me09 608(p)_ thermal lab ii

8/10/2019 ME09 608(P)_ Thermal Lab II

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Department of Mechanical Engineering

LABMANUAL

FOR

THERMAL LAB II

DEPT. OF MECHANICAL ENGINEERING

Downloaded from Official website of Ammini College of Engineering, Palakkad

http://ammini.edu.in/content.aspx?pageid=362


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Wher

SAM

RES

Best c

INFE

,

TFC = T

CV = Cal

BP = Bra

LE GR

LT

ooling wa

ENCE

tal fuel c

orific val

e power

PH

ter tempe

nsumpti

e of dies

in watts

rature (fr

3

n in kg/h

l = 45.2

m graph)

r

106J/K

=

.K




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4

2.RETARDATION TEST ON SINGLE CYLINDER

VERTICAL

DIESEL ENGINE

AIM

To determine the frictional horse power of the engine by conducting retardation

test and Plot the graph Rated Speed Vs Retardation Time.

ENGINE DETAILS

Brake Power = 6HP =6 x 736 W

Rated Speed = 650 rpmStroke length = 139.7 mm

Bore diameter = 114.3 mm

Brake drum Radius = 197 mm

APPARATUS REQUIRED

Stop watch

MAXIMUM LOAD CALCULATION

Maximum brake power; B.P max = 2NT/60 Watts.Where B.P max = 4.416 KW = 4416 Watts,

N = speed of the engine = 650 rpm,

T = torque on the engine shaft in Nm = WR in Nm

Where W = load on the engine in Kg

R = radius of the Brake drum = 197 mm

Maximum load in Kg W max = B.P max x60/ (2NR*9.81)

PROCEDURE

Calculate the load to be applied for the maximum output. Take the following

precautions before starting the engine.

1. Check the fuel level

2. Check the lubricating oil level.

3. Check the cooling water circulation.

4. Check whether the engine is on no load.

Engine is started at No- load condition and is run at rated speed. The fuel is thencut- off using fuel cut off lever and the time taken for the speed to drop to a




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lower

lever i

experi

load a

and fr

FOR

Fricti

Wher

Mech

speed is

s engage

ment is r

d the ab

ctional p

ULAE

nal powe

,

nical Effi

Where, I

oted usin

] and th

peated w

ve proce

wer is de

, F.P. =

N= s

Fricti

Where,

ciency,

dicated P

a stop

engine i

ith vario

ure is re

termined.

N TF /

eed = 66

nal Torq

t3 = Ret

t2 = Reta

TL = Loa

TL = W

Where,

ech =

ower, IP

5

atch. The

again b

s lower s

eated an

0 Watts

rpm

e, TF =

rdation ti

dation ti

d Torque

* 9.81 N

W =

R = R

P/IP

BP + F

fuel is a

ought ba

peeds. Th

noted t

L x [t3 /(

me at hal

e at No-

oad on e

dius of b

ain turne

k to the

e engine

e reading

2-t3)] N

load [fro

load [fro

gine

rake dru

on [fuel

ated spe

is loaded

s. Plot th

m the gra

the gra

cut off

d. The

to half

graph

h]

h]




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7

3.DETERMINATION OF EFFECTIVENESS OF

PARALLEL FLOW AND COUNTER FLOW HEAT

EXCHANGER

OBJECTIVE:

To determine the logarithmic mean temperature difference (LMTD),

effectiveness and overall heat transfer coefficient for parallel and counter flow

heat exchanger.

EQUIPMENT:

1.

The apparatus consists of a concentric tube heat exchanger.

2.

The hot fluid namely hot water is obtained from the Geyser (heatercapacity 3 kW), it flows through the inner tube.

3.

The cold fluid i.e. cold water can be admitted at any one of the ends

enabling the heat exchanger to run as a parallel flow or as a counter flow

exchanger.

4. Rotameters are used for measuring flow rate of cold and hot water.

5.

This can be adjusted by operating the different valves provided.

6.

Temperature of the fluid can be measured using thermocouples with

digital display indicator. The outer tube is provided with insulation tominimize the heat loss to the surroundings.

