Turbo Machines Lab manual

1

AKSUM UNIVERSITY

COLLEGE OF ENGINEERING AND TECHNOLOGY

DEPARTMENT OF MECHANICAL ENGINEERING

TURBO MACHINES & MACHINES LAB MANUAL

Compiled by

A.Syed Bava Bakrudeen,

Associate Professor,

Mechanical Department

Aksum University, Ethiopia

LIST OF EXPERIMENTS

1. Pelton turbine

2. Reaction turbine (Efficiency Test)

3. Axial turbine

4. Reaction turbine (Velocity Test)

5. Centrifugal pump

6. Piston pump

7. Gear pump

8. Axial fan

9. Radial fan

10. Tubular heat exchanger (Parallel flow)

11. Tubular heat exchanger (Counter flow)

12. Air compressor

13. Multipurpose air duct.

2

1. PELTON TURBINE

THEORY: Pelton wheel turbine is an impulse turbine, which is used to act on

high loads and for generating electricity. All the available heads are classified in to

velocity energy by means of spear and nozzle arrangement. Position of the jet

strikes the knife-edge of the buckets with least relative resistances and shocks.

While passing along the buckets the velocity of the water is reduced and hence an

impulse force is supplied to the cups which in turn are moved and hence shaft is

rotated.

AIM: To determine the efficiency of the given pelton turbine using constant

volume method.

APPARATUS REQUIRED:

1. Pelton turbine setup.

2. Collecting tank

3. Stop watch

4. Meter scale

FORMULA:

1. Mechanical Power = Pmech = Mω = Mx 2xπxn/60

M=Moment in N/m.

n = Speed in rpm.

2. Hydraulic Efficiency = Phyd = pxV

p = Pressure in N/m2.

V= Volume flow rate in m3/Sec (Discharge)

(1liter =1000 cm3 = 0.001 m

3) (1bar = 1x10

5 N/m

2)

(1liter/min =16.67x10-6

m3/Sec) (1liter = 1kg)

3. Efficiency η = Pmech x100/ Phyd

PROCEDURE:

1. Tare values in the system diagram

2. Select "Measurement Diagram" in the program.

3

3. Enable new series of measurements. Make settings for the measurements file.

4. Open the brake fully with the adjusting screw(8).

5. Open valve (4) fully to create the maximum volume flow.

6. Change the pump to "Pressure Control" and specify a control pressure. This

corresponds to the constant height difference of a real turbine to the storage

lake. With the pumps used it is possible to achieve pressures up to about 3.8 bar

with the needle nozzle fully opened. Higher pressures are only possible by

throttling the volume flow. We recommend using this pressure range. The

measured values obtained by GUNT reach 4 bar.

7. Wait until the operating point is reached. In this case the observation of the

water jet from the blade is particularly relevant for the subsequent explanation.

Then record measurements (the current measurement data set is written to the

measurements file). The program is now ready for the next measurement.

8. a) Use the adjusting screw on the brake (8) to increase the brake torque. The

turbine's rotational speed and torque change. The change depends on the desired

number of measurement points. Meaningful characteristics are often obtained

with 5 to 6 measurement points.

b) The volume flow can still be varied at each braking position. To do this, the

nozzle is closed gradually from the initial fully open position.

9. Repeat steps 7 and 8 as often as needed until the turbine's rotational speed has

fallen to zero and tabulate the value.

GRAPHS:

i Torque (M) Vs Speed (n) (Ordinary Graph)

RESULT:

1. Maximum efficiency of the turbine = ……………………… %

2. Speed corresponding to maximum efficiency = ………….. rpm

3. Power corresponding to maximum efficiency = …………..W.

4. Discharge corresponding to maximum efficiency = ………….. L/min.

4

TABULATION:

S.

No

Torque

(M)

Ncm

Pressure

(p) bar

Speed(n)

1/min

Volume

Flow V

(L/min)

Mechanical

Power

(Pmech)

W

Hydraulic

Power

(Phyd)

W

Efficiency

(η) %

1. 22.8 3.0 2260 22 53.96 110 49.05

MODEL CALCULATIONS:

1. Phyd = pxV = 3x105x22x0.001/60 = 110 W

2. Pmech = Mω = Mx 2xπxn/60 = 22.8x10-2

x2xπx2260/60 = 53.96 W.

3. η = Pmech x100/ Phyd= 53.96/110x100 =49.05

Fig1.Pelton Turbine

5

2. REACTION TURBINE (EFFICIENCY TEST)

THEORY: Reaction turbines are acted on by water, which changes pressure as it

moves through the turbine and gives up its energy. They must be encased to

contain the water pressure (or suction), or they must be fully submerged in the

water flow. Newton's third law describes the transfer of energy for reaction

turbines.Most water turbines in use are reaction turbines and are used in low

(<30 m or 100 ft) and medium (30–300 m or 100–1,000 ft) head applications. In

reaction turbine pressure drop occurs in both fixed and moving blades. It is largely

used in dam and large power plants

AIM: To determine the efficiency of the given reaction turbine under constant

pressure condition.

APPARATUS REQUIRED:

1. Francis turbine setup.

2. Collecting tank

3. Stop watch

4. Meter scale

FORMULA:

1. Mechanical Power = Pmech = Mω = Mx 2xπxn

M=Moment in N/m.

n = Speed in rpm.

(1liter =1000 cm3 = 0.001 m

3) and (1bar = 1x10

5 N/m

2)




3. Efficiency = Pmech x100/ Phyd

PROCEDURE:




https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Newton.27s_third_law

6

4. Open the brake fully with the adjusting screw(8).

5. Open valve (4) fully to create the maximum volume flow.

6. Change the pump to "Pressure Control" and specify a control pressure. Pressure

up to about 3 bar can be achieved with the pump used. Higher pressure cannot

be maintained over the entire operating range due to increasing the volume

flow.

7. Wait until the operating point is reached. Then record measurements (the

current measurement data set is written to the measurements file). The program

is now ready for the next measurement.

8. The brake torque is increased via the adjusting screw (7). The torque is varied

depend up on the desired number of measurement points. Meaningful

characteristics are often obtained with 10 measurement points.

9. Repeat steps 7 and 8 as often as needed until the turbine's rotational speed has

fallen to zero and tabulate the value.

