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 =815kg/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
R d x D 224
4
π
π
=
(1)
so
R D d x 2
??
?
??= (2)
and hydrostatic equilibrium gives following relationship
g R g x p g R p A c c ρρρ++=+21 (3)
so
g R g x p p c A c )(21ρρρ-+=- (4)
substituting the equation (2) for x into equation (4) gives
g R g R D d p p c A c )(2
21ρρρ-+??
?
??=- (5)
（a ）when the change in the level in the reservoirs is neglected,
()Pa g R g R g R D d p p c A c A c 26381.98151000145.0)()(2
21=?-=-≈-+??
?
??=-ρρρρρ
（b ）when the change in the levels in the reservoirs is taken into account
()Pa g R g R D d g
R g R D d p p c A c c A c 8.28181.98151000145.081.9815145.0515.6)()(22
2
21=?-+?????
? ??=-+??
?
??=-+??
?
??=-ρρρρρρ
error=
％＝7.68
.281263
8.281-
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 .
Solution: There is a gaseous mixture in the U-tube manometer meter. The densities of fluids are denoted by Hg O H g ρρρ,,2, respectively. The pressure at point A is given by hydrostatic equilibrium
g R R g R g R p g Hg O H A )(32232+-+=ρρρ
g ρis small and negligible in comparison with Hg ρand ρH2O , equation above can be simplified
c A p p ≈=232gR gR Hg O H ρρ+
=1000×9.81×0.05+13600×9.81×0.05
=7161N/m2
1gR p p p Hg A D B ρ+=≈=7161+13600×9.81×0.4=60527N/m
Figure for problem 1.4
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:
2
22
2
222111
u gz p u gz p ++=++ρρ Where p 1=0, p 2=0, and u 1=0, simplification of the equation 1
The relationship between the velocity at outlet and velocity u o at throat can be derived by the continuity equation:
2
2??
? ??=???? ??D d u u o 2
2??
?
??=d D u u o 2
Bernoulli equation is written between the throat and the station 2-2 3
Combining equation 1,2,and 3 gives
22
2
u Hg =222200u u p =+ρ
Solving for H
H=1.39m
1.6 A liquid with a constant density ρ kg/m3 is flowing at an unknown velocity V1 m/s through a horizontal pipe of cross-sectional area A1 m2 at a pressure p1 N/m2, and then it passes to a section of the pipe in which the area is reduced gradually to A2 m2 and the pressure is p
2. Assuming no friction losses, calculate the velocities V1 and V2 if the pressure difference (p1 - p2) is measured.
Solution:
In Fig1.6, the flow diagram is shown with pressure taps to measure p1and p2. From the mass-balance continuity equation , for constant ρ where ρ1 = ρ2 = ρ, 2
1
1
2A
A
V
V=
For the items in the Bernoulli equation , for a horizontal pipe,
z1=z2=0
Then Bernoulli equation becomes, after substituting
2
1
1
2A
A
V
V=for V2,
ρ
ρ
2
2
1
2
1
2
1
1
2
1
2
2
p
A
A
V
p
V
+
+
=
+
+
()＝
＝
＝
1
44
.2
81
.9
2
1000
81
.9
1000
2
1
25
.1
1
1
1
24
4
2
-
?
?
?
-
-
?
?
?
?
?
=
=
ρ
ρg
h
d
D
u
Hg
Rearranging,
2
)
1(21
2
121
21-=
-A A V p p ρ ??
?
?????-????
??-122
2
12
11A A
p p V ρ
＝
Performing the same derivation but in terms of V 2,
???
????????
?
??--2
1
2
2
1212A A p p V ρ
＝
1.7 A liquid whose coefficient of viscosity is μ 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
L p ? is 232d
V μ. 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
From velocity profile equation for laminar flow 2 substituting equation 2 for u into equation 1 and integrating 3
rearranging equation 3 gives
?
?==R R rdr u R udA A V 020211ππ???
