9.3 Solutions
405
Volume flux, Q = av = a
√
2gx
Volume flux is also equal to Q = A
dx
dt
We then have a
√
2gx = A
dx
dt
t =
dt =
A
a
√
2g
h 1
h 2
dx
√
x
=
A
a
2
g
h 1 −
h 2
9.21 Pressure at the bottom due to water column = (3 − 1) atm = 2 atm = 2 ×
10 5 Pa.
P = hρg
∴ h =
P
ρg
=
2 × 10 5
1000g
=
200
g
v =
2gh =
2g
200
g
= 20 m/s
Second method
Apply Bernoulli’s equation
P 1 +
1
2
ρv
2
1 = P 2 +
1
2
ρv
2
2
where the left side refers to the point inside the tank and right side to a point
outside the tank.
3 × 10
5
+ 0 = 1 × 10
5
+
1
2
× 1000v
2
2
∴ v 2 = 20 m/s
9.22 Q = v 1 A 1 = v 2 A 2
2gh 1
(2A 2 ) =
2gh 2
A 2
∴
h 1
h 2
=
1
4
9.23 Apply Bernoulli’s equation to a point just outside the hole and a point at the
top of the kerosene surface. If P is the atmospheric pressure, h 1 and h 2 the
heights of water and kerosene columns, respectively, ρ 1 and ρ 2 the respective
densities,
405
Volume flux, Q = av = a
√
2gx
Volume flux is also equal to Q = A
dx
dt
We then have a
√
2gx = A
dx
dt
t =
dt =
A
a
√
2g
h 1
h 2
dx
√
x
=
A
a
2
g
h 1 −
h 2
9.21 Pressure at the bottom due to water column = (3 − 1) atm = 2 atm = 2 ×
10 5 Pa.
P = hρg
∴ h =
P
ρg
=
2 × 10 5
1000g
=
200
g
v =
2gh =
2g
200
g
= 20 m/s
Second method
Apply Bernoulli’s equation
P 1 +
1
2
ρv
2
1 = P 2 +
1
2
ρv
2
2
where the left side refers to the point inside the tank and right side to a point
outside the tank.
3 × 10
5
+ 0 = 1 × 10
5
+
1
2
× 1000v
2
2
∴ v 2 = 20 m/s
9.22 Q = v 1 A 1 = v 2 A 2
2gh 1
(2A 2 ) =
2gh 2
A 2
∴
h 1
h 2
=
1
4
9.23 Apply Bernoulli’s equation to a point just outside the hole and a point at the
top of the kerosene surface. If P is the atmospheric pressure, h 1 and h 2 the
heights of water and kerosene columns, respectively, ρ 1 and ρ 2 the respective
densities,