Specimen material - Copper Tube

Size of specimen- diameter 12.5mm, length -1500mm

Outer shell material G I

Size of outer shell diameter- 40 mm

BASICS:

LOGARITHMIC MEAN TEMPERATURE DIFFERENCE (LMTD): LMTD

LMTD = (2 1)/ ln(2/ 1)0C

where, 1= Thi -Tci, and2 = Tho. -Tco for parallel flow heat exchanger1= Tho -Tci , and2= Thi, -Tco for counter flow heat exchanger

This is defined as that temperature difference which, if constant, would give the

same rate of heat transfer as usually occurs under variable conditions oftemperature difference.




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8

PRECAUTION:

Switch ON the heater only after starting water supply.

PROCEDURE:

1.

Switch ON the unit panel.

2. Start the flow of cold water through the annulus and maintain the

exchanger as counter flow or parallel flow.

3.

Switch ON the geyser provided on the panel & allow water to flow

through the inner tube by regulating the valve.

4. Adjust the flow rate of hot water and cold water by using rotameters &

valves.

5.

Keep the flow rate same till steady state conditions are reached.

6.

Note down the temperatures on hot and cold water sides. Also note the

flow rate.

7.

Repeat the experiment for different flow rates and for different

temperatures. The same method is followed for parallel flow also.

OBSERVATIONS:

SI. No Hot

water

flow

rate

cc/s

Cold

water

flow

rate

cc/s

Temperature

of cold water

in C

Temperature

of hot water in

C

Tci Tco Thi Tho

Parallel

flow

Counter

flow

CALCULATION:

Heat transfer from hot water Qh= mcph (Thi Tho) W

where mh - mass flow rate of hot water kg/s.

Cph - Specific heat of hot water = 4186.8 J/kgK




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Heat gain by the cold fluid Qc= mCpc (Tco Tci) W

where me- Mass flow rate of cold fluid, kg/s

Cpc - Specific heat of cold fluid=4186.8 J/kgK

Q = (Qh + Qc ) /2 W

LMTD = (2 1)/ ln(2/ 1)0C

Where 1= Thi -Tci, and2 = Tho. -Tco for parallel flow heat exchanger1= Tho -Tci , and2= Thi, -Tco for counter flow heat exchanger

Overall heat transfer coefficient based on outside surface area of inner tube

U0= Q / A0LMTD W/m2K

where, Area, A0= d0L m2

d0 - Outer diameter of the tube = 0.0125 m

L - length of the tube = 1.5 m

Effectiveness, = ( Thi Tho) / ( Thi - Tci ) if ch < cc

Effectiveness, = ( Tco Tci) / ( Thi - Tci ) if cc < ch

This is applicable for both parallel and counter flow heat exchanger.

RESULT:

INFERENCE:




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10

4.DETERMINATION OF COP OF THE REFRIGERATION

TEST RIG

AIM

To determine the actual, theoretical and relative COP of refrigerating plant.

APPARATUS

The given Refrigeration Test Rig with refrigerant R-132.

a) Digital indicator b) compressor c) condenser

d) Expansion device e) evaporator f) waterchiller

THEORY

The system works on vapour compression refrigeration cycle.

Theoretical COP = Refrigeration effect = (h1-h4)

Work done (h2-h4)

Where,

h1= Enthalpy corresponding to P1, and T1, kJ/kg



P1and T1- Pressure and Temperature of Refrigerant at inlet of

compressor.

P2and T2- Pressure and Temperature of Refrigerant at exit of compressor.

T3-Temperature of Refrigerant at the exit of condenser.

Actual COP = Heat removed

Actual workdone

Heat removed = mCp dT

m- mass of water taken in the chiller in kg

Cp - specific heat of water

dt - drop in temperature of water

Actual work input = V x IV- Voltage

I - Current

Relative COP = Actual COP

Theoretical COP

PROCEDURE

1) Fill the chiller with water

2)

Switch- On the power




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Theoretical COP = Refrigeration effect = (h1-h4)

Work done (h2-h4)

Where,

h1= Enthalpy corresponding to P1, and T1, kJ/kg from p-H chart.



