-
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
?
4
so
D
2
x
?
?
4
2
d
2
R
(1)
?
d
?
x
?<
/p>
?
?
R
(2)
?
D
?
and
hydrostatic equilibrium gives following
relationship
p
1
?<
/p>
R
?
c
g
?
p
2
?
x
?
c
g
?
R
?
A
g
(3)
so
p
1
?
p
2
?
x
?
c
g<
/p>
?
R
(
?
A
?
?
c
)
g
(4)
substituting the
equation (2) for x into equation (4) gives
?
d
?
p
1
?
p
2
?
?
?
R
?
c
g
< br>?
R
(
?
A
?
?
c
)
g
(5)
?
D
?
（
a
）
wh
en the change in the level in the reservoirs is
neglected,
2
2
?
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
?
（
b
）
when the change in the levels in the
reservoirs is taken into account
?
d
?
p
1
?
p
2
?
?
?
R
?
c
g
?
R
(<
/p>
?
A
?
?
c
)
g
?
D
?
?
d
?
?
?
?
R
?
c
g
?
R
(
?
A<
/p>
?
?
c
)
g
?
D
?
?
6
.
5
?
?
?
?
?
0
.
145
?
815
?
9
.
81
?
0
.
145
?
1000
?
815
?
?
9
.
81
?
< br>281
.
8
Pa
51
?
?
error=
2
2
2
281
.
8
?
263
＝
6
.
7
％
281
.
8
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
=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
equilibrium
?
g
,
?
H
2
O
,
?
Hg
, respectively. The pressure at
point A is given by hydrostatic
p
A
?
?
H
< br>2
O
R
3
g
?
?
H
g
R
2
g
?
?
g
(
R
2
?
R
3
)
g
?
g
is
small
and
negligible
in
comparison
with
?<
/p>
H
g
and
ρ
H2O
,
equation
above
can
be
p
A
?
p
c
=
?
H
2
O<
/p>
gR
3
?
?
H
g
gR
2
simplified
=1000×
9.81×
0.05+136
00×
9.81×
0.05
=7161N/m?
p
B
?
p
D
< br>?
p
A
?
?
H
g
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,
p
a
tank
A
and
the
exit
of
drainpipe are all open to
air.
H
d
D
h
p
a
Figure for problem 1.5
A
Solution:
Bernoulli equation
is written between stations 1-1 and 2-2, with
station 2-2 being reference plane:
p
1
2
u
1
2
p
2
u
2
?
gz
< br>1
?
?
?
gz
2
?
?
2
?
2
Where
p
1
=0,
p
2
=0, and
u
1
=0, simplification of the
equation
2
u
2
Hg
?
1
2
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
?
D
?
u
o
?
< br>u
2
?
?
2
?<
/p>
d
?
Bernoulli
equation is written between the throat and the
station 2-2
2
2
p
u
u
0
?
0
?
3
2
2
?
Combining equation 1,2,and
3 gives
2
u
1
h
?
g
< br>1
2
?
1000
?
9
.
81
< br>2
?
9
.
81
Hg
?
?
＝
＝
＝
4
4
2
?
D
?
?
1000
2<
/p>
.
44
?
1
?
1
.
25
?
?
1
?
?
?
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
Then
Bernoulli equation
becomes, after substituting
V
2
?
V
1
A
1
for
V
2
,
A<
/p>
2
A
1
2
V
2
V
1
p
1
A
1
2
p
2
0
?
?
?
0
?
?
2
?<
/p>
2
?
2
1
Rearranging,
A
1
2
?
V
(
2
?
1
)
A
1
p
1
?
p
2
?<
/p>
2
2
1
V
1
＝
p
1
?
p
2
2
?
?
A
1
?
?
?
A
?
?
?
2
?<
/p>
?
?
?
?
1
?
?
?
?
2
?
Performing the same derivation but in
terms of
V
2
,
V
2
＝
p
1
?
p
2
< br>2
?
?
A
2
?
1
?
?
?
A
?
?
?
1
?
?
?
?
2
?
?
?
?
?
< br>
1.7 A liquid whose
coefficient of viscosity is
?
flows below the critical
velocity for laminar flow
in a circular
pipe of diameter
d
and with
mean velocity
V
. Show that
the pressure loss in a length
of pipe
?
p
32
?
V
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
R
R
1
1
V
?
2
?
rdr
1
udA
?
2
u
A
0
?
R
0
From
velocity profile equation for laminar flow
2
?
