-
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.
p
a
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
?
gz
1
?
u
1
2
2
?
?
p
2
?
?
gz
2
?
u
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
u
o
?
D
?
?
< br>u
2
?
?
2
?<
/p>
d
?
Bernoulli
equation is written between the throat and the
station 2-2
?
?
3
?
2
2
Combining equation 1,2,and 3 gives
p
0
u
0
2
u
2
< br>1
h
?
g
1
2
?
1000
?
9
.
81
2
?
9
.
8
1
Hg
?
?
＝
＝
＝
4
4
2
?
1000<
/p>
2
.
44
?
1
?
1
.
25
?
?
1
?
D
?
?
?
?
1
?
d
?
Solving for H
u
2
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
1
A
1
A
2
V
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
1
2
2
2
?
V
1
(
p
1
?
p
2
?
2
?
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
?
2
V<
/p>
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
R
R
1
1
V
?
u
2
?
1
udA
?
rdr
2
?
A
0
?
R
?
0
From velocity profile
equation for laminar flow
2
?
u
?
0
L
R
2
?
1
r
?
?
2
?
<
/p>
?
?
?
4
?
L
?
R
?
?
?
?
p
?
p
?
?
substituting equation 2
for
u
into equation 1 and
integrating
p
0
?
p
L
2
V
?
D
3
32
?
L
rearranging equation 3 gives
2
L
d
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
greater than that at A.
Assuming
the
losses
in
the
pipe
between
A
and
B
can
be
expressed
as
k
V
2
?
p
?
32
?
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
?
V
A
?
?<
/p>
4
d
A
V
A
Q
?
2
2
0
.
02
0
.
785
?
0
.
15
2
?
4
?
1
.
132
m
/
s
d
A
Q
?
V
B
?<
/p>
?
4
d
B
V
B
Q
?
2
B
2
0
.
02
0
.
< br>785
?
0
.
< br>075
2
?
4
< br>?
4
.
529
< br>m
/
s
d
When the fluid flows down,
writing mechanical balance equation
p
A
?
z
A
g
?
V
A
2
2
?
?
< br>p
B
?
?
z
B
g
?
V
B
2
2
?
k
V
A
2
2
2
.
5
?
9
.
< br>81
?
1
.
13
2
2
?
14715
1000
?
4
.
53
2
2
?
k
1
.
< br>13
2
2
24
.
525
?
< br>0
.
638
?
< br>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
H
g
?
?
g
?
?
< br>14715
?
2
.
5
?
1000
?
9
.
81
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
?
g
,
h
fab
,
?
p
cd
?<
/p>
g
and
h
fcd
,
respectively.
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
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
p
b
?
lg
?
?
h
fab
?
?
hence
p
a
?
p
b
?
lg
?
h
fab
2
?
from station
c
to station
d
p
p
c
?
d
?
h
fcd
?
?
hence
p
?
p
c
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
（
?
?
?
?
）
g
?
l
?
g
1
?
lg
?
h
fab
?
therefore
?
?
?
?
h
fab
?
R
g
6
1
?
Substituting equation 5 in equation 3
，
then
?
?
?
?
h
fcd
?
R
2
g
7
?
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.
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
3
，
respecti
vely
；
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
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
?
＝
0
.
004
u
?
?
max
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
< br>?
p
2
?
?
?
h
f
,
1
—
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
u
1
?
0
< br>Z
2
?
0
2
When the gate valve is fully closed,
the height of water level in the
reservoir can be related
?
H
g
(
Z
1
?
h
)
?
?
Hg
gR
to h (the distance
between the center of pipe and the meniscus of
left arm of U tube).
2
O
（
b
）
where h=1.5m
R=0.6m
Substitute the known
variables into equation b
Z
1
?
13600
?
0
.
6
1000
l
d
?
1
.
5
?
6
.<
/p>
66
m
?
K
c
)
V
2
?
h
f
,
1
_
2
< br>?
(
?
2
?
(
0
.
