1.

Example Problem Velocity & Discharge Channel geometry known Depth of flow known Determine
the flow velocity and discharge ?

Given required
Bed slope of 0.002 ft/ft Q = va

2/3 1/2
Manning’s n of 0.04 V=(1.5/n) R S

A = width x depth

R= a/P

Solution

2
A= BY = 20 x 1.5 ft A=30 ft

P= B+ 2Y P= 20 + 1.5 + 1.5 ft P= 23 ft.

R= a/P R= 30/23 R= 1.3 ft

S = 0.002 ft/ft (given) and n = 0.04 (given)

2/3 1/2 2/3 1/2

V=(1.5/n) R S v = (1.5/0.04)(1.3) (0.002) V= 2 ft/s

3
Q = va =2x30 Q = 60 ft /s or 60 cfs

Answer: the velocity is 2 ft/s and the discharge is 60 cfs

2. Example Problem Velocity & Discharge Discharge known Channel geometry known
Determine the depth of flow?

35ft

? ft
Discharge is 200 cfs
 Bed slope of 0.005 ft/ft
Stream on a plain, clean, winding, some pools and stones

1. q = va
2. v =(1.5/n) R2/3 S1/2
3. R= a/P Guess a depth! Lets try 2 ft
a = width x depth
= 35 x 2 ft = A=70 ft2
P= 35 + 2 + 2 ft P = 39 ft.
R= 70/39 R = 1.8 ft
S = 0.005 ft/ft (given)
n = 0.033 to 0.05 (Table 7.1) Consider deepest depth
v = (1.5/0.05)(1.8)2/3(0.005)1/2
V = 3.1 ft/s
q = va=3.1 x 70 Q= 217 ft3/s or 217 cfs
Answer: The flow depth is about 2 ft for a discharge of 200 cfs
3. Example of Pascal’s Law
Let us understand the working principle of Pascal’s law with an example.
A pressure of 2000 Pa is transmitted throughout a liquid column due a force being applied on a piston. If the
piston has an area of 0.1 m2, what is the force applied?
This can be calculated using Pascal’s Law formula.
F = PA Pa = N/m2
given required solution
A = 0.1 m2 F=? F = PA
P = 2000 = 2000Nx0.1 m2
F = 200 N
Substituting values, we arrive at F = 200 N
4. Example Determine normal discharge for a 200 mm inside diameter common clay drainage tile
running half-full if the slope drops 1m over 1000 m.

Given
S = 1/1000 = 0.001
n for clay tile = 0.013
R = 0.05 m Solution
A= ½ IID2 /4 = 0.5*π*(0.2)2/4
A= 0.0157 m2
WP = (1/2) * (π D) = 0.5*π*0.2 = 0.3141 m
Flow Regimes
Re =VDρ/μ NR = PvD/µgc    
where:
NR = Reynolds number (unitless)
v = average velocity (ft/sec)
D = diameter of pipe (ft)
µ = absolute viscosity of fluid (lbf-sec/ft2 )
ρ = fluid mass density (lbm/ft3 )
gc = gravitational constant (32.2 ft-lbm/lbf-sec2 )
For practical purposes, if the Reynolds number is less than 2000, the flow is laminar. If it is greater than
3500, the flow is turbulent. Flows with Reynolds numbers between 2000 and 3500 are sometimes referred to
as transitional flows. Most fluid systems in nuclear facilities operate with turbulent flow
Flow-through pipes are classified into three main flow regimes.
1.Laminar flow – R < 2000
2.Critical flow – R > 2000 and R < 4000
3.Turbulent flow – R > 4000
Formula to calculate discharge rectangular channel
Discharge:
𝑄 = 𝑉𝐴
Where, in normal flow,

5. Example
The discharge in a rectangular channel of width 6 m with Manning’s 𝑛 = 0.012 m−1⁄3 s is 24 m3s–1
If the stream wise slope is 1 in 200 find:
(a) the normal depth;
(b) the Froude number at the normal depth;
(c) the critical depth
State whether the normal flow is subcritical or supercritical.
Given = 6 m
𝑛 = 0.012 m−1⁄3s
𝑄 = 24 m3s−1
𝑆 = 0.005
Discharge: 𝑄 = 𝑉𝐴 Where, in normal flow
Calculating discharge of trapezoidal channel flow

Geometry: trapezoidal cross-section with base width b, surface width 𝑏 + 2 × (2ℎ) and two sloping side lengths
Area and wetted perimeter:

Channel Geometry Characteristic

• Depth( y)
• Area,( A)
• Wetted perimeter,( P)
• Top width(, T)
Hydraulic Radius, Rh = Area / Wetted perimeter
Hydraulic Depth, Dh = Area / Top width
Trapezoidal Channel

