q1 = 2/3 Cd1 √2g L (H1 – H2 )3/2
Submerged discharge:
__________
Q2 = Cd2 L H2 √2g(H1 – H2 )
--(9.a)
--(9.b)
Total discharge = Q1 + Q2
Where:
H1 = upstream water surface above weir top,
H2 = downstream water surface above weir top,
Cd1 & Cd2 = discharge coefficients.
Example (9):
A 60cm high submerged weir in a 2m width channel. Upstream water depth
is 1.1m and downstream water depth = 0.8m. Calculate the discharge if
Cd1=0.6 and Cd2 = 0.8.
Solution:
Free discharge:
__
Q1 = 2/3 Cd1 √2g L (H1 – H2 )3/2
____
Q1 = 2/3 (0.6) √19.62 × 2 × (0.5 – 0.2)3/2
= 0.582m3/s.
8
Submerged discharge: __________
Q2 = Cd2 L H2 √2g(H1 – H2 )
________
Q2 = 0.8 × 2 × 0.2 √19.62(0.3)
= 0.776m3/s.
Total discharge:
Q = Q1 + Q2
= 0.582 + 0.776
= 1.358m3/s.
************
g. Streamlined Triangular Weir) ‫ (سذ غاطس مثلث بخطوط أالنسياب‬:
V2/2g
Fig. (10) Streamlined Triangular Weir
3/2
Q = Cd (2/3)
_
√g L E3/2
-- (10)
Where:
L = weir length
E = total head
= h1 + V2/2g
h1 = upstream head
h2 = downstream water depth
Example (10):
Find the discharge over a streamlined triangular weir in a rectangular canal.
If weir height = 2ft, canal width = 5ft, upstream water level = 0.5ft above
weir crest, and Cd = 0.99.
9
Solution:
_
3/2
Q = Cd (2/3) √g L E3/2
___
3/2
Q = 0.99(2/3) √32.2 × 5 × [0.5 + Q2/2(32.2)(5×2.5)2]3/2
= 15.29(0.5 + 9.94 × 10-5 Q2)3/2
By trial & error,
Q = 5.45cfs.
**********
3. Sluice & Tainter Gate (Underflow Gates):
They used for controlling water in canals, flumes, & on spillways.
Fig. (11) Underflow Gates
____
Q = Cd L Y √2gH
-- (11)
Where:
Cd = discharge coefficient, from Fig.
L = gate width
H = upstream water depth
y = gate opening
y3 = submergence depth
Fig.(12): Flow coefficients for Sluice gate.
Free: find H/y =? → Free curve → Cd
Submergence: find H/y, & y3 /y → H/y cross y3 /y → Cd
10
H/y
Fig.(12) Discharge coefficients for sluice gate.
Example (11):
Estimate the flow under a sluice gate 3m wide. The upstream depth is
5.50m, & the gate opening is 60cm with no downstream submergence.
Solution:
H/y = 5.5/0.6 = 9.17
Fig.(12) free flow curve
Cd = 0.575
____
Q = Cd L y √2gH
________
Q = 0.575 × 3 × 0.6 √19.62×5.5
= 10.75m3/s.
************
11
4. Venture Flume (Parshall Flume))‫(قناة فنجوري‬:
Fig.(13) Venture Flume
0.026
Q = 4B H1.522B
(EI)
-- (12)
Where: Q = flow rate, cfs
B = throat width, ft
H = upper head, ft.
Example (12):
What is the flow through a Parshall flume with a throat width of 2ft & max.
free flow head of 2.5ft?
Solution:
0.026
Q = 4B h1.522B
0.026
1.522(2)
Q = 4 × 2 × (2.5)
= 33.4cfs.
Current meter
Used to measure flow velocity in rivers & canals.
Fig.(14) price current meter
Reference:
- Roberson, J.A., et.al., “ Hydraulic Engineering”, 2nd Ed, John Wiley &
Sons. Inc., New York
12
-
Chow, V. T., 1959, “ Open-channel Hydraulics” Japan.
