EXPERIMENT No.1 FLOW MEASUREMENT BY ORIFICEMETER 1.1 AIM: To determine the co-efficient of discharge of the orifice meter 1.2 EQUIPMENTS REQUIRED: Orifice meter test rig, Stopwatch 1.3 PREPARATION 1.3.1 THEORY An orifice plate is a device used for measuring the volumetric flow rate. It uses the same principle as a Venturi nozzle, namely Bernoulli's principle which states that there is a relationship between the pressure of the fluid and the velocity of the fluid. When the velocity increases, the pressure decreases and vice versa. An orifice plate is a thin plate with a hole in the middle. It is usually placed in a pipe in which fluid flows. When the fluid reaches the orifice plate, with the hole in the middle, the fluid is forced to converge to go through the small hole; the point of maximum convergence actually occurs shortly downstream of the physical orifice, at the so-called vena contracta point. As it does so, the velocity and the pressure changes. Beyond the vena contracta, the fluid expands and the velocity and pressure change once again. By measuring the difference in fluid pressure between the normal pipe section and at the vena contracta, the volumetric and mass flow rates can be obtained from Bernoulli's equation. Orifice plates are most commonly used for continuous measurement of fluid flow in pipes. This experiment is process of calibration of the given orifice meter. 1
Transcript
EXPERIMENT No.1
FLOW MEASUREMENT BY ORIFICEMETER
1.1 AIM: To determine the co-efficient of discharge of the orifice meter
1.2 EQUIPMENTS REQUIRED: Orifice meter test rig, Stopwatch
1.3 PREPARATION
1.3.1 THEORY
An orifice plate is a device used for measuring the volumetric flow rate. It uses the same principle as a Venturi nozzle, namely Bernoulli's principle which states that there is a relationship between the pressure of the fluid and the velocity of the fluid. When the velocity increases, the pressure decreases and vice versa. An orifice plate is a thin plate with a hole in the middle. It is usually placed in a pipe in which fluid flows. When the fluid reaches the orifice plate, with the hole in the middle, the fluid is forced to converge to go through the small hole; the point of maximum convergence actually occurs shortly downstream of the physical orifice, at the so-called vena contracta point. As it does so, the velocity and the pressure changes. Beyond the vena contracta, the fluid expands and the velocity and pressure change once again. By measuring the difference in fluid pressure between the normal pipe section and at the vena contracta, the volumetric and mass flow rates can be obtained from Bernoulli's equation. Orifice plates are most commonly used for continuous measurement of fluid flow in pipes. This experiment is process of calibration of the given orifice meter.
Fig.1. Orifice Plate
1.3.2 PRE-LAB QUESTIONS
1.3.2.1 Write continuity equation for incompressible flow?
1.3.2.2 What is meant by flow rate?
1.3.2.3 What is the use of orifice meter?
1.3.2.4 What is the energy equation used in orifice meter?
1.3.2.5 List out the various energy involved in pipe flow.
N.B.: Keep the delivery valve open while start and stop of the pump power supply.
1.4.1 Switch on the power supply to the pump
1.4.2 Adjust the delivery flow control valve and note down manometer heads (h1, h2) and
time taken for collecting 10 cm rise of water in collecting tank (t). (i.e. Initially the
delivery side flow control valve to be kept fully open and then gradually closing.)
1.4.3 Repeat it for different flow rates.
1.4.4. Switch off the pump after completely opening the delivery valve.
1.5 OBSERVATIONS
1.5.1 FORMULAE / CALCULATIONS
1.5.1.1 The actual rate of flow, Qa = A x h / t (m3/sec)
Where A = Area of the collecting tank = lengh x breadth (m2 )
h = Height of water(10 cm) in collecting tank ( m),
t = Time taken for 10 cm rise of water (sec)
1.5.1.2 The Theoretical discharge through orifice meter,
Qt = (a1 a2 2g H ) / (a12 – a2
2 ) m3/sec
Where, H = Differential head of manometer in m of water
= 12.6 x hm x 10 -2 (m)
g = Acceleration due to gravity (9.81m/sec2)
Inlet Area of orifice meter in m2 , a1 = d12/ 4 ,
Area of the throat or orifice in m2 , a2 = d22/ 4
1.5.1.3 The co-efficient of discharge,
Cd = Actual discharge / Theoretical discharge = Qa/Qt
1.5.2 TABULATION
Size of Orifice meter :
Inlet Dia. d1 = 25 mm ,
Orifice dia d2 = 18.77 mm, Measuring area in collecting tank A = 0.3 x 0.3 m2
2
Sl.
No. Manometer Reading
(cm)
Manomet
er Head
H
Time for
10 cm rise
t
Actual
Discharge
Qa
Theoretical
Discharge
Qt
Co-eff. of
discharge
Cd
h1 h2 hm = h1 - h2 m sec m3/sec m3/sec
1.
2.
3.
4.
5.
6
Average Cd value
1.5.3 GRAPH:
Draw Qa Vs Qt .
Find Cd value from the graph and compare it with calculated Cd value from table.
1.6 POST-LAB QUESTIONS
1.6.1 How do you find actual discharge?
1.6.2 How do you find theoretical discharge?
1.6.3 What do you meant by co-efficient of discharge?
1.6.4 Define vena-contracta?
1.6.5 List out the Bernoulli’s applications.
1.7 INFERENCES
1.8 RESULT
The co-efficient of discharge of orifice meter = ……………. From Calculation
The co-efficient of discharge of orifice meter = ……………. From Graph
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EXPERIMENT No.2
FLOW MEASUREMENT BY VENTURIMETER
2.1 AIM: To determine the co-efficient of discharge of the venturimeter
2.2 EQUIPMENTS REQUIRED: Venturimeter test rig, Stopwatch
2.3 PREPARATION
2.3.1 THEORY
Fig.2. Venturimeter
In a Venturi meter there is first a converging section in which the cross sectional area for flow is reduced. Then there is a short section at the reduced diameter, known as the throat of the meter. Then there is a diverging section in which the cross sectional area for flow is gradually increased to the original diameter. The velocity entering the converging section is where the pressure is P1. In the converging section the velocity increases and the pressure decreases. The maximum velocity is at the throat of the meter where the minimum pressure P2 is reached. The velocity decreases and the pressure increases in the diverging section. There is a considerable recovery of pressure in the diverging section. However, because of frictional effects in the fluid, the pressure leaving the diverging section is always less than P1, the pressure entering the meter.
2.3.2 PRE-LAB QUESTIONS
2.3.2.1 Differentiate mass and volume flow rate?
2.3.2.2 Which property is remains same in the incompressible flow?
2.3.2.3 What is meant by discharge?
2.3.2.4 What is the use of venturimeter?
2.4 PROCEDURE:
N.B.: Keep the delivery valve open while start and stop of the pump power supply.
2.4.1. Switch on the power supply to the pump
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2.4.2. Adjust the delivery flow control valve and note down manometer heads (h1, h2) and
time taken for collecting 10 cm rise of water in collecting tank (t). (i.e. Initially the
delivery side flow control valve to be kept fully open and then gradually closing.)
2.4.3. Repeat it for different flow rates.
2.4.4. Switch off the pump after completely opening the delivery valve.
2.5 OBSERVATIONS
2.5.1 FORMULAE / CALCULATIONS
2.5.1.1 The actual rate of flow, Qa = A x h / t (m3/sec)
Where A = Area of the collecting tank = lengh x breadth (m2 )
h = Height of water(10 cm) in collecting tank ( m),
t = Time taken for 10 cm rise of water (sec)
2.5.1.2 The Theoretical discharge through venturimeter,
Qt = (a1 a2 2g H ) / (a12 – a2
2 ) m3/sec
Where, H = Differential head of manometer in m of water
= 12.6 x hm x 10 -2 (m)
g = Acceleration due to gravity (9.81m/sec2)
Inlet Area of venturimeter in m2 , a1 = d12/ 4 ,
Area of the throat in m2 , a2 = d22/ 4
2.5.1.3 The co-efficient of discharge,
Cd = Actual discharge / Theoretical discharge = Qa/Qt
2.5.2 TABULATION:
Inlet Dia. of Venturimeter (or) Dia of Pipe d1 = 25 mm
Throat diameter of Venturimeter d2 = 18.79 mm
Area of collecting tank , A = Length x Breadth = 0.3 x 0.3m2
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Sl.No.
Manometer Reading
(cm)
Mano-meter Head
H
Time for 10 cm rise
t
Actual Discharge
Qa
TheoreticalDischarge
Qt
Co-eff. of discharge
Cd
h1 h2 hm = h1 - h2 m sec m3/sec m3/sec1.2.3.4.5.
Average Cd value
2.5.3 GRAPH:
Draw Qa Vs Qt .
Find Cd value from the graph and compare it with calculated Cd value from table.
2.6 POST-LAB QUESTIONS
2.6.1 How do you find actual and theoretical discharge?
2.6.2 What do you meant by throat of the venturimeter?
2.6.3 List out the practical applications of Bernoulli’s equation?
2.6.4 What is the use of U-tube manometer?
2.7 INFERENCES
2.8 RESULT
The co-efficient of discharge of Venturi meter = ……………. From Calculation
The co-efficient of discharge of Venturi meter = ……………. From Graph
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EXPERIMENT No.3
VERIFICATION OF BERNOULLIS THEOREM
3.1 AIM: To verify the Bernoulli’s theorem
3.2 EQUIPMENTS REQUIRED: Bernoulli’s Theorem test set-up, Stopwatch
3.3 PREPARATION
3.3.1 THEORY
Bernoulli’s Theorem
According to Bernoulli’s Theorem, in a continuous fluid flow, the total head at any point along the flow is the same. Z1 + P1/ g +V1
2/2g= Z2 + P2/ g +V22/2g , Since Z1 –Z2 = 0
for Horizontal flow, h1 +V12/2g= h2 +V2
2/2g ( Pr. head, h = P1/ g ). Z is ignored for adding in both sides of the equations due to same datum for all the positions.
3.3.2 PRE-LAB QUESTIONS
3.3.2.1 State Bernoulli’s theorem?
3.3.2.2 What is continuity equation?
3.3.2.3 What do you meant by potential head?
3.3.2.4 What do you meant by pressure head?
3.3.2.5 What do you meant by kinetic head?
3.4 PROCEDURE
N.B.: Keep the delivery valve open while start and stop of the pump power supply.
3.4.1 Switch on the pump power supply. 3.4.2 Fix a steady flow rate by operating the appropriate delivery valve and drain valve 3.4.3. Note down the pressure heads (h1 – h8) in meters3.4.4. Note down the time taken for 10 cm rise of water in measuring (collecting) tank.3.4.5. Switch off the power supply.
3.5 OBSERVATIONS
3.5.1 FORMULAE / CALCULATIONS
3.5.1.1 Rate of flow Q = Ah /t. Where A: Area of measuring tank = Length x Breadth (m2)
h: Rise of water in collecting tank (m) .. (i.e. h = 10 cm )t: Time taken for 10 cm rise of water in collecting tank (sec)
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3.5.1.2 Velocity of flow, V = Q/a ,
Where a – Cross section area of the duct at respective peizometer positions (a1 - a8)
3.5.1.3 Hydraulic Gradient Line (HGL): It is the sum of datum and pressure at any point
HGL = Z + h
3.5.1.4 Total Energy Line (TEL): It is the sum of Pressure head and velocity head
TEL = Z + h +V2/2g
3.5.2 TABULATIONS Area of measuring tank = 0.3 x 0.3 m2
Assume Datum head Z = 0
Diameter at the sections of the
channeld
Cross –SectionArea
a = d2/ 4
Time for 10 cm rise t
Discharge
Q=Ah/t
VelocityV=Q/a
VelocityHeadV2/2g
PiezometerReading i.e.
Pr. Head(h=P/g )
Total HeadZ +h+V2/2g
m x10-3 m2 sec m3/sec m/sec m m md1 = 0.04295
1.448
d2 = 0.03925
1.209
d3= 0.03555
0.992
d4= 0.03185
0.796
d5 = 0.02815
0.622
d6= 0.02445
0.469
d7= 0.02075
0.338
d8= 0.01705
0.228
3.5.3 GRAPH
Draw the graph: Distance of channel (Locations 1-8) Vs HGL, TEL
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3.6 POST-LAB QUESTIONS
3.6.1 What do you meant by velocity head?
3.6.2 What do you meant by HGL?
3.6.3 What do you meant by datum head?
3.6.4 What is the use of piezometer?
3.6.5 Define TEL?
3.6.6 What is the reason for the slight decrease in the total energy head between the
successive locations in the duct?
3.7 INFERENCES
3.8 RESULT
The Bernoulli’s theorem is verified.
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EXPERIMENT No.4
DETERMINATION OF PIPE FRICTION FACTOR
4.1 AIM: To determine the friction factor for the given pipe.
The major loss in the pipe is due to the inner surface roughness of the pipe. There
are three pipes (diameter 25 mm, 20 mm and 15 mm) available in the experimental set
up. The loss of pressure head is calculated by using the manometer. The apparatus is
primarily designed for conducting experiments on the frictional losses in pipes of
different sizes. Three different sizes of pipes are provided for a wide range of
experiments.
4.3.2 PRE-LAB QUESTIONS
4.3.2.1. What do you meant by friction and list out its effects?
4.3.2.2 What do you meant by major loss in pipe?
4.3.2.3 Write down the Darcy-Weisbach equation?
4.3.2.4 What are the types of losses in pipe flow?
4.4 PROCEDURE
N.B.: Keep the delivery valve open while start and stop of the pump power supply.
4.4.1. Switch on the pump and choose any one of the pipe and open its corresponding inlet and exit valves to the manometer.
4.4.2. Adjust the delivery control valve to a desired flow rate. (i.e. fully opened delivery valve position initially)
4.4.3 Take manometer readings and time taken for 10 cm rise of water in the collecting tank
4.4.4 Repeat the readings for various flow rates by adjusting the delivery valve. (i.e. Gradually closing the delivery valve from complete opening)
4.4.5 Switch of the power supply after opening the valve completely at the end.
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4.5 OBSERVATIONS
4.5.1 FORMULAE / CALCULATIONS
4.5.1.1 The actual rate of flow Q = A x h / t (m3/sec)
Where A = Area of the collecting tank = lengh x breadth (m2 )h = Height of water(10 cm) in collecting tank ( m),
t = Time taken for 10 cm rise of water (sec)
4.5.1.2 Head loss due to friction, hf = hm ( Sm – Sf)/ (Sf x 100) in m
hf = hm (13.6 – 1 ) x 10 -2 (m) Where Sm = Sp. Gr. of manometric liquid , Hg =13.6 ,
Sf = Sp. Gr. of flowing liquid, H2O = 1 hm = Difference in manometric reading = (h1-h2) in cm
4.5.1.3 The frictional loss of head in pipes (Darcy-Weisbach formula)
hf = 4f L V 2 2 g d
Where f = Co-efficient of friction or friction factor for the pipe (to be found)L = Distance between two sections for which loss of head is measured = 3 mV = Average Velocity of flow = Q/a (m/s), Area of pipe a= d2/ 4 (m2),
d = Pipe diameter = 0.015 mg = Acceleration due to gravity = 9.81 m/sec2
4.5.2 TABULATION
Length between Pressure tapping, L = 3 mPipe Diameter, d = 0.015 m,Measuring tank area, A= 0.6 x 0.3m2 ,
Sl.No. Pipe Dia
Manometer Reading
Head loss
Time for 10 cm rise
Discharge
Velocity
Frictional factor
d h1 h2 hm = (h1-h2)
hf t Q V=Q/a f
m cm m sec m3/s m/s12345 Average friction factor, f
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4.5.3 GRAPH
Draw the graph: Q Vs hf
4.6 POST-LAB QUESTIONS
4.6.1 What is the relationship between friction head loss and pipe diameter?
4.6.2 What is the relationship between friction head loss and flow velocity?
4.6.3 What is the relationship between friction head loss and pipe length?
4.6.4 How is the flow rate and head loss related?
4.6.5 If flow velocity doubles, what happen to the frictional head loss?
4.7 INFERENCES
4.8 RESULT
The friction factor for the given pipe diameter of ……… m is = __________
12
EXPERIMENT NO.5
PERFORMANCE TEST ON GEAR PUMP
5.1 AIM: To study the performance of gear oil pump.
5.2 EQUIPMENTS REQUIRED: Gear pump test rig, Stopwatch
5.3 PREPARATION
5.3.1 THEORY
A gear pump uses the meshing of gears to pump fluid by displacement. They are one of the most common types of pumps for hydraulic fluid power applications. Gear pumps are also widely used in chemical installations to pump fluid with a certain viscosity. There are two main variations; external gear pumps which use two external spur gears, and internal gear pumps which use an external and an internal spur gear. Gear pumps are positive displacement (or fixed displacement), meaning they pump a constant amount of fluid for each revolution. Some gear pumps are designed to function as either a motor or a pump. The gear oil pump is works based on the squeezing action of the two meshing gears (internal or external gears). The gear pump is one of the positive displacement pump and the reduction in volume inside the pump results in increase in pressure of fluid.
As the gears rotate they separate on the intake side of the pump, creating a void and suction which is filled by fluid. The fluid is carried by the gears to the discharge side of the pump, where the meshing of the gears displaces the fluid. The mechanical clearances are small— in the order of 10 μm. The tight clearances, along with the speed of rotation, effectively prevent the fluid from leaking backwards. The rigid design of the gears and houses allow for very high pressures and the ability to pump highly viscous fluids.
Fig.3. External Gear pump
5.3.2 PRE-LAB QUESTIONS
5.3.2.1 What is the purpose of gear pump?
5.3.2.2 What do you meant by internal and external gears?
N.B. : NEVER operate the pump with closed delivery valve during start and stop of the pump power supply. The violation leads to damage of the pipe line and the pump.
5.4.1 Ensure the complete opened position of delivery valve. 5.4.2 Measure height of the pressure gauge above the vacuum gauge.5.4.3 Switch on the pump.5.4.4 Vary the flow rate (discharge) by closing the delivery valve.5.4.5 Adjust the delivery valve accordingly the pressure gauge reading of 1kg/cm2.5.4.6 Note down vacuum gauges reading. 5.4.7 Note down time taken for ‘h’ cm rise of oil (10 cm) in collecting tank. 5.4.8 Note down the time taken for ‘n’ revolutions for energy meter disc (3 rev). 5.4.9 Repeat the procedure for 1 kg/cm2 incremental by closing the delivery valve
gradually, (i.e. 1.0, 1.5, 2.0, 2.5 and 3.0 kg/cm2 ). 5.4.10 Switch off the power supply after opening the delivery valve completely.
5.5 OBSERVATIONS
5.5.1 FORMULAE / CALCULATIONS
5.5.1.1 Total head H = [ P + (V/760 ] x 105/( g) + Z (m)
Where P = Pressure gauge reading in kg/cm2, V = Vacuum gauge reading in mm Hg, N.B.: Unit Conversion: For V, 1 mm Hg/ 760 = 1 bar & for P, 1 bar = 1 kg/ cm2 .
5.5.1.2 Discharge, Q = (A x h ) / t (m3/s) ,
Where A = Area of tank in m2 , h = Rise oil level in collecting tank (m), t = Time taken for the rise of oil 10 cm in collecting tank (sec)
5.5.1.3 Output in Watts, OP = g Q H / 1000 (kW)
Where = Density of oil = 860 kg/m3
g = Acceleration due to gravity = 9.81 m/sec2
5.5.1.4 Input in kWatts , IP = (n x 3600 x m ) / (Ec x T) (kW)
Where Ec = Energy meter constant in Rev /kWh = 1200 Rev / kWhn = Number of revolution taken in energy meter = 3
T = Time required to complete ‘n’ revolution in secm = Efficiency of motor = 0.80
5.5.1.5 Efficiency, = (Output / Input ) x 100% = (OP/IP) x 100%
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1.5.2 TABULATION
Measuring Area in collecting tank = 0.3 x 0.3 m2 Datum head Z = 0.3 m. Density of oil, = 860 kg/m3
Sl.No. P V Z H
Time for 10 cm rise
(t)
Flow rateQ
Time for 3 rev of Energy
meter (T)
Input
IP
Output
OP
Efficiency
kg/cm2 mm Hg
M m sec m3/sec sec kW kW %
1 1.02 1.53 2.04 2.55 3.06 3.5
5.5.3 GRAPHDraw the graph: Discharge vs Head, Output Power, Efficiency.
5.6 POST-LAB QUESTIONS
5.6.1. What is the squeezing in gear pumps?
5.6.2 What is the type of gears used in gear pumps?
5.6.3 List out the different pressure heads used in gear pump?
5.6.4 What is mechanical efficiency of pump?
5.7 INFERENCES
5.8 RESULT
The performance characteristics of the given gear oil pump is studied.
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EXPERIMENT No.6
PERFORMANCE TEST ON SUBMERSIBLE PUMP
6.1 AIM: To study the performance characteristics of submersible pump.
6.2 EQUIPMENTS REQUIRED: Submersible pump test rig, Stop watch
6.3 PREPARATION
6.3.1 THEORY
A submersible pump (or electric submersible pump (ESP)) is a device which has a hermetically sealed motor close-coupled to the pump body. The whole assembly is submerged in the fluid to be pumped. The main advantage of this type of pump is that it prevents pump Cavitation, a problem associated with a high elevation difference between pump and the fluid surface. Submersible pumps push fluid to the surface as opposed to jet pumps having to pull fluids. Submersible pumps are more efficient than jet pumps
6.3.2 PRE-LAB QUESTIONS
6.3.2.1 What is the submersible pump?
6.3.2.2 What is the working principle of submersible pump?
6.3.2.3 What are applications of submersible pump?
6.3.2.4 What is the value of suction head in submersible pump?
6.3.2.5 Is priming necessary in submersible pump?
6.4 PROCEDURE
N.B. : NEVER operate the pump with closed delivery valve during start and stop of the pump. The violation leads to damage of the pipe line and the pump.
6.4.1 Start the pump and run it at particular head on it. 6.4.2 Ensure the complete opening of the delivery valve provided. 6.4.3 The pressure gauge reading to be noted for the pressure of 0.5 kg/cm2.6.4.4 The time is to be noted for collecting 10 cm rise of water in the collecting tank. 6.4.5 The time is to be noted for 3 revolution of Energy meter disc.6.4.6 Repeat the procedure for 0.5, 1.0, 1.5, 2.0, 2.5, 3.0 kg/cm2 delivery pressures.
by closing the delivery valve. 6.4.7 Open the delivery valve completely and then Switch off the power supply.
6.5 OBSERVATIONS
6.5.1 FORMULAE / CALCULATIONS
6.5.1.1 Total Head, H = P x 105/ ( g) …… in m Where P – Pressure gauge reading in kg/cm2,
6.6.3 List out the different heads used in submersible pump?
6.6.4 What is advantage of submersible pump over jet pump?
6.6.5 What is mechanical efficiency?
6.6.6 What do you meant by cavitaion?
6.6.7 How do you prevent cavitaion in pump?
6.6.8 What is advantage of submersible pump over centrifugal pump?
6.6.9 Is priming necessary in submersible pump?
6.7 INFERENCES
6.8 RESULT
The performance characteristics of the given submersible pump are studied.
EXPERIMENT No.7
PERFORMANCE TEST ON RECIPROCATING PUMP
7.1 AIM: To study the performance of Reciprocating pump.
7.2 EQUIPMENTS REQUIRED: Reciprocating pump test rig, Stop watch
7.3 PREPARATION
7.3.1 THEORY
18
A reciprocating pump is a positive plunger pump. It is often used where relatively small quantity of liquid is to be handled and where delivery pressure is quite large. Reciprocating pumps are now typically used for pumping highly viscous fluids including concrete and heavy oils, and special applications demanding low flow rates against high resistance.
7.3.2 PRE-LAB QUESTIONS
7.3.2.1 What is the working principle of reciprocating pump?
7.3.2.2 What is type of suction and delivery valves in reciprocating pump?
7.3.2.3 What do you meant by slip in reciprocating pump?
7.3.2.4 What do you meant by single and double acting pump?
7.3.2.5 List out the parts of reciprocating pump.
7.4 PROCEDURE
N.B. : NEVER operate the pump with closed delivery valve during start and stop. Never operate the pump above the gauge pressure of 3 kg/cm2.
7.4.1 Start the pump and run it at particular head on it. 7.4.2 Ensure the complete opened position of delivery valve. 7.4.3 Vary the flow rate (discharge) by closing the delivery valve in order to maintain
certain pressure gauge reading i.e.0.5 kg/cm2.7.4.4 Note down pressure gauge reading (0.5 kg/cm2) and vacuum gauges readings.7.4.5 Measure height of the pressure gauge above the vacuum gauge (Datum level)7.4.6 Note down time taken (t) for ‘h’ cm rise of water (10 cm) in collecting tank. 7.4.7 Note down the time taken (T) for ‘n’ revolutions for energy meter (3 rev) disc. 7.4.8 Repeat the procedure for 0.5, 1.0, 1.5, 2.0, 2.5 kg/cm2 in the Pressure gauge
reading by closing the delivery valve. 7.4.9 Switch off the power supply after opening the delivery valve completely.
7.5 OBSERVATIONS
7.5.1 FORMULAE / CALCULATIONS
7.5.1.1. Total head, H = [P + (V/760) ] x 105 / ( g ) + Z (m)
Where, P – Pressure gauge reading in kg/cm2, V- Vacuum gauge reading in mm Hg, Z – Datum level between Pressure gauge and Vacuum gaugeg = Gravitational acceleration = 9.81 m2/s = Density of fluid (water) = 1000 kg/m3
N.B.: For V, 1 mm Hg/ 760 = 1 bar & for P, 1 bar = 1 kg/ cm2
Where A- Collecting tank area = l x b in m2, t - time for 10 cm rise of water level in the collecting tank (sec)h – Rise of water level in the collecting tank = 0.10 m
7.5.1.3. Output in kW, OP = g QH / 1000
7.5.1.4. Input in kW, IP = (‘n’ rev of energy meter x 3600 x Efficiency of motor) / (Energy meter constant in Rev/kW-hr x Time for n revolutions)
= (n x 3600 x m ) / (Ec x T)
Where Ec = Energymeter constant in Rev /kW – hr = 1200 rev / kWh n = Number of revolution taken in energy meter = 3
T = Time required to complete ‘n’ revolution in sec m = Efficiency of motor = 0.8
7.5.1.5. Efficiency of Pump, = (Output / Input) x 100 %
1.5.3 TABULATION:
Area of Measuring tank A : 0.3 x 0.3 m2 ,
Motor Efficiency m : 0.8
Energymeter Constant : 1200 Rev/kW-hr
Z – Datum level between Pressure gauge and Vacuum gauge = 0.75 m
Sl.No.
Pressure Gauge reading
P
Vacuum Gauge reading
V
Total Head
H
Time for
10 cm riseT
DischargeQ
x10-4
Time for 3 rev of
EM discT
Input Power
IP
Output Power
OP
Efficiency
kg/cm2 mm Hg m Sec m3/sec sec kW kW %1. 0.52. 13. 1.54. 25. 2.56. 3
20
7.5.3 GRAPH: Disharge Vs head, Output, Efficiency.
7.6 POST-LAB QUESTIONS
7.6.1 List out the different heads used in reciprocating pump?
7.6.2 What is the use of air vessel?
7.6.3 Why the delivery valve should be kept open always?
7.6.4 What do you meant by indicator diagram and its use?
7.6.5 What do you meant by percentage slip and negative slip?
7.6.6 In what situations, the reciprocating pump is desired?
7.7 INFERENCES
7.8 RESULT
The performance test on reciprocating pump is conducted and the performance characteristics are drawn.
EXPERIMENT No.08
PERFORMANCE TEST ON JET PUMP
8.1 AIM: To study the performance test on the jet pump
8.2 EQUIPMENTS REQUIRED: Jet pump test rig, stop watch
8.3 PREPARATION
8.3.1 THEORY
Jet pumps, also referred to as ejectors, are devices for the conveyance, compression or mixing of gases, vapours, liquids or solids in which a gaseous or liquid medium serves as the motive force. They operate by the conversion of pressure energy into velocity in suitable nozzles. They are "pumps without moving parts." The basic
21
principle of jet pumps consists in the liquid or gas jet being emitted by a nozzle at high speed entraining and accelerating the surrounding liquid, gas or solid matter. The result of this action is a mixture of the driving and entrained (sucked) fluids, the velocity of which is reduced and the pressure increased in a second nozzle.
Fig.5 Jet Pump
8.3.2 PRE-LAB QUESTIONS
8.3.2.1 What is the purpose of jet pump?
8.3.2.2 What is the working principle of jet pump?
8.3.2.3 What are applications of jet pump?
8.3.2.4 List out the parts of jet pump.
8.4 PROCEDURE
8.4.1 Start the pump and run it at particular head on it.
8.4.2 Ensure the complete opened position of delivery valve.
8.4.3 Vary the flow rate (discharge) by closing the delivery valve in order to maintain
certain pressure gauge reading i.e.0.5 kg/cm2.
8.4.4 Note down pressure gauge reading (0.5 kg/cm2) and vacuum gauges readings.
8.4.5 Measure height of the pressure gauge above the vacuum gauge (Datum level)
8.4.6 Note down time taken (t) for ‘h’ cm rise of water (10 cm) in collecting tank.
8.4.7 Note down the time taken (T) for ‘n’ revolutions for energy meter (3 rev) disc.
22
8.4.8 Repeat the procedure for 0.5, 1.0, 1.5, 2.0, 2.5 kg/cm2 in the Pressure gauge
reading by closing the delivery valve.
8.4.9 Switch off the power supply after opening the delivery valve completely.
8.5 OBSERVATIONS
8.5.1 FORMULAE / CALCULATIONS
8.5.1.1 Total Head, H = P x 105/ ( g) …… in m Where P – Pressure gauge reading in kg/cm2,
g = Gravitational acceleration = 9.81 m2/s = Density of fluid (water) = 1000 kg/m3
8.5.1.2 Discharge, Q = (A x h ) / t (m3/sec) Where A = Area of tank in m2 A = l x b
h = Rise of water level in collecting tank = 0.10 mt = Time taken for 10 cm rise of water in the collecting tank
8.5.1.3 Output in kW, OP = g QH / 1000
8.5.1.4 Input in kW, IP = (n x 3600 m ) / (N x T)
Where N = Energymeter constant in Rev /kW – hr = 1200 rev / kWhn = Number of revolution taken in energymeter
T = Time required to complete ‘n’ revolution in secm = Efficiency of motor = 0.8
8.5.1.5 Efficiency, = Output / Input x 100%
8.5.2 TABULATIONEnergy meter constant N = 1200 Rev/kW-hr
Measuring Area in collecting tank=0.3 x0.3m2
Sl.
No.
Pressure
gauge
Total
Head
Time for 10
cm rise
Discharge Time for 3
rev of EM
Input Output Efficiency
P H t Q T IP OP
kg/cm2 M sec m3/sec sec kW Kw %
1 0.5
2 1.0
3 1.5
4 2.0
23
5 2.5
8.5.3 GRAPH:
Draw Q Vs H, OP and
8.5 POST-LAB QUESTIONS
8.6.1 How do you measure actual flow rate?
8.6.2 What is the output power of the pump?
8.6.3 List out the different pressure heads used in jet pump?
8.6.4 What is mechanical efficiency?
8.6.5 What is overall efficiency?
8.7 INFERENCES
8.8 RESULT
The performance of the given jet pump is studied.
EXPERIMENT No.9
DETERMINATION OF MINOR LOSSES DUE TO PIPE FITTINGS
9.1 AIM: To study the various losses due to the pipe fittings
9.2 EQUIPMENTS REQUIRED: Minor losses test rig, Stopwatch
9.3 PREPARATION
9.3.1 THEORY
The various pipe fittings used in the piping applications are joints, bends, elbows, entry, exit and sudden flow area changes (enlargement and contraction) etc. The energy losses associated with these types of pipe fittings are termed as
24
the minor losses due its lesser values compared to the major loss (pipe friction) in the pipe. The loss of head is indicated by the manometer connected across the respective pipe fitting.
9.3.2 PRE-LAB QUESTIONS
9.3.2.1 List out the various types of pipe fittings?
9.3.2.2 What do you meant by minor losses?
9.3.2.3 What are the types of losses in pipe flow?
9.3.2.4 What do you meant by entry loss?
9.3.2.5 What do you meant by exit loss?
9.4 PROCEDURE
N.B.: Keep the delivery valve open while start and stop of the pump power supply.
9.4.1 Switch on the pump. Adjust the delivery valve to a desired steady flow rate.9.4.2 Note down the time taken for 10 cm rise of water level in the collecting tank.9.4.3 Choose any one of the pipe fittings (2 bends, one enlargement and one
contraction). e.g. Bend-1 9.4.4 Open the levers (cocks) of respective pipe fitting to the manometer. Ensure other
fitting levers should be closed. e.g. Open the entry and exit levers of Bend-1( left & right side cocks at the top of the panel)
9.4.5 Note down the manometer head levels (e.g. h1 & h2 for bend-1) 9.4.6 Now open the other two entry and exit levers of next pipe fitting. Then close the
levers of first chosen pipe fitting. (e.g. Open the 2nd left & right levers for Bend-2 and close the top levers of Bend-1)
9.4.7 Note down the manometer for the second pipe fitting. (e.g. h1 & h2 for bend-2)
9.4.8 Repeat this procedure by opening the respective levers of sudden enlargement fitting after closing other levers( i.e. for sudden enlargement by opening the next down left & right cocks of sudden enlargement and then close the previous left & right cocks of Bend-2).
9.4.9 Repeat this procedure by opening the respective levers of sudden contraction fitting after closing other levers( i.e. for sudden contraction by opening the next down left & right cocks of sudden contraction and then close the previous left & right cocks of Sudden enlargement).
9.4.10 Ensure the readings taken for all pipe fittings and then switch off the pump. 9.5 OBSERVATIONS
9.5.1 FORMULAE / CALCULATIONS
25
9.5.1.1. Discharge, Q = (A x h ) / t ….. (m3/s) A = Area of tank in m2 , h = 0.10 m // Rise water level in collecting tank (m),
t = Time taken for the 10 cm rise of water in collecting tank (sec) 9.5.1.2. Velocity, V = Discharge / Area of pipe = Q/A… (m/s)
Where A = d2/4 , d – Dia of pipe in m
9.5.1.3. Actual loss of head, hf = ( h1 – h2 ) x 12.6 x 10-2 … (m)
9.5.1.4. Theoretical Velocity loss heads for pipe fittings
Velocity head loss for bend and elbow hv = V2 / (2g)
Velocity head loss for sudden enlargement hv = ( V1 – V2 )2 / (2g)
Velocity head loss for sudden contraction hv = 0.5 (V2)2 / (2g)
Where V2= velocity of smaller pipe
9.5.1.5. Loss co-efficient K = Theoretical Velocity head /Actual loss of head = hv / hf
9.5.2 TABULATION
Collecting Tank area, A = 0.6 m x 0.3 m, Pipe Diameter , d = 0.02 m
Pipe fittings
Manometer Reading
(cm)
Time for 10 cm rise(sec)
Discharge
(m3/s)
Velocity
(m/s)
Actual Loss of head, (m)
Loss of head(Theoretical)
(m)
Loss co-efficient
K
h1 h2 hm t Q V hf hv hv / hf
Bend-1Bend-2Sudden
26
Enlarge(20-40 mm)Sudden Contraction (40-20 mm)
9.5 POST-LAB QUESTIONS
9.6.1 What is the equation for head loss due to sudden enlargement?
9.6.2 What is the equation for head loss due to sudden contraction?
9.6.3 What is the equation for head loss due to bend?
9.6.4 What is the equation for head loss at entry of pipe?
9.6.5 What is the equation for head loss at exit of pipe?
9.6.6 Which Newton’s law is applicable to impulse turbine?
9.7 INFERENCES
9.8. RESULT
The various minor losses in pipe fittings are determined.
EXPERIMENT No.10
PERFORMANCE TEST ON PELTON TURBINE
10.1 AIM: To study the performance of Pelton turbine.
The Pelton wheel is an impulse turbine which is among the most efficient types of water turbines. The Pelton wheel extracts energy from the impulse (momentum) of moving water, as opposed to its weight like traditional overshot water wheel. The water flows along the tangent to the path of the runner. Nozzles direct forceful streams of water against a series of spoon-shaped buckets mounted around the edge of a wheel. As water flows into the bucket, the direction of the water velocity changes to follow the contour of the bucket. When the water-jet contacts the bucket, the water exerts pressure on the bucket and the water is decelerated as it does a "u-turn" and flows out the other side of the bucket at low velocity. In the process, the water's momentum is transferred to the turbine. This "impulse" does work on the turbine.
For maximum power and efficiency, the turbine system is designed such that the water-jet velocity is twice the velocity of the bucket. Often two buckets are mounted side-by-side, thus splitting the water jet in half. This balances the side-load forces on the wheel, and helps to ensure smooth, efficient momentum transfer of the fluid jet to the turbine wheel. Because water and most liquids are nearly incompressible, almost all of the available energy is extracted in the first stage of the hydraulic turbine. Therefore, Pelton wheels have only one turbine stage, unlike gas turbines that operate with compressible fluid. Pelton turbine is preferred for water source has relatively high hydraulic head at low flow rate.
Fig.4 Pelton Turbine
10.3.2 PRE-LAB QUESTIONS
10.3.2.1 What are the types of hydraulic turbines?
10.3.2.2 What is the purpose of turbine?
10.3.2.3 What is the use of nozzle?
10.3.2.4 Define water hammer.
10.3.2.5 What is the purpose of surge tank?
10.3.2.6 Classify the hydraulic turbines.
10.3.2.7 Differentiate impulse and reaction turbines.
10.4.2.6. Increase the weights from 0 kg as 0, 2 , 4, 6, 8 kg and note down the speed and
spring balance readings. No need to vary the flow rate and head.
10.4.2.7. First remove all the dead weights on the hanger. Close the delivery valve and
then switch off the pump power supply in order to stop the turbine.
10.5 OBSERVATIONS
10.5.1 FORMULAE / CALCULATIONS
10.5.1.1 Input Power, IP = g Q H (W)
Where g = Specific weight of water = 9.81 kN/m3
Q = Discharge in m3/sec Q = K √h = 3.183 x 10-3 √h (h in m )Size of Venturimeter : 50mm and Throat Diameter : 29.58mmWhere ‘K’ Value = (a1 a2 2g ) / (a1
2 – a2 2 = 3.183 x 10-3
Where h = Venturimeter head in m of water h = p x105/ g = (p1 – p2) x105/ (g) (m)
H = Supply head in m = Input total head in m H = P x105/ (g )
10.5.1.2. Output (Brake) Power, OP = 2 N T / 60 (W)
Where N = Turbine speed in RPM., T = W Re g = Torque in N-m
Brake drum dia D =0.2m ,Rope dia d =0.015mEffective radius of Brake Drum, Re = (D/2 )+d) = 0.115m
Weight of rope & hanger = 1kg.Brake drum Net load W =(W1 + weight of rope hanger) –W2 kg = ( 5+1) – 2 = 4 kg
10.5.1.3 Efficiency of turbine, = (OP/IP ) x 100%
10.5.2 TABULATION
SlN
Pressure gauge
readingP
Pressure Gauge reading Venturi -meter
Headh
Discharge
Q
Weight on
hangerW1
Spring balance reading
W2
Net Load
W
Speed
N
Input
IP
Output
OP
Efficiency
p1 p2 P = p1-p2
30
kg/cm2 kg/cm2 m m3/sec kg kg kg rpm kW kW %1.2.3.4.5.6.
10.5.3 GRAPH: Speed Vs Output, Efficiency
10.6 POST-LAB QUESTIONS
10.6.1 What do you meant by impulse turbine?
10.6.2 How do you regulate the flow of water to the turbine?
10.6.3 What is the energy conversion from nozzle to turbine?
10.6.4 Why is braking jet used?
10.6.5 What is the spear and nozzle?
10.6.6 What is the pressure inside the turbine casing?
10.6.7 What is mechanical efficiency?
10.6.8 What do you meant by volumetric efficiency?
10.7 INFERENCES
10.8 RESULT: The performance of Pelton turbine is studied.
31
EXPERIMENT No.11
PERFORMANCE TEST ON FRANCIS TURBINE
11.1 AIM: To study the performance of Francis turbine.
11.2 EQUIPMENTS REQUIRED: Francis turbine test rig, Stop watch, Weights,
tachometer.
11.3 PREPARATION
11.3.1 THEORY
32
The Francis turbine is an inward-flow reaction turbine that combines radial and axial flow concepts. Francis turbines are the most common water turbine in use today. They operate in a head range of 10 m – 650 m and are primarily used for electrical power production. The power output ranges from 10 to 750MW. Runner diameters are between 1 and 10 meters. The speed range of the turbine is from 83 to 1000 rpm. Medium size and larger Francis turbines are most often arranged with a vertical shaft. The Francis turbine is a reaction turbine, which means that the working fluid changes pressure as it moves through the turbine, giving up its energy.
The inlet is spiral shaped. Guide vanes direct the water tangentially to the turbine wheel, known as a runner. This radial flow acts on the runner's vanes, causing the runner to spin. The guide vanes (or wicket gate) may be adjustable to allow efficient turbine operation for a range of water flow conditions.
11.3.2 PRE-LAB QUESTIONS
11.3.2.1 What are the types of hydraulic turbines?
11.3.2.2 What is the use of draft tube?
11.3.2.3 What is the flow direction in reaction turbine?
11.3.2.4 What is the use of guide vanes?
11.3.2.5 Which Newton’s law is applicable to reaction turbine?
3. Before switching off the supply pump set, first remove all the dead weights on the
hanger.
11.5 OBSERVATIONS
11.5.1 FORMULAE / CALCULATIONS
11.5.1.1 Input Power, IP = g Q H (W)
Where g = Specific weight of water = 9.81 kN/m3
Q = Discharge in m3/sec Q = K √h = 3.183 x 10-3 √h (h in m )Size of Venturimeter : 50mm and Throat Diameter : 29.58mmWhere ‘K’ Value = (a1 a2 2g ) / (a1
2 – a2 2 = 3.183 x 10-3
Where h = Venturimeter head in m of water h = p x105/ g = (p1 – p2) x105/ (g) (m)
H = Supply head in m = Input total head in m H = P x105/ (g )
11.5.1.2. Output (Brake) Power, OP = 2 N T / 60 (W)
Where N = Turbine speed in RPM., T = W Re g = Torque in N-m
Brake drum dia D =0.2m ,Rope dia d =0.015mEffective radius of Brake Drum, Re = (D/2 )+d) = 0.115m
Weight of rope & hanger = 1kg.Brake drum Net load W =(W1 + weight of rope hanger) –W2 kg = ( 5+1) – 2 = 4 kg
11.5.1.3 Efficiency of turbine, = (OP/IP ) x 100%
Input Power = g QH in kW
Where = Density of water = 1000 kg/m3
g = Acceleration due to gravity (9.81m/sec2) Q = Discharge in m3/sec.
H = Supply head in meters.
2 NReW x 9.81
Brake Power =------------------- kW
60000
Output
Efficiency = ------------- x 100%
34
Input
Where N = Turbine speed in RPM.
T= Torque in kgm, (effective radius of the brake drum in meters (0.165m)
Sl.
No
.
Pressure
gauge
reading
Pressure
Gauge
Reading
Orifice
meter
head
Disch
arge Speed
Weight
on
hanger
Spring
balance
reading
Net load Output Input Efficiency
P.S kg/cm2 h Q N W1 W2 W OP IP
kg/cm2 p1 p2 dp m m3/
sec
RPM kg kg kg kW kW %
1
2
3
4
5
FRANCIS TUBRINE
Orificemeter Head h in m of water h=(p1-p2)x 10m of water
Effective Radius of = (D/2 + t) Discharge Q = Kh (h in m of water)
Input power IP = x H x Q kW (H in m of water) Brake drum Re = 0.115m
Brake Drum net load W = (W1 + weight of rope & hanger) – W2 kg
Inlet diameter of Orificemeter: 80mm , Orificemeter diameter : 60 mm
Meter constant for Orificemeter: K=9.11 x10-3 h , h in m of water
Brake drum dia D=0.23m Rope Dia t = 0.015m
Input total head H in m of water = Pressure gauge reading in kg/cm2x 10
11.5.3 GRAPH: Discharge Vs Speed, output, Input, Efficiency
35
11.6 POST-LAB QUESTIONS
11.6.1 What is the energy conversion across the turbine?
11.6.2 Differentiate stator and rotor vanes or blades?
11.6.3 What do you meant by runner?
11.6.4 What is the pressure inside the turbine casing?
11.6.5 What do you meant by reaction turbine?
11.6.6 List out different component efficiencies used?
11.6.7 What do you meant by brake power?
11.6.8 What do you meant by cavitation?
11.6.9 What are the methods to avoid cavitation?
11.7 INFERENCES
11.8 RESULTS
The performance test on Francis turbine is conducted.
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EXPERIMENT No.12
PERFORMANCE TEST ON CENTRIFUGAL PUMP
12.1 Aim: To study the performance of centrifugal pump at constant speed.
12.2 EQUIPMENTS Required: Centrifugal pump test rig, Stop watch
12.3 PREPARATION
12.3.1 THEORY
A centrifugal pump is a rotodynamic pump that uses a rotating impeller to increase the pressure and flow rate of a fluid. Centrifugal pumps are the most common type of pump used to move liquids through a piping system. The fluid enters the pump impeller along or near to the rotating axis and is accelerated by the impeller, flowing radially outward or axially into a diffuser or volute chamber, from where it exits into the downstream piping system. Centrifugal pumps are typically used for large discharge through smaller heads. The priming of the pump is necessary in order to get discharge. The priming can be done by filling the impeller with water in order to provide continuous water column from the sump water level to pump.
12.3.2 PRE-LAB QUESTIONS
12.3.2.1 What is the purpose of pump?
12.3.2.2 What do you meant by centrifugal force?
12.3.2.3 What is type of flow in centrifugal pump?
N.B.: If the pump is not delivering water output (discharge), prime the pump and then start the motor.
12.4.1 Ensure the complete opened position of delivery valve. 12.4.2 Start the pump power supply.12.4.3 Vary the flow rate (discharge) by closing the delivery valve.12.4.4 Note down pressure gauge reading for 0.5 kg/cm2 and vacuum gauges readings.12.4.5 Measure height of the pressure gauge above the vacuum gauge. (Z)12.4.6 Note down time taken (t) for ‘h’ cm rise of water (10 cm) in collecting tank. 12.4.7 Note down the time taken (T) for ‘n’ revolutions for energy meter (3 rev) disc. 12.4.8 Repeat the procedure for 0.5, 1.0, 1.5, 2.0, 2.5 kg/cm2 in the pressure gauge
reading by gradual closing of delivery valve.12.4.9 Switch off the power supply after opening the delivery valve completely.
12.5 OBSERVATIONS
12.5.1 FORMULAE / CALCULATIONS
12.5.1.1. Total head, H = [ P + (V/760) ] x 105 / ( g ) + Z (m)Where P – Pressure gauge reading in kg/cm2,
V- Vacuum gauge reading in mm Hg, Z – Datum level between Pressure gauge and Vacuum gaugeg = Gravitational acceleration = 9.81 m2/s = Density of fluid (water) = 1000 kg/m3
N.B.: For V, 1 mm Hg/ 760 = 1 bar For P, 1 bar = 1 kg/ cm2
12.5.1.2. Discharge, Q = (A h )/ t (m3/sec)
Where A- Collecting tank area = l x b in m2, t - time for 10 cm rise of water level in the collecting tank (sec)h – Rise of water level in the collecting tank = 0.10 m
12.5.1.3. Output in kW, OP = g Q H / 1000
12.5.1.4. Input in kW, IP = (‘n’ rev of energy meter x 3600 x Efficiency of motor) / (Energy meter constant in Rev/kW-hr x Time for n
revolutions) = (n x 3600 x m ) / (Ec x T)
Where Ec = Energymeter constant in Rev /kW – hr = 1200 rev / kWh n = Number of revolution taken in energymeter
T = Time required to complete ‘n’ revolution in sec m = Efficiency of motor = 0.8
12.5.1.5. Efficiency of Pump, = (Output / Input) x 100 %
12.5.2 TABULATIONMotor Efficiency m : 0.8, Energy meter Constant = 200 rev/kWh, Hight between the pressure gauges, Z = 500mm, Collecting tank Area, A = 0.5 x 0.5 m2
38
Sl.No.
