Chapter 12
Bernoulli and Energy Equations
Review Problems
1275
Water discharges from the orifice at the bottom of a pressurized tank. The time it will take for half of
the water in the tank to be discharged and the water level after 10 s are to be determined.
Assumptions
1
The flow is uniform and incompressible, and the frictional effects are negligible.
2
The tank
air pressure above the water level is maintained constant.
Properties
We take the density of water to be 1000 kg/m
3
.
Analysis
We take point 1 at the free surface of the tank, and point 2 at the exit of orifice. We take the
positive direction of
z
to be upwards with reference level at the orifice (
z
2
= 0). Fluid at point 2 is open to
the atmosphere (and thus
P
2
=
P
atm
) and the velocity at the free surface is very low
(
V
1
0). Then,
/
2
2
2
2
2
gage
,
1
1
2
2
2
1
1
2
2
2
2
1
2
1
1
P
gz
g
g
P
z
g
P
z
g
g
P
z
g
g
P
atm
V
V
V
V
or,
/
2
2
gage
,
1
2
P
gz
V
where
z
is the water height in the tank at any time
t
. Water surface moves down
as the tank drains, and the value of
z
changes from
H
initially to
0
when the tank is emptied completely.
We denote the diameter of the orifice by
D
, and the diameter of the tank by
D
o
.
The flow rate of
water from the tank is obtained by multiplying the discharge velocity by the orifice crosssectional area,
/
2
2
4
gage
,
1
2
2
orifice
P
gz
D
A
V
V
Then the amount of water that flows through the orifice during a differential time interval
dt
is
dt
P
gz
D
dt
V
dV
/
2
2
4
gage
,
1
2
(1)
which, from conservation of mass, must be equal to the decrease in the volume of water in the tank,
dz
D
dz
A
dV
4
)
(
2
0
tank
(2)
where
dz
is the change in the water level in the tank during
dt
. (Note that
dz
is a negative quantity since the
positive direction of
z
is upwards. Therefore, we used
–dz
to get a positive quantity for the amount of water
discharged). Setting Eqs. (1) and (2) equal to each other and rearranging,
/
2
2
1
4
/
2
2
4
gage
,
1
2
2
0
2
0
gage
,
1
2
dz
P
gz
D
D
dt
dz
D
dt
P
gz
D
The last relation can be integrated since the variables are separated. Letting
t
f
be the discharge time and
integrating it from
t
= 0 when
z
=
z
0
to
t
=
t
when
z
=
z
gives
t
D
D
g
P
g
z
g
P
g
z
gage
gage
2
2
0
2
,
1
2
,
1
0
2
2
2
2
where
2
2
2
2
3
2
2
gage
,
1
s
274
.
7
kN
1
m/s
kg
1000
)
m/s
81
.
9
)(
kg/m
1000
(
kN/m
)
100
450
(
2
2
g
P
The time for half of the water in the tank to be discharged (
z
=
z
0
/2) is
t
2
2
2
2
2
2
m)
2
(
m)
1
.
0
(
s
274
.
7
m/s
81
.
9
m)
5
.
1
(
2
s
274
.
7
m/s
81
.
9
m)
3
(
2
t =
22.0 s
(
b
) Water level after 10s is
s)
10
(
m)
2
(
m)
1
.
0
(
s
274
.
7
m/s
81
.
9
2
s
274
.
7
m/s
81
.
9
m)
3
(
2
2
2
2
2
2
2
z
z
=
2.31 m
Discussion
Note that the discharging time is inversely proportional to the square of the orifice diameter.
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 Spring '08
 Chung
 Fluid Dynamics, Energy Equations

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