1. 49
Fluid Mechanics Laboratory
Observation Note Book
By
Mr.B.Ramesh, M.E.,(Ph.D),
Associate professor,
Department of Mechanical Engineering,
St. Joseph’s College of Engineering,
Jeppiaar Trust, Chennai-119
Ph.D. Research Scholar, College of
Engineering Guindy Campus, Anna
University, Chennai.
2. 50
Observation:
Constant speed of the pump , N = 1440 rpm
Area of the collecting tank , A = 0.3 x 0.3 ,m2
Energy meter constant , EMC = 1200 , rev / kwhr
Distance between the centres of
vacuum gauge and pressure gauge , X= 0.2 ,m
Lubricating oil used = SAE 40
Sl.No.
Pressure
gauge,G
Vacuum
gauge,V
TotalHead
H
Timefor
h=10cm
rise,t
Actual
discharge,
Qact
Timefor5
revolutions
Tn.
Input
Power,Pi
Output
Power,Po
ηpump
Units↓
Kgf/
cm2
mm
Hg
m of
oil
sec
m3
/s x
10-4 sec w w %
1 0.4
2 0.8
3 1.2
4 1.6
5 2.0
6 2.4
3. 51
Exp. No. :
Characteristic tests on gear oil
pump at constant speed
Date :
Aim:
To study the characteristics of the gear pump at constant speed.
Apparatus required:
i) Gear oil pump set up and ii) Stop watch.
Description:
The gear pump is a positive displacement type of pump and consists of a pair of
helical or spur gears, housed closely in a casing. The pressure gauge is fitted to the
delivery side and a vacuum gauge to the suction side. The energy input to the pump can
be measured through an energy meter. There is a collecting tank with a level indicator
and a gate valve at the drain.
Procedure:
i) Keeping the gate valve in the delivery side fully open the experiment is
started.
ii) The pressure gauge reading ,vacuum gauge reading ,the time taken for 5
revolutions of the energy meter disc , time taken for 10 cm rise of oil level
are noted.
iii) By closing the delivery valve gradually, the flow rate is varied.
iv) For each valve setting the above readings are noted and tabulated.
Formulae:
a) Total head, H = [ G x 11.33 ] + [ V x 0.0179 ] + X , m of oil
where,
G = Pressure head ,kg/cm2
V = Vacuum head ,mmHg
X = distance between pressure gauge and vacuum
gauge ,m
b) Actual discharge, Qact = [ Ah ] / t ,m3
/s
where,
A = Area of collecting tank ,m2
h = Rise of oil in the collecting tank = 0.1 m
t = Time for 10 cm rise of oil in collecting tank , sec
c) Input power, pi = [ 3600 x n x ηmotor x 1000 ] / [ Tn x EMC ] ,w
5. 53
where,
n = Number of revolutions of energy meter disc = 5
ηmotor = 0.75
Tn = Time for 5 revolutions of energy meter disc , sec
EMC = Energy meter constant , rev / kwhr
d) Output power , Po = ρgQact H , w
where,
ρ = Density of the oil = 882 ,Kg / m3
g = Acceleration due to gravity = 9.81 , m/s2
e) Efficiency of the gear oil pump , ηpump = [ Po / Pi ] x 100 ,%
Graphs:
The following graphs are drawn taking head(H) on X axis:
i) Head vs Actual discharge
ii) Head vs Efficiency of pump
iii) Head vs Output power
7. 55
Result:
The characteristic test was conducted on the gear oil pump and the following
graphs were drawn:
i) H vs Qact ii) H vs ηpump and iii) H vs Po
i) Maximum efficiency of gear oil pump, ηpump = ,%
ii) Actual discharge , Qact = ,m3
/s
iii) Output power from the pump , Po = ,w
iv) Total head, H = ,m of oil
8. 56
Observation:
Diameter at inlet of venturimeter ,d1 = 25.4 ,mm
Diameter at throat of venturimeter ,d2 = 12.5 ,mm
Sl.No
Manometer
readings
Time
taken
for 20
litre ,t
Total
head ,H √H
Qact Qtheo
Cd
h1 h2
units
→
cm cm sec
m of
water
m of
water
m3
/sec
x 10-4
m3
/sec
x 10-4
1
2
3
4
5
6
7
Mean cd =
9. 57
Exp. No. :
Venturimeter
Date :
Aim:
To find the co-efficient of discharge of the given venturimeter.