0
25
?
15
0
.
1
?
0
.<
/p>
5
)
V
2
?
2
.
13
V
2
2
Substitute the known variables equation a
9.81×
6.66=
V
2
2
?
< br>39630
1000
?
2
.
13
V
2
the velocity is
V =3.13m/s
the flow rate of water is
?
2
V
h
?
3600
?<
/p>
4
d
V
?
3600
?
?
4
?
0
.
12
?
3
.
13
?
88
.
5
m
/
h
3
2
）
the
pressure of the point where pressure is measured
when the gate valve is wide-open.
Write mechanical energy balance
equation between the stations
1
—
1
’
and 3-3?
，
then
gZ
1<
/p>
?
V
1
2
2
?
p
1
?
?
gZ
3
?
V
3
2
< br>2
?
p
3
?
?
?
h
f
,
1
—
3
（
c
）
since
Z
1
?
6
.
66
m
Z
3
?
0
u
1
?
0
p
1
?
p
3
?
h
f
,
< br>1
_
3
?
(
?
l
?
l
e
d
35
?<
/p>
K
c
)
V
2
2
V
2
?
[
0
.
025
(
?<
/p>
4
.
81
V
2
0
.
1
?
15
)
?
0
.
5
]
2
input the above data
into equation c
，
9.81
?
6
.
66
?
V
2
2
?
4
.
81
V
2
the velocity is:
V
=3.51 m/s
Write
mechanical
energy
balance
equation
between
thestations
1
—
1
’
and
2
——2’
,
for
the
same
situation of water level
gZ
1
?<
/p>
V
1
2
2
2
?
p
1
?
?
gZ
2
?
V
2
2
< br>?
p
2
?
?
?
h
f
,
1
—
2
（
d
）
since
Z
1
?
6
.
66
m
Z
2
?
0
u
1
?
0
u
2
?
3.51
m
/
s
p
1
?
0(page pressure
）
?
h
f
,
1
_
2
?
(
?
l
d
?
K
c
)
V<
/p>
2
2
?
(
0
.
025
?
15
0
.
1
?
0
.
5
)
3
.
51
2
2
?
26
< br>.
2
J
/
kg
input the above data
into equation d
，
9.81×
6.66=
3
.
51
2
2
?
p
2
1000
?
26
.
2
the pressure is:
p
2
?
32970
1.14
Water at
20
℃
passes through a steel
pipe
with an inside diameter of 300mm
and 2m long.
There
is
a
attached-
pipe
(Φ60
?
3.5mm)
which
is
parallel
with
the
main
pipe.
The
total
length
including
the
equivalent
length
of
all
form
losses
of
the
attached-pipe
is
10m.
A
rotameter
is
installed
in
the
branch
pipe.
When
the
reading
of
the
rotameter
is
2.72m
3
/h, try
to calculate
the
flow
rate
in
the
main
pipe
and
the
total
flow rate,
respectively.
The frictional
coefficient
of
the
main pipe
and the attached-pipe is 0.018 and 0.03,
respectively.
Solution
：
The variables of main pipe are denoted
by a subscript1, and branch pipe by subscript 2.
The friction loss for parallel
pipelines is
?
h
f
1
?
?
h
f
2
V
s
?
V
S
1
< br>?
V
S
2
The energy loss in the branch pipe is
l
2
?
?
h
f
2
?
?
2
?
l
d
2
e
2
< br>u
2
2
2
In the equation
?
2
?
0
.
03
l
2
?
?
l
e
2
?
10
m
d
2
?
0
.
053
u
2
?
3600
?<
/p>
2
.
72
?
4
?
0
.
343
m
/
s
2
?
0
.
053
input the data into
equation c
?
h
f
2
?
0
.
03
?
10
0
.
053
?
0
.
343
2
2
?
0
.
333
J
/
kg
The
energy loss in the main pipe is
?
h
f
1
?
< br>?
h
f
2
?
?
1
l
1
u
1
d
1
2
2
?