H.W: 05-06
M
Ex.1: d = 15 + 1(no.) = 16cm
E
H = 1 + 0.1(no.) = 1.1m
Ex.2:
v = 3 + 0.05(no.) = 3.05ft/s
y = 1 + 0.1(no.) = 1.1ft
Ex.3:
Q = 0.25 + 0.1(no.) = 0.26cfs
H = 0.2 + 0.01 = 0.21m
Ex.4:
A = 100+5no. x 100+5no.m2
= 105 + 105m2
L = 2 + 0.1(no.) = 2.1m
Ex.5:
H = 0.22 + 0.01(no.) = 0.23m
Ex.6:
A = 25 x 15m2
= 26 x 26m2
θ = 60o
H = 0.22 + 0.01(no.) = 0.23m
θ = 60o
A = 25 x 15m2
= 26 x 16m2
Ex.7:
H = 16 + 1(no.) = 17cm
θ = 90o
H = 16 + 1(no.) = 17cm
Ex.8:
L = 1 + 0.1(no.) = 1.1ft
H = 0.5 + 0.01(no.) = 0.51ft
Ex.9:
60 – 1(no.) = 59cm
60 → 70cm
y = 1.1 + 0.01(no.) = 1.11cm
13
Ex.10: P = 2 + 0.1(no.) = 2.1ft
Ex.11:
5.5 + 0.1(no.) = 5.6m
Ex.12:
B = 2 + 0.1(no.) = 2.1ft
h = 0.5 + 0.1(no.) = 0.6ft
0.6 + 0.01(no.) = 0.61m
H = 2.5 + 0.1(no.) = 2.6ft
H.W: 06-07
M
Ex.1: d = 16 + 1(no.) = 17cm
E
H = 2 + 0.1(no.) = 2.1m
Ex.2:
v = 4 + 0.05(no.) = 4.05ft/s
y = 2 + 0.1(no.) = 2.1ft
Ex.3:
Q = 0.3 + 0.01(no.) = 0.31cfs
H = 0.3 + 0.01(no.) = 0.31m
Ex.4:
A = 110+5(no.) x 110+5(no.)
= 115 + 115m2
L = 2.5 + 0.1(no.) = 2.6m
Ex.5:
H = 0.25 + 0.01(no.) = 0.26m
θ = 60o
H = 0.25 + 0.01(no.) = 0.26m
Ex.6:
A = 30+1(no.) x 20+1(no.)
= 31 x 21m2
θ = 60o
A = 30+1(no.) x 20+1(no.)
= 31 x 21m2
Ex.7:
H = 18 + 1(no.) = 19cm
θ = 90o
H = 18 + 1(no.) = 19cm
Ex.8:
L = 2 + 0.1(no.) = 2.1ft
H = 0.7 + 0.01(no.) = 0.71ft
Ex.9:
50 – 1(no.) = 49cm
P : 60 → 70cm
y = 1.5 + 0.01(no.) = 1.51cm
14
Ex.10: P = 2.5 + 0.1(no.) = 2.6ft
2
Ex.11: 6 + 0.1(no.) = 6.1m
1.1
h = 0.7 + 0.1(no.) = 0.8ft
0.5
y = 0.8 + 0.01(no.) = 0.81m
Ex.12:
H = 3 + 0.1(no.) = 3.1ft
B = 2.5 + 0.1(no.) = 2.6ft
H.W: 07-08
E
Ex.1: d = 16 + 1(no.) = 17cm
M
H = 2 + 0.1(no.) = 2.1m
Ex.2:
v = 4 + 0.05(no.) = 4.05ft/s
y = 2 + 0.1(no.) = 2.1ft
Ex.3:
Q = 0.3 + 0.01(no.) = 0.31cfs
H = 0.3 + 0.01(no.) = 0.31m
Ex.4:
A = 110+5(no.) x 110+5(no.)
= 115 + 115m2
L = 2.5 + 0.1(no.) = 2.6m
Ex.5:
H = 0.25 + 0.01(no.) = 0.26m
θ = 60o
H = 0.25 + 0.01(no.) = 0.26m
Ex.6:
A = 30+1(no.) x 20+1(no.)
= 31 x 21m2
θ = 60o
A = 30+1(no.) x 20+1(no.)
= 31 x 21m2
Ex.7:
H = 18 + 1(no.) = 19cm
θ = 90o
H = 18 + 1(no.) = 19cm
Ex.8:
L = 2 + 0.1(no.) = 2.1ft
H = 0.7 + 0.01(no.) = 0.71ft
Ex.9:
50 – 1(no.) = 49cm
Ex.10: P = 2.5 + 0.1(no.) = 2.6ft
2
Ex.11: 6 + 0.1(no.) = 6.1m
P : 60 → 70cm
y = 1.5 + 0.01(no.) = 1.51cm
1.1
h = 0.7 + 0.1(no.) = 0.8ft
0.5
y = 0.8 + 0.01(no.) = 0.81m
Ex.12:
H = 3 + 0.1(no.) = 3.1ft
B = 2.5 + 0.1(no.) = 2.6ft
15
16