Page 132 - Water Engineering Hydraulics, Distribution and Treatment

110
Chapter 5
Water Hydraulics, Transmission, and Appurtenances
where
P = ( h )A = I A
f
H = energy at section A
A
where I is the intensity of pressure at center of gravity of
p
H = energy at section B
B
the area. The line of action of the force passes through the
H = energy added
center of pressure, which can be determined by the following
a
H = energy lost
equation:
l
H = energy extracted. (The units of H , H , H , H ,
(5.8)
= [I ∕y A] + y
y
A
B
e
cp
cg
and H are feet or meters of the fluid.)
e
where
2
P , P = pressures at sections A and B (lb/ft or kN/m
A
B
= moment of inertia of the area about its center of
I
2
cg
[1,000 Newton/m ])
gravity axis
3
3
= specific weight of water (lb/ft or kN/m ) a 2 l The engineering equation is cg cg p cg (5.7)
y cp = distance of the center of pressure measured along
g = acceleration due to gravity the plane from an axis located at the intersection
2
2
(32.2 ft/s or 9.81 m/s ) of the plane and the water surface, extended if
V , V = velocities at sections A and B (ft/s or m/s) necessary
A
B
Z , Z = heights of stream tube above any assumed datum y cg = distance of the center of gravity measured along
B
A
plane at sections A and B (ft or m) the plane from an axis located at the intersection
h = head loss (ft or m) of the plane and the liquid surface, extended if
f
necessary
It is important to determine the magnitude, direction,
The moment of inertia of the most common rectangu-
and sense of the hydraulic force exerted by fluids in order
lar shape (width b and height h) can be calculated by the
to design the constraints of the structures satisfactorily. The
following equation:
force P (lb or kg) exerted by a fluid on a plane area A I = (1∕12)Ah 2
f
2
2
(ft or m ) is equal to the product of the specific weight cg 2 (5.9)
3
3
(lb∕ft orkN∕m ) of the liquid, the depth (ft and m) of the = (1∕12)(bh)h
3
2
2
center of gravity h cg of the area, and the area A (ft or m ). = (1∕12)bh
EXAMPLE 5.1 EQUATION OF CONTINUITY
For incompressible fluids, such as water, the equation of continuity shown in Eq. (5.4) can be used for all practical purposes. Derive
the equations of continuity for compressible fluids, such as air, carbon dioxide, and chlorine, which are frequently used in water and
wastewater treatment plants.
Solution:
The following equations of continuity should be used:
A v = A v (5.4a)
1 1
2 2 2
1
A v ∕g = A v ∕g (5.4b)
2 2 2
1
1 1
A v = A v (5.4c)
2 2 2
1 1
1
where , , A , A , v , v , , ,and g have been defined in Eqs. (5.2)–(5.4). The above equations of continuity result from the
2
1
1
1
2
2
2
1
basic principle of conservation of mass. For a steady flow, the mass of fluid passing any and all sections in a stream of fluid per unit
of time is the same.
EXAMPLE 5.2 SIZING AIR PIPE DIAMETER
Determine the minimum diameter of an air pipe to carry 0.55 lb/s (0.25 kg/s) of air with a maximum velocity of 19 ft/s (5.8 m/s). The
◦
◦
◦
2
air is at 82 F (27.8 C) and under an absolute pressure of 34.5 lb/in. (239.4 kPa). Gas constant for air is 53.3 ft/ R (0.287 kJ/kg-K).
Solution 1 (US Customary System):
2
2
P = P + P = 34.5 lb/in. = 34.5 × 144 lb/ft .
a gauge atm
◦
R = 53.3 ft/ R.
g