Apparatus required:
i) Venturimeter pipe set up ii) Mercury manometer and
iii) Stop watch.
Description:
i) The arrangement is of closed type.
ii) Water is circulated through the venturimeter from reservoir to collecting
tank by means of a monoblock pump.
iii) The collecting tank of the venturimeter is connected to a mercury
manometer.
Procedure:
i) The pump is primed and started.
ii) Keeping the gate valve fully open the experiment is started.
iii) The manometer readings and the time taken for 20 litre of water are noted.
iv) The gate valve is gradually closed; for each valve setting the readings are
noted and the values are tabulated.
Formulae:
a) Co-efficient of discharge ,Cd = Qact /Qtheo
b) Actual discharge , Qact = Volume of water collected / time
taken for collection of 20 litres of
water , m3
/sec.
c) Theoretical discharge ,Qtheo = [ a1a2√2gH ] / [ √ a1
2
- a2
2
] , m3
/sec.
where,
a1 = cross sectional area of inlet = π d1
2
/4 ,m2
a2 = cross sectional area of throat = π d2
2
/4 ,m2
d1 = diameter at inlet of the venturimeter ,m
d2 = diameter at throat of the venturimeter ,m
g = acceleration due to gravity = 9.81 ,m/sec2
.
d) Total head ,H = [ (h1- h2) /100 ] × [ ( SH / SL ) – 1 ] ,m of water
= [ (h1- h2) /100 ] × [ ( 13.6 / 1 ) – 1 ] ,m of water
where,
h1- h2 = difference of mercury level in the manometer.
SH = specific gravity of mercury = 13.6
SL = specific gravity of water = 1
13. 61
Graphs:
The following graph is drawn:
i) Qact vs Qtheo
Result :
The co-efficient of discharge( cd ) of the given venturimeter is :
i) Experimentally =
ii) Graphically =
14. 62
Observation:
Galvanized iron pipe (GI) : d = 12.5 mm
Sl.No.
Manometer
readings
Timetakenfor
20litre,t
Lossofhead
,Hf
Actual
discharge,
Qact
Velocityof
water,v
v2
i
Co-efficientof
friction,
f1
Co-efficientof
friction,
f2
h1 h2
units↓
cm cm sec m of water
m3
/ s
x 10-4 m/s
1
2
3
Mean f1= f2=
Copper (Cu) : d = 12.5 mm
Sl.No.
Manometer
readings
Timetakenfor
20litre,t
Lossofhead
,Hf
Actual
discharge,
Qact
Velocityof
water,v
v2
i
Co-efficientof
friction,
f1
Co-efficientof
friction,
f2
h1 h2
units↓
cm cm sec m of water
m3
/ s
x 10-4 m/s
1
2
3
Mean f1= f2=
15. 63
Exp. No. :
Losses due to pipe friction
Date :
Aim:
To determine the co-efficient of friction for flow of water through the given pipes.
Apparatus required:
i) Pipe line set up and ii) Stop watch.
Description:
The given arrangement is closed type fitted with a reservoir and a collecting tank.
A monoblock pump which is fitted on the reservoir can pass water through any one of the
four pipes of different materials ( Galvanized iron, Aluminium, Copper and Stainless
steel). Two tapping at a distance of 60 cm are connected to a water manometer.
Procedure:
i) The pump is primed & started.
ii) The discharge valve of the required pipe is fully opened and values in the
manometer are noted.
iii) Time taken for 20 litre of water collection is also noted.
iv) By closing the discharge valve gradually the flow rate is varied.
v) For each position of the discharge valve, the above readings are noted.