? ????? ??--=2
2014R r R L p p u L μ2032D L p p V L μ-=
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
expressed as g
V k 22
where V is the velocity at A, find the
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
s m d Q
V V d Q A
A A
A /132.115.0785.002
.04
4
2
22=?==
=
ππ s m d Q
V V d Q B
B B
B /529.4075.0785.002
.04
4
2
22=?==
=
ππ When the fluid flows down, writing mechanical balance equation
2
222
22A B B B A A A
V k V g z p V g z p +++=++ρρ 2
13.1253.4100014715213.181.95.2222k ++=+?
Figure for problem 1.8
232d V L p μ=?Pa d VL p 115201.01206.005.0323222=???==?μ
k 638.0260.10715.14638.0525.24++=+
=k 0.295
making the static equilibrium
g
R g x g l p g R g x p Hg A B ρρρρρ+?++=+?+()()
mm g
g
l p p R g
H A B 7981
.91260081
.910005.214715-=???-=
---=
ρρ
ρ
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
g
p ab ρ?, h fab , g p cd
ρ? and h fcd , respectively.
（2）the relationship between readings R 1and R 2 in the U tube.
Solution:
(1) From Fanning equation
and
so
Fluid flows from station a to station b , mechanical energy conservation gives
hence
2
from station c to station d
Figure for problem 1.9
22
V d l h fab λ=2
2
V d l h fcd λ
=fcd
fab h h =fab b
a h p p +=+ρρlg fa
b b a h p p =+-lg ρfcd d
c h p
p +=ρρ
hence
3
From static equation
p a -p b =R 1（ρˊ-ρ）g -l ρg 4
p c -p d =R 2（ρˊ-ρ）g 5 Substituting equation 4 in equation 2 ，then
therefore
6
Substituting equation 5 in equation 3 ，then
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 2O column?
2) Find the maximum velocity and point r at which
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 kg/m 3，respectively ；and the viscosity
and density of kerosene are 0.003 Pas and 800
kg/m 3，respectively ）
solution: 1）1600001
.01000
004.04.0Re =??=
=
μ
ρ
ud
fcd d c h p p =-ρfab h g l g R =+--'lg 1ρ
ρρρ）（g
R h fab ρρ
ρ-'=1g R
h fcd ρρ
ρ-'=2
Figure for problem 1.10
from Hagen-Poiseuille equation
1600004.0001
.024.032322
2=???==
?d uL P μ m g p h 163.081
.910001600=?=?=
ρ 2）maximum velocity occurs at the center of pipe, from equation 1.4-19 max
0.5V
u =
so u max =0.4×2=0.8m
3）when u=V=0.4m/s Eq. 1.4-17
2
max
1???
? ??-=w
r
r u u 5.0004.01max
2
=???
??-u V r ＝
m r 00284.071.0004.05.0004.0=?==
4) kerosene:
427003
.0800
004.04.0Re =??=
=
μ
ρ
ud
Pa p
p 4800001
.0003.01600=='?='?μμ m g p h 611.081
.98004800=?=''?=
'ρ
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 m3/h)
b)When the gate valve is widely open, calculate the static pressure at the tap (in gauge pressure, N/m2). l e/d≈15 when the gate valve is widely open, and the friction coefficient λ is still 0.025.
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
∑
+
+
+
=
+
+
2
1,
2
2
2
2
1
2
1
12
2—
f
h
p
u
gZ
p
u
gZ
ρ
ρ
（a）
In the equation 0
1
=
p(the gauge pressure)
2
2
2
/
39630
4.1
81
.9
1000
4.0
81
.9
13600m
N
gh
gR
p
O
H
Hg
=
?
?
-
?
?
=
-
=ρ
ρ
2
1
=
≈
Z
u
When the gate valve is fully closed, the height of water level in the reservoir can be related to h (the distance between the center of pipe and the meniscus of left arm of U tube).
gR
h
Z
g
Hg
O
H
ρ
ρ=
+)
(
1
2
（b）
where h=1.5m
R=0.6m
Substitute the known variables into equation b
2
2
2
2
_
1,
1
13
.2
2
)5.0
1.0
15
025
.0(
2
)
(
66
.6
5.1
1000
6.0
13600
V
V
V
K
d
l
h
m
Z
c
f
=
+
?
=
+
=
=
-
?
=
∑λ
Substitute the known variables equation a
9.81×6.66=2
2
13
.2
1000
39630
2
V
V
+
+
the velocity is V =3.13m/s
Figure for problem 1.12
the flow rate of water is
h m V d V h /5.8813.312.04
36004
360032=???