GRAPHS:


RESULT:





7

TABULATION:

S.

No

Torque

(M)

Ncm

Pressure

(p) bar

Speed

(n)

1/min

Volume

Flow

(V)

(L/min)

Mechanical

Power

(Pmech)

W

Hydraulic

Power

(Phyd)

W

Efficiency

(η) %

1. 7.9 3 8848 28.3 73.2 141.5 51.73

MODEL CALCULATIONS:

1. Phyd = pxV = 3x105x28.3x0.001/60 = 141.5 W

2. Pmech = Mω = Mx 2xπxn/60 = 7.9x10-2

x2xπx8848/60 = 73.2 W.

3. η = Pmechx100/Phyd = (73.2/141.5)x100 = 51.73%.

Fig.2 Reaction Turbine

8

3. AXIAL TURBINE

THEORY: If the water flows parallel to the axis of the rotation of the shaft, the

turbine is known as axial flow turbine. If the head at the inlet of the turbine is the

sum of pressure energy and kinetic energy and during the flow of water through

runner a part of pressure energy is converted into kinetic energy, the turbine is

known as reaction turbine.

AIM: To determine the efficiency of the given axial flow axial turbine under

constant speed and pressure.

APPARATUS REQUIRED:

1. Kaplan turbine setup.

2. Collecting tank

3. Stop watch

4. Meter scale

FORMULA:

1. Mechanical Power = Pmech = Mω = Mx 2xπxn

M=Moment in N/m.

n = Speed in rpm.

(1 Millibar = 0.001 bar)




3. Efficiency = Pmech x100/ Phyd

PROCEDURE:




4. Open the brake fully with the adjusting screw.

5. Open control valve V1 to obtain maximum upstream pressure

6. Switch on the pump.

9

7. Wait until the operating point is established. Then record the measuring values

(the current measurement data set is written to the measurement file). The

program is now ready for next measurement.

8. Increase the turbine torque using the adjusting screw. The torque variation is

depending on the number of measuring points chosen. Meaningful

characteristics are often obtained with 5 to 6 measuring points.

10. Repeat steps 7 and 8 as often as needed until the adjusting screw is at the stop.

11. Save the measurement file.

12. Alternatively, further series of measurement can be recorded at reduced

upstream pressure. To do so, reduce the pressure at control valve V1 and repeat

the measurement from step 6 onwards.

GRAPHS:


RESULT:





10

TABULATION:

S.

N

o

Torque (M)

(Ncm)

Pressure

(p)

(mbar)

Speed

(n)

(1/min)

Volume

Flow (V)

( L/min)

Mechanical

Power

(Pmech) (W)

Hydraulic

Power

(Phyd) (W)

Efficiency

(η) (%)

1. 34.8 890 3490 125 127.2 185.42 68.6

MODEL CALCULATIONS:

1. Phyd = pxV = 0.89x105x125x0.001/60 = 185.42 W

2. Pmech = Mω = Mx 2xπxn/60 = 34.8x10-2

x2xπx3490/60 = 127.2 W.

3. η = Pmechx100/ Phyd = (127.2 /185.42)x100 =68.6%

Guided Wheel Impeller

Fig3. Axial Turbine

11

4. REACTION TURBINE ( VELOCITY TEST)

THEORY:

Absolute velocity (c): The velocity that the water jet has at the outlet from the

turbine compared to the environment.

Circumferential velocity (u): The velocity of the impeller. Since this depends on

the flow of the exiting water jet, it is the velocity at the diameter of the outlet

nozzle.

Relative velocity (w): It is the velocity corresponds to the velocity of the flow

relative to the nozzle. It is calculated by adding circumferential velocity (u) and

absolute velocity (c).

AIM: To find the relative velocity and circumferential velocity of the francis

turbine and determine the correlation between relative velocity and volume flow

rate of the given turbine.

APPARATUS REQUIRED:

1. Francis turbine setup.

2. Collecting tank

3. Stop watch

4. Meter scale

FORMULA:

1. Circumferential Velocity = u2 = πdn/60

d= Diameter of nozzle = 52 mm.

n = Speed in rpm.

2. Relative velocity = W2 = √(((Pux2)/ρ)+ u22)

pu = Net Pressure in N/m2.

ρ = Density of water = 1000 kg/m3.

PROCEDURE:




4. Loosen off adjusting screw to open the brake.

12

5. Let the pump run to100% power (This ensures that air in the flow section does

not affect the measured values).

6. Wait until the operating point is reached. Then record measurements (the

current measurement data set is written to the measurements file). The program

is now ready for the next measurement.

7. The capacity of the pump is retracted one step. The power is varied depending

on the desired number of measurement points. Meaningful characteristics are

often obtained with 5 to 6 measurement points

8. Repeat steps 6 and 7 as many times as needed until there is no more rotational

speed at the turbine.

9. Save the measurements file.

GRAPHS:

i Relative Velocity Vs Volume flow.

OBSERVATION: Diameter of nozzle (d) = 52 mm.

RESULT:

1. The maximum relative velocity = ________________ m/s.

2. Net pressure at inlet = _____N/m2 at maximum relative velocity

3. The relative velocity is directly proportional to volume flow rate.

13

TABULATION:

S.

No

Pressu

re (p)

bar

Speed

(n)

1/min

Volume

Flow (V)

L/min

Circumferential Velocity

(u2) (m/s)

Relative velocity

(W2)

(m/s)

1 3.18 19894 46.7 54.16 59.8

MODEL CALCULATIONS:

1. u2 = πdn = πx0.052x19894/60 = 54.16 m/s

2. W2 = √(((Pux2)/ρ)+ u22) = √(((3.18x10

5x2)/1000)+ 54.2

2)= 59.8 m/s

W

C

u2

Fig 4. Velocities in reaction turbine

C- Absolute velocity at outlet (m/s)

W= Relative velocity (m/s)

u2= Circumferential velocity (m/s)

14

5. CENTRIFUGAL PUMP

THEORY: Centrifugal pumps are a sub-class of dynamic axisymmetric work-

absorbing turbomachinery. Centrifugal pumps are used to transport fluids by the

conversion of rotational kinetic energy to the hydrodynamic energy of the fluid

flow. The rotational energy typically comes from an engine or electric motor. The

fluid enters the pump impeller along or near to the rotating axis and is accelerated

by the impeller, flowing radially outward into a diffuser or volute chamber

(casing), from where it exits.

