Post graduate exam in hydraulics.pptx

‫يا‬‫لعل‬‫ا‬ ‫اسات‬‫ر‬‫ادل‬ ‫امتحان‬ ‫حل‬
2018

A rectangular swimming pool 1.0 m deep at one end and increases uniformly
in depth to 2.6 m at the other end. The pool is 8 m wide and 32 m long and is
emptied through an orifice of area 0.225 m2, at the lowest point in the side of
the deep end (see the shown sketch). Taking Cd for the orifice as 0.60, find
from the first principles:
 The time for the depth to fall by 1.0 m,
 The time to empty the pool completely.
Question No 1 (10 Marks)

Atank
H1
H2
dh
H
g
2
A
C
Q orifice
d
out 


0
Qin 
dh
A
dt
.
Q tank
out 


dh
.
Q
A
dt
t
0
H
H
tank
2
1
 


dh
.
h
g
2
A
.
C
A
dt
t
0
H
H
5
.
0
orifice
d
tank
2
1
 




Qout



Given:
Area of the tank = 32 × 8.0 = 256 𝑚2, where width of the tank
= 8.0 𝑚
Area of the orifice = 0.225 𝑚2
Coefficient of discharge of the orifice = 0.60
dh
.
h
g
2
A
.
C
A
dt
t
0
H
H
5
.
0
orifice
d
tank
2
1
 


 ……………… (1)
Reference Datum

 
  S
6
.
297
348
.
0
22
.
856
6
.
2
6
.
1
22
.
856
T
h
h
81
.
9
2
225
.
0
60
.
0
256
2
T
1
1
2
1
















Substituting into Eq. (1) gives:
 The time required to fall the depth by 1.0 m = 297.6 S



dh
.
h
g
2
A
.
C
A
dt
t
0
H
H
5
.
0
orifice
d
tank
2
1
 


 ……………… (2)
Reference Datum
h
X
𝑋 ℎ = 32 1.60 = 20
or 𝑋 = 20 ℎ
∴ 𝑑𝐴 = 𝑋 𝑑ℎ = 20 ℎ. 𝑑ℎ
dh

 
 
.
min
71
.
5
S
7
.
342
1
.
45
6
.
297
T
T
completely
pool
the
emptying
for
time
The
S
1
.
45
02
.
2
3
.
22
6
.
1
0
3
.
22
T
h
h
81
.
9
2
225
.
0
60
.
0
3
20
2
T
2
1
2
3
2
3
2
2
3
1
2
3
2
2























Substituting into Eq. (2) gives:
 The time required for emptying the pool completely =
342.7S= 5.71 min.

“3-b” Drop Inlet (Shaft or Morning Glory) Spillways

“3-d” Morning-glory Spillway (from U.S. Bureau of Reclamation 1987)

Maximum water surface
Crest
Circular conduit
Top of the dam
D
It is often:
 selected for the projects where space limitations or topographic features do not
permit instilling other types of spillways.
 used with embankment dams because it is not safe for standard chutes to be
constructed on the embankment.

The major components of a shaft spillway are:
 Circular cross section, a vertical shaft, an elbow in a vertical plane, a tunneling
section, and a terminal structure.
 Used with embankment dams because it is not safe for standard chutes to be
constructed on the embankment.
 For large structures, the various components are usually constructed from
concrete, with the tunnel and much of the vertical curve tunneled through rock.

Nature of Discharge Characteristic of
Morning-glory Spillway

There are three types of possible flow regimes in vertical shaft spillway.
 For Low Flows “First case”: the flow-rate is governed by the relationship for weir
flow at the crest
𝑄 = 𝐶 2𝑔 𝐿 𝐻3 2
Where:
C = is a function of both H/ R and P/ R, where R is the radius of the spring point,
L = the length (the circumference of the crest measured at the spring point,
H = the head on the weir.

