Uniform Flow Calculation May 2015

Presented by:

TEODORO M. CERALDE Bureau of Design

Contents

Types of Open Channel Flows Common Types of Flood Control Channel Applications of Uniform Flow Calculation Uniform Flow Sample Calculations

Steady flow

Unsteady flow

E.L. E.L. E.L.

E.L. E.L.

W.S. W.S.

W.S.

W.S.

W.S.

UNIFORM GRADUALLY VARIED RAPIDLY VARIED

GRADUALLY VARIED RAPIDLY VARIED

V

V

SOME TYPES OF OPEN CHANNEL FLOW

STEADY UNIFORM FLOW

FLOW THAT IMPLIES THAT THE DEPTH, WATER AREA, VELOCITY AND DISCHARGE DONT CHANGE WITH DISTANCE ALONG THE CHANNEL. IT ALSO IMPLIES THAT THE ENERGY GRADE LINE, WATER SURFACE AND CHANNEL BOTTOM ARE PARALLEL.

IT IS THE FUNDAMENTAL TYPE OF FLOW TREATED

IN OPEN CHANNEL HYDRAULICS.

COMMON TYPES OF FLOOD CONTROL CHANNEL CROSS SECTION

1. TRAPEZIODAL CROSS SECTION - have sloped sides and are formed by excavating in situ

material - usually the most economical channel when ROW is

available 2. RECTANGULAR CROSS SECTION - have vertical or near vertical sides - may be required for channels located in urban areas

where the ROW is severely restricted.

GEOMETRIC PROPERTIES OF COMMON OPEN

CHANNEL SHAPES SHAPE SECTION FLOW WETTED HYDRAULIC

AREA, A PERIMETER, P RADIUS, R

Trapezoidal y(b+ycot) b+ 2y y(b+ycot)

Triangular y2cot 2y ycos sin 2

Rectangular by b+2y by

y b + 2y

Wide flat by b y

sin b+(2y/sin)

y

b

y

b

y

b>>y

VALUES OF MANNINGS COEFFICIENT, n

SURFACE/DESCRIPTION RANGE

MIN. MAX.

1. Natural stream channels (top flood width < 30 m.)

(i) Fairly regular section

a. Some grass and weeds, little or no brush 0.030 0.035

b. Dense growth of weeds, depth of flow

materially greater than the weed height 0.035 0.05

c. Some weeds, light brush on banks 0.035 0.05

d. Some weeds, heavy brush on banks 0.05 0.07

e. Some weeds, dense trees 0.06 0.08

f. For trees within channel, with branches

submerged at high flood, increase all

above values by 0.01 0.02

(ii) Irregular sections, with pools, slight

channel meander; increase values given

above about 0.01 0.02

VALUES OF MANNINGS COEFFICIENT, n

SURFACE/DESCRIPTION RANGE

MIN. MAX.

(iii) Mountain streams, no vegetation in channel,

banks usually steep, trees and brush along

banks submerged at high flood:

a. Bottom of gravel, cobbles and few boulders 0.040 0.05

b. Bottom of cobbles, with large boulders 0.050 0.07

2) Larger stream channels (top flood width >30m)

Reduce smaller stream coefficient by 0.010

3) Flood plains (adjacent to stream beds)

Pasture, short grass, no brush 0.030 0.035

Pasture, tall grass, no brush 0.035 0.050

Cultivated land- no crop 0.030 0.040

Cultivated land- nature field crop 0.045 0.055

Scrub and scattered bush 0.050 0.070

Wooded 0.12 0.16

VALUES OF MANNINGS COEFFICIENT, n

SURFACE/DESCRIPTION RANGE

MIN. MAX.

4) Man made channels and ditches

Earth, straight and uniform 0.017 0.025

Grass covered 0.035 0.050

Dredged 0.025 0.033

Stone lined and rock cuts, smooth and uniform 0.025 0.035

Stone lined and rock cuts, rough and irregular 0.035 0.045

Lined- metal corrugated 0.021 0.024

Lined- smooth concrete 0.012 0.018

Lined- grouted riprap 0.017 0.030

UNIFORM FLOW FORMULA Applicable for rivers with the following conditions:

a. When there are no points of abrupt change of riverbed gradients

b. When there are no structures/obstruction that impede the flow discharge

c. When the cross- sectional area of the river is almost the same longitudinally

d. When there is relatively long straight river reach.