?
p
?
p
r
?
?
2
u
?
0
L
?
R
1
?
?
?
?
2
?
?
R<
/p>
?
?
4
?
L
?
?
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
?
V
?
L
d
2
32
?
VL
32
?
0
.<
/p>
05
?
0
.
6
?
120
?
p
?
?
?
11520
Pa
2
2<
/p>
d
0
.
1
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
A and when the flow rate down the pipe
is 0.02 m
3
/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
V
2
expressed
as
k
where
V
is
the
velocity
at
A,
find
the
2
g
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
?
?
4
2
d<
/p>
A
V
A
V
A
?
Q
?
4
2
d
A
0
.
02
< br>?
?
1
.
132
m
/
s
0
.
785
?
0
.
15
2
Q
?
?
4
2<
/p>
d
B
V
B
V
B
?
Q
?
4
2
d
B
0
.
02
?
?
4
.
529
m
/
s
0
.
785
?
0
.
075
2
When the fluid flows down, writing
mechanical balance equation
p
A
2
2
V
A
p
B
V
B
2
V
A
?
z
A
g
?
?
?
z
< br>B
g
?
?
k
?
2
?
2
2
1
.
13<
/p>
2
14715
4
.
53
2
1
.
13
2
2
.<
/p>
5
?
9
.
81
?
?
?
?
k
2
1000
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
?
H
g
g
R
?
?
p
B<
/p>
?
p
A
?
?
l
?
g
?
?
H
g
?
?
g
?
?
14715
?
2
.
5
?
1000
?
9
.
81
?
?
79
mm
12600
?
9
.
81
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
?
p
cd
,
h
fab
,
and
h
fcd
,
respectively.
?
g
?
g
Figure for problem 1.9
（
2
）
the
relationship between readings
R
1
and
R
2
in the U tube.
Solution:
(1)
From Fanning equation
2
l
V
h
fab
?
?
d
2
and
l
V
2
h
fcd
?
?
d
2
so
h
?
h
fab
fcd
Fluid flows from station
a
to station
b
, mechanical energy
conservation gives
p
a
p
b
?
lg
?
?
h
fab
?
?
hence
p
?
p
b
a
?
lg
?
h
fab
2
?
from station
c
to station
d
p
p
c
?
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
c
-p
d
=R
2
（<
/p>
ρ
ˊ
-
ρ
）
g
5
Substituting equation 4 in
equation 2
，
then
?
R
（
1
?
?
?
）
g
?
l
?
g
?
lg
?
h
fab
?
therefore
?
?
?
?
h
fab
?
R
g
6
1
?
Substituting equation 5 in equation 3
，
then
?
?
?
?
7
h
fcd
2
g
?
R
?
Thus
R
1
=R
2
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.
r
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
kg/m
3
，
respectively
；
and
the
viscosity
Figure for problem 1.10
and
density
of
kerosene
are
0.003
Pas
and
800
kg/m
3
，
respectiv
ely
）
solution:
1
）
Re
?
ud
?
?
?
0
.
4
?
0
.
004
?
1000
?
1600
0
.
001
from Hagen-Poiseuille equation
< br>?
P
?
32
uL
?
32
?
0
.
4
?
2
?
0
.
00
1
?
?
1600
2
2
d
0
.
004
h
?
?
p
1600
?
?
0
.
16
3
m
?
g<
/p>
1000
?
9
.
81
2
）
ma
ximum velocity occurs at the center of pipe, from
equation 1.4-19
so
u
max
=
0.4×
2=0.8m
3
）
when u=V=0.4m/s
Eq. 1.4-17
V
?
0.5<
/p>
u
max
u
u<
/p>
max
?
r
?<
/p>
?
1
?
?
?
r
?
?
?
w
?
2
2
V
?
r
?
1
?
?
?
0
.
5<
/p>
?
＝
0
.
004
u
?
?
max
r
?
0
.
004
0
.
5
?
0
.
004
?
0
.
71
?
0
.
00284
m
4) kerosene:
Re
?<
/p>
ud
?
?
?
0
.
4
?
0
.
004
?
800
?
427
0
.
003
?
p
?
?
?
p
?
?
0
.
003
?
1600
?
4800
Pa
<
/p>
?
0
.
001<
/p>
h
?
?
?
p
?
4800
?
?
0<
/p>
.
611
m
<
/p>
?
?
g
800<
/p>
?
9
.
81
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
-
-
-
-
-
-
-
-
-
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