-
1.1 What will be the (a) the gauge pressure
and (b) the absolute pressure of water at depth
12m
below the surface?
ρ
water
= 1000
kg/m
, and
P
atmosphere
=
101kN/m
.
Solution:
Rearranging the equation 1.1-4
< br>p
b
?
p
a
?
?
gh
3
2
Set the
pressure of atmosphere to be zero, then the gauge
pressure at depth 12m below the surface
is
p
b
?
p
a
?
?
gh
?
0
?
1000
?
9
.
81
?
12
?
117
.
72<
/p>
kPa
Absolute
pressure of water at depth 12m
p
b
?
p
a
?
?
gh
?
101000
?
1000
?
9
.
81
?
12
?
218720<
/p>
Pa
?
218
.
72
kPa
1.3
A
differential
manometer
as
shown
in
Fig.
is
sometimes
used
to
measure
small
pressure
difference. When the reading is zero,
the levels in two reservoirs are equal. Assume
that fluid B is
methane
（甲烷）
, that
liquid C in the reservoirs is kerosene (specific
gravity = 0.815), and that
liquid
A
in the U tube is water.
The inside diameters of the reservoirs and U tube
are
51mm and
6.5mm ,
respectively. If the reading of the manometer
is145mm., what is the pressure difference
over
the
instrument
In
meters
of
water,
(a)
when
the
change
in
the
level
in
the
reservoirs
is
neglected, (b) when the change in the
levels
in the reservoirs is taken into
account? What is the
percent error in
the answer to the part (a)?
Solution
：
p
a
=1000kg/m
3
p
c
=81
5kg/m
3
p
b
=0.77kg/m
3
D/d=8
R=0.145m
When the pressure
difference between two
reservoirs is
increased, the volumetric changes in the
reservoirs and U tubes
?
?
2
2
D
x
?
d
R
< br>
(1)
4
4
so
?
d
?
x
?
?
?
R
(2)
?
D
?
2
and hydrostatic
equilibrium gives following relationship
p
1
?
R
?
c
g
?
p<
/p>
2
?
x
?
c
g
?
R
?
A
g
(3)
so
p
1
?
p
2
?
x
?
c
g
?
R
(
?
A
?
?
c
)
< br>g
(4)
substituting the equation (2) for x
into equation (4) gives
?
d
?
p
1
?
p
2
?
?
?
R
?
< br>c
g
?
R
(
?
A
?
?
c
)
g
(5)
?
D
?
2
（<
/p>
a
）
when the change
in the level in the reservoirs is neglected,
?
d
?
p
1
?
p
2
?
?
?
R
< br>?
c
g
?
R
(
?
A
?
?
c
)
g
?
R
(
?
A
?
?
c
)
g
?
0
< br>.
145
?
1000
?
815
?
?
9
.
81
?
263
Pa
?
D
?
2
（
b
）
when the change in the
levels in the reservoirs is taken into account
?
d
?
p
1
?
p
2
?
?
?
R
?
c
g
?
R
(
?
A
?<
/p>
?
c
)
g
?
D
?
?
d
?
?
?
?
R
?
c
g
?
R
(
?
A
?
?
c<
/p>
)
g
?
D
?
?
6
.
5
?
?
?
?
?
0
.
145
?
815
?
9
.
81
?
< br>0
.
145
?
< br>1000
?
815
?
?
9
.
81
?
281
.
8
Pa
?
51
?
2
2
2
< br>error=
281
.
8
?
263
281
.<
/p>
8
＝
6
.
7
％
1.4
There are two
U-tube manometers fixed
on the fluid bed reactor, as shown in the figure.
The
readings of two U-tube
manometers are
R
1
=400mm
，
R
2
< br>=50mm, respectively. The indicating liquid
is mercury. The top of the manometer is
filled with the water to prevent from the mercury
vapor
diffusing into the air, and the
height
R
3
=50mm.
Try to calculate the pressure at point
A
and
B
.
Figure for problem 1.4
Solution:
There is a gaseous
mixture in the U-tube manometer meter. The
densities of fluids
are denoted by
?
g
,
?
H
2
O
,
< br>?
Hg
, respectively. The
pressure at point A
is given by
hydrostatic
equilibrium
p
A
?
?
H
2
O
R
3
g
?
?
Hg
R
2
g
?
?
g
(
R
2
?
R
3
)
g
?
g
is
small
and
negligible
in
comparison
with
?<
/p>
Hg
and
ρ
H2O
,
equation
above
can
be
simplified
p
A
?
p
c
=
?