Formulae:
a) Darcy-weisbach’s formula:
co-efficient of friction , f1 = [ Hf g d ] / [ 2Lv2
]
where,
Loss of head ,Hf = [ (h1- h2) /100 ] × [ ( SH / SL ) – 1 ] ,m of water
= [ (h1- h2) /100 ] × [ ( 13.6 / 1 ) – 1 ] ,m of water
where,
h1- h2 = difference of mercury level in the manometer.
SH = specific gravity of mercury = 13.6
SL = specific gravity of water = 1
d = diameter of pipe line ,m
L = length of the pipe = 0.6 ,m
g = acceleration due to gravity = 9.81 ,m/s2
velocity of flow ,v = Actual discharge / area of cross section of the pipe ,m/s
= Qact / a
Actual discharge , Qact = Volume of water collected / time
taken for collection of 20 litres of
water , m3
/sec.
19. 67
b) Chezy’s formula:
co-efficient of friction , f2 = [ miρg ] / [ 4v2
]
where,
Hydraulic radius , m = area of flow / wetted perimeter
= [ πd2
/ 4 ] / [ πd ] = d / 4 ,m
Loss of head per unit length of pipe ,i = Hf / L
ρ = density of water = 1000 ,kg/m3
Graphs:
The following graphs (for both GI & Cu) are drawn taking (velocity of flow)2
on
X axis:
i) Velocity of flow2
vs Loss of head and
ii) Velocity of flow2
vs Loss of head per unit length of pipe.
Result :
The test was conducted on the given pipe lines and the following graphs were
drawn:
i) v2
vs Hf and ii) v2
vs i
Pipe Experimentally Graphically
GI
f1
f2
Cu
f1
f2
20. 68
Observation:
Constant speed of the pump , N = 1200 rpm
Area of the collecting tank , A = 0.7 x 0.7 ,m2
Energy meter constant , EMC = 750 , rev / kwhr
Distance between the centres of
vacuum gauge and pressure gauge , X= 0.31 ,m
Sl.No.
Pressure
gauge,G
Vacuum
gauge,V
Total
Head,H
Timefor
h=10cm
rise,t
Actual
discharge,
Qact
Timefor
10
revolutions
Tn.
Input
Power,Pi
Output
Power,Po
ηpump
Units↓
Kgf/
cm2
mm
Hg
m of
water
sec
m3
/s x
10-3 sec w w %
1 0.2
2 0.4
3 0.6
4 0.8
5 1.0
6 1.2
7 1.4
21. 69
Exp. No. :
Characteristic tests on centrifugal
pump at constant speed
Date :
Aim: To study the characteristics of the centrifugal pump at constant speed.
Apparatus required:
i) Centrifugal pump setup & ii) Stop watch
Description:
i) The pump is run by a single phase motor.
ii) The pressure gauge is fitted to the delivery side and a vacuum gauge to the
suction side.
iii) The energy input to the pump can be measured through an energy meter.
iv) There is a collecting tank with a level indicator.
Procedure:
i) Prime the pump with water.
ii) Close the gate valve.
iii) Start the motor.
iv) Note:
a) The pressure gauge reading , G.
b) The vacuum gauge reading, V
c) Time for 10 revolutions in the energy meter by means of stopwatch.
d) Time for 10cm. rise in the collecting tank by means of stopwatch
e) Difference of level between the pressure and vacuum gauge.
v) Take atleast 6 sets of readings by varying the head from minimum when
the gate valve is fully open to maximum at shut off. This can be done by
throttling the delivery valve.