Example 7 An undershot sluice controls the flow in a long rectangular channel of width 2.5 m, Manning’s
roughness coefficient 𝑛 = 0.012 m−1⁄3 s and stream wise slope 0.002. The depths of parallel flow upstream and
downstream of the gate are 1.8 m and 0.3 m, respectively.
(a) Assuming no losses at the sluice, find the volume flow rate, Q.
(b) Find the normal and critical depths in the channel.
(c) Compute the distance from the sluice gate to the hydraulic jump, assuming normal
depth downstream of the jump. Use two steps in the gradually-varied-flow equation
GIVEN 𝑏 = 2.5 m
ℎ1 = 1.8 m
ℎ2 = 0.3 m
𝑛 = 0.012 m−1⁄3s
𝑆 = 0.002
(a) Assuming the same total head on either side of the gate

(b)
Normal depth
𝑄 = 𝑉A

Calculation of pipeline flow rate or discharge

Formula for calculation of the pipeline flow rate formula is based on the diameter of pipe (for circular pipe):
Q = (Πd²/4)·v Where
Q – flow rate of pumped fluid, m3/s
d – pipeline diameter, m
v – flow velocity, m/s
The flow rate is most often a set quantity in problems on pipeline design. In such case the unknown quantities
are only a pipeline diameter and flow velocity. Comprehensive technical and economic calculation may be
very labor-intensive and complicated, so optimal values of pumped medium velocity, taken from reference
materials, drawn up on the basis of experimental findings, are used in practice:
Example No 8 What are the head losses for local resistances in horizontal pipeline having diameter of 20 x 4
mm, through which water is pumped from open reservoir to reactor with pressure of 1.8 bar? Distance between
reservoir and reactor is 30 m. Water flow rate is 90 m3/h. Total head equals to 25 m. Friction coefficient is
taken equal to 0.028.

Solution:
Water flow velocity in pipeline equals to:
v=(4·Q) / (π·d2) = ((4·90) / (3,14·[0,012]2))·(1/3600) = 1,6 m/s
We find head friction losses in the pipeline:
HТ = (λ·l) / (dэ·[w2/(2·g)]) = (0,028·30) / (0,012·[1,6]2) / ((2·9,81)) = 9,13 m
Total losses are:
hп = H - [(p2-p1)/(ρ·g)] - Hг = 25 - [(1,8-1)·105)/(1000·9,81)] - 0 = 16,85 m
Losses on local resistance fall within: 16,85-9,13=7,72 m
Example No. 9 The pipeline with the inside diameter of 42 mm is given. It is connected to the water pump
with flow rate of 10 m3/h and creating head of 12 m. Temperature of the pumped medium is 20° C. Pipeline
configuration is given in the figure below. It is necessary to calculate the head losses and check whether this
pump is capable of pumping water at pipeline set parameters. Absolute roughness of pipes is taken as equal to
0.15 mm.
Solution: We calculate the velocity of the fluid flow in the pipeline:
v = (4·Q) / (π·d2) = (4·10) / (3,14·0,0422)·1/3600 = 2 m/s
Velocity head corresponding to the found velocity will equal to:
v2/(2·g) = 22/(2·9,81) = 0,204 m
Friction coefficient should be found before the calculation of friction losses in pipes. In the first place we
determine pipe relative roughness:
e = Δ/dЭ = 0,15/42 = 3,57·10-3 mm
Reynolds criterion for water flow in the pipeline (water dynamic viscosity at 20° C is 1·10-3 Pa·s, and density
is 998 kg/m3):
Re = (V·dЭ·ρ) / μ = (2·0,042·998) / (1·10-3) = 83832
We find out the water flow mode:
10/e = 10/0,00357 = 2667
560/e = 560/0,00357 = 156863
The found value of Reynolds criterion falls within the range of 2667<83,832<156,863 (10/e < Re < 560/e),
hence, friction coefficient should be calculated by the following formula:
λ=0,11·(e+68/Re)0,25 = 0,11·(0,00375+68/83832)0,25 = 0,0283
Head friction losses in the pipeline will equal to:
HТ = (λ·l)/dэ · [V2/(2·g)] = (0,0283·(15+6+2+1+6+5))/0,042 · 0,204 = 4,8 m
 2
r
PI
L2 V 2 V 2 V
F = = = ⇒ F r=
Froude Number ( Fr )
P V gL 3 gL √ gL

Inertia P I ρV 2 L2 ρVL VL
= = = =
Reynolds number (Re): R = Vis cos ity P V μVL μ υ
e

State of flow can be classified depend on the value of Froude Number as:

1. Critical flow
Fr =1
 Unstable flow state conditions
 Mostly occurred on control volumes

2. Sub-critical flow
Fr <1
 Flow has low velocity and larger depth of flow than super-critical sate
 Often described as tranquil and streaming
 Gravity force played more pronounced role

3. Super-critical flow
Fr >1
 Flow has high velocity and low flow depth compared to sub-critical flow
 Inertia forces become dominant
 Usually described as rapid, shooting, and torrential flow
Variation in velocity and pressure in open channel flow causes by
 Presence of free surface
 friction along the channel wall
Equation 3.1 is a very useful tool for the computation and analysis of critical flow in an open channel. When
the discharge is given, the equation gives the critical section factor Z c and, hence the critical depth y c. On the
other hand when the depth and, hence the section factor are given, the critical discharge can be computed by
the following form.

Q= Z √g

Q = √ g
α

Assignments 1 solve the following problem each 5 point ?