P1and T1- Pressure and Temperature of Refrigerant at inlet of

compressor.

P2and T2- Pressure and Temperature of Refrigerant at exit of compressor.

T3-Temperature of Refrigerant at the exit of condenser.

Relative COP = Actual COP

Theoretical COP

Result

Theoretical COP of refrigerator =

Actual COP of refrigerator =

Relative COP refrigerator =

Inference

Blockdiagramofvapourcompressioncycle

Expansion

Device

T 2 P 3

3

4 1

Entropy Enthalpy

2

41

Condenser

Compressor

Evaporator




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5.MORSE TEST ON 4 CYLINDER PETROL ENGINE

AIM

To determine the frictional horse power in each cylinder of a 4 cylinder petrol

engine. Also determine the indicated power and mechanical efficiency of the

engine.

ENGINE DETAILS

Brake Power = 7.36 KW

Rated Speed = 1500 rpm

Stroke length = 75 mm

Bore diameter = 68mm

Radius of dynamometer wheel = 0.125 m

APPARATUS REQUIRED

Cylinder cut-off arrangement


Maximum brake power; B.Pmax = 2NT/60 Watts.Where B.Pmax = 7.36 KW = 7360 Watts,




R = radius of the dynamometer wheel = 0.125 m


PROCEDURE

Calculate the load to be applied on the eddy current dynamometer for the

maximum output. Take the following precautions before starting the engine.

1. Check the fuel level.




The test is conducted at constant speed with constant fuel supply (the throttle

valve is not adjusted). Start the engine using self starter. Engage the engine with

dynamometer using the clutch. Allow the engine to run for a few minutes at the

rated speed (1500rpm) to attain steady conditions. The engine is loaded to about




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50% of the maximum load and throttle valve is adjusted to maintain constant

speed.

Now the voltage to the spark plug of 1st cylinder is cut-off and now the engine

is running on the expense of 2nd and 3rd cylinders. The speed is maintained

constant by reducing load and the load is noted. Close the circuit of 1st cylinder

and the 2nd cylinder is short circuited. Repeat the procedures for other cylinders

and note the load on the dynamometer on each case. After completion of the

experiment, bring the engine to no load conditions and stop the engine by

switching off ignition key. Maintain cooling water circulation for some more

time.

FORMULAE

Brake power, 1st cylinder cut-off = 2NT/60 WattsWhere N= speed = 1500 rpm

T = WR* 9.81 Nm

B.P = ...........................Watts

Indicated power of 1st cylinder;

I.P. 1 = B.P. total B.P. 1stcylinder cut-off

Brake power, 2nd cylinder cut-off = 2NT/60 WattsWhere N= speed = 1500 rpm

T = WR* 9.81 Nm




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B.P = ...........................Watts

Indicated power of 2nd cylinder;

I.P. 2 = B.P. total B.P. 2nd

cylinder cut-off

Brake power, 3rd cylinder cut-off = 2NT/60 WattsWhere N= speed = 1500 rpm

T = WR* 9.81 Nm

B.P = ...........................Watts

Indicated power of 3rd cylinder;

I.P. 3 = B.P. total B.P. 3rd

cylinder cut-off

Indicated power of 4th cylinder;

I.P. 4 = B.P. total B.P. 4thcylinder cut-off

Total indicated power;

I.P. total = I.P. 1 + I.P. 2 + I.P. 3

Mechanical efficiency;

Mech = B.P. total X 100I.P. total

RESULT

Total indicated power = .. W

Mechanical efficiency=...................%

INFERENCE




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6.LOAD TEST ON 4 STROKE PETROL ENGINE

AIM

To determine the total fuel consumption, specific fuel consumption, brake mean

effective pressure and brake thermal efficiency of the petrol engine at various

loads and to plot the following graphs.

Brake power output Vs T.F.C

Brake power output Vs S.F.C

Brake power output Vs B.M.E.P

Brake power output Vs Br.Th.efficiency

ENGINE DETAILS

Brake Power = 7.36 KW

Rated Speed = 1500 rpm

Stroke length = 75 mm

Bore diameter = 68mm

Radius of dynamometer wheel = 0.125 m

APPARATUS REQUIRED

Stopwatch

PROCEDURE

Calculate the load to be applied on the eddy current dynamometer for the

maximum output. Take the following precautions before starting the engine.

