HY D-RAULIC
PROPE.RTlES
BOXES,
AND
OF- PIPE,
RECTANGULAR CHANNELS
BUREAU OF ENGINEERING
City of Leg Ar?ge!es
LYALL A. PAROLE:
City Engineer
OFFICE
STANDARDS
STORM
DRAIN
No. 116 and 117
DESIGN
DIVISION
STORM DRAIN DZSIGN DIVISION
OFFICE STAXD.;IRD
NO. 116
rnaly of the hydraulic properties
of part-full pipe us:ially found
The
in individual tables.
tabular values range from j.
depth/diameter rd.ti.0of .COIito
0.549 in increments of O.Wl.
The tables dre b;ised upon the following formula:
The vdlues for tl-ieconvedace
reciprocal v.IIi~c
*.w,
lJ
factor
Aii2'3Z.rldt'r:e
Da/3
are btlsed upon the crlssumptionthat
I .486~R'~
uniform flow condition=; exist.
5h
813
Tr, %cili.t,titte
tht: calculations, vdiues for D , D
,
D
‘13
9
form after tne T;;rbleof I",ydrstiiic
D", dre included in t~~;iijular
D2,
PropertiG:s.
The
viilui.:;;v dr'y
i'?orn t
inches to 12 Incne3 iii
Increments oi' 2 inches; from 13 iriches to ij4Inches in incremerlts
Oi’
:j
inches; and from >jG inche;; to 144 inches I:i increments of
6 inches.
-
l-
D
d
A
R
s"
TW
dc
hvc
%
P
Discharge, cubic feet per second (cfs.)
Pipe Diameter, feet
Depth of flow, feet
Area of water, square feet
Hydraulic radius, feet
Mannings roughness coefficient
Slope of energy gradient, feet per foot
Top width of water surface, feet
Critical depth, feet
Velocity head where d = d,
Critical slope where d = d,
Static pressure, cubic feet
EXPLANATION OF TABLES
Column 1
$;
ratio of depth to diameter
Column 2
-$
value x D2 = Area
Column 3
;;
value x D = Hydraulic Radius
Column 4
K, =
(1)
AR""; Q = K, (l.;,,)
e/,'
D
Given:
$3
$2
Q, D, S, n
Required:
d, A, V
Solve for K,.
From Table read $, -.k,
D2
Solve for d, A, V
(2)
Given:
d, D, S, n
Required:
Q, A, V
Determine $.
From Table read K,, A
D2
Solve for Q, A, V
-2-
Column 5
K2 =
Dti . S
1.486AR"'
Given:
d, D, n, Q
Required: S, A, V
Determine g.
Solve
Column 6
(1)
for S, A, V
value x D = Top width
%
Column '7
From Table read K,, A
D2
xD
2.
value x D6k= Q (d = dc)
Given:
Q, D
Required:
d,, A V,
Determine 3;
D
dc
From Table read D,
A
D'
Solve for d,, A, V,
(2)
Given:
d,, D
Required:
Q, A, Vc
Determine g;
From Table read &,
D=/2
Solve
Column 8
hVC
7);
4
D-d
for Q, A, V,
value x D = Velocity head (d = dc)
-3-
Column 9
S,D'jS value x n2
yz--;
D‘13
Given:
=
S,(d = dc)
Q, D, n
Required:
S,
Determine Q
-%*
D
d,
From Table read D,
Solve for S,
Column 10
value x D3 = Pressure
IRC
12-l-66
-4-
S,D"3
n
HYDRAULIC PROPERTIES OF PART-FULL PIPE
dD
v
A
R
0
AR%
m=Kc
0%
,_a6AR%=K2
TW
7
Q
DH
hi
D
ScD”
n2
P
7
.ooooo
.0013
.0015
.OOl?
