WATER
Torrential rain, because of global warming, will be common in the coming days. Thus one should equip oneself with some knowledge on hydraulic and water engineering. The requirement for this essay is that of a text book on Physics.
1. Not Containment but Division, further division, and diversion.
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Many many years ago, China was troubled by floods. The king then assigned a person
to tackle this problem.
He tackled this problem by building levees. But floods still occurred. The king was
angry, and the king killed him and put his son in charge instead.
His son took another approach, he did not build levees, but adopt the tactics of
division and diversion. He channeled away water all along the river. He built canals
as well as tunnels. And the problem was solved.
Another benefit from this is that the water diverted may be used in irrigation; and
if the canal is wide, may be used for transport as well as for leisure.
2. Have cities and towns on upper course of river where the elevation is high and not
on lower course of river. Move to higher ground.
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( ) (Upper-stream) , ( ) (Lower-stream) are words used by Chinese to describe
noble people and base people. This originated from the fact that prudent people
would choose upper-stream to live, and foolish people would choose lower stream.
3. BERNOULLI'S EQUATION.
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Two equations, Bernoulli's Equation and Manning's Equation are very important.
IT IS ALSO THE ONLY 2 EQUATIONS WE NEED IN HYDRAULIC COMPUTATION.
Here we describe Bernoulli's equation first. Also we use MKS system.
We write Bernoulli's Equation in this form:
Hence we see that the first term is kinetic energy per unit volume, and the second
term, potential energy per unit volume, and the third term, energy per unit volume
arising from pressure.
You may wonder how Bernoulli's Equation is derived. This is derived from a set of
Partial Differential Equation resulting from Newton's Law of Motion, when applied
to a volume of water under direct and shear force (direct stress and shear stresses).
Many simplifications are made: e.g. No vicuous force, i.e. no viscosity and hence
friction free; irrotational (curl velocity-field = 0) and imcompressible
(density is constant) .... etc. You may learn about them later, but for the time
being, just learn to know how to apply the equation.
Notice that friction is assumed to be 0. Hence we CANNOT APPLY BERNOULLI'S EQUATION
TO FLUID WHERE FRICTION IS PRESENT. In reality, Energy will sure be lost to
friction, which is to be accounted by Manning's equation to be described later.
Exercise. 1. There is a water fall. The drop in height is 4 m. Assume that the river
above is rectangular in cross section, of width 1 m, height 15 cm. Assume
that velocity of flow is 0.5 m/sec.
What is the discharge rate in m^3/sec?
How much power may be gotten from this fall, assuming no friction and
the outward velocity of flow is again 0.5 m/sec. Express this in
horse-power too (1 h.p. = 746 watt, and 1 watt = 1 Joule/sec = 1 Newton-m/sec,
density of water = 1 gram/c.c. = 1000 kg/m^3.)
(Ans: discharge rate Q = 1 m x 15/100 m x 0.5 m/sec = 0.075 m^3/sec;
power = density . g . h . Q
= 1000 kg/m^3 * 9.81 m/s^2 * 4 m * 0.075 m^3/s
= 2943 Newton-m/s
= 2943 Joule/s
= 2943 Watt
= 2943/746 h.p. = 3.945 h.p.)
Exercise. 2. We are to pump water to a height of 5 m. The diameter of pipe is 35 cm,
and water velocity is 2.5 m/sec.
What is the discharge rate in m^3/sec?
How much power is required, expressed first in watt, then in h.p.
(Ans: discharge rate Q = 3.1416 * (35/2/100)^2 m^2 * 2.5 m/s
= 0.240528 m^3/s
power = density . g . h . Q
= 1000 kg/m^3 * 9.81 m/s^2 * 5 m * 0.240528 m^3/s
= 11797.898 Watt = 15.8 h.p. )
Exercise. 3. Water is emerging from a jet-nozzle of diameter 25 cm at a velocity
of 14 m/sec. If the water is totally stopped (i.e. velocity = 0 m/s)
after hitting the blade of a water wheel. How much power is delivered
to the wheel, express in watt and in h.p.