=?
=π
π
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
∑+++=++31,32331
21122—f h p V gZ p V gZ ρ
ρ （c ）
since m Z 66.61=
3
11300
p p u Z =≈= 2
2
2
3_1,81.4 2
]
5.0)151.035(025.0[ 2
)(V V V K d l l h c e f =++=++=∑λ input the above data into equation c ，
9.8122
V 81.42
66.6+=?V 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
∑+++=++21,2
222121122—f h p V gZ p V gZ ρ
ρ
（d ）
since m Z 66.61=
21210
03.51/0(page pressure Z u u m s
p =≈≈=）
kg J V K d l h
c f /2.262
51.3)5.01.015025.0(2)(222
_1,=+?=+=∑λ
input the above data into equation d ，
9.81×6.66=2.261000
251.322
++p
the pressure is: 329702=p
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 subscript 1, and branch pipe by subscript 2.
The friction loss for parallel pipelines is 2
12
1
S S s f f V V V h h
+==∑∑
The energy loss in the branch pipe is
2
222222
2
u d l l h
e f ∑∑+=λ In the equation 03.02=λ
s
m u d m
l l e /343.0053.04
360072
.2053
.0102
2222=??
=
==+∑π
input the data into equation c kg J h
f /333.02
343.0053.01003.02
2
=??=∑
The energy loss in the main pipe is
333.02
211112
1
===∑∑u d l h h
f f λ
So s m u /36.22
018.02
3.0333.01=???=
The water discharge of main pipe is
h m V h /60136.23.04
3600321=???
=π
Total water discharge is
h m V h /7.60372.26013≈+=
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
R 5.2 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 Venturi meter
f o o h
g z V p g z V p +++=++2212
11
2
2ρρ 1 rearranging the equation above, and set (z 2-z 1)=x
f o o
h xg V V p p ++-=-2
2
121ρ
2
from continuity equation
112
2
1
125.625V V d
d V V o o =??? ??=???
? ??= 3 substituting equation 3 for V o into equation 2 gives
()
f
f
f f o
h xg R h xg R
h V h xg V V p p ++=++=+=++-=-94.1185.203.1903.192
06.392
2121211ρ
4
from the hydrostatic equilibrium for manometer
g x g R p p Hg o ρρρ+-=-)(1 5
substituting equation 5 for pressure difference into equation 4 obtains
f H
g
h xg R g
x g R ++=+-94.118)(ρ
ρρρ 6
rearranging equation 6
kg J R R R R g
R h Hg f /288.267.494.11861.12394.118)(==-=--=
ρ
ρρ
Figure for problem 1.16
1.17.Sulphuric 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)
2.050
10
10==D D =?-=-=-81.9)130013600(1.0)(21g R p p Hg ρρ12066.3pa
s kg V D m /268.0130063.201.04
4
2220=???=
=
π
ρπ
b) approximate pressure drop
=?-=-=-81.9)130013600(1.0)(21g R p p Hg ρρ12066.3Pa
pressure difference due to increase of velocity in passing through the orifice
Pa D D V V V V p p o
8.44882
)2.01(63.213002
24
24
122222
12221=-=????
??-=-=-ρ
ρ
pressure drop caused by friction loss
Pa p f 5.75778.44883.12066=-=?
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 3/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.
()s m p p D D C V o /63.231.461.056.1861.0130081.9)130013600(1.022.0161.021*******=?=≈?-?-=-???? ??-=ρ
Solution:
Write the mechanical energy balance equation between the suction connection and discharge connection
2_1,22
22121122f H g
p g u Z H g p g u Z +++=+++ρρ
where
m Z Z 4.012=-
(Pa 1052.1(Pa 1047.22_1,215241≈=?=?-=f H u u pressure gauge p pressure gauge p ））
total heads of pump is m H 41.1881
.9100010247.01052.14.05
5=??+?+
= efficiency of pump is N N e /=η since kW g QH N e 3.13600
81
.9100041.18263600=???==
ρ N=2.45kW
Then mechanical efficiency
%1.53%10045
.23
.1=?=
η The performance of pump is
Flow rate ，m3/h 26 Total heads ，m 18.41 Shaft power ，kW 2.45 Efficiency ，%
53.1
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 φ57×
3.5 mm , the orifice coefficient of C o and orifice diameter d o are 0.62 and 25 mm, respectively. Frictional
coefficient λ 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)
Mass flow rate
s kg S V m o o /02.21000025.04
14
.312.42=???