1. Check the fuel level.




Start the engine using self starter. Engage the engine with dynamometer using

the clutch. Allow the engine to run for a few minutes at the rated speed

(1500rpm) to attain steady conditions. Observe time for 10cc fuel consumption.

Now load the engine keeping the speed constant. Again wait for a few minutes

to attain steady conditions at that load. Observe time for 10cc fuel consumption

and actual load acting on the engine. Repeat the procedure for six different

loads (from no load to full load). Care must be taken not to overload the engine.




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After completion of the experiment, bring the engine to no load conditions

before stopping. Maintain cooling water circulation for some more time.

OBSERVATION AND TABULAR COLUMN


Maximum brake power; B.Pmax = 2NT/60 Watts.Where B.Pmax = 7.36 KW = 7360 Watts,




R = radius of the dynamometer wheel = 0.125 m


SAMPLE CALCULATIONS (SET NO....)

Brake power output; B.P = 2NT/60 WattsWhere N= speed = 1500 rpm

T = WR* 9.81 Nm

B.P = ...........................Watts

Total fuel consumption; T.F.C = (10/t)*(3600/1000)*0.75 Kg /hr

Where t = time for 10cc fuel consumption =.................sec




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0.75 = sp. weight of petrol

T.F.C = ............................. Kg/hr

Specific fuel consumption; S.F.C = T.F.C/ B.P Kg/W hr

Where T.F.C = Total fuel consumption in Kg/hr

B.P = Brake power in Watts

S.F.C =.............................. Kg/W hr

Brake mean effective pressure; B.M.E.P = (B.P*60)/ LA(N/2)*n in N/m2

Where B.P = brake power in watts

L = stroke length = 0.75 m

D = diameter of the cylinder (bore) = 0.68m

A = area of the cylinder in m2 = /4* D2N = speed = 1500 rpm (N/2 is because of four stroke engine)

n = no: of cylinders = 3Nos

B.M.E.P =.......................... in N/m2

Brake thermal efficiency; Br.th.= (B.P*3600)/(T.F.C*C.V) X 100 %Where T.F.C = total fuel consumption in Kg/hr

C.V = calorific value of petrol = 43.5 J/Kg K

B.P = brake power in Watts

Br.Th.=...............................%

RESULT

Maximum brake thermal efficiency =...................%

INFERENCE




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7.LOAD TEST ON TWO STAGE AIRCOMPRESSOR.

AIM

To conduct load test on the two stage reciprocating air compressor and to

determine the volumetric

efficiency and isometric efficiency at various delivery pressure. Also plot the

following graphs.

Delivery pressure Vs Volumetric efficiency.

Delivery pressure Vs Isothermal efficiency.

TEST RIG DETAILS:

Working pressure = 12 kgf/cm2.

Motor power = 3HP

Low pressure cylinder bore diameter, D1 = 90mm.

High pressure cylinder bore diameter, D2 = 63mm.

Stroke length L = 89.5mm

Speed, Nc = 925rpm

Diameter of orifice, d = 0.015m

Coefficient of discharge of the orifice meter, Cd = 0.6

Energy meter constant, K = 1200 impulse/kwh.

Number of impulse on energy meter, n = 10

THEORY:

During the downward motion of the piston the pressure inside the cylinder falls

below the atmospheric pressure and the inlet valve is opened due to this

pressure difference. The air is sucked into the cylinder until the piston reaches

the BDC (Bottom dead centre). As the piston starts moving upwards the inlet

valve closed and the pressure starts building up continuously until the pressure

inside the cylinder is above the pressure of the receiver. Then the delivery valve

opens and the air is delivered during the remaining upward motion of the piston

to the receiver.

At the end of the delivery stoke, small volume of high pressure air left in the

clearness space. The high pressure left in the clearness space expands as the

piston moves downwards and the pressure of the air falls, until the pressure is

just below the atmosphere and then the inlet valve opens and fresh air is sucked

in and whole process will repeat.