Applications of the uniform flow calculation

Case 1. For selecting the type of revetment

Case 2. For calculating existing channel capacity (Qcap)

Case 3. For determining the discharge (Qmax) corresponding to the max. experienced flood level

Case 4. For determining the water levels corresponding to design discharges.

Case 1. For selecting the type of revetment

In selecting the type of revetment, it is necessary to calculate the flow velocity of the channel.

The flow velocity shall be calculated under the following conditions:

1. Design water level

2. Design cross section

Case 2. For calculating existing channel capacity (Qcap) In order to determine any countermeasures along the channel,

it is essential to determine the existing capacity of the channel. The discharge capacity shall be compared with the maximum experienced flood (Qcap) and the calculated discharges for a specific return periods.

Existing Capacity of a Simple Cross Section Channel

Qcap = AV = 1/n(A/P)2/3(S)1/2 x A = A5/3S1/2/nP2/3

=(A1+A2+A3+.An)5/3xS1/2/n(P1+P2+P3+Pn)2/3

Case 2. For calculating existing channel capacity (Qcap) Existing Capacity of a Compound Cross Section Channel

Qcap = A1V1 + A2V2 +A3V3

V1 = (1/n1)(A1/P1)2/3(S)1/2

V2 = (1/n2)(A2/P2)2/3(S)1/2

V3 = (1/n3)(A3/P3)2/3(S)1/2

High Water Channel

Low Water Channel

High Water Channel

Case 3. For determining the discharge (Qmax) corresponding to the maximum experienced flood level

Based on the flood mark of the maximum experienced flood, corresponding discharge shall be determined by the uniform flow calculation under two (2) considerations:

1. No overtopping of floodwater of the riverbanks

2. Overtopping of floodwater of the riverbanks.

Case 3. For determining the discharge (Qmax) corresponding to the maximum experienced flood level

1. No overtopping of floodwater of the riverbanks

2. Overtopping of floodwater of the riverbanks.

Max Water level Left bank Right bank

Left bank

Left bank

Right bank

Right bank

Case 4. For determining the water levels corresponding to the design discharges The uniform flow calculations will determine the water levels

corresponding to the calculated discharges of a specific return periods.

The maximum experienced flood level will serve as the reference level in plotting the water levels of the calculated discharges

MEFL

WL25yr

WL10yr

EXAMPLE PROBLEM NO. 1

The flow is uniform in a trapezoidal channel at a depth of 1.5 m. The bottom width is 15.0 m., side

slopes of 1:1, bottom slope is 0.0001 and the

Manning's n is 0.02. Determine the discharge

capacity of the channel.

Figure:

d = 1.5 m

b = 15 m

1

1

water level

Solution

Given:

b =15 m

d =1.5 m

s =0.0001

n =0.02

From the eq.

V = 1/n(R^2/3)(s^1/2) for metric system

It is known that Q = AV, so

Q = (1/n)A(R^2/3)(s^1/2)

Knowing that R = A/P and referring to the Figure,

A = (0.5)(15+18)(1.5) = 24.75 sq. m.

P = (1.5)(2^0.5)(2)+(15) =19.24264 m.

so

R = A/P = 1.286206 m

therefore

Q = (1/0.02)(24.75)(1.29^2/3)(0.0001^0.5) =14.66465

cu. m./sec

EXAMPLE PROBLEM NO. 2

If the channel shown below has the following properties,

n(a,b), n(b,c), n(f,g), n(g,h) = 0.03 and n(c,d), n(d,e),

n(e,f)= 0.025; slope= 1/2000.

Determine the discharge in the channel.

Figure:

d1

d2

30 m 38 m 44 m

1 2 1

1

a

b c

d e

f g

h water level

Solution

Given: b1 = 30m. b2 = 38m. b3 = 44m.

d1 = 3 m. = d2 s = 0.0005

From the eq., V = 1/n(R^2/3)(s^1/2) for metric system

It is known that Q = AV, so Q = (1/n)A(R^2/3)(s^1/2)

Knowing that R = A/P and referring to the Figure,

A1 = say the area of the high water channel

A1 = (0.5)(30+36)(3) + (0.5)(44+50)(3) = 240 sq.m.