0
.
333
So
u
1
?
0
.
333
?
0
.
3
?
2
0
.
018
?
2
?
2
.
< br>36
m
/
s
The water discharge of main pipe is
?
2
3
V
h
1
?
3600<
/p>
?
?
0
.
3
?
2
.
36
?
601
m
/
h
4
Total water discharge is
V
h
?
601
?
2
.
72
?
603
.<
/p>
7
m
/
h
3
1.16
A
Venturimeter
is
used
for
measuring
flow
of
water
along
a
pipe.
The
diameter
of
the
Venturi throat is two fifths the
diameter of the pipe. The inlet and throat are
connected by water
filled tubes to a
mercury U-tube manometer. The velocity of flow
along the pipe
is found to be
2
.
5
R
m/s, where
R
is
the manometer reading in metres of mercury.
Determine the loss of head
between
inlet and throat of the Venturi when
R
is 0.49m. (Relative
density of mercury is 13.6).
Solution:
Writing mechanical energy balance
equation between the inlet
1
and throat o for V
enturi meter
< br>p
1
?
V
1
2
2
?
2
?
z
1
g
?
p
o
?
?
V
o
2
?
z
2
g
< br>?
h
f
1
rearranging
the equation above, and set (
z
2
-
z
1
)=
x
p
1
?
p
o
?
V
o
?
V
1
2
2
2
?
?
xg
?
h
f
2
Figure for problem 1.16
from continuity equation
?
d
1
?
?
5
?
?
?
?
?
V
1
< br>?
6
.
25
V
1
3 <
/p>
V
o
?
V
1
?
?
d
?
?
2
?
?
o
?
2
2
substituting equation 3 for
V
o
into equation
2 gives
p
1
?
p
o
?
?
< br>39
.
06
V
< br>1
?
V
1
2
2
2
2
?
xg
?
h
f<
/p>
?
19
.
03<
/p>
V
1
?
h
f
?
19
.
03
2
.
5
R
?
?
2
?
xg
?
h
< br>f
4
?
118
.
94
R
?
x
g
?
h
f
fr
om the hydrostatic equilibrium for manometer
p
1
?
p
o
?
R
(
?
Hg
?
?
)
g
?
x
?
g
5
substituting equation 5
for pressure difference into equation 4 obtains
R
(
?
Hg
?
?
)
g
?
x
?
g
?
118
.
94
R
?
xg
?
h
f
6
?
rearranging equation 6
h
f
?
R
(
?
Hg
?
?
)
g
?
118
.
94
R
?
123
.
61
R
?
118
.
94
R
?
4
.
67
R
?
2
.
288
J
/
kg
?
ric acid of
specific gravity 1.3 is flowing through a pipe of
50 mm internal diameter. A
thin-lipped
orifice, 10mm, is fitted in the pipe and the
differential pressure shown by a mercury
manometer is 10cm. Assuming that the
leads to the manometer are filled with the acid,
calculate
(a)the
weight
of
acid
flowing
per
second,
and
(b)
the
approximate
friction
loss
in
pressure caused by the orifice.
The coefficient of the orifice may be
taken as 0.61, the specific gravity of mercury as
13.6, and the
density of water as 1000
kg/m
3
Solution:
a)
D
0
D
1
?
10
50
?
0
.
2
p
< br>1
?
p
2
?
R
(
?
H
g
?
?
)
g<
/p>
?
0
.
1
(
13600
?
13
00
)
?
9
.
81
?
V
2
?
m
?
C
o
?
D
0
?
?
1
?
?
< br>?
D
?
?
1
?
4
2
?
p
1
?
p
2
?
?
?
0
.
61
1
?
0
.
2
4
2
?
0
.
1
(
13600
?
1300
)
?
9
.
81
1300
?
0
.
61
18
.
56
?
0
.
61
?
4
.
31
?
2
.
63
m
< br>/
s
?
4
D
0
V
2
?
?
2
?
4
?