AIM:

To determine the characteristics of a centrifugal pump under constant speed

and varying the discharge method.

APPARATUS REQUIRED:

1. Centrifugal pump setup.

2. Collecting tank

3. Stop watch

4. Meter scale

FORMULA:

1. Hydrualic Power Phyd= ∆P xV in Watts.

∆P = P2-P1=Difference in pressure in N/m2.

V = Volume flow rate (Discharge or Actual discharge) in m3/sec.

(1L/min = 0.001 m3/min) (1 min = 60 sec) (1 L/min = 16.66x10

-6 m

3/sec)

(1bar = 1x105 N/mm

2)

2. Head H = ∆P/ ρxg in meter

ρ= Density of water=1000 kg/m3.

g = Specific gravity =9.81 m/s2.

(1L/min = 0.001 m3/min)

3. Efficiency = Phyd/Pel.

Phyd = Hydrualic Power in W

Pel = Electrical Power in W

4. Specific speed Ns = n√V/ (H)3/4

(unitless)

https://en.wikipedia.org/wiki/Turbomachinery

https://en.wikipedia.org/wiki/Volute_(pump)

15

PROCEDURE:

1. Bleed the pump demonstrator.

2. Open valve V2 fully.

3. Use the Tare button to calibrate to zero.

4. Leave pump to turn to n=_________ rev/min.

5. Record measuring values of the suction pressure (P1), the pump outlet

pressure (P2), hydraulic and electrical power and volume flow (V).

6. Reduce the volume flow bit by bit by gradually closing valve V2 and take

the measurements according to point 5.

7. Repeat steps 5 and 6 until the volume flow is completely throttled.

8. Additional curves can be recorded with different rotational speed.

GRAPHS:

i Volume flow rate (V2) Vs Hydraulic Power (Phyd)

ii. Volume flow rate (V2) Vs Electrical Power (Pel)

iii. Volume flow rate (V2) Vs Efficiency (η)

RESULT:

1. Maximum efficiency of the pump = ……………………… %

2. Discharge corresponding to maximum efficiency = ………….. m3/Sec.

3. Input power corresponding to maximum efficiency = …………..W.

4. Head corresponding to maximum efficiency = ………….. meter.

16

TABULATION: S

.

N

o

Rotation

al speed

(n) (Rev

/ min)

Volume

flow

(V)

(L/min)

Suction

Pressure

(P1) (bar)

Discharg

e

Pressure

(P2) (bar)

Temp

eratu

re (T)

(0C)

Electric

al

Power(

Pel)

(W)

Pressure

Differen

ce (∆P)

(bar)

Head

(H)

(M)

Hydrau

lic

Power

(Phyd)

(W)

Efficien

cy (η)

(%)

Specifi

c speed

(Ns)

-

1 3300 25.2 -0.0056 0.865 25 343 0.8706 8.875 36.56 10.66 13.15

MODEL CALCULATIONS: 1. Phyd= ∆P xV = 0.8706X10

5 x 25.2x0.001/60 = 36.56

2. Head H = ∆P/ ρxg = 0.8706X105/(1000X9.81) = 8.875

3. Efficiency = Phydx100/Pel = 36.56x100/343=10.66%.


= 3300 x √(25.2x0.001/60 )/ (8.875)3/4

= 13.15

Fig6. Schematic diagram of centrifugal pump

EI1 Energy input Pel of the pump B1 Water tank

FI1 Volume flow P1 Centrifugal pump

PI1 Pressure p1 upstream of the pump V1 Valve to throttle the suction side

PI2 Pressure p2 downstream of the pump V2 Valve to throttle the pressure side

TI1 Water temperature V3 Outlet valve

17

6. PISTON PUMP

THEORY: Reciprocating is a positive displacement pump in which the liquid is

sucked and then it is actually pushed or displaced due to the thrust exerted on it by

a moving member, which results in lifting the liquid to the required height These

pumps usually have one or more chambers which are alternatively filled with the

liquid to be pumped and then emptied again As such the discharge of liquid

pumped by these pumps almost wholly depends on the speed of the pump. It is

widely used in Automobile Service Stations and Chemical Industries.

AIM:

To conduct the performance test on piston pump

APPARATUS REQUIRED:

1. Piston pump setup.

2. Collecting tank

3. Stop watch

4. Meter scale

FORMULA:

1. Hydrualic Power Phyd= ∆P xV in Watts.




-6 m

3/sec)

(1bar = 1x105 N/mm

2.)

2. Indexed Work Pind = Wind x60 / n

Pind = Indexed work in joules.







5. Qt = Theoretical discharge = 2xLxAxn/60 in m3/sec

18

Where Area of the piston A= (π/4) x d2 in m

2.

d= diameter of the piston =32mm

L= Stroke length of the piston = 15mm

n = Motor speed in rpm.

6. Percentage of slip = (Qt – V)x100/Qt.

PROCEDURE:

1. Open the throttle valve and suction-side throttle valve fully.

2. Check whether the air cushion in the air vessel fills approx. half the useable volume.

3. Tare values in the system diagram.

4. Set overflow valve to maximum set point.

5. Switch on the piston pump, set speed to 100min-1

.

6. Gradually close the throttle valve (7) until the supply pressure p2 is approx.

2 bars.

7. Save a screenshot of the system diagram in a file.

GRAPHS:




OBSERVATIONS: d= diameter of the piston =32mm

L= Stroke length of the piston = 15mm

RESULT:





5. Percentage of slip corresponding to maximum efficiency=………..

19

TABULATION: S

.