==ρ 2) Fluid flow through the pipe from the reactor to tank, the Bernoulli equation is as follows for V 1=V 2
f H z g
p p H +?+-=
ρ1
2 ?z=10m
Pa p 7570710013.1760
200
1081.95.054=??+
??=? ?p/ρg=7.7m
The relation between the hole velocity and velocity of pipe
Friction loss
so
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,
s m Rg D d C V f /12.444.69375.062
.01000)100013600(81.9168.025025162.02144000
=?=
-?????
??-=-??
? ??-=ρρρ）（s m D d V V /12112.42
200=???
???=??? ??=m g u d l f
H f 1.581
.92105.0200025.0242
2=??==NPSH H g
p p H f v
o g ---=
ρ
Where
P o =760-640=120mmHg P v =760-710=50mmHg
Use of the equation will give the minimum height H g as
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 ? Viscosity of acid 25×10-3 Pas
Solution:
Velocity of acid in the pipe:
s
m d m d m
pipe of area tional cross flowrate volumetric u /32.3025.01840785.03785.04
sec 2
22=??===-=
ρπρ
Reynolds number:
610910
2532
.31840025.0Re 3
=???=
=
-μ
ρu
d from Fig.1.22 for a smooth pip
e when Re=6109, f=0.0085 pressure drop is calculated from equation 1.4-9
kg J u d l f p
h f /4502
32.3025.0600085.04242
2=?==?=ρ
kPa p 5.8271840450=?=?
or friction factor is calculated from equation1.4-25
kg
J u d l u d l f p
h f /4262
32.3025.0606109046.042Re 046.042422
.02
2.02=???==?=--＝ρkPa p 84.7831840426=?=?
if the pressure drop falls to 783.84/2=391.92kPa
m
NPSH H g
p p H f v
o g
55.335.181.9100081.913600)05.012.0-=--???-=---=（ρ
8
.18
.12.12
.038
.12
.12
.022
.0012.089.1079`2
025.060102518401840046.042046.042Re 046.043919202u u u d
l u d l p p =??
? ?????=???? ?????==?='?----ρμρρ＝
so
s m u /27.236.489
..1079012.03919208
.18
.1==?=
new mass flowrate=0.785d 2u ρ=0.785×0.0252×2.27×1840=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 Viscosity of acid 25×10-3 Pa Friction factor 32
.0Re
500
.00056.0+=f for hydraulically smooth pipe Solution:
Write energy balance equation:
f h g
u z g p H g u z g p +++=+++222
2
222111ρρ g
u d l g p h H f 22
λρ=?==
34
2=ρπ
u d
s m d u /32.31840
025.014.3124322=??=?=
ρπ 611510251840
025.032.3Re 3
=???=
-
0087.061155
.00056.0Re 500.00056.032
.032.0=+=+=f
92.4681
.9232.3025.0600087.0422
2=??==?==g u d l g p h H f λρ
Δp=46.92×1840×9.81=847.0kpa
2.6 The fluid is pumped through the horizontal pipe from section A to B with the φ38?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 C o ＝0.6
3. the permanent loss in pressure is
3.5×104
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 ：
∑+++=+++f A
A A A A
A h u p g z w u p g z 2
222ρρ
ρ
λρ
022p u d l h p p f B
A ?+==-∑
247.0334.162
=??
?
??=A A o ()()s m gR C u /5.8870
870136006.081.9297.063.02247.012
00=-??=''--=
ρρρ
∴u = (16.4/33)2×8.5=2.1m/s
∴242
/76855105.32
1.2033.030870
024.0m N h p p f B A =?+?==-∑ρ (2)
W u d p Wm 1381.2033.0785.0768554
Ne 22
=???=?=
=ρπρ so
the ratio of power obliterated in friction losses in AB to total power supplied to the fluid
％％＝461006
.0500138
??