0
.
01
?
2
.
63
?
1300
?
0
.
268
kg
/
s
2
b)
approximate pressure drop
p<
/p>
1
?
p
2
?
R
(
?
Hg
?
?
)
g
?
0
.
< br>1
(
13600
?
1300
)
?
9
.
81
?
12066.3
Pa
pressure difference due to increase
of velocity in passing through the orifice
V
2
?
?
2
p
1
?
p
2
?
?
V<
/p>
2
?
V
1
2
2
2
?
D
o
?
2
?
?
V
2
?
?
D
?
?
1
?
2
4<
/p>
?
1300
2
.
63
(
1
?<
/p>
0
.
2
)
2
2
4
?
4488
.
8
Pa
pressure drop
caused by friction loss
?
p
f
?
12066
.
3
?
448
8
.
8
?
75
77
.
5
Pa
2.1
Water
is
used
to
test
for
the
performances
of
pump.
The
gauge
pressure
at
the
discharge
connection is 152 kPa
and the reading of vacuum gauge at the
suction connection of the pump is
24.7
kPa
as
the
flow
rate
is
26m
/h.
The
shaft
power
is
2.45kw
while
the
centrifugal
pump
operates
at
the speed
of
2900r/min.
If
the
vertical
distance
between
the
suction connection
and
discharge connection is 0.4m, the
diameters of
both the suction and
discharge line are the same.
Calculate
the
mechanical
efficiency
of
pump
and
list
the
performance
of
the
pump
under
this
operating condition.
3
Solution:
Write
the
mechanical
energy
balance
equation
between
the
suction
connection
and
discharge
connection
Z
1
?
where
Z
2
?
Z
1<
/p>
?
0
.
4
m
p
1
?
?
2
.
47
?
10
Pa
(
gauge
p
2
?
1
.
52
?
10
Pa
(
gauge
u
1
?
u
2
H
f
,
1
_
2
5
4
u
1
2
2
g
?
p
1
?
g
?<
/p>
H
?
Z
2
?
u
2
2
2
g
?
p
2
?
g
?
H
f
,
1
_
2
pressure
）
pressure
）
?
0
total heads of
pump is
H
?
0
.
4
?
< br>1
.
52
?
10
5
?
0
.
247
?
10
5
1000
?
9
< br>.
81
?
18
< br>.
41
m
efficiency of pump is
?
?
N
e
/
N
since
N
e
?
QH
?
g
3600
?
26
?
18
.
41
?
10
00
?
9
.
8
1
3600
?
1
.
3
kW
N=2.45kW
Then mechanical
efficiency
?
?
1
.
3
2
.
45
?
100
%
?
53
.
1
%
The performance of
pump is
Flow rate
，
m?
/h
Total heads
，
m
Shaft power
，
kW
Efficiency
，
%
2.2 Water is transported by
a
pump from reactor, which has
200
mm
Hg
vacuum,
to
the
tank,
in
which
the
gauge
pressure
is
0.5
kgf/cm
2
,
as
shown
in
Fig.
The
total
equivalent
length
of
pipe
is
200
m
including
all
local
frictional loss. The pipeline is
?
57
×
3.5
mm
,
the
orifice
coefficient
of
C
o
and
orifice
diameter
d
o
are
0.62
and
25
mm,
respectively.
Frictional
26
18.41
2.45
53.1 <
/p>
2
10m
1
co
efficient
?
is 0.025.
Calculate: Developed head H
of pump, in m (the reading
R
of
U
pressure
gauge in orifice meter is 168 mm Hg)
Solution:
Equation(1.6-9)
V
?
0
?
C
0
?
d
?
1
?
?
0
?
?
< br>D
?
0
.
62
0
.
9375
< br>4
2
Rg
（
?
f
?
?
）
?
?
0
.<
/p>
62
?
25
?<
/p>
1
?
?
?
50
?
?
4
2
?
0
.
168
?
9
.
81
(
13600
?
1000
)
1000
?