N

o

Rotat

ional

speed

(n)

(Rev

/ min)

Volume

flow

(V)

(L/min)

Suctio

n

Pressu

re (P1)

(N/m2)

Disch

arge

Pressu

re (P2)

(N/m2)

Elect

rical

Powe

r (Pel)

(W)

Index

ed

powe

r (Pin)

(W)

Index

work

in (J)

Wind

Pressur

e

Differe

nce

(∆P)

(bar)

Hea

d

(H)

(M)

Hydrau

lic

Power

(Phyd)(

W)

Effic

ienc

y (η)

(%)

Theoretica

l

Discharge

(Qt) (m3/s)

% of

Slip

1 94 1.03

-0.08 1.93 55.2 3.6 2.23 2.01 20.5 3.45 6.25 3.8 x10-5

53.4

MODEL CALCULATIONS:

1. Phyd= ∆P xV = 2.01x105x1.03x0.001/60 =3.45 W

2. Wind = Pind x60 / n = 3.6x60/94=2.23 J

3. H = ∆P/ (ρxg) =2.01x105/ (1000x9.81) = 20.5 m

4. η=Phyd x100/Pel = 3.45x100/55.2 = 6.25%

5. Qt = 2xLxAxn/60 = 2x 0.015 x(π/4)x 0.0322x 94/60 = 3.8 x10

-5 m

3/Sec.

6. % slip = (Qt – V)x100/Qt.

= (3.8x10-5

– (1.03x16.67x10-6

))x100/ 3.8x10-5

= 53.46%

Fig 6. Piston pump

20

7. GEAR PUMP

THEORY: A rotary gear pump consists essentially of two intermeshing spur gears

which are identical and which are surrounded by a closely fitting casing. One of

the pinions is driven directly by the prime mover while the other is allowed to

rotate freely. The fluid enters the spaces between the teeth and the casing and

moves with the teeth along the outer periphery until it reaches the outlet where it is

expelled from the pump. Each tooth of the gear acts like a piston or plunger of on

reciprocating pump and hence the pump can be termed a positive displacement

pump. Gear pump is widely used for cooling water and pressure oil to be supplied

for lubrication to motors, turbine, machine tools etc.

AIM:

To determine the characteristics of a gear pump under constant speed and

varying the discharge and obtain the best-driven conditions by drawing the

performance curves.

APPARATUS REQUIRED:

1. Gear pump setup.

2. Collecting tank

3. Stop watch

4. Meter scale

FORMULA:

1. Hydrualic Power Phyd= HxV in Watts.




-6 m

3/sec)

(1bar = 1x105 N/mm

2)







21


(unit less)

PROCEDURE:

1. Bleed the pump demonstrator.

2. Open valve V2 fully.

3. Use the Tare button to calibrate to zero.

4. Leave pump to turn to n=_________ rev/min.

5. Record measuring values of the suction pressure (P1), the pump outlet

pressure (P2), hydraulic and electrical power and volume flow (V).

6. Reduce the volume flow bit by bit by gradually closing valve V2 and take

the measurements according to point 5.

7. Repeat steps 5 and 6 until the volume flow is completely throttled.

8. Additional curves can be recorded with different rotational speed.

GRAPHS:




RESULT:





5. Specific speed corresponding to the flow =……………………

22

TABULATION:

S

.

N

o

Rotati

onal

speed

n

(Rev

/ min)

Volu

me

flow

V

(L/mi

n)

Suctio

n

Pressu

re (P1)

(N/m2)

Discha

rge

Pressu

re (P2 )

(N/m2)

Temp

eratu

re

(T1)

(0C)

Electr

ical

Power

(Pel)

(W)

Pressure

Differen

ce (∆P)

(bar)

Head

(H)

(M)

Hydraulic

Power

(Phyd) (W)

Effici

ency

(η)

(%)

Speci

fic

speed

(Ns)

-

1 600 8.7 -0.09 0.98 23.9 147 1.07 17.33 15.515 10.55 0.85

MODEL CALCULATIONS:

1. Phyd= ∆PxV = 1.07x105x 8.7x0.001/60 =15.515 W

2. H = ∆P/ (ρxg) = 1.07x105/(1000x9.81) = 17.33m

3. Efficiency = Phyd x100/Pel = 15.515x100/147= 10.55%


= 600 x √(8.7x16.67x10-6

)/ (17.33)3/4

= 0.85

Figure. 7 Gear Oil Pump

23

8. AXIAL FAN

THEORY: An axial fan is a type of a compressor that increases the pressure of the

air flowing through it. The blades of the axial flow fans force air to

move parallel to the shaft about which the blades rotate. In other words, the flow is

axially in and axially out, linearly, hence their name. The design priorities in an

axial fan revolve around the design of the propeller that creates

the pressure difference and hence the suction force that retains the flow across the

fan. The main components that need to be studied in the designing of the propeller

include the number of blades and the design of each blade. Their applications

include propellers in aircraft, helicopters, hovercrafts, ships and hydrofoils. They

are also used in wind tunnels and cooling towers.

AIM:

To identifying characteristics data, to investigate of typical dependencies

and recording the fan characteristics are three aims of our experiment

APPARATUS REQUIRED:

1. Axial fan apparatus setup.

2. Computer system.

FORMULA:

1. ρ =ρ0 x (T0/T1)x(pamb/p0)

ρ0 = 1.293 kg/m3 is air density at reference temperature at T0 = 273.15 K.

p0= 1,013 mbar,in

T1= Temperature of intake air in Kelvin. (1K = X0C +273.15)

2. Air velocity c = √((2/ρ) x dpN) in m/sec.

dpN = Dynamic pressure in N/m2.

3. Suction volume flow Vs = c x A in m3/sec.

A = Area of the intake pipe in m2.

A= (π/4) x d2 where d = diameter of intake pipe = 110 mm.

4. Power hydraulic Phyd = dpF x Vs in Watts

5. Efficiency = Phydx100/Pel.



https://en.wikipedia.org/wiki/Gas_compressor

https://en.wikipedia.org/wiki/Pressure

https://en.wikipedia.org/wiki/Axial_flow

https://en.wikipedia.org/wiki/Parallel_(geometry)

https://en.wikipedia.org/wiki/Propeller

https://en.wikipedia.org/wiki/Pressure

https://en.wikipedia.org/wiki/Force

https://en.wikipedia.org/wiki/Aircraft

https://en.wikipedia.org/wiki/Helicopters

https://en.wikipedia.org/wiki/Hovercrafts

https://en.wikipedia.org/wiki/Ships

https://en.wikipedia.org/wiki/Hydrofoils

https://en.wikipedia.org/wiki/Wind_tunnels

https://en.wikipedia.org/wiki/Cooling_towers

24

PROCEDURE:

1. Tare values and enter the ambient pressure in the system diagram.

2. Select ‘Measurement Diagram’ in the program.

3. Enable new series of measurements. Make any setting for the measurement

file.