(Ans: discharge rate Q = 3.1416 * (25/2/100)^2 m^2 * 14 m/s
= 0.68722 m^3/s
power = 1/2 . density . V^2 . Q
= 0.5 * 1000 kg/m^3 * (14 m/s)^2 * 0.68722 m^3/s
= 67347.56 Watt = 90.27 h.p. )
Exercise. 4. Write simple computer programs using BASIC (Qbasic) to automate these
computations.
Exercise. 5. Find further use of Bernoulli's equation in Physics Textbooks.
Exercise. 6. A storage reservoir is built on the top of a small hill. The height of
that hill is 50 m. Water is delivered to houses at the foot of the hill
by galvanized steel pipes. What is the static pressure in the pipe?
(Ans: At the reservoir, V = 0 m/s, h = 50 m, P = atmospheric pressure.
At the foot of the hill, V = 0 m/s, h = 0 m, Pressure is unknown,
let it be x Newton/m^2.
Applying Bernoulli's equation,
0 + density.g.h + P = 0 + 0 + x
x - P = density.g.h = 1000 kg/m^3 * 9.81 m/s^2 * 50 m
= 490500 Newton/m^2
= 5 kg/(cm)^2 = 70.97 lb/(in)^2
But Atmospheric pressure P = 1.02 kg/(cm)^2 or 14.7 lb/(in)^2, therefore
x = 6.02 kg/(cm)^2 or 85.67 lb/(in)^2.
Therefore, when buying pipes from hardware store, be sure to ask the shop-keeper
whether the pipe will withstand the pressure. Often, we may find such details
from manufacturer's spec.)
4. Manning's Equation:
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In reality, friction will always be there, and water flow is usually turbulent.
Unlike Bernoulli's equation, which may be rigorously derived from Newton's Law
of Motion under various assumptions, turbulent flow is extremely hard to solve
theorectically, (perhaps one day you will be able to derive an acceptable theory
using statistics etc. Or, perhaps, no solution will ever be found.)
Hence an empirical formula is needed. Around 1930, Nikuradse at Gottingen performed
experiments with water flow in pipes roughened to various degree of roughness.
Water flow will be peaceful (laminar flow) at first, but becomes turbulent when
velocity increased beyond certain Reynold number, which is a function of velocity
as well as viscosity.
From the experiments, a formula is derived for turbulent flow (incidentally, laminar
flow may be solved theorectically. You may try to learn them from textbooks. But
river flow, water flow in domestic water systems are all turbulent flow).
Manning's formula:
Explanation of terms of the Equation:
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(a) R = hydraulic radius (not that radius of circle !) is defined as
= cross-sectional area / perimeter wetted by water.
(Hence length around the perimeter which is DRY will not be counted. This
is because only surface in contact with water will cause friction to flow.)
Example: A channel with water height = 35 cm, width = 60 cm.
Area = 35/100 m x 60/100 m = .21 m^2.
Wetted perimeter = 2 x 35/100 m + 60/100 m = 1.3 m
R = hydraulic radius = .21 m^2/ 1.3 m = 0.1615 m
Exercise: Find hydraulic radius for a pipe of diameter D m totally filled with water.
(Ans: D/4 )
Exercise: Find hydraulic radius of a pipe of diameter D m filled only to half height.
(Ans: = ((1/2)*PI*D*D/4 )/(1/2*PI*D) = again D/4 )
Exercise: Find hydraulic radius of a square channel of width = a meter.
(Ans: a/3 m)
Exercise: Find hydraulic radius of a channel with 3 equal sides of length
a meter, but inclined at 60 degree, (see figure)
(Ans: area = (a + (a + 2a.cos(60))) . a sin(60) /2
wetted perimeter = 3a.
R = a . (1+cos(60)) . sin(60) / 3.
