Page 155 - Water Engineering Hydraulics, Distribution and Treatment

Solution 2 (SI System):
Apply the Bernoulli equation from A to B and use A as the datum:
2
2
v
P
A
A
B
+
+ 0.9144.
2g

2
v
P
A
B
B
−
=
2g

From continuity equation
Q = A v = A v .
B B
A A
2
2
( ∕4)(0.3048) v = ( ∕4)(0.1524) v .
A
4v = v . P + 0 = v 2g P B − + 2 A v 2g + 0.9144. B 5.2 Fluid Transport 133
B
A
v = 0.25 v .
B
A
Apply the Bernoulli equation for L to R (s.g. mercury, Hg = 13.6):
P A P B
+ z + 0.4191 = + 0.9144 + z + 0.4191 × 13.6.

P P
A B
− = 6.195 m of water.

v 2 (0.25v ) 2
6.195 = B − B + 0.9144.
2g 2g
Since g = 9.81 m/s 2
v = 10.51 m∕s.
B
2 3
Q = A v = (0.1524) ( ∕4)(10.51) = 0.192 m ∕s.
B B
EXAMPLE 5.28 BUOYANCY AND FLOTATION
A body, floating or immersed in a liquid, is acted upon by a buoyant force equal to the weight of the liquid displaced. The buoyancy
center is the point through which this buoyant force acts. It is located at the center of gravity of the displaced liquid. If the center of
gravity of the body lies directly below the center of buoyancy (gravity) of the displaced liquid, the submerged body is stable. If the
two points coincide, the submerged body is in neutral equilibrium position.
Determine the volume of a metal body and its specific gravity if the body weighs 100 lb (45.4 kg) in air and weighs 60 lb
(27.2 kg) when immersed in water.
Solution 1 (US Customary System):
Buoyant force = (Weight in air) − (Weight in water)
= 100 lb − 60 lb = 40 lb.
Buoyant force = Weight of displaced liquid.
3
40 lb = V = (62.4 lb/ft )V.
3
V = 0.641 ft .
Specific gravity = (Weight of metal body)∕(Weight of an equal volume of water)
= 100 lb∕40 lb
= 2.5.
Solution 2 (SI System):
Buoyant force = (Weight in air) − (Weight in water)
= 45.4kg − 27.2kg = 18.2kg.
Buoyant force = Weight of displaced liquid.
18.2 kg = V = (1 kg/L)V.
V = 18.2 L.
Specific gravity = (Weight of metal body)∕(Weight of an equal volume of water)
= 45.4kg∕18.2kg
= 2.5.