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The suction, compression and delivery of air take place within two strokes of

the piston or one revolution of the crankshaft. The compression of air from

initial pressure to the final pressure in more than one cylinder is known as

multistage compression.

DESCRIPTION:

The compressor basically consists of an electric motor (prime mover), two

cylinders namely HP cylinder and LP cylinder. The system is intercooled.

Pressure gauges are provided at the both of the HP cylinder and LP cylinder

outlets to read the pressures. The AC motor gives input power to the

compressor.

APPARATUS

Manometer, Digital rpm indicator, stopwatch

PROCEDURE:

1. The water present if any in the receiver is drained out using the drainage

cock.

2. The outlet valve of the receiver is kept open to facilitate starting and then the

motor is switched on.

3. When the compressor reaches its normal speed the outlet valve of the receiver

is closed and the compressor is allowed to build the required pressure.

4. When the pressure reaches the desired valve, the outlet valve is adjusted so

that the delivery pressure remains constant at that pressure. At this point

manometer reading, speed of the motor and energy meter readings is noted

down.

5. The experiment is repeated for different values of pressures and the above set

of reading are noted down.

After completing the experiment, switch of the motor and release the air from

the receiver.




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FORMULAE:

1) Initial pressure, P1 = Atmospheric pressure = 1,01,325 N/m2 .

2) Final pressure, P2 = (Gauge pressure x 10 5 ) + Atmospheric pressure

N/m2

Where gauge pressure = Pressure gauge reading in Kgf/cm2.

3) Actual volume of air intake per second , Va = Cd A 2ghaWhere Cd = Coefficient of discharge of orifice meter = 0.60

A = Area of orifice = d2/4 m2.g = Acceleration due to gravity = 9.81 m/s2

ha= Head difference in terms of air column in meters.

= h x (density of water /density of air at RTP.)Where, h = Difference in level of water in manometer in meters.Density of water w = 1000 Kgf/m3.

Density of air at RTP a = 1.293 X 273 / (273 +t ) Kgf/m3.Where t = Room temperature in

0C.




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4) Theoretical air intake per second, Vth = /4 x D12x L x Nc / 60.Where D1 = Diameter of low pressure cylinder in metres.

L = Stroke length in metres

Nc= Speed of the compressor in rpm.

5) Volumetric efficiency , vol = (Va / Vth ) x 100 %Where Va = Actual volume of air delivered in m3/s.

Vth = Theoretical air intake in m3/s.

6) Input power , P = (3600 x n) / (K x t) kw

Where , K = Energy meter constant = 200 rev/kwh.

n = Number of revolutions of energy meter disc.

t = Time taken for n revolution of energy meter disc.

Assuming transmission and mechanical losses as 20 %. i.e. Total Input = 0.8 P

7) Isothermal work done = P1 V1 loge P2 /P1 x 10 -3 kJ/s

Where P1 = initial pressure or atmospheric pressure in N/ m2 .

P2 = final pressure in N/ m2.

V1 = Va = Actual air intake in m3/s.

8) Isothermal efficiency , iso= (Isothermal work done ) / ( Total Input) x100 %.

RESULT

1) Maximum Volumetric efficiency of the compressor.

2) Maximum Isothermal efficiency of the compressor.

INFERENCE




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8.PERFORMANCE TEST ON BLOWER

AIM:-

To conduct and evaluate the performance test on the blower by plotting the

following performance curve:

a) Total head Vs discharge

b) Efficiency Vs discharge

c) Input Vs discharge

SPECIFICATIONS

a) Power (P) : 7.5HP or 5.5KW

b) Speed (N) : 2900rpm

c) Pipe Diameter : 125mm

d) Throat Diameter : 75mm

e) Co-efficient of Discharge (Cd) : 0.6

f) Impeller Diameter : 500mm

g) Motor : 7.5hp or 5.5kw, AC,3phase,440V,

50Hz, Squirrel cage Inductionh) Delivery Size : 125x80mm

i) Inlet Diameter : 200mm

APPARATUS REQUIRED

a) Centrifugal Air blower test rig

b) Stop watch

c) Tachometer

PRINCIPLE

The main components of air blower are the impeller and the diffuser. The fresh

air enters into the eye of the of impeller. Because of the high rotational speed of

the impeller, the air contained in the rotational passage is subjected to

centrifugal force which causes air to flow radially outwards. All the mechanical

energy driving the impeller is transmitted to fluid stream in the impeller, where

it is converted into kinetic energy with a slight pressure rise. Blower is used to

discharge higher volume of air at a lower pressure it is used in blast furnace,cupolas, air conditioning plant, etc.