4. Switch on radial fan, select speed of ____ %

5. For the first measurement, close the throttle valve completely.

6. Wait until the displayed measurement stable. Then record measurements


program is now ready for the next measurement.

7. Open the throttle valve a little bit. The position of the throttle valve is

dependent on the desired number of measurement points. Meaningful

characteristics are often obtained with 5 to 6 measurement points.

8. Repeat the steps until throttle valve is fully open.

9. Repeat the steps with the newly selected speed of 100%, save the

measurements and plot the characteristics curve using system.

OBSERVATION:

Diameter of intake pipe ……110…….. mm

GRAPHS:

Fan speed Vs Efficiency

RESULT:


2. Hydraulic power corresponding to maximum efficiency = …………..W.

3. Electrical power corresponding to maximum efficiency = …………..W

4. Air velocity corresponding to maximum efficiency = …………..m/s

5. Suction volume flow corresponding to maximum efficiency = …….. m3/Sec

25

TABULATION:

S.

N

o

Fan

speed

(n)

(Rev/

min)

Differe

ntial

Pressu

re flow

(dpN)

(N/m2)

Pressu

re

increas

e (dpF )

(N/m2)

Tempe

rature

of

intake

air (T1)

(0C)

Elect

rical

Powe

r (Pel)

(W)

Ambie

nt

pressur

e (pamb)

(mbar)

Density

of

intake

air (ρ)

(Kg/m3)

Air

Velo

city

(c)

(m/

sec)

Suction

volume

flow

(Vc)

(m3/

sec)

Power

hydru

alic

(Phyd)

(W)

Effic

ienc

y (η)

(%)

1 9602 364 394 20.4 78.4 1013 1.203 7.8 0.7412 29.2 37.2

MODEL CALCULATIONS:

1. ρ =ρ0 x (T0/T1)(pamb/p0) = 1.293x (273.15/293.55)x(1013/1013) =1.203kg/m3.

2. c = √((2/ρ) x dpN) = √((2/1.203) x 36.4) =7.8 m/s

3. Vs = c x A =7.8 x (π/4) x 0.112 =0.07412 m

3/s = 266.8 m

3/hr .

4. Phyd = dpF x Vs = 394x0.07412 =29.2 W

5. Efficiency = Phydx100/Pel = 29.2/78.4 = 37.2%

Main components

M = Drive motor dpF = Differential pressure, radial fan

V-R = Axial fan dpN = Differential pressure, inflow

V1 = Throttle valve n = Speed

Pel = Electrical power of the drive motor T1= Temperature of the intake air

26

9. RADIAL FAN

THEORY: A mine fan (or radial flow fan) in which the air enters along the axis

parallel to the shaft and is turned through a right angle by the blades and

discharged radially. There are three main types with (1) backwardly inclined

blades; (2) radial blades; and (3) forward curved blades. In (2) and (3) the blades

are made of sheet steel, while in (1) the present tendency is to replace curved

sheet-steel blades by blades of aerofoil cross section. The aerofoil bladed radial-

flow fan has an efficiency of about 90%.

AIM:

To determine the efficiency of the radial fan in constant speed condition and

plot the necessary chart.

APPARATUS REQUIRED:

3. Radial fan apparatus setup.

4. Computer system.

FORMULA:

1. Density of intake air ρ =ρ0 x (T0/T1)x(pamb/p0)

ρ0 = 1,293 kg/m3 is air density at reference temperature at T0 = 273.15 K.

p0= 1,013 mbar.

T1= Temperature of intake air in Kelvin. (1K = X0C +273.15)

2. Air velocity c = √((2/ρ) x dpN) in m/sec.

dpN = Dynamic pressure in N/m2.

3. Suction volume flow Vs = c x A in m3/sec.

A = Area of the intake pipe in m2.

A= (π/4) x d2 where d = diameter of intake pipe = 90 mm.

4. Power hydraulic Phyd = dpF x Vs in Watts




27

PROCEDURE:

1. Tare values and enter the ambient pressure in the system diagram.

2. Select ‘Measurement Diagram’ in the program.

3. Enable new series of measurements. Make any setting for the measurement

file.

4. Switch on radial fan, select speed of ____ %

5. For the first measurement, close the throttle valve completely.

6. Wait until the displayed measurement stable. Then record measurements


program is now ready for the next measurement.

7. Open the throttle valve a little bit. The position of the throttle valve is

dependent on the desired number of measurement points. Meaningful

characteristics are often obtained with 5 to 6 measurement points.

8. Repeat the steps until throttle valve is fully open.

9. Repeat the steps with the newly selected speed of 100%, save the

measurements and plot the characteristics curve using system.

GRAPHS:

Speed in rpm Vs suction volume.

OBSERVATION: Diameter of intake pipe (d) = ……90…….. mm

RESULT:


2. Hydraulic power corresponding to maximum efficiency = …………..W.


4. Air velocity corresponding to maximum efficiency = …………..m/s

5. Suction volume flow corresponding to maximum efficiency = …….. m3/Sec

28

TABULATION:

S.

N

o

Fan

speed

(n)

(Rev/

min)

Differe

ntial

Pressu

re flow

(dpN)

(N/m2)

Pressur

e

increas

e (dpF )

(N/m2)

Tempe

rature

of

intake

air (T1)

(0C)

Elect

rical

Powe

r (Pel)

(W)

Ambie

nt

pressur

e (pamb)

(mbar)

Density

of intake

air (ρ)

(Kg/m3)

Air

Velo

city

(c)

(m/

sec)

Suction

volume

flow

(Vc)

(m3/

sec)

Power

hydru

alic

(Phyd)

(W)

Effici

ency

(η)

(%)

1 2640 22.2 250 25 55.1 1013 1.184 6.12 0.04 10 18.15

MODEL CALCULATIONS:

1. ρ =ρ0 x (T0/T1)(pamb/p0) = 1.293x (273.15/298.15)x(1013/1013) =1.184kg/m3.