6
.
44
?
4
.
12
m<
/p>
/
s
Mass flow rate <
/p>
m
?
V
o
S
o
?
?
4
.
12
?
3
.
14
4
?
0
.
025
2
?
1000
?
2
.
02
kg
/
s
2) Fluid
flow through the pipe from the reactor to tank,
the Bernoulli equation is as follows
for
V
1
=
V
2
H<
/p>
?
p
2
?
p
1
?
?
z
?
H
f
?
g
?
z=10m
?
p
?
0
.
5
?
9
.
81
?
10
4
?
200
760
?
1
.
013
?
10
5
?
75707
Pa
?
p/
?
g=
7.7m
The relation between the hole
velocity and veloc
ity of pipe
?
d
?
?
1
?
V
?
V
0
?
0
?
?
4
< br>.
12
?
?
?
?
1
m
/
s
?
2<
/p>
?
?
D
?
Friction loss
2
2
l
u
200
1
?
0
.
025
?
H
f
?
4
f
so
d
2
g
2
2
0
.
05
2
?
9
.
81<
/p>
?
5
.
1
m
H=7.7+10+5.1=22.8m
2.3 . A
centrifugal pump is to be used to extract water
from a condenser in which the vacuum is
640
mm
of
mercury,
as shown
in
figure.
At
the
rated
discharge,
the
net
positive
suction
head
must
be
at
least
3m
above
the
cavitation
vapor
pressure
of
710mm
mercury
vacuum.
If
losses
in
the
suction
pipe
accounted for a head of 1.5m. What must
be the least height of the liquid
level
in the condenser above the pump inlet?
Solution
:
From an
energy balance,
p
?
p
v
H
g
?
o
?
H
f
?
NPSH
?<
/p>
g
H
g
Where
P
o
=76
0-640=120mmHg
P
v
=760-710=50mmHg
Use of the equation will give the
minimum height H
g
as
p
o
?
p
v
?
H
f
?
NPSH
H
g
?
?
g
?
（
0
.
12
?
0
.
05
)
?
13600
?
9
.
81
< br>?
1
.
5
?
3
?
?
3
.
55
m
10
00
?
9
.
8
1
2.4 Sulphuric
acid is pumped at 3 kg/s through a 60m length of
smooth 25 mm pipe. Calculate the
drop
in pressure. If the pressure drop falls by one
half, what will the new flowrate be ?
?
Density of acid
1840kg/m
3
-3
?
V
iscosity of acid
25
×
10
Pas
Solution:
V
elocity of acid in the
pipe:
m
u
?
< br>volumetric
flowrate
cross
?
sec
tional
area
of
pipe
?
?
?
4
d
2
?
m
0
< br>.
785
?
d
< br>2
?
3
0
.
785
?
1840
?
0
.
025
2
?
3
.
32
m
/
s
Reynolds number:
Re
?
d
?
u
?
0
.
025
?
1840
?
3
.
< br>32
25
?
10
?
3
?
?
6109
from Fig.1.22 for a
smooth pipe when Re=6109, f=0.0085
pressure drop is calculated from
equation 1.4-9
h
f
?
?
p
?
4<
/p>
f
l
u
2
2
?
d
2
?
4
?
0
.
0085
60
0
.
025
3
.
32
2
?
450
J
/
kg
?
p
?
450
?
1840
?
827
.
5
kPa
or friction factor is calculated from
equation1.4-25
h
f
?
?
p
?
4<
/p>
f
l
u
2
?
d
2
?
4
?
0
.
046
Re
?
0
.
2
l
u
< br>2
d
2
＝
4
?
0
.
0
46
?
6109
?
0
.
2
60
0
.
025
3
.
32
2
2
?
426
J
/
kg
?
p
?
4
26
?
1840
?
783
.
84
kPa
if the pressure drop falls to
783.84/2=391.92kPa
?
?
?
?
0
.
2
?
p
?
?
?
391920
?
< br>4
?
0
.