Formulae:
a) Efficiency of the pump , ηpump = [Output power / Input power ] x 100
= [ Po / Pi ] x 100
b) Output power , Po = ρgQactH ,w
where,
Density of water , ρ = 1000 ,kg / m3
Acceleration due to gravity ,g = 9.81 ,m / s2
Actual discharge , Qact = Ah / t ,m3
/ s
Where,
A = area of collecting tank ,m2
h = rise of water level in collecting tank = 0.1 ,m
t = time taken for 10 cm rise in collecting tank , sec
23. 71
c) Total head , H = [ G x 10 ] + [ V x 0.0136 ] + X ,m of water
where,
G = pressure head ,kg/cm2
V = vacuum head ,mmHg
X = distance between pressure gauge and vacuum
gauge ,m
d) Input power, pi = [ 3600 x n x ηmotor x 1000 ] / [ Tn x EMC ] ,w
where,
n = Number of revolutions of energy meter disc = 10
ηmotor = 0.75
Tn = Time for 10 revolutions of energy meter disc,sec
EMC = Energy meter constant , rev / kwhr
Graphs:
The following graphs are drawn taking head(H) on X axis:
i) Head vs Actual discharge
ii) Head vs Efficiency of pump
iii) Head vs Output power
25. 73
Result:
The characteristic test was conducted on the centrifugal pump and the following
graphs were drawn:
i) H vs Qact ii) H vs ηpump and iii) H vs Po
i) Maximum efficiency of centrifugal pump, ηpump = ,%
ii) Actual discharge , Qact = ,m3
/s
iii) Output power from the pump , Po = ,w
iv) Total head, H = ,m of water
26. 74
Observation:
Constant speed of the pump , N = 1440 rpm
Area of the collecting tank , A = 0.495 x 0.495 ,m2
Energy meter constant , EMC = 1200 , rev / kwhr
Distance between the centres of
vacuum gauge and pressure gauge , X= 0.35 ,m
Sl.No.
Pressure
gauge,G
Vacuum
gauge,V
TotalHead
H
Timefor
h=10cm
rise,t
Actual
discharge,
Qact
Timefor10
revolutions
Tn.
Input
Power,Pi
Output
Power,Po
ηpump
cd % slip
Units↓
Kgf/
cm2
mm
Hg
m of
water
sec
m3
/s x
10-4 sec w w % %
1 0.4
2 0.8
3 1.2
4 1.6
5 2.0
6 2.4
27. 75
Exp. No. :
Characteristic tests on
reciprocating pump at constant speed
Date :
Aim:
To study the characteristics of the reciprocating pump at a constant speed.
Apparatus required:
i) Reciprocating pump set up & ii) Stop watch.
Description:
The reciprocating pump is a displacement type of pump and consists of a piston or
a plunger working inside a cylinder. The cylinder has got two valves, one allowing water
into the cylinder from the suction pipe and the other allowing water from the cylinder
into the delivery pipe.
During the suction stroke, a petrol vacuum is created inside the cylinder,
the suction valve opens and water enters into the cylinder. During the return stroke the
suction valve closes and the water inside the cylinder is displaced into the delivery pipe
through the delivery valve. In case of double acting pump two sets of delivery and
suction valves are provided. So for each stroke one set of valves are operated and there is
a continuous flow of water.
Specification of the pump :
Type of the pump : Double acting cylinder
Piston stroke : 4.5 cm
Piston diameter : 5 cm
Suction pipe : 1”
Delivery pipe : 3/4 ”
An energy meter is provided for determination of input to the motor. The
pump is belt driven by a A.C motor .The pump can be run at three different speeds by the
use of V- belt and differential pulley system. The belt can be put in different grooves of
pulleys for different speeds. A set of pressure gauge are provided and the required pipe
lines are also provided.
Procedure:
i) Select the required speed.
ii) Open the gate valve in the delivery pipe fully.
iii) Start the motor.
iv) Throttle the gate valve to get the required head.
v) Note the following :
a) Pressure gauge (G) and vacuum gauge (V) readings.
b) Time taken for : 10 cm rise of water in the collecting tank and
10 revolutions of the energy meter.
vi) Repeat the experiment for different heads. Take atleast 6 set of readings.
29. 77
Formulae:
a) Actual discharge of water , Qact = [ Ah ] / t ,m3
/ s
where,
A = Area of collecting tank ,m2
h = Rise of water in collecting tank = 0.1 ,m
t = Time taken for 0.1 m rise of water ,sec
b) Theoretical discharge of water , Qtheo = [ kalNp ] / 60 ,m3
/ s
where,
K = No. of strokes of the pump = 2
l = Stroke length = 0.045 ,m
d = Diameter of cylinder = 0.05 ,m
a = Cross sectional area of cylinder = [ πd2
] / 4 ,m2
Np= Pump speed = 300 ,rpm
c) Co-efficient of discharge ,cd = Qact / Qtheo
d) Slip = Qtheo - Qact
% Slip = [ Qtheo - Qact ] / [Qtheo ] x 100 ,%
e) Power input to the pump , Pi = [ 3600 x n x ηmotor x 1000 ] / [ EMC x Tn ]
where,
n = No. of revolutions of the energy meter disc = 10
ηmotor = 1
EMC = Energy meter constant ,rev / kwhr
Tn = Time taken for n revolutions of the energy meter disc ,sec
f) Power output from the pump , Po = ρgQactH ,w
where,
ρ = Density of water = 1000 kg/m3
g = Acceleration due to gravity = 9.81 ,m / s2
H = Total head , m of water
g) Total head , H = [ G x 10 ] + [ V x 0.0136 ] + X , m of water
where,
G = Pressure gauge reading , kgf / cm2
V = Vacuum gauge reading , mmHg
X = Distance between the centres of pressure gauge and
vacuum gauge , m
h) Efficiency of the pump , ηpump = [ Po / Pi ] x 100 ,%
31. 79
Graphs :
The following graphs are drawn taking Total head on X axis:
i) Total head vs Actual discharge
ii) Total head vs Efficiency of the pump
iii) Total head vs Power output
iv) Total head vs % Slip
Result :
The characteristic test was conducted on the reciprocating pump and the
following graphs were drawn:
i) H vs Qact ii) H vs ηpump iii) H vs Po and
iv) H vs % S
i) Maximum efficiency of reciprocating pump, ηpump = ,%
ii) Actual discharge , Qact = ,m3
/s
iii) Output power from the pump , Po = ,w
iv) Total head, H = ,m of water
v) % Slip, S = ,%
vi) Co-efficient of discharge ,cd =
32. 80
Observation:
No P1 P2 Gauge
Pr.
G
Total
Head
H
Dead
Wt.
T1
Sp.
Wt
To
Kg
T1 -T2 +T0
Kg
Speed
N
Q
m3
/s
Input
Power
watts
Output
Power
watts
η
%
1
2
3
4
5
6
7
33. 81
Exp. No. :
Pelton wheel
Date :
Aim :
To conduct a test on the pelton wheel (turbine) at constant head.
Apparatus required:
i) Pelton wheel set up & ii) Stop watch.
Description:
Pelton turbine is an impulse turbine, which is used to utilise high heads for
generation of electricity. All the available head is converted into velocity energy by
means of spear and nozzle arrangement. The water leaves the nozzle in jet formation. The
jet of water then strikes the buckets of the pelton wheel runner. The buckets are in the
shape of double cups, joined at the middle portion. The jet strikes the knife edge of the
buckets with least resistance and shock. Then the jet glides along the path of the cup, and
the jet is deflected through more than 160° to 170°. While passing along the buckets, the
velocity of the water is reduced and hence an impulsive force is supplied to the cups
which in turn are moved and hence the shaft is rotated. The specific speed of the pelton
wheel changes from 10 to 100.
Procedure :
i) Keep the nozzle opening at the required position. i.e. full opening or 3/4
opening.
ii) Start the pump.
iii) Allow water in to the turbine, then the turbine rotates.
iv) Note the speed of the turbine.
v) Take readings in manometer.
vi) Note the pressure of water in the pressure gauge.
vii) Load the turbine by putting weights.
viii) Note dead weight T1 and spring weight T2
ix) Note the head.
x) Repeat the experiment for different loadings.
xi) Tabulate the readings.
Formulae:
To find discharge
The venturimeter and the manometer has been calibrated.