= 0.433 a )
Notice that only perimeter in contact with water will cause friction.
(b) S = slope = sin(...).
Exercise: If the river has fallen 3 m after traversing 1000 m, what is the slope?
(Ans: S = 3/1000 = 0.003 )
Exercise: If a pipe is of length 70 cm, and one end is down 15 cm, what is S?
(Ans: 15/70 = 0.214 )
(c) n is computed from 26n = k^(1/6). k must be in meter.
Exercise: If roughness of a pipe is 2.5 mm, what is n?
(Ans: n = (2.5/1000)^(1/6)/26 = 0.014169)
Exercise: If roughness of a river is 3 cm, what is n?
(Ans: n = (3/100)^(1/6)/26 = 0.021439)
Exercise: If roughness of a river is 15 cm, what is n?
(Ans: n = (15/100)^(1/6)/26 = 0.0280355)
We are now to compute various things using Manning's equation.
Exercise. 7: A drainage pipe is of diameter 25 cm, length 1.2 m, and is sloped at 15o.
Assume roughness = 3 mm. Water is flushed level at the upper end, what
is the emerging velocity?
(Ans: R = D/4 = (25/100)/4 = 0.0625
S = sin(15o) = 0.258819
n = (3/1000)^(1/6)/26 = 0.014606
V = R^(2/3)*S^(1/2)/n = (0.0625)^(2/3)*(0.258819)^(1/2)/0.014606
= 5.485549 m/s
)
Exercise. 8: A drainage pipe is of diameter 15 cm, length 7 m. Roughness = 2 mm.
Assume that we require water velocity to be 1.5 m/s in order to avoid
sedimentation. At what angle should the pipe be sloped?
(Ans: R = D/4 = (15/100)/4 = 0.0375
n = (2/1000)^(1/6)/26 = 0.01365
S^(1/2) = nV/R^(2/3)
S = (nV/R^(2/3))^2 = (0.01365*1.5/0.0375^(2/3))^2
= 0.03339946
angle = 1.9145 degree
)
Exercise. 9: The same pipe as in Ex. 7 (Diameter = 15 cm, length = 7 m, roughness = 2 mm,
Velocity = 1.5 m/s) But the pipe is now level, see diagram. How high should
the water be at the other end to achieve this velocity ?
(Ans: height = 7 m * sin(1.9145) = 23.38 cm.
)
Exercise. 10: Refer back to Ex.2, (Diameter = 35 cm, Velocity = 2.5 m/s,
height to be raised = 5 m), assume the total distance to be transported
is 1 km (i.e. length of pipe = 1000 m), and the roughness of pipe is 4 mm.
What is the power required ?
(Answer: R = D/4 = (35/100)/4 = 0.0875
n = (4/1000)^(1/6)/26 = 0.0153239
S^(1/2)= nV/R^(2/3)
S = (nV/R^(2/3))^2
= (0.0153239*2.5/0.0875^(2/3))^2
= 0.03778
As the length of pipe is 1000 m, this is equivalent to a height = 1000*S
= 37.78 m.
Therefore, because of friction, it is equivalent to raising the water to
a total height of = (5 + 37.78) m = 42.78 m
Power = density . g . h . Q
= 1000 kg/m^3 * 9.81 m/s^2 * 42.78 m *
(3.1416 * (35/2/100)^2 m^2 * 2.5 m/s)
= 100942.89 Watt
= 100942.89/746 h.p. = 135.3 h.p.
)
Exercise. 11: There is a tank filled to a height of 1.5 m. If a hole is punched near the
bottom, what is the velocity of water running out? If we solder a tube
of diameter 4 mm, length 1.3 m, roughness 0.7 mm, what is the velocity
of water running out then?
(Ans: Case 1:
Pressure is the same at the top and at the hole, i.e. atmospheric pressure.