 The second type of flow 𝐻 𝑅 𝑎𝑏𝑜𝑣𝑒 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒𝑙𝑦 0.45 “: the weir becomes
partially submerged. As 𝐻 𝑅 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ𝑒𝑠 𝑢𝑛𝑖𝑡𝑦, the entrance becomes
completely submerged, and orifice control, occurs, where
𝑄 ∝ 𝐻1 2
 The third type of flow will occur when the downstream tunnel becomes full and
full pipe flow relationship apply:
𝑄 ∝ 𝐻𝑃𝑖𝑝𝑒
1 2
Where:
𝐻𝑝𝑖𝑝𝑒= is a full head for full-pipe flow.
(it is common practical to avoid designing for full- pipe flow, shaft spillways are
usually designed so that the flow depth in the downstream tunnel is not more than
75% of the diameter).

Surge tanks are usually provided in high or medium-head plants when
there is a considerable distance between the water source and the power
unit, necessitating a long penstock. The main functions of the surge tank
are:
1. When the load decreases, the water moves backwards and gets
stored in it.
2. When the load increases, additional supply of water will be provided
by surge tank.
In short, the surge tank mitigates pressure variations due to rapid changes
in velocity of water.

Surge tank (or surge chamber) is a device introduced within a hydropower
water conveyance system having a rather long pressure conduit to absorb the
excess pressure rise in case of a sudden valve closure.
The surge tank is located between the almost horizontal or slightly inclined
conduit and steeply sloping penstock and is designed as a chamber
excavated in the mountain.
It also acts as a small storage from which water may be supplied in case of a
sudden valve opening of the turbine. In case of a sudden opening of turbine
valve, there are chances of penstock collapse due to a negative pressure
generation, if there is no surge tank.

Surge Tank Function
When the valve in a hydroelectric power plant is suddenly completely closed,
because of its small inertia the water in the penstock stops almost at once. The
water in the pipeline, with large inertia retards slowly. The difference in flows
between pipeline and penstock causes a rise in the water level in the surge tank.
The water level rises above the static level of the reservoir water, producing a
counter-pressure so that water in the pipeline flows towards the reservoir and
the level of water in the surge tank drops. In the absence of damping, oscillation
would continue indefinitely with the same amplitude.

The flow into the surge tank and water level in the tank at any time during the oscillation
depends on the dimension of the pipeline and tank and on the type of valve movement. The
main functions of a surge tank are:
1. It reduces the amplitude of pressure fluctuations by reflecting the incoming pressure
waves
2. It improves the regulation characteristic of a hydraulic turbine.
The surge tank dimensions and location are based on the following considerations
1. The surge tank should be located as close to the power or pumping plant as possible;
2. The surge tank should be of sufficient height to prevent overflow for all conditions of
operation;
3. The bottom of surge tank should be low enough that during its operation the tank is
drained out and admit air into the turbine penstock or pumping discharge line; and
4. The surge tank must have sufficient cross sectional area to ensure stability.

Surge Tank Types
There are different types of surge tanks that are possible to be installed. Some of
the most common types of surge tanks which are as follows:
1. Simple Surge Tank: A simple surge tank is a shaft connected to pressure
tunnel directly or by a short connection of cross-sectional area not less than
the area of the head race tunnel.
2. Restricted Orifice Surge Tank: A simple surge tank in which the inlet is
throttled to improve damping of oscillations by offering greater resistance
and connected to the head race tunnel with or without a
connecting/communicating shaft
3. Differential Surge Tank: Differential Surge tank is a throttled surge tank with
an addition of a riser pipe may be inside the main shaft, connected to main
shaft by orifice or ports. The riser may also be arranged on one side of
throttled shaft.

AT
AS
VT VP
Ao

Communicating
Shaft
AS
Ao

AT
VT VP
Restricted Orifice Surge Tank: A simple surge tank in which the inlet is
throttled to improve damping of oscillations by offering greater resistance and
connected the head race tunnel with or without a connecting/ communicating
shaft.