.0019
.0021
.0024
.0026
.0029
.0032
.0034
.0037
.0066
.0072
.0079
.0086
.0092
.0099
.0105
00112
.0118
.0125
.0132
.00004
.00005
.00006
.00008
.00009
.OOOll
.00012
.00014
.00016
.00018
.00020
14360.1690
11687.7140
9685.1971
8147.9219
6943.3754
5982.8586
5205.2172
4567.2235
4037.6460
3593.4682
3217.4274
.1989
.2086
.2177
.2265
.2349
.2431
.2509
.2585
.2659
.2730
.2799
.0006
.0007
.0008
.OOlO
.0012
.0013
r0015
.0017
.0019
.0022
.0024
.0033
.0036
.0040
.0043
.0046
.0050
.0053
.0056
.0060
.0063
.0066
76.7099
74.3963
72.3505
70.5229
68.8763
67.3819
66.0169
64e7632
63.6061
62.5334
61.5352
.021 .0040
.022 .0043
.023 .0046
.024 .0049
.025 .0052
.026 .0055
cO27 .0058
.028 .0061
.029 .0065
.030 .0068
.0138
.0145
.0151
.0158
.0164
.0171
.Ol77
.0184
.0190
.0197
.00023
.00025
.00028
.00030
.00033
.00036
.00039
.00043
.00046
.00050
2896.4142
2620.2916
2381.1385
2172.7068
1989.9971
1828.9900
1686.4119
1559.5804
1446.2835
1344.6802
.2867
.2933
.2998
.3060
.3122
.3182
.3241
.3299
.3356
.3411
.0027
.0029
.0032
.0035
.0038
.0041
.0044
.0048
.0051
.0055
.0070
.0073
.0077
.0080
.0083
.0087
.0090
.0093
.0097
.OlOO
60.6030
59.7297
58.9092
58.1363
57.4066
56.7159
56.0611
55.4389
54.8468
54.2823
.00003
.00003
.00004
.00004
.00005
.00005
.00006
.OOOOh
.00007
.00008
.031
,032
.033
.034
.035
.036
.037
.038
.039
.040
.0203
.0210
.@216
.0223
.0229
.0235
.0242
.02fi8
.0255
.0261
.00053
.00057
.00061
.00065
.00@69
.00074
.00078
.00083
.00087
.00092
1253.2308
1170.6396
1095.8092
1027.8064
965.8336
909.2057
857.3313
809.6985
765.8606
725.4307
.3466
.3519
.3572
.3624
.3675
.3725
.3775
.3823
.3871
.3919
.0058
.0062
.0066
.0070
.0075
.0079
.0083
.CO88
.0093
.0098
.0103
.0107
.OllO
.0114
.0117
.0120
.0124
.0127
.0131
.0134
53.7434
53.2282
52.7350
52.2622
51.8@85
51.3726
50.9534
50.5498
50.1610
49.7860
.00008
.00009
.OOOlO
.OOOll
.00012
.00013
.00013
.00014
.00015
.00016
.0268
.0274
.0280
.0287
.0293
.0300
.0306
.0312
.03?9
.0325
.00097
.00102
.GO108
.00113
.00119
.00125
.00130
.0,0137
.00143
.00149
688.0646
653.4675
621.3739
591.5488
563.7867
537.9026
513.7333
491.1320
469.9670
450.1210
.3965
.4011
.4057
.4101
.4146
.4189
.4232
.4275
.4317
.4358
.0102
.0108
.0113
.0118
.0123
.0129
.0135
.ol.40
.0146
.0152
.0137
.0141
.0144
.0148
a0151
.0154
.0158
.0161
.0165
.0168
49.4240
49.0743
48.7363
48.4092
48.0927
47.7859
47.4886
47.2003
46.9204
46.6487
.00017
.odo19
.00020
.0@021
.00022
r00023
.00025
.00026
.00028
.00029
.OlO
.011
.012
.013
.014
.015
,016
.017
.018
.019
.020
.0072
.0075
.0079
.0082
.0086
.0090
.0093
.0097
.0101
.0105
.041 .0109
.042 .0113
.043 .0117
.044 .0121
.045 .0125
.046 .01.?9
.047 .0133
.048 .0138
.049 .0142
.050 .0146
-5-
.00000
.ooooo
.00001
.00001
.00001
.00001
.00002
.00002
.00002
.00003
STORM DRAIN DESIGN DIVISION
OFFICE STANDARDNO. 117
HYDRAULIC PROPERTIES
OF
RECTANGULAR CHANNELS
AND
BOX STRUCTURES
This table combines many of the hydraulic properties of
rectangular channels and box structures usually found In individual
tables. The tabular values range from a depth/width ratio of 0.020
to 1.500 1n Increments of 0.001.
The structures have a 'V" shape Invert with a slope
toward the center of 0.04.
The
usual fillets In the upper cor-
ners of the box section have been ignored.
.
b
b
w
.
t-
d
s = 0.04
S
Box
Channel
The tables are based upon the following formula:
For the open channel, the values for the conveyance
Y3
bQ3
are based upon the
factor AR and the reciprocal value
1 . 486A~~~
bW3
assumption that uniform flow conditions exist.
The box structure
is assumed to be flowing with a full wetted perimeter.