046
Re
?
＝
4
?
0
.
046
?
?
?
?
?
2
d
2<
/p>
?
?
?
?
p
l
u
?
1840
?
?
4
?
0
.
046
?
1840
?
?
?
3
?
25
?
10
?
?
0
.
2
2
?
0
.
2
?
l
d
1
.
2
u
1
.<
/p>
8
2
60
0
.
025
1
.
2
u
1079<
/p>
.
89
1
.
8
?
`
u
2
0
.
012
1
.
8
so
u
?
1
.
8
391920
?
0
.
012
1079
.
.
89
?
1
.
8
4
.
36<
/p>
?
2
.
27
m
/
s
new mass flowrate=0.785d
2
u
ρ
=0.785×
0.025
2
×
2.27×
1840=2.05kg/s
2.4 Sulphuric acid is pumped at 3 kg/s
through a 60m length of smooth 25 mm pipe.
Calculate the
drop in pressure. If the
pressure drop falls by one half on assumption that
the change of friction
factor is
negligible, what will the new flowrate be ?
Density of acid
1840kg/m
3
-3
V
iscosity of acid
25
×
10
Pa
Friction factor
f
?
0
.
0056
< br>?
Solution:
Write energy
balance equation:
p
1
< br>?
z
1
?
u
1
2
2
0
.
500
Re
0
.
32
for hydraulically smooth pipe
?
g
2
g
?
p
?
H
?<
/p>
p
2
?
g
2
?
z
2
?
u
2
2
g
?
h
f
H
?
h
f
?
?
g
?<
/p>
?
l
u
d
2
g
?
4
d
u
?
?
3
2
u
?
3
?
4
?
d
?
2<
/p>
?
12
3
.
14
?
0
.
025
2
?
1840
?
3
.
32<
/p>
m
/
s
Re
?
3
.
32
?
0
.
025
?
1840
25<
/p>
?
10
?
3
?
6115
f<
/p>
?
0
.
0056
?
0
.
500
Re
?
?
0<
/p>
.
32
?
0
.
0056
?
2<
/p>
0
.
5
6115
0
.
32
?<
/p>
0
.
0087
3
.
32
2<
/p>
H
?
h
f
?
?
p
l
u
?
g
d
2
g
?
4
?
0
.
0087
< br>60
0
.
025
2
?
9
.
81
?
46
.
92
Δ
p=46.92
×
1840
×
9.81
=847.0kpa
2.6 The fluid
is pumped
through the horizontal pipe from section
A
to
B with the
φ
38
?
2.5mm
diameter and length of 30 meters, shown
as figure. The orifice meter of 16.4mm diameter is
used
to
measure
the
flow
rate.
Orifice
coefficient
< br>C
o
＝
0.63.
the
permanent
loss
in
pressure
is
3.5
×
10
4
N/m
2
, the friction coefficient
λ
=0.024. find:
(1) What is the pressure drop along the
pipe AB?
(2)What is the
ratio of power obliterated in pipe AB to total
power supplied to the fluid when the
shaft work is 500W
,
60%efficiency? (The density of fluid is
870kg/m
3
)
solution
：
z
A
g
?
p
A
?
u
A
2
2
?
?<
/p>
w
?
z
A
g
?
2
p
A
?
?
u
A
2
2
?
?
h
f
p
A
?
p
B<
/p>
?
A
o
?
?
2
h
f
?
?
l
u
d
2
?
?
p
0
?
?
16
.
4
?
?
?
?
?
0
.
247
A
33
?
?
u
0
?
C
0
1
?
0
.
247
2
2
gR
?
?
?
< br>?
?
?
?
?
?
0
.
6
3
0
.
97
2
?
9
.
81<
/p>
?
0
.
6
?
13600
?
87
0
870
?
?
8
.
5
m
/<
/p>
s
∴
u
= (16.4/33)
2
×
8.5=2.1m/s
∴
p
A
?
p
B
?
?
?
h
f
?