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OBSERVATION AND TABULATION

FORMULA USED

Atmospheric pressure (Pa) = 1.013 x 105N/m

2

Diameter of pipe (d1) = 125mm

Diameter of throat (d2) = 75mm

Co-efficient of discharge (Cd) = 0.6

a) Static Head (Hs)

Hs =(hsxw)/ a (m)

Where , hs= Static pressure manometer (m)

w=density of water ,1000 (Kg/m3)

a =density of air ,1.18 (Kg/m3)

b) Head causing flow

Ha = (Hwxw)/a (m)

SlNo

ManometerReading

(m)

Venturimeterreading

Timefor

5 pulse

on e/mt

Headcausing

flow

h1

DischargeQ

Velocity

of air

v

Staticpressu

re

headHs

Dynamichead

Hd

Total

head

H

Poweroutput

Po

Powerinput

Pi

Efficiency

h

1

h

2

h1

h2

h

1

h

2

h1

h2sec m m

3/s m/s m m m w w %




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Where, Hw = Veturimeter reading (m)

w = density of water ,1000 (Kg/m3)

a = density of air ,1.18 (Kg/m3)

mercury = density of mercury (Kg/m3)

c) Discharge (Q)

Q=

(m3/sec)

Where, Cd =Co-efficient of discharge, 0.6

a1 =Area of cross-section of pipe (m2

)

=0.012

a2 =Area of cross-section of throat (m2)

=0.004

g = Acceleration due to gravity (m/s2)

=9.81

Ha = Head causing flow

d) Velocity of air in pipe (V)

V= (m/s)

Where, Q = Discharge (m3/s)

A = Area of cross-section of pipe (m2)

e) Dynamic Head (Hd)

Hd =V

(m)

f) Total Head (H)

H = Hs+Hd+Z (m)

Where, Z= Datum height from suction to delivery (m)

=1




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g) Power output (Po)

Po = axgxQxH (w)Where, a = Density of air =1.18 (kg/m3)

g = Accelaration due to gravity =9.81 (m/s2)

Q= Discharge (m)

H = total head (m)

h) Power input (Pi)

Pi =

(w)

Where, n = no of pulse on energy meter

m = motor efficiency = 85 (%)t = Time for 5 pulse of energy meter (sec)

k = Energy meter constant ,1600 pulse/KWH

i) Efficiency ()

= 100 (%)

PRECAUTIONS

The following precautions were taken before starting the test

a) Check the test rig is under no load

b) Check the level of water in the manometer

PROCEDURE

a) Start the blower at no load condition by keeping the delivery valve closed

position

b)

Open the delivery valve in full open conditionc) Take the manometer readings and the time taken for n number of pulses

of energy meter

d) Repeat the experiments by closing the delivery valvegradually

e) Finally take the reading at closed condition of delivery valve

f) Close the delivery valve and switch off the blower




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RESULT

The following graph are plotted :

a) Total head Vs discharge

b)

Efficiency Vs discharge

c) Input Vs discharge

Maximum efficiency obtained :

Maximum output :

Maximum value of air discharge :

INFERENCE

The efficiency increases gradually with increase in discharge

The input power increases

Gradually with increase in discharge

The head decreases

Gradually with increase in discharge




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9.HEAT BALANCE TEST ON SINGLE CYLINDER

DIESEL ENGINE

AIMTo study the variation of heat losses with load and to plot heat balance chart.

TEST RIG DETAILS

BP of the engine = 6HP = 6 x 736 W

Bore diameter of the engine = 114.3mm.

Stroke length of the engine = 139.7mm.

Speed of the engine = 650rpm.

Orifice diameter = 20 mm

APPARATUS REQUIRED

Stop watch

PROCEDURE

Calculate load to be applied on the engine corresponding to the maximum

output. Take all necessary precautions before starting the engine.

Open the cooling water supply valve to the engine and dynamometer. Start the

engine by cranking. Allow the engine to run for few minutes at no load to attain

steady conditions. After this condition reached, note the following readings.

1. Load (Kg)

2. Time for 10cc fuel consumption (s).

3. Manometer reading (m of H2o).

4. Time for 10 litres of water circulation through the engine jacket (s).

5. Temperature of exhaust gas (0C).6. Inlet and outlet temperatures of cooling water (0C).

Now change the load to full load. Allow the engine to run for few minutes to

attain steady condition and note the above set of readings.

After completion of experiment, bring the engine back to no load condition and

then stop.




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OBSERVATIONS AND TABULATION


Brake power, BP = (2NT)/60. Watt

Where,

N = Speed of the engine in rpm.

T = Torque on the brake drum in Nm

= (W1 W2) R x 9.81Nm.

W1 = weight on hanger + hanger weight in kg.

W2 = spring balance reading in kg.

R = Radius of brake drum + thickness of rope in meters.

FORMULAE

1. Brake power output

Brake power output = Wmax X N/ 2.71 Watts

Where N = speed = 1500 rpm

Wmax = load applied on the engine = .................Kg

2. Heat input

Heat input = T.F.C X CV Watts

Where T.F.C = Total Fuel Consumption = 10 X 3600 X 0.83 Kg/hr

SlNo

LoadApplied

W

Speedof the

Engine

N

Time takenfor 10cc of

fuelconsumption

t

Coolingwater

flow rate

ManometerReadings

Temperatureof cooling

water

Temperature of

exhaustgas

t3

Temperature of

inlet air

t4

h1 h2 t1 t2

kg r.p.m. sec m3/s cm cm 0C 0C 0C 0C




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30

t X 1000

CV = Calorific value of diesel = 45.2 X 106 J/Kg

Heat input may be taken as 100%

3. Heat carried away by cooling water

Heat carried away by cooling water = m CP T WattsWhere m = mass flow rate of cooling water in Kg/s [for 10 Ltrs of water]

m = 10/t Kg/s

Where t = time for 10 ltrs of water circulation

CP = Specific heat of water at constant pressure = 4.186 X 103 J/Kg K

T = (T1-T2) = difference of inlet and outlet cooling water temperature in K

% of heat loss through cooling water =

Heat carried away by cooling water x100%

Heat Input

4. Heat loss through exhaust gas

Heat carried away by exhaust gas = m CP T WattsWhere m = mass flow rate of exhaust gas in Kg/s

m = mass flow rate of fuel in Kg/s + mass flow rate of air in Kg/s

Mass flow rate of fuel = T.F.C/3600 Kg/s

Mass flow rate of air = Volume of air in m3/s (Va) X density of air at RTP (a)Volume of air in m3/s (Va) = Cd .a. 2gha m3/sWhere Cd = 0.62

A = area of orifice = d2 /4 m2D = diameter of orifice = mm

ha = Manometric head of air column = hw w/a m of airhw = Manometer difference of water column

w = density of water = 1000 kg/m3

a = density of air at R.T.P = 1.293 X 273 Kg/m3273+t

t = Ambient temperature

CP = specific heat of exhaust gas = 1.005 X 103 J/KgK

T = (T1-T2) = difference of exhaust gas and room temperature in K

% of heat loss through exhaust gas = Heat loss through exhaust gas X 100%

Heat input




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5. Unaccounted heat loss (radiation and other losses)

Unaccounted heat loss = heat input (work output + cooling water loss+

exhaust gas loss) watts

RESULT

Heat equivalent of brake power =

Heat carried away by cooling water =

Heat carried away by exhaust gas =

Unaccounted heat loss =

Energy Input = 100 %

INFERENCE

me09 608(p)_ thermal lab ii

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