At the top V = 0 m/s, h = 1.5 m. At the bottom, h = 0, V is unknown,
therefore, using Bernoulli's equation,
0 + density . g . (1.5) + P = 1/2 . density . V^2 + 0 + P
V = (2gh)^(1/2)
= (2*9.81 m/s^2 * 1.5 m)^(1/2)
= 5.425 m/s
Case 2: Now some energy is dissipated because of friction along the tube,
the equivalent drop in height is then
S^(1/2) = (nV/R^(2/3))
S = (nV/R^(2/3))^2
= (0.01146/0.001^(2/3))^2 * V^2
because n = k^(1/6)/26
= (0.7/1000)^(1/6)/26
= 0.01146
R = D/4 = (4/1000)/4 = 0.001
S = 1.313316 V^2 meter
But S = drop-in-height/length
= drop-in-height/1.3
Hence drop-in-height = 1.3 * S m
= 1.3 * 1.313316 * V^2 m
= 1.7073 * V^2 m
Applying Bernoulli's equation, (P = atmospheric pressure)
(0 + density.g.h + P) = { (1/2).density.V^2 + 0 + P } + density.g.(1.7073 * V^2)
9.81 * 1.5 = (1/2 + 9.81*1.7073) * V^2
V = (9.81*1.5/(1/2+9.81*1.7073))^(1/2)
= 0.92364 m/s
Exercise 12: There is a square plot of land, of side = 1000 meter (= 1 kilometer).
Rain is falling at a rate of 7 cm per hour. A drainage channel of
slope = (1 in 200), and of square section, is to be built. Roughness = 2 cm.
What is the dimension of that square channel if flooding is not to
happen?
(Ans: Assume the side of the square channel is x meter. Then
R = x / 3
n = (2/100)^(1/6)/26 = 0.020038
V = (x/3)^(2/3) . (1/200)^(1/2) / n
= x^(2/3) . (1.69648)
Now the discharge rate is = 1000 m . 1000 m . 7/100 m / 3600 s
= 19.4444 m^3/s
But discharge is also = cross-sectional-area * velocity, hence
(x^2) . x^(2/3).(1.69648) = 19.4444
x = 2.495 meter.
)
5. CENTRIFUGAL PUMP
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The following gives a basic understanding of the working of centrifugal pump.
Question: If the rotor is rotating at M r.p.m., what is the angular velocity in
radian per second?
(Ans: = M*2*PI/60)
Question: If the length of the rotor blade is L m, what is the tangential
velocity at the tip of blade?
(Ans: velocity V = L . angular velocity = L . ( M * 2 * PI /60) )
Question: If the kinetic energy is wholly converted into potential energy,
what is the height attained? (Use Bernoulli's Equation)
(Ans: 1/2 . V^2 = g . H
H = 1/(2g)* V^2
= 1/(2g)* (L*M*2*PI/60)^2
Hence H is directly proportional to L^2 (which is the radius of rotor blade), and M^2.
(Note that in reality, the water at the tip of the rotor of the pump has tangential
velocity as well as radially outward velocity. But the order of magnitude agrees
with the simplication here.)
Therefore it is very important to consult the specification sheet of the manufacturer
of the pump. Each pump is designed for certain discharge rate, and for certain height.
If there is a mis-match in M (r.p.m.), or in L , much energy is wasted, and even more,
we may not be able to pump water to the required height.
If there is a mis-match in M, usually some gears (or V-belts pulley) are used to
increase/decrease M, or the radius of the impeller is changed by machining.
6. If you find these interesting, and wish to pursue further, you may find the following
books useful:
College Physics (Sears F.W., Zemansky M.W. Addison-Wesley Publishing Co. Ltd.)
Physics (Parts I and II) (Halliday, D., Resnick R. John Wiley & Sons, Inc.)
Water Engineering (Willcock D.B., Evans Basic Technology)
Fluid Mechanics for Civil Engineers (Webber N.B. Chapman and Hall Ltd.)
Also, if you are a tropical fish hobbyist, you will find water filtration used
in an aquarium is very similar to that used in a water treatment plant.
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