To facilitate the calculations, values for b4"",
b6',b'3,
b2, b3, are Included in tabular form after the Tables of Hydraulic
Properties. The values vary from 4.00 to 25.00 In increments of
0.,25.
-
26 -
Formulas for the hydraulic properties included in the
tables follow the section titled "Explanation of Tables".
Symbol
Q
b
d
A
R
n
S
de
Sc
P
Discharge, cubic feet per second (cfs.)
Width of channel or box, feet
Depth of flow, or depth of box, feet
Area of water, square feet
Hydraulic radius, feet
Manning's roughness coefficient
Slope of energy gradient, feet per foot
Critical depth, feet
Critical slope, when d=d,
Static pressure, cubic feet
EXPLANATION OF TABLES
Channel and Box
Column 1
d.
->
b
Column 2
F'
ratio of depth to width
A.
'value x b2 = Area
Channel
Column 3
R
-_;
b
Column 4
Kl
(1)
value x b = Hydraulic Radius
=AR2/3.
b93 ’
Given:
Q
Q, b, S, n
Required:
d, A, V
From Table read d* A
b' F
Solve for K,.
Solve for d, A, V
(2)
Given:
d, b, S, n
Required:
Q, A, V
A
From Table read K,, i;z-
Determine g.
Solve for Q, A, V
-
27
-
Column 5
value x be - Q
(1)
Given:
Q, b
Required:
d,, A, Vc
From Table read d , $
+
Determine Q
i?=
Solve for dc, A, Vc
(2)
Given:
d,, b
Required:
Q, A, V,
%
Determine b.
From Table read p?F
Q
A
Solve for Q, A, V,
Column 6
&b"=_ value x n* -_S, (d = dc)
7
b‘h
Given:
Q,
b, n
Required: S,, A, V,
dc
From Table read b
De.terml.neQ
35'
S,d/"
* Solve for S,, A, V,
-7n
Column 7
Column tj
P.
value x b3 = Pressure
i7
K2 =
be/, .
1.486AR'
Given:
S
d, b, n, Q
Required: S, A, V
From Table read K2, A
7
Determine $.
Solve for S, A, V
-
28
-
Box
value x b = Hydraulic radius
Column 9
Column 10
K3
(1)
=
AR'/"..
Q_K
-p’
Given:
Q, b, S, n
Required:
d, A, V
From Table read d A
6' p
Solve for K3.
Solve for d, A, V
(2)
Given:
b, d, S, n
Required: Q, A, V
Determine g.
From Table read K,, -!&
Solve for Q, A, V
-
29
-
FORMULAS
s= 0.04
t
S
3
Channel
Box
Channel
1.
Area
A zbd-$
- $b2= b2 (x-0.01)
qb2
$=
2.
x-o.01
Netted Perimeter
'JJ.P.
= 2 (d-3) + 2 [($
+ ($br]"
= 2 (xb-gb) + 2 @)(1+sll"
r=b [(2x-S) + (l+S2)""]
Assume 1+S2 -1
W.P. = b (2x-S+l)
= b (2x+0.96)
3.
Hydraulic Radius
R=
$k
b2 (
1
= b (2:i::;fi)
=: b(“-0.01)
(2x+0.96)
x-o.01
-2!xr$E
- 30 -
=x
4.
Conveyance
Fat tor
*k’/3 = b2(x-0.01)
5.
Reciprocal
1
Conveyance
1.4&
6.
Factor =
2x+0. g6)2’3
(bvs)(X-.Ol)a
Plow At Critical Depth
&=
A3 “2
f\
\
bl
= [gb6(x-b0.CUf]v2
= 5.t5'75b5'*
(x-O.Olp
$,*=
‘7
l
5.675
(x-O.Olp
Critical Slope
S = (1.4~~AR~2where
6.
&=
5.675
($r
Pressure
P = “j;s; _ (bd) (;) - $
- 31 -
[d; (q)]
Sb2d + $
t
b3x2
= - 2 - 0.01b3x + .oooo667b3
P
F
=-- X2
2
0.01x
+
0.0000667
Box
9.
Area
A = b2(x-0.01)
A = x-o.01
i7
10.
Netted Perimeter
W.P.
= 2d-Sb+2b
= 2xb-Sb+2b = b(2x+l.$h)
11. Hydraulic Radius
R=
=
R
-=
b
12.
x-o.01
'Yf5zXqT
Conveyance Factor
ARe'3
= b*(x-0.01)
IRC
12-l-66
-
32
-