AS
Ao

AT
VT VP
Differential Surge Tank: is a
throttled surge tank with an
additional of a riser pipe may
be inside the main shaft,
connected to main shaft by
orifice or ports as shown in
the figure.


ZSp
Ar
Reference Datum Z

AS
Ao

AT
VT VP
The riser may also be
arranged on one side of
throttled shaft as shown in
the figure.

Reference Datum Z
Z1

In an underground development of hydropower system, tail race surge tanks are
usually provided to protect tail race tunnel from water hammer effect due to
fluctuation in load. These are located downstream of turbines which discharge
into long tail race tunnels under pressure. The necessity of tail race surge tank
may be eliminated by insuring free-flow condition in the tunnel but in case of long
tunnels this may become uncommercial than a surge tank.

Successive division and rejoining of flow around islands.
Most common in mountain reaches, on alluvial fans and glacial
outwash.
Requires:
 High bed load
 Steep gradient (aids in higher bed load)
 Relative ease of bank erosion
Grain Size:
 Gravels most common,
 Can braid fines if discharge is higher and banks are weak.

Braiding occurs when stream discharge is insufficient to transport the
available load. Conditions that favor braiding over meandering are high
erosion rates (high gradient, proximity to scour, non-cohesive channel
margins), coarse grain size, and low or variable discharge.

A round jet of water is discharging upward from near the bottom of a bay 100 ft
deep. The jet is 1 ft diameter, and the discharge rate is 30 cfs.
 What is the velocity at the points (shown by the Xs) in the diagram shown
below?
“Assume the density of the jet is the same as the density of the water un the bay”
6 ft
20 ft
20 ft
X
X
Mid-Year Exam.

The figures show the maximum unit tractive forces in terms of (𝜌𝑔 𝑦 𝑆𝑜 =
𝛾 𝑦 𝑆𝑜) for different (𝐵 𝑦 = 𝑏 𝑦) ratios
Maximum unit tractive forces in terms of (𝜌𝑔 𝑦 𝑆𝑜 = 𝛾 𝑦 𝑆𝑜)

A meander: is a bend in a sinuous watercourse.
Meandering is the process when the faster-moving water in a river
erodes the outer banks and widens its valley, and the slower-moving
water on the inner side of the bend becomes a place where sediments
are deposited (point bars).
As a result, rivers tend to constantly change their course over a
floodplain over time (See the following figure).
An oxbow is a crescent lake on .
a stream floodplain formed when a
meandering stream channel is cut off and
isolated by changes in a stream channel

How meanders grow: laterally through erosion (outside bend) and
sediment deposition (inside bend, point bar). When the loops get too
large and consume too much energy (friction), the river will eventually
find a less energetically "taxing" shortcut, and a part of the old channel
will be abandoned and becomes an oxbow lake.

Stream channel flow results in meandering forming cut
banks and point bars

A pump on test at its designed speed of 1450 rpm gave the following results:
The pump is to run continuously at its design speed and is to deliver water
through 40 ft of pipe against a static lift of 5 ft. The only pipes available are 6in,
8in and 10in diameter all with =0.028.
i. Which will be the most suitable pipe for this duty?
ii. Estimate the horse power required to drive the pump and calculate the
specific speed of the pump at maximum efficiency?
iii. Plot on the square paper an estimated head-quantity (Q) curve for the pump
running at 1000 rpm.
H (ft) 27 25 21.5 17.5 13.5 8 3
Q (gal/min) 0 250 500 750 1000 1250 1375
 (%) 0 20 42 60 69 66 61
Question No 2 (15 Marks)

5
2
2
5
2
D
Q
028
.
0
Q
D
2
.
32
40
028
.
0
8






For a 6-in pipe line
 
2
5
2
Q
902
.
0
12
6
Q
0282
.
0
H 


 
2
5
2
Q
214
.
0
12
8
Q
0282
.
0
H 


 
2
5
2
Q
070
.
0
12
10
Q
0282
.
0
H 


Worked Example
 The characteristic data for the pump and the calculated values of
“H” required from pump (HS +hL) For each pie line are shown in
the following table:
 and H for pipe line 6-in, 8-in and 10-in against Q are plotted on
superimpose on them the pump characteristic curves (see the
next graphs):
“H“
for 10-in
“H“
for 8-in
“H"
for 6-in