Pressure gauge
reading
P
Vacuum gauge
reading
V
Time for 10cm rise of water
t
Time for 3 rev of energy
meter discT
Head
H
Discharge
Q
Input power
IP
Output power
OP
Efficiency
kg/cm2 mm of hg Sec Sec m m3/sec kW kW %
1. 0.52 13. 1.54 25 2.5
12.5.3 GRAPH: Disharge Vs Head, Output, Efficiency
12.6 POST-LAB QUESTIONS
12.6.1 What is the energy conversion across the impeller?
12.6.2 List out the different heads used in centrifugal pump?
12.6.3 What do you meant by manometric head?
12.6.4 What is the pressure inside the pump casing?
12.6.5 What do you meant by manometric efficiency?
12.6.6 What do you meant by cavitation, its effects and its prevention?
12.7 INFERENCES:
12.8 RESULTThe performance test on centrifugal pump is completed and the performance characteristics are studied.
39
EXPERIMENT No.13
IMPACT OF JET OF WATER ON VANES
13.1 AIM: To determine the coefficient of impact of water jet on different vanes
13.2 EQUIPMENTS REQUIRED: Jet on vane apparatus, Weighing machine, Flat
vane, Flat vane with oblique impact, Conical vane, stop watch
13.3 PREPARATION
13.3.1 THEORY
Water turbines are widely used throughout the world to generate power. In the type of water turbine referred to as a Pelton wheel, one or more water jets are directed tangentially on to vanes or buckets that are fastened to the rim of the turbine disc. The impact of the water on the vanes generates a torque on the wheel, causing it to rotate and to develop power. Although the concept is essentially simple, such turbines can generate considerable output at high efficiency. To predict the output of a Pelton wheel, and to determine its optimum rotational speed, we need to understand how the deflection of the jet generates a force on the buckets, and how the force is related to the rate of momentum flow in the jet. In this experiment, we measure the force generated by a jet of water striking a flat plate or a hemispherical cup, and compare the results with the computed momentum flow rate in the jet.
40
Fig. Impact of Water jet on flat vane
13.3.2 PRE-LAB QUESTIONS
13.3.2.1 What is the water jet?
13.3.2.2 What is the effect of water jet on vanes?
13.3.2.3 What do you meant by impact?
13.3.2.4 List out different types of vanes?
13.4 PROCEDURE
13.4.1 Switch on the power supply.
13.4.2 Open the gate valve and note down the reading from the weight balance.
13.4.3 Then note the time for ‘h’ m rise in collecting tank.
13.4.4 Repeat the procedure for different gate valve openings.
13.4.5 Take readings for different vanes and nozzles also.
13.5 OBSERVATIONS
13.5.1 FORMULAE / CALCULATIONS
13.5.1.1 Actual discharge, Q = Volume of collecting tank/ time taken = A h / t
Where, A - Area of collecting tank = length x breadth
h - Water level rise in the collecting tank = 10 cm
t - Time taken for ‘h’ cm rise of water in the tank in sec
13.5.1.2 Theoretical force Ft = ( AN V 2)/ g
Density of water, = 1000 kg/m3
Area of nozzle, AN = d2/4
Gravity, g = 9.81 m/s2
41
13.5.1.3 Velocity V = Q/ [Cc. AN]
13.5.1.4 Co-efficient of Impact, Ci = Fa / Ft Where Fa = Actual force acting on the Disc shown from Dial Gauge.
13.5.2 TABULATION
Measuring area in tank = 0.5 x 0.4 m2
Dia of jet = 15mm
Type of vane = Flat vane / Conical vane
Co-efficient of Contraction, Cc = 0.97
Sl.
No. Type of Vane
Time for 10 cm
rise of water
(sec)
Actual
force,
Fa in kg
Theoretical
Force,
Ft in kg
Co-efficient
of impact,
Ci
1
2
3
4
5
6
13.5 POST-LAB QUESTIONS
13.6.1 How do you compare different vanes?
13.6.2 What do you meant by co-efficient of impact?
13.6.3 How do you measure the force of the jet?
13.6.4 How do you measure actual flow rate?
13.6.5 How do you measure theoretical flow rate?
42
13.7 INFERENCES
13.8 RESULT
The co-efficient of impact of the given vane = ___________
EXPERIMENT No.14
FLOW VISUALIZATION - REYNOLDS APPARATUS
14.1 AIM: To demonstrate the flow visualization – laminar or turbulent flow.
14.2 EQUIPMENTS REQUIRED: Reynolds Experimental set up, stop watch
14.3 PREPARATION
14.3.1 THEORY
The flow of real fluids can basically occur under two very different regimes namely laminar and turbulent flow. The laminar flow is characterized by fluid particles moving in the form of lamina sliding over each other, such that at any instant the velocity at all the points in particular lamina is the same. The lamina near the flow boundary move at a slower rate as compared to those near the center of the flow passage. This type of flow occurs in viscous fluids , fluids moving at slow velocity and fluids flowing through narrow passages. The turbulent flow is characterized by constant agitation and intermixing of fluid particles such that their velocity changes from point to point and even at the same point from time to time. This type of flow occurs in low density fluids, flow through wide passage and in high velocity flows.
43
Fig. Reynolds Experimental Set-up
Reynolds conducted an experiment for observation and determination of these regimes of flow. By introducing a fine filament of dye in to the flow of water through the glass tube ,at its entrance he studied the different types of flow. At low velocities the dye filament appeared as straight line through the length of the tube and parallel to its axis, characterizing laminar flow. As the velocity is increased the dye filament becomes wavy throughout indicating transition flow. On further increasing the velocity the filament breaks up and diffuses completely in the water in the glass tube indicating the turbulent flow. There are two different types of fluid flows laminar flow and Turbulent flow. The velocity at which the flow changes laminar to Turbulent is called the ‘Critical Velocity’.
Fig. Types of internal (pipe) flow
Reynolds number determines whether any flow is laminar or Turbulent. Reynolds
number corresponding to transition from laminar to Turbulent flow is about 2,300.
14.3.2 PRE-LAB QUESTIONS
14.3.2.1 What do you meant by fluid?
14.3.2.2 What are the types of flow?
14.3.2.3 Define Reynolds number?
14.3.2.4 What is laminar flow?
14.3.2.5 What is turbulent flow?
44
14.4 PROCEDURE
1. Switch on the power supply. Adjust the water inflow slowly by flow control
valve ( delivery valve).
2. Inject a filament of dye into the water stream by opening the value from dye
tank.
3. When the flow is laminar, the colored stream of dye does not mix with the
stream of water and is apparent long the whole length of the pipe. Increase the
velocity of the stream gradually by opening the flow control valve, to see the
turbulent flow. The stream of dye begins to oscillate and then diffused. This
velocity of water in the pipe is ‘Critical Velocity’.
14.5 OBSERVATIONS
14.5.1 FORMULAE / CALCULATIONS
Discharge, Q = (A h )/ t (m3/sec)
Where A- Collecting tank area = l x b in m2, t - time for 10 cm rise of water level in the collecting tank (sec)h – Rise of water level in the collecting tank = 0.10 m
Reynolds number for pipe flow, Re = ( V D)/
Where V= Velocity of the fluid (m/s),
D= diameter of the pipe (m)
= Kinetic viscosity of the fluid (m2/s)
14.5.2 TABULATION
Internal plan area of collecting tank = 0.3 x 0.3m2
Diameter of pipe D = 32 mm , Kinematics viscosity of fluid (water) = 1.01 x 10-6 m2/sec
Sl. No. Time taken for 10 cm rise t sec
Discharge Qm3/sec
Velocity V m/s
Reynolds number Re
Remarks (Laminar/ Turbulent flow)
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14.5 POST-LAB QUESTIONS
45
14.6.1 What do you meant by stream and streak lines?
14.6.2 Mention the Reynolds no for laminar and turbulent flow?
14.6.3 What do you meant by steady and unsteady flow?
14.6.4 What do you meant by path line?
14.6.5 What do you meant by uniform and non-uniform flow?
