thermodynamics lab manual_s

DHANALAKSHMI SRINIVASAN ENGINEERING COLLEGE

(Approved by AICTE, New Delhi & Affiliated to Anna University , Chennai – 600 025)

(NBA Accredited and ISO 9001:2008 Certified Institution)

PERAMBALUR - 621 212.

DEPARTMENT OF AERONAUTICAL ENGINEERING

AE 6311 Thermodynamics Laboratory

MANUAL NOTE BOOK

Name :…………………………………………………………….

Reg. No. :…………………………………………………………….

Semester :…………………………………………………………….

Academic Year :…………………………………………………………….

CONTENTS

Sl. No. Date Name of the Experiment Page Marks

Sign

Completed / Not Completed

Signature of Staff in Charge

Average

Marks

Dhanalakshmi Srinivasan Engineering College, Perambalur – 621 212. AUC R2013

Gurunath K – AE 6311 Thermodynamics Laboratory | 1

EXPERIMENTS



VALVE TIMING DIAGRAM OF FOUR STROKE DIESEL ENGINE

S. No.

Description

Angle in degree Events

Before TDC/BDC

After TDC/BDC



AIM

To study the salient points of operation and to draw the valve timing diagram

for a four stroke engine.

APPARATUS REQUIRED

1. Tape

2. Chalk

3. Spanner

4. Screw driver

5. Diesel engine

Marking BDC and TDC position

The flywheel is made to rotate slowly in the direction of rotation. When the

bottom edge of the piston coincides with the bottom edge of the cylinder, make a mark

on the flywheel rim against pointer. Further rotