9.3 Solutions
401
Comparing (1) and (2)
v =
2ghρ L
ρ g
9.13 Apply Bernoulli’s equation at the sections A 1 and A 2 :
P 1 +
1
2
ρv
2
1 = P 2 +
1
2
ρv
2
2
(1)
∴ p 2 − p 1 = p = hρg =
1
2
ρ(v
2
1 − v
2
2 )
∴ 2gh = v
2
1 − v
2
2
(2)
Q = v 1 A 1 = v 2 A 2
(3)
v 2 =
v 1 A 1
A 2
(4)
Using (4) in (2)
2gh = v
2
1
(A 2
2 − A 2
1 )
A 2
2
v 1 = A 2
2gh
A 2
2 − A 2
1
Q = A 1 v 1 = A 1 A 2
2gh
A 2
2 − A 2
1
9.14 Volume of water flowing out per second
Q = sv
(1)
where v is the speed and s is the cross-sectional area.
Volume flowing out
V = Qt = svt
(2)
1
2
ρv
2
= P =
F
A
=
F L
AL
=
W
V
(3)
where L is the length of the cylinder and W is the work done.
∴ W =
1
2
ρV 3
s 2 t 2
(4)
where we have used (2).