Example 1

For a trapezoidal channel with base width b = 6.0 m and side slope m = 2, calculate the critical depth of flow if
Q = 17 m3/s.

Solution
Example 2

A circular channel 0.91 m in diameter conveys a flow of 0.71 m3/s; estimate the critical depth of flow.

Solution

Example 3

A trapezoidal channel with b = 6.0 m and m = 1.5 conveys a flow 17 m3/s; estimate the critical depth of flow.

Solution

Example 4

Given a trapezoidal channel with a bottom width of 3m, side slope 1.5, a longitudinal slope of 0.0016 and
estimated on of 0.13, find the normal depth of flow for a discharge of 7.1m3/5.

Solution:
Example #10: Water is flowing 1.5 feet deep in a 4 foot wide, open channel of
rectangular cross section, as shown in the diagram below. The channel is made of concrete
(made with steel forms), with a constant bottom slope of 0.003.
a) Estimate the flow rate of water in the channel. b) Was the assumption of
turbulent flow correct ?
Solution: a) Based on the description, this will be uniform flow. Assume that
the flow is turbulent in order to be able to use equation (1), the Manning equation.
All of the parameters on the right side of equation (1) are known or can be calculated:
From Table 1, n = 0.011. The bottom slope is given as: S = 0.003.
From the diagram, it can be seen that the cross-sectional area perpendicular to flow is 1.5
ft times 4 ft = 6 ft2. Also from the figure, it can be seen that the wetted perimeter is 1.5 +
1.5 + 4 ft = 7 ft. The hydraulic radius can now be calculated: Rh = A/P = 6 ft2/7 ft =
0.8571 ft
Substituting values for all of the parameters into Equation 1:
Q = (1.49/0.011)(6)(0.85712/3)(0.0031/2) = 40.2 ft3/sec = Q
b) Since no temperature was specified, assume a temperature of 50o F. , ρ = 1.94
slugs/ft3, and μ = 2.730 x 10-5 lb-s/ft2. Calculate average velocity, V:
V = Q/A = 40.2/6 ft/sec = 6.7 ft/sec
Reynold’s number (Re = ρVRh/μ) can now be calculated:
Re = ρVRh/μ = (1.94)(6.7)(0.8571)/( 2.730 x 10-5) = 4.08 x 105
Since Re > 12,500, this is turbulent flow
Example #11: A triangular flume has 10 ft3/sec of water flowing at a depth of 2 ft
above the vertex of the triangle. The side slopes of the flume are: horiz : vert = 1
1. The bottom slope of the flume is 0.004. What is the Manning roughness coefficient, n, for
this flume?
Solution: From the problem statement: y = 2 ft and z = 1, substituting into
Equation (5): Rh = 22(1)/(2[22(1 + 12)]1/2 ) = 0.250 ft
The cross-sectional area of flow is: A = y2z = (22)(1) = 4 ft2
Substituting these values for Rh and A along with given values for Q and S into
equation (1) gives:
10 = (1.49/n)(4)(0.252/3)(0.0041/2)
Solving for n: n = 0.015
Example #12: Determine the bottom slope required for a 12 inch diameter circular storm sewer made of
centrifugally spun concrete, if must have an average velocity of 3.0 ft/sec when it’s flowing full.
Solution: Solving Equation (6) for S, gives: S = {(nV)/[1.49(Rh
2/3)]}2. The velocity, V, was specified as 3 ft/sec. From Table 1, n = 0.013 for centrifugally spun concrete.
For the circular, 12 inch diameter sewer, Rh = D/4 = ¼ ft.
Substituting into the equation for S gives:
S = {(0.013)(3.0)/[1.49(1/4)2/3]}2 = 0.00435 = S
Example #13: A reach of channel for a stream on a plain is described as clean, straight, full stage, no rifts or
deep pools. The bottom slope is reasonably constant at 0.00025 for a reach of this channel. Its cross-section is
also reasonably constant for this reach, and can be approximated by a trapezoid with bottom width equal to 7
feet, and side slopes, with horiz : vert equal to 3:1. Using the minimum and maximum values of n in the above
table for this type of stream, find the range of volumetric flow rates represented by a 4 ft depth of flow.
Solution to Example #13: From the problem statement, b = 7 ft, S = 0.00025, z = 3, and y = 4 ft. From the
above table, item 1. a. (1) under “Natural Stream”, the minimum expected value of n is 0.025 and the
maximum is 0.033. Substituting values for b, z, and y into equation (4) for a trapezoidal hydraulic
radius gives:
Rh = [(7)(4) + 3(42)]/[7 + (2)(4)(1 + 32)1/2 ] = 2.353 ft
Also A = (7)(4) + 3(42) = 76 ft2
Substituting values into the Manning Equation [Q = (1.49/n)A(Rh
2/3)S1/2] gives the following results:
Minimum n (0.025): Qmax = (1.49/0.025)(76)(2.3532/3)(0.00025)1/2
Qmax = 126.7 ft3/sec
Maximum n (0.033): Qmin = (1.49/0.033)(76)(2.3532/3)(0.00025)1/2
Qmin = 95.99 ft3/sec