0
< br>.
024
?
870
(2)
Ne
?
Wm
?
?
p
?
d
u
?
?
76855
?
0
.
785
?
0
.
033
2
2
30
0
.
033
2
.
1
2
2
?
3
.
5
< br>?
10
4
?
76855
N
/
m
2
?
4
?
2
.
1
?
138
W
so
the ratio of power
obliterated in friction losses in AB to total
power supplied to the fluid
138
500
?
0
< br>.
6
?
100
< br>％＝
46
％
3.2
A
spherical quartzose
particle
（颗粒）
with
a density of 2650 kg/m?
settles
（沉淀）
freely
in
the
20
℃
air,
try
to
calculate
the
maximum
diameter
obeying
Stocks
’
law
and
the
minimum
diameter obeying
Newton’
s law.
Solution:
The
gravity settling is followed
Stocks
’
law, so maximum
diameter of particle settled can be
calculated from Re that is set to 1
Re
?
d
c
u
t
?
?
1
, then
t
?
u
t
?
?
d
c
?
equation 3.2-16 for the terminal
velocity
2
?
d
c
?
?
d
c
(
?
S
?
?
)
g
18
?
solving for critical diameter
d
c
?
1
.
224
3
2
?
(
?
S
?
?
)
g
Check up
the appendix
-3
2
The density of
20
℃
air
ρ
=1.205 kg/m?
and
viscosity ?
=1.81×
10
N
·
s/m
d
c
?
1
.
224
?
5
.
73
?
10
?
57
.
3
?
m
3
(
1
.
81
?
10
?
3
)
2
(
2650
?
1
.
205
)
?
1
.
< br>205
?
5
m
< br>
when Reynolds number
≥
1000, the flow pattern
follows Newton
’
s law and
terminal velocity can
be calculated by
equation 3.2-19
u
t
?
1
.
75
gd
p
?
?
?
p
?
?
?
1
critical
Reynolds number is
Re
t
?
?
u
< br>t
?
d
c
?
1000
，
2
?
rearranging the equation 2
gives
u
t
?
< br>1000
?
?
?
d
c
3
combination of equation 1
with equation 3
1000
?
?
?
d
c
?
1
.
74
?
(
?
S
?
?
)
g
d
c
?
solving for critical diameter
?
?
32
.
3
3
d
c
(
?
S
?
?
)
?
?
2
?
?
32<
/p>
.
3
3
d
c
(
1
.
81
?
10
?
3
?
3
)
2
(
2650
?
1
.
205
)
?
1
.
205
m
?
1
.
< br>512
?
10
?
1512
um
3.3
It is desired to remove dust particles
50 microns in diameter from 226.5m
/min
of air, using
a settling chamber for
the purpose. The temperature and pressure are
21
o
C and 1 atm. The particle
density
is
2403kg/m
3
.
What
minimum
dimensions
of
the
chamber
are
consistent
with
these
conditions? (the
maximum permissible velocity of the air is 3m/s)
solution:
to calculate terminal velocity from the
equation 3.2-16
u
t
?
d
p
?
?
p
?
?
?
g
2
3
18
?
-5
2
The density of
21
℃
air
ρ
=1.205 kg/m?
and
viscosity ?
=1.81×
10
N
·
s/m
u
t
?
d
p
?
?
p
?<
/p>
?
?
g
2
18
?
＝
(
50
?
10
?
6
)
(
2403
?
1
.
205
)
?
9
.
81
?
5
2
18
?
1
.
81
?
10
?
0
.
181
m
/
s
Q
?
BLu
t
so
Q
u
t
226
.
5
6
0
?
0
.
18
1
2
BL
?
?
?
20
.
86
m
1
from
equation3.3-4
L
u
?
H
u
t
the maximum permissible velocity of the
air is 3m/s
L
3
?
H
0
.
181
L
?
16
.
58
H
2
set
B
to be
3
m
, then from equation 1
L
=7
m
And
H
=0.42m
-
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