2. c = √((2/ρ) x dpN) = √((2/1.184) x 22.2) =6.12 m/s

3. Vs = c x A =6.12 x (π/4) x 0.092 =0.04 m

3/s = 140.16 m

3/hr .

4. Phyd = dpF x Vs = 250x0.04 =10 W

5. Efficiency = Phydx100/Pel = 10x100/55.1 = 18.15%.

Figure 9. Radial fan

Main components

M = Drive motor dpF = Differential pressure, radial fan

V-R = Radial fan dpN = Differential pressure, inflow

V1 = Throttle valve n = Speed

Pel = Electrical power of the drive motor T1= Temperature of the intake air

29

10. TUBULAR HEAT EXCHANGER (PARRALLEL FLOW)

THEORY: A heat exchanger is a device used to transfer heat between one or more

fluids. The fluids may be separated by a solid wall to prevent mixing or they may

be in direct contact.[1]

They are widely used in space heating, refrigeration, air

conditioning, power stations, chemical plants, petrochemical plants, petroleum

refineries, natural-gas processing, and sewage treatment. The classic example of a

heat exchanger is found in an internal combustion engine in which a circulating

fluid known as engine coolant flows through radiator coils and air flows past the

coils, which cools the coolant and heats the incoming air. There are several types

heat exchanger shell and tube heat exchanger and plate type heat exchanger etc.

AIM: Parameter determination of the tubular heat exchanger in parallel flow of

water.

APPARATUS REQUIRED:

1. Tubular heat exchanger setup.

2. Water tank.

FORMULA:

1. LMTD = Logarithmic Mean Temperature Difference.

LMTD = [Thi –Tci]- [Tho – Tco] / {ln [(Thi- Tci)/(Tho- Tco)]}

Where Tci = T6 = Entry temperature of cold fluid in Kelvin.

Thi = T1 = Entry temperature of hot fluid in Kelvin.

Tco = T4 = Exit temperature of cold fluid in Kelvin.

Tho = T3 = Exit temperature of hot fluid in Kelvin.

2. Qh = Heat transfer rate from hot water in KJ = mh x Cph [Thi – Tho]

Where mh = Mass flow rate of hot water [Kg/s]

Cph = Specific heat of hot water [KJ/KgK] = 4.187 KJ/KgK

3. Qc = Heat Transfer rate to the cold water = mc x Cpc [Tco- Tci]

Where mc = Mass flow rate of cold water [Kg/s]

Cpc = Specific heat of cold water [KJ/KgK] =4.187 KJ/KgK

4. Q = Heat transfer rate in Watts = [Qh + Qc] / 2

5. U = Overall Heat transfer co-efficient W/m2K = Q/(A x[ΔT]M)

https://en.wikipedia.org/wiki/Heat_exchanger#cite_note-1

https://en.wikipedia.org/wiki/Space_heating

https://en.wikipedia.org/wiki/Refrigeration

https://en.wikipedia.org/wiki/Air_conditioning



https://en.wikipedia.org/wiki/Power_station

https://en.wikipedia.org/wiki/Chemical_plant

https://en.wikipedia.org/wiki/Petrochemical

https://en.wikipedia.org/wiki/Oil_refinery



https://en.wikipedia.org/wiki/Natural-gas_processing

https://en.wikipedia.org/wiki/Sewage_treatment

https://en.wikipedia.org/wiki/Internal_combustion_engine

https://en.wikipedia.org/wiki/Engine_coolant

https://en.wikipedia.org/wiki/Radiator

https://en.wikipedia.org/wiki/Air

https://en.wikipedia.org/wiki/Air

30

Where [ΔT]M = LMTD

A = Area = πdl

6. Cr = Cmin/ Cmax

Ch = Cph x mh

Cc = Cpc x mc

In Cc , Ch which is minimum called Cmin and which is maximum called Cmax.

7. NTU = No of transfer units = Ux A/ Cmin

1- exp [ - NTU x (1+ Cr)]

8. Effectiveness E = ------------------------------------

1+ Cr

PROCEDURE:

1. Give the necessary connection to the set up.

2. Heat the water in the setup using heater.

3. Give the flow of hot water and cold water using the valve according to the

diagram. Note down the flow rate of hot and cold water.

4. Now the change in temperatures take place, note down the temperatures

after the change in temperatures reaches a steady value

5. Repeat the process of for other flow rates.

6. Tabulate the value and plot the graph.

GRAPHS: Heat transfer rate Vs Effectiveness.

Observation:

Overall length (L) = ……..560…….. mm.

Diameter (D) = ………..7…….. mm.

RESULT:

1. Heat transfer rate (Q)= _______________ W

2. Overall heat transfer coefficient (U)= _____________ W/m2k.

3. Effectiveness (E) = ___________ .

31

TABULATION:

S

.

N

o

Flow rate

of Hot

water

Flow rate

of Cold

water

Inlet

temp of

hot

water

(Thi)

(T1)

outlet

temp of

hot water

(Tho) (T3)

Inlet

temp of

cold

water

(Tci)

(T6)

Outlet

temp of

hot water

(Tco)

(T4)

LMT

D

Hea

t

tran

sfer

rate

(Q)

Ove

r all

heat

tran

sfer

coef

ficie

nt

(U)

Effe

ctiv

enes

s

(E)

L/hr Kg /s L/mi

n Kg /s

0C K

0C K

0C K

0C K

1

144

.04

120

.0333

56

329

45

318

34

307

39

312

12.31

1.27

8.4

.403

MODEL CALCULATIONS:

LMTD = [Thi – Tci] - [Tho – Tco] / ln [Thi – Tci/Tho – Tco]

= [329 – 307] – [318 – 312] / ln [(329 – 307) / (318 – 312)]= 12.31 K.

Qh = mh x Cph [Thi – Tho] = 0.04 x 4.187 x [329 – 318] = 1.842 KJ/sec.

Qc = mc x cpc [Tco –Tci]= 0.0333 x 4.187 [312 – 307] = 0.691 KJ/sec.

Q = [Qh + Qc] / 2 = [1.842 + 0.691] / 2 = 1.27 KJ/sec.

A = π x D x L= π x 0.007 x 0.56= 0.0123 m2.