HP
Q
Q
ft
ft
ft
%
ft
ft3/sec
gallon/min
5.00
5.00
5.00
0
19.1
0.00
0
5.02
5.07
5.28
21
19.1
0.56
250
5.09
5.27
6.12
42
18.8
1.11
500
5.20
5.60
7.52
60
17.0
1.67
750
5.35
6.06
9.47
69
13.2
2.23
1000
5.54
6.66
11.99
64
8. 0
2.78
1250
Worked Example
 The characteristic data for the pump and the calculated values of
“H” required from pump (HS +hL) For each pie line are shown in
the following table:
“H“
for 10-in
“H“
for 8-in
“H"
for 6-in

HP
Q
Q
ft
ft
ft
%
ft
ft3/sec
gallon/min
5.00
5.00
5.00
0
19.1
0.00
0

System Curve for D = 6 in
H (ft)

Q (ft3/s)
Characteristic
Curve
Efficiency Curve

From the graph:
A. Pipe line with diameter “D = 6 in“ is better, since the pump performance
around a higher efficiency,
B. From the economical point
 With N = 1450 rpm, Q = 2.52 ft3/s and  = 72%
 Horsepower required to drive the pump
ℎ𝑝 = 𝜌 𝑔 𝑄 ∙ 𝐻 (𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 × 550) =
1.94 × 32.2 × 2.52 × 11 0.72 × 550 = 4.37 𝐻𝑝 ≈ 5 𝐻𝑝

Using Affinity Law for a constant impeller diameter (DA = DB):
𝐻𝐵 = 𝐻𝐴 ×
𝑁𝐵
𝑁𝐴
2
& 𝑄𝐵 = 𝑄𝐴 ×
𝑁𝐵
𝑁𝐴
3
, we obtain
H (ft) 27 25 21.5 17.5 13.5 8 3
Q (gal/min) 0.00 250 500 750 1000 1250 1375
H (ft) for N = 1000 rpm 12.8 11.9 10.2 8.3 6.4 3.8 1.4
Q (gal/min) for N = 1000 rpm 0.00 82 164 246 328 410 451
Q (ft3/sec) for N = 1000 rpm 0 0.56 1.11 1.67 2.23 2.78 3.06

0
2
4
6
8
10
12
14
0 0.5 1 1.5 2 2.5 3 3.5
H (ft)
Characteristic Curve for N
100 rpm
Using Affinity Law for a constant impeller diameter (DA = DB):
𝐻𝐵 = 𝐻𝐴 ×
𝑁𝐵
𝑁𝐴
2
& 𝑄𝐵 = 𝑄𝐴 ×
𝑁𝐵
𝑁𝐴
3
, we obtain

When the rate of flow of a fluid passing down a pipe-line changes, there is a change in
pressure. The severity of this effect depends upon the rate of change in the flow rate, the
length of the pipe and its diameter. In small bore pipes there is no real problem other than
maybe an annoying hammering sound. In large water mains the rate of change in flow is
carefully controlled to avoid damage to pipes and valves. Unfortunately this solution will
not work with Turbines where a sudden change in load requires a rapid change in water
demand.
Here Surge Tanks or Stand Pipes are used to reduce the pressure surges. When the flow
to the turbine is reduced, water flows into the surge tank and conversely for increased
load , the initial extra water required is from the surge tank. The size of the tank should
be such that water will not overflow when the turbine is suddenly shut down, nor allow air
to be drawn into the system following a sudden increase in demand. In addition it must be
sited as close to the turbines as possible to avoid surges in the length of pipe between
the surge tank and the turbine.

Post graduate exam in hydraulics.pptx

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