35. 83
d
Venturimeter , = 0.6
D
a1
= 0.36
a2
K. 0.36 a1 √2g √ H1
Qa =
√ 1 - (0.36)
= K. 1.72 a1√ H1
Where, K = Coefficient of the venturimeter
D = 0.065 m
a1 = 0.00332 m2
Qa = K 0.0057 √H1
K = 0.962
-----------------------------------------------
Qa = 0.0055 √H1 m3
/s
-----------------------------------------------
Manometer Reading:
The height of mercury column in left arm, h1 = m
The height of mercury column in right arm, h2 = m
Difference of level , h = h1 -h2 m
Equivalent water column, H1 = 12.6 h m
Calculation of input power :
Discharge = Q m3
/s
Head = H m
Input power = 1000 X gQH
Calculation of output power :
Dead weight, T1 = kg
Spring weight , T2 = kg
Weight of hanger T0 = kg
Resultant load W = T1 - T2 + T0 kg
Speed, N = RPM
Dia. of brake drum = 0.40 m
Thickness of pipe = 0.015 m
Resultant dia. D = 0.415 m
2 π N (W*R)*9.81
output. power =
60
37. 85
Result:
A test is conducted on Pelton wheel (turbine) and the following graphs were drawn.
i) Output power Vs N & ii) Efficiency Vs N
38. 86
Observation :
Gate Opening: Weight of Hanger T0 = 1Kg.
Sl.
No
G V Head
G+V
Pr.Gauge
P1 P2
H1
`m’
Discharge
Q m3
/s
I P
(W)
Speed
N
T1
K
g
T2
Kg
W O P
(W)
η
%
39. 87
Exp. No. :
Francis turbine
Date :
Aim:
To study the characteristics of Francis Turbine at constant head.
Apparatus required:
i) Francis turbine set up & ii) Stop watch.
Description:
Francis turbine is a prime mover. It converts the hydraulic energy (head of water)
into mechanical energy, which in turn can be transformed into electrical energy by
coupling a generator to the turbine. Francis Turbine is a radial inward flow reaction
turbine. This has the advantage of centrifugal force acting against the flow, thus reducing
the tendency of the wheel to race. The turbine consists essentially of runner (G.M.), a
ring of adjustable guide vanes, a volute casing(spiral casing) , draft tube. Francis turbines
are best suited for medium heads, say 40m to 300m. The specific speed ranges from 25 to
300.
Procedure:
i) Keep the guide vanes fully opened or 6/8 opening.
ii) Prime the pump
iii) Start the pump
iv) Vent the manometer
v) Note the pressure gauge reading (G) and vacuum gauge reading (V).
vi) Adjust the gate valve so that G +V reads = 15m.
vii) Note the readings in the pressure gauge
Left limb reading = P1 m
Right limb reading = P2 m.
viii) Measure the speed of the turbine by tachometer
ix) Load the turbine by putting weights in the weights hanger. Take all
readings.
x) Repeat the experiments for various loadings and take 6 readings.
xi) Experiment can be repeated for different guide vane opening.
Formulae:
I. Discharge:
Pressure gauge readings:
Left limb reading = P1m.
Right limb reading = P2m.
Difference of levels H1 = (P1 -P2) 10 m of water
41. 89
Venturimeter equation Q = 0.0131 ÖH1 m3
/s
II. Head:
Pressure gauge = G m.
Vaccuum gauge = V m.
Total Head = G + V + X ,(X = Difference of levels pressure & vacuum
gauge)
= H m.
III. Input to the turbine:
I. H. P. = 1000 QH
75
IV. Output:
Brake drum diameter = 0.30m.
Rope diameter = 0.015m
Equivalent drum diameter = 0.315m.
Hanger weight = T0 Kg = 1 kg.
Dead Weight = T1 kg.
Spring Load = T2 kg.
Resultant load = T1 -T2 + T0 = T kg.
Speed of the turbine = N RPM
Output power = 2πN
(w x R) watts
60
Output
Efficiency = x 100
Input
43. 91
Result:
The characteristics of Francis turbine at constant head is studied and the
following graphs were drawn.
i) Output power Vs Speed ii) Output power Vs Input power and
iii) Output power Vs Efficiency
Calculate specific speed.