H
2
O
gR
3
?
?
Hg
gR
2
=10
00×
9.81×
0.05+13600×
9.81×
0.05
=7161N/m?
p
B
?
p
D
< br>?
p
A
?
?
Hg
gR
1
=7161+13600×
9.81×
0.4=60527N
/m
1.5 Water discharges from the reservoir
through the drainpipe, which the throat diameter
is
d.
The
ratio
of
D
to
d
equals 1.25. The vertical
distance
h
between the tank
A
and axis of the drainpipe
is
2m. What height
H
from the centerline of the
drainpipe
to the water level in
reservoir is required
for
drawing
the
water
from
the
tank
A
to
the
throat
of
the
pipe?
Assume
that
fluid
flow
is
a
potential
flow.
The
reservoir,
tank
A
and
the
exit
of
drainpipe are all open to air.
Solution:
Bernoulli equation
is written between stations 1-1 and 2-2, with
station 2-2 being reference plane:
p
1
?
gz
1
?
u
1
2
2
p
a
H
D
d
h
p
a
Figure for problem 1.5
A
?
?
p
2
?
?
gz
2
?
u
2
2
< br>2
Where
p
1
=0,
p
2
=0, and
u
1
=0, simplification of the
equation
2
u
2
Hg
?
2
1
The
relationship
between
the
velocity
at
outlet
and
velocity
u
o
at
throat can
be
derived
by
the
continuity equation:
?
u
2
?
?
< br>u
?
o
?
?
d
?
?
?
?
?
?
?
?
D
?
2
2
u
o
?
D
?
?
< br>u
2
?
?
2
?<
/p>
d
?
Bernoulli
equation is written between the throat and the
station 2-2
2
2
0
?
0
?
3
2
2
?
p
u
u
Combi
ning equation 1,2,and 3 gives
< br>2
u
1
h
?
g
1
2
?
1000
?
9
.
81
2
?
9
.
81
Hg
?
?
＝
＝
＝
4
4
2
?
1000
2
.
44
?
1
?
1
.
25
?
?
1
?
D
?
?
1
?
?
?
d
?
Solving for H
H=1.39m
1.6
A
liquid with a constant density
ρ
kg/m
3
is flowing at an
unknown velocity
V
1
m/s through a
horizontal pipe of cross-sectional area
A
1
m
2
at a pressure
p
1
N/m
2
, and then it passes to
a section
of the pipe in which the area
is reduced gradually to
A
2
m
2
and the pressure is
p
2
. Assuming no
friction losses, calculate the
velocities
V
1
and
V
2
if the
pressure difference
(
p
1
-
p
2
) is measured.
Solution
:
In Fig1.6, the flow diagram is shown
with pressure taps to measure
p
1
and
p
2
.
From the mass-balance continuity
equation
, for constant
ρ
where
ρ
1
=
ρ
2
=
ρ
,
V
2
?
V
1
A
1
A
2
For the
items in the Bernoulli equation , for a horizontal
pipe,
z
1
=
< br>z
2
=0
V
< br>1
A
1
A
2
Then
Bernoulli
equation
becomes, after sub
stituting
V
2
?
for
V
2
,
0<
/p>
?
V
1
2
2
?
p
1
V
1
?
0
?
2
A
1
A
1
2
2
2
?
?
p
2<
/p>
?
Rearranging,
A
1
A
2
2
?
V
1
(
p
1
?
p
2
?
2
2
1
?
1
)<
/p>
V
1
＝
p
1
?
p
2
2
?
?
A
?
?
1
?
?
?
?
A
?
?
1
?<
/p>
?
?
?
?
2
?
?
2
?
Performing the same
derivation but in terms of
V
2
,
p<
/p>
1
?
p
2
2
?
?
1
?
?
?
?
A
2
?
?
?
?
?
A
?
?
?
1
?<
/p>
?
?
2
V
2
＝
?
1.7 A liquid whose
coefficient of viscosity is
?
flows below the critical
veloc
ity for laminar flow
in
a circular pipe of diameter
d
and with mean velocity
V
. Show that the pressure
loss in a length
?
p
32
?
V
of
pipe
is
.
2
L
d
Oil
of viscosity 0.05 Pas flows through a pipe of
diameter 0.1m with a average velocity of 0.6m/s.
Calculate the loss of pressure in a
length of 120m.
Solution
:
The
average velocity
V
for a
cross section is found by summing up all the
velocities over the cross
section and
dividing by the cross-sectional area
1
1
V
?
u
2
?