U = Q/(A x[ΔT]M)= 1.27 / (0.0123 x 12.31)= 8.4 W/m2K.

Cr = Cmin/ Cmax = 0.14/0.167= 0.8383

Ch = Cph x mh = 4.187 x 0.04= 0.167 = Cmax

Cc = Cpc x mc = 4.187 x 0,0333= 0.140 = Cmin

32

NTU = Ux A/ Cmin = 8.4x0.0123/ 0.14 = 0.737.

1- exp [ - NTU x (1+ Cr)] 1- exp [- 0.737x 1.8383]

9. E = ---------------------------------- = --------------------------------- = 0.4036

1+ Cr 1.8383

Fig 12.Parrallel Flow Tubular Heat Exchanger

33

11. TUBULAR HEAT EXCHANGER (COUNTER FLOW)

THEORY: A heat exchangers are classified parallel flow and counter flow based

on the direction of both the fluids flow. In parallel flow heat exchanger both the

fluids (hot and cold) are flowing in the same direction. But in counter flow heat

exchanger both the fluids are flowing in the opposite direction.

AIM: Parameter determination of the tubular heat exchanger in counter flow of

water.

APPARATUS REQUIRED:

1. Tubular heat exchanger setup.

2. Water tank.

FORMULA:

1. LMTD = Logarithmic Mean Temperature Difference.

LMTD = [Thi –Tci]- [Tho – Tco] / {ln [(Thi- Tci)/(Tho- Tco)]}

Where Tci = T4 = Entry temperature of cold fluid in Kelvin.

Thi = T1 = Entry temperature of hot fluid in Kelvin.

Tco = T6 = Exit temperature of cold fluid in Kelvin.

Tho = T3 = Exit temperature of hot fluid in Kelvin.

2. Qh = Heat transfer rate from hot water in KJ = mh x Cph [Thi – Tho]

Where mh = Mass flow rate of hot water [Kg/s]

Cph = Specific heat of hot water [KJ/KgK] = 4.187 KJ/KgK

3. Qc = Heat Transfer rate to the cold water = mc x Cpc [Tco- Tci]

Where mc = Mass flow rate of cold water [Kg/s]

Cpc = Specific heat of cold water [KJ/KgK] =4.187 KJ/KgK

4. Q = Heat transfer rate in Watts = [Qh + Qc] / 2

5. U = Overall Heat transfer co-efficient W/m2K = Q/(A x[ΔT]M)

Where [ΔT]M = LMTD

A = Area = πdl

6. Cr = Cmin/ Cmax

Ch = Cph x mh

34

Cc = Cpc x mc

In Cc , Ch which is minimum called Cmin and which is maximum called Cmax.

7. NTU = No of transfer units = Ux A/ Cmin

1- exp [ - NTU x (1- Cr)]

8. Effectiveness E = -------------------------------------------

1- {Cr xexp [ - NTU x (1- Cr)]}

PROCEDURE:

1. Give the necessary connection to the set up.

2. Heat the water in the setup using heater.

3. Give the flow of hot water and cold water using the valve according to the

diagram. Note down the flow rate of hot and cold water.

4. Now the change in temperatures take place, note down the temperatures

after the change in temperatures reaches a steady value

5. Repeat the process of for other flow rates.


GRAPHS: Heat transfer rate Vs Effectiveness.

Observation:

Overall length (L) = ……..560…….. mm.

Diameter (D) = ………..7…….. mm.

RESULT:

1. Heat transfer rate (Q)= _______________ W

2. Overall heat transfer coefficient (U)= _____________ W/m2k.

3. Effectiveness (E) = ___________ .

35

TABULATION:

S.

N

o

Flow rate

of Hot

water

Flow rate

of Cold

water

Inlet

temp of

hot

water

(Thi)

(T1)

outlet

temp of

hot

water

(Tho)

(T3)

Inlet

temp of

cold

water

(Tci)

(T4)

Outlet

temp of

hot

water

(Tco)

(T6)

LMT

D

Hea

t

tran

sfer

rate

(Q)

Over

all

heat

trans

fer

coeff

icien

t (U)

Effe

ctiv

enes

s

L/h

r Kg /s L/hr Kg /s

0C K

0C K

0C K

0C K

1

288

.08

191

.053

82

355

57

330

35

308

51

324

26.24

5.95

18.43

0.44

MODEL CALCULATIONS:

LMTD = [Thi – Tci] - [Tho – Tco] / ln [Thi – Tci/Tho – Tco]

= [355 – 324] – [330 – 308] / ln [(355 – 324) / (330 – 308)]= 26.24 K.

Qh = mh x Cph [Thi – Tho] = 800 x 10-4

x 4.187 x [355 – 330] = 8.347 KJ/sec.

Qc = mc x cpc [Tco –Tci]= 0.053 x 10-4

x 4.187 [32 4 – 308] = 3.551 KJ/sec.

Q = [Qh + Qc] / 2 = [8.347 + 3.551] / 2 = 5.95 KJ/sec.

A = π x D x L= π x 0.007 x 0.56= 0.0123 m2.

U = Q/(A x[ΔT]M)= 5.95 / (0.0123 x 26.24)= 18.4352 W/m2K.

Cr = Cmin/ Cmax = 0.335/0.222 = 1.51

Ch = Cph x mh = 4.187 x 0.08 = 0.335 = Cmax

Cc = Cpc x mc = 4.187 x 0.053 = 0.222 = Cmin

NTU = Ux A/ Cmin = 18.4352x0.0123/ 0.222 = 1.02.

36

1- exp [ - NTU x (1- Cr)] 1- exp [- 1.02 x -0.51]

E = -------------------------------------- = -------------------------------------- = 0.44

1- {Cr xexp [ - NTU x (1- Cr)]} 1- {1.51xexp [- 1.02 x -0.51]}

Fig 13. Counter Flow Tubular Heat Exchanger

37

12. AIR COMPRESSOR

THEORY: An air compressor is a device that converts power (using an electric

motor, diesel or gasoline engine, etc.) into potential energy stored in pressurized air

(i.e., compressed air). This air compressor is a two stage reciprocating type. The air

is sucked from atmosphere and compressed in the first cylinder. The compressed

air then passes through an inter cooler into the second stage cylinder, where it is

further compressed. The compressed air then goes to a reservoir through a safety

valve. This valve operates an electrical switch that shuts off the motor when the

pressure exceeds the set limit.