P1 = [(3)(5^0.5)+(30)] + [(44)+ (3)(5^0.5)] =87.41641m

so

R1 = A1/P1 =2.74548 m, therefore

Q1 = (1/0.03)(240)(2.74548^2/3)(0.0005^0.5)

= 350.7415 cu. m./sec

A2 = say the area of the low water channel

A2 = (0.5)(38+44)(3) + (44)(3) = 255 sq.m.

P2 = [(3)(2^0.5)(2)+(38)] = 46.48528 m.

so

R 2= A2/P2 = 5.485607 m.

therefore

Q2 = (1/0.025)(255)(5.485607^2/3)(0.0005^0.5)

= 709.4171 cu. m./sec

Total Q = Q1 + Q2

Total Q = 1060.159 cu.m./sec

EXAMPLE PROBLEM NO. 3

Determine the design flood level of the channel (trapezoidal

section) from the given data below:

upstream elev = 58.612 m

downstream elev = 58.598 m

length of stream, L = 50.0 m

Figure:

DFL Elev. 70.453

Elev. 60.3 d

22.0 m

1 1

Solution

Given:

Q = 137 cum./sec s = 0.00028 b = 22 m.

ss = 1H:1V n = 0.03 arb = el. 60.3

Try depth of flow, d1 = 4.0 m

From the eq.

V = 1/n(R^2/3)(s^1/2) for metric system

It is known that Q = AV, so, Q = (1/n)A(R^2/3)(s^1/2)

Knowing that R = A/P and referring to the Figure,

A1 = (0.5)(22+30)(4) = 104 sq.m.

P1 = (4)(2^0.5)(2)+(22) = 33.31371 m.

so

R1 = A1/P1 = 3.121838 m.

therefore

Q1 = (1/0.03)(104)(3.121838^2/3)(0.00028^0.5)

= 123.9076 cu. m./sec < 137 cu. m./sec

Try depth of flow, d2 = 4.5 m

A2 = (0.5)(22+31)(4.5) = 119.25 sq. m.

P2 = [(4.5)(2^0.5)(2)+(22)] = 34.72792 m.

so

R 2= A2/P2 = 3.433836 m.

therefore

Q2 = (1/0.03)(119.25)(3.433836^2/3)(0.00028^0.5)

= 151.39186 cu. m./sec > 137 cu. m./sec

By interpolation:

y/(Q2-Q) = (d2-d1)/(Q2-Q1)

y/(151.39-137) = (4.5-4.0)/(151.39-123.91)

y = 0.261827 m.

therefore

d = d1 + y

d = 4.261827 m.

DFL = 60.3+ 4.26187

DFL = 64.56187 m

EXAMPLE PROBLEM NO. 4

Given the properties of the earth canal shown, find the discharge.

Figure:

d1

b1 b2 b3 b4

d3

d2

n = 0.023 n = 0.030

s = 0.00042 s = 0.00042

water level

Solution

Given:

b1 = 6 m. b2 = 20 m. b3 = 4 m.

b4 = 150 m. d1 = 4 m. d2 = 3 m.

d3 = 1 m. n1 = 0.023 n2 = 0.03

From the eq.

V = 1/n(R^2/3)(s^1/2) for metric system

It is known that Q = AV, so

Q = (1/n)A(R^2/3)(s^1/2)

Knowing that R = A/P and referring to the Figure,

A1 = say the area of the main section

A1 = (0.5)(4)(6) + (20)(4) + (0.5)(1+4)(4) = 102 sq.m.

P1 = ((6^2+4^2)^0.5)+(20)+((4^2+3^2)^0.5) =32.2111 m

so

R1 = A2/P2 = 3.16661 m.

therefore

Q1 = (1/0.023)(102)(3.16661^2/3)(0.00042^0.5)

= 195.987 cu. m./sec

A2 = say the area of the overflow section

A2 = (150)(1) = 150 sq.m.

P2 = (150+1) = 151 m.

so

R2= A2/P2 = 0.9934 m.

Q2 = (1/0.030)(150)(0.9934^2/3)(0.0002^0.5)

= 102.0181 cu. m./sec

Total Q = Q1 + Q2

Total Q = 298.005 cu.m./sec

Thank You and God Bless