1
udA
?
2
rdr
?
?
A
0
?
R
0
R
R
From velocity
profile equation for laminar flow
2
?
?
p
?
p
r
?
< br>?
u
?
0
L
R
2
?
1
?
?
?
?
2
?
4
?<
/p>
L
?
R
?
?
?
?
substituting
equation 2 for
u
into
equation 1 and integrating
p
0
?
p
L
2
V
?
D
3
32
?
L
rearranging equation 3 gives
?
p
32<
/p>
?
V
?
2
L
d
32
?
VL
d
2
?
p
?
?
32
?
0
.
05
?
0
.
6
?
< br>120
0
.
1
< br>2
?
11520
Pa
1.8.
In
a
vertical
pipe
carrying water,
pressure
gauges
are
inserted
at
points
A
and
B
where
the
pipe
diameters
are
0.15m and 0.075m respectively. The
point B is 2.5m below
3
A
and
when
the flow
rate
down
the
pipe
is
0.02
m
/s,
the
pressure at B is 14715
N/m
2
greater than that at A.
Assuming
the
losses
in
the
pipe
between
A
and
B
can
be
expressed
as
k
V
2
2
g
where
V
is
the
velocity
at
A, find
the
Figure for problem 1.8
value
of
k
.
If
the
gauges
at
A
and
B
are
replaced
by
tubes
filled
with
water
and
connected
to
a
U-tube
containing mercury of
relative density 13.6, give a sketch showing how
the levels in the two limbs
of the
U-tube differ and calculate the value of this
difference in metres.
Solution:
d
A
=0.15m;
d
B
=0.075m
z
A
-
z
B
=
l
=2.5m
Q
=0.02
m
3
/s,
p
B
-p
A
=14715
N/m
2
Q
?
?
2
4
d<
/p>
A
V
A
V
Q
A
?
?
?
0
.
02
2
0
.
785
?
0
.
15
2
?
1
.
132
m
/
s
4
d
A
Q
?
?
2
4<
/p>
d
B
V
B
V
Q
B
?
?
?
0
.
02
2
0
.
< br>785
?
0
.
< br>075
2
?
4
< br>.
529
m
/
< br>s
4
d
B
When the fluid flows down, writing
mechanical balance equation
p
2
2
2
A
?
z
A
p
B
A
?
A
g
?
V
2
?
?
?
z
B
< br>g
?
V
B
2
?
k
V
2
2
2
2
.
5
?
9
.
81
?
1
.
13
2
?
14715
1000
?
4<
/p>
.
53
2
?
k
1
.
13
2
2
24
.
525
?
0
.
638
?
14
.
715
?
10
.
260
?
0
.
638
k
k
?
0.295
making the static equilibrium
p
B
?
?
x
?
g
?
R<
/p>
?
g
?
p
A
?
l
?
g
?
?
x
?
g
?
R
?
Hg
g
R
?
?
p
B
?
p
?
A
?
?
l
?
g
?
2
.
5
?
1000
?
9
.
81
?
H
g
?
?
?
< br>g
?
14715
12600
?
9
.
81
?
?
79
mm
1.9
．
The
liquid
vertically
flows
down
through
the
tube
from
the
station
a
to
the station
b
,
then
horizontally
through
the
tube
from
the station
c
to the station
d
, as shown in figure. Two
segments of
the tube, both
ab
and
cd
，
have the same
length, the diameter and
roughness.
Find:
（
1
）
the
expressions of
?
p
ab
cd
?
g
,
h
fab
,
?
p
?
g
and
h
fcd
,
respectively.
（
2
）
the
relationship between readings
R
1
and
R
2
in the U tube.
Solution:
(1)
From Fanning equation
h
l
V
2
fab
?
?
d
2
Figure for problem 1.9
and
2
l
V
h
fcd
?
?
d
2
so
h
fab
?
h
fcd
Fluid flows from station
a
to station
b
, mechanical energy
conservation gives
p
?
a
?
lg
?
p
?
b
?
h
fab
hence
p
a
?
p
b
?
lg
?
h
fab
2
?
from station
c
to station
d
p
c
p
?
?
d
?
?
h
fcd
hence
p
c
?
p
d
?
h
?
fcd
3
From static equation
p
a
-p
b
=R
1
（
ρ
ˊ
-
ρ
）
g -l
ρ
g
4 <
/p>
p
c
-p
d
=R
2
（
ρ
ˊ
-
ρ
）
g
Substituting
equation 4 in equation 2
，
then
R
（
1
?