AIM: To conduct a performance test on a two stage air compressor and determine

its volumetric efficiency.

APPARATUS REQUIRED:

1.Two stage air compressor apparatus.

2. Stop watch

FORMULA:

1. Volume V0 = Ad x√(2x∆p/ρ) ) in m3.

Ad = Area of the duct= 1.131x10-4

m2.

ρ = Density of air =1.293 kg/m3.

(1bar = 1x105N/m

2) (1mbar = 1x10

-3bar) (1Pa = 1N/m

2)

2. Isothermal power Piso = p1V0 ln (p4/p1)

3. Efficiency = Piso/Pel.


1 p1 - Inlet pressure

2 T1 - Inlet temperature

3 p2 - Pressure after 1st compressor stage

4 T2-Temperature after 1st compressor stage

5 p4-Pressure vessel pressure

6 T3-Temperature before 2nd compressor stage

7 ∆p-Differential pressure across Venturi nozzle

8 T4-Temperature after 2nd compressor stage

38

PROCEDURE: 1. Close the outlet valve.

2. Switch on compressor (see Section 2.4) If it does not start up, it is possible

that the over-current protection switch may have cut out directly on the

motor - restart.

3. Allow the system to run, until a constant pressure p3 has built up, set the

desired final pressure with the bleeder valve and record the measured values.


GRAPHS:

Pressure Vs Volume

OBSERVATION: Area of duct inlet= ……1.131x10-4

m2…….

RESULT:

1. Maximum efficiency of the compressor = ……………………… %

2. Isothermal power corresponding to maximum efficiency = …………..W.


4. Volume flow rate V0 at maximum efficiency = …………………l/min

39

TABULATION: S.

No

p1 T1 p2 T2 T3 p4 T4 ∆p V0 Pelec Tim

e

Piso η

Un

it

bar 0C bar

0C

0C bar

0C mba

r

m3/ Sec W min W %

1 0.99 23 3.4 127.1 58.8 11.7 151.6 7.4 38.26x10-4

2550 3 935.43 36.7

MODEL CALCULATIONS:

1. V0 = Ad x√(2x∆p/ρ) ) = 1.131X10-4

x(√(2x7.4/1.293) ) = 38.26x10-4

m3/s.

2. Piso = p1V0 ln (p4/p1) = 0.99x105x38.26x10

-4xln(11.7x10

5/0.99x10

5) =935.43W

3. η = Pisox100/Pel = 935.43x100/2550 =36.7%.

Fig 12 Air compressor

40

13. MULTI PURPOSE AIRDUCT

THEORY: There are various forms of heat transport: Convection is the transport

of heat by a moving fluid. Example: In forced convection, a conveying unit (pump,

blower) moves the fluid to be heated or cooled along the surfaces of a heat

exchanger. Thermal radiation is energy emitted by electromagnetic waves.

Example: Thermal radiation from the sun. Conduction is kinetic energy being

transported between two neighboring atoms or molecules. Example: A refrigerator

is insulated to prevent conduction.

Convective heat transfer takes part in heat exchangers and plays a large role in

many areas of industry. There are many different forms of heat exchangers which

transfer heat from one medium to another. Convective heat transfer in heat

exchangers can take place according to different principles: Parallel flow,

Counterflow, Cross-flow.

The Multipurpose Air Duct and Heat Transfer Unit WL 312 offers an excellent

supplement for calculations. It can be used to determine convective heat transfer on

an experimental basis. It provides a view of industrial applications with the

possibility of installing different types of heat exchangers with different heat

transfer media.

AIM: To determine the flow velocity and volume flow rate in multipurpose air

duct.

APPARATUS REQUIRED:

1. Air duct compressor apparatus.

2. Stop watch

FORMULA:

1. Velocity c = √(2xpdyn/ρ) ) in m/sec.

pdyn = Differential pressure between the ambient air and air duct (PD1) in mbar.

ρ = Density of air =1.293 kg/m3.

(1bar = 1x105N/m

2) (1mbar = 1x10

-3bar) (1Pa = 1N/m

2)

2. Volume flow rate V = c xAd in m3/sec.

Ad= Area of duct inlet in m2 = hdx bd.

hd = Height of the duct in mm.

bd = Width of the duct in mm.

41

PROCEDURE:

1. Place the throttle valve in a position that is vertical (90°) to the air flow. This

ensures that the maximum possible flow rate of the fan is achieved, since the

resistance is at its lowest level on the pressure side.

2. Switch on the fan.

3. Read the dynamic pressure (which is a proportion of the flow velocity in the air

duct) on the digital display with the differential pressure sign (flow).

4. Change the position of the throttle valve in order to obtain a different dynamic

pressure (flow rate).

GRAPHS:

Pressure Vs Velocity

OBSERVATION: hd = Height of the duct …290…. mm.

bd = Width of the duct …150…mm.

RESULT:

1. Maximum Velocity of air in duct = ……………. m/s

2. Maximum flow rate of air in duct=……………... m3/s

3. Angle of opening of throttle valve at Maximum flow rate of air in

duct=……………... 0

42

TABULATION:

S.No Throttle valve

Position in °

Dynamic

pressure

pdyn

in mbar

Flow velocity

c

in m/s

Flow rate

V in m³/Sec

1 90 0.873 12.1 0.505

MODEL CALCULATIONS:

4. Ad = hdx bd = 0.29x0.15= 0.0435 m2.

5. c = √(2xpdyn/ρ) ) = (√(2x0.873x105x10

-3/1.293) ) = 11.62 m/s.

6. V = c xAd = 11.62x0.0435=0.505 m3/sec.

Switch Cabin Test section

Throttle valve Fig 13. Air duct Air inlet

43

IMPORTANT TERMS

1. 1liter =1000 cm3 = 0.001 m

3

2. 1bar = 1x105 N/m

2

3. 1liter/min =16.67x10-6

m3/Sec

4. 1liter = 1kg

5. 1Pa = 1N/m2.

6. 1mbar =0.001bar

Turbo Machines Lab manual

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