?
?
?
）
g
?
l
?
g
?
lg
?
h
fab
?
therefore
h
?
?
fab
?
R
?
?
1
g
?
6
Substituting equation 5 in equation 3
，
then
h
?
?
?
?
fcd
?
R
2
g
7
?
Thus
R
1
=R
2
5
1.10
Water
passes
through
a
pipe
of
diameter
d
i=0.004
m with
the
average
velocity
0.4 m/s,
as
shown in Figure.
1)
What is the pressure drop
–
?
P
when water flows through
the pipe length
L
=2 m, in m
H
2
O
column?
2) Find the maximum velocity and point
r
at which
L
it occurs.
3)
Find
the
point
r
at which
the
average
velocity
equals the local velocity.
4
）
if kerosene
flows through this pipe
，
how
do the
variables above
change
？
（
the viscosity and density
of
Water are 0.001 Pas
and
1000
< br>kg/m
，
respectively
< br>；
and
the
viscosity
and
density
of
kerosene
are
0.003
Pas
and
800
kg/m
3
，
respectiv
ely
）
solution:
1
）
Re
?
ud
?
?
0
.
4
?
0
.
004
?
1000
0
.
001
?
1600
Figure for problem 1.10
3
r
?
from Hagen-Poiseuille
equation
32
uL
?
32
?
0
.
4
?
2
?
0
.
001
?
P
?
d
2
< br>?
0
.
004
< br>2
?
1600
h
?
?
p
?
g
?
1600
< br>1000
?
9
.
81
?
0
.
< br>163
m
2
）
maximum
veloc
ity occurs at the center of pipe,
from equation 1.4-19
V
u
m
ax
?
0.5
so
u
max
=0.4×
2=0.8m
3
）
when u=V=0.4m/s
Eq. 1.4-17
u
u
max<
/p>
?
r
?
?
?
1
?
?
?
r
?
?
w
?
2
2
V
?
r
?
1
?
?
?<
/p>
0
.
5
?
＝
u
max
?
0
.
004
?
r
?
0
.
004
0
.
5
?
0
.
004
?
0
.
71
?
0
.
< br>00284
m
4)
kerosene:
Re
?
ud
?
?
0
.
4
?
0
.
004
?
800
0
.
003
0
.
003
0
.
001
?
427
?
?
p
?
?
?
p
?
?
?
?
1600
?
4800
Pa
h
?
?
?
p
?
?
?
g
?
4800
800
?
9
.
81
?
0
.
611
m
1.12 As
shown in the figure, the water level in the
reservoir keeps constant. A
steel
drainpipe (with
the inside diameter of
100mm) is connected to the bottom of the
reservoir. One arm of the U-tube
manometer
is
connected
to
the
drainpipe
at
the
position
15m
away
from
the
bottom
of
the
reservoir,
and the other is opened to the air, the U tube is
filled with mercury and the left-side arm
of the U tube above the mercury is
filled with water. The distance between the
upstream tap and
the outlet of the
pipeline is 20m.
a)
When the gate
valve is closed, R=600mm, h=1500mm; when the gate
valve is opened partly,
R=400mm,
h=1400mm.
The friction coefficient λ
is
0.025, and the loss coefficient of
the entrance
is 0.5. Calculate the flow
rate of water when the gate valve is opened
partly. (in m?
/h)
b)
When the gate
valve is widely open, calculate the static
pressure at the tap (in gauge pressure,
N/m?
).
l
e
/
d
≈15
when the gate valve is widely
open,
and the friction coefficient λ is still
0.025.
Figure for problem 1.12
Solution
：
(1) When the gate valve is opened
partially, the water discharge is
Set
up
Bernoulli
equation
between
the
surface
of
reservoir
1
—
1
and
the
section
of
pressure point
2
—2’
，
and take the
center of section
2
—
2
’
as the referring plane, then
gZ
1
?<
/p>
u
1
2
2
’
?
p
1
?
?
gZ
2
?
u
2
2
2
?
p
2
?
?
?
h
f
,
1
—<
/p>
2
（
a
）
In the equation
p
1
?
0
(the gauge
pressure)
p
2
?
?
Hg
gR
?
?
H
2
O
gh
?
13600
?
9
.
81
?
0
.
4
?
1000
?
9
.
81
?
1
.
4
?
39630
N
/
m
2
-
-
-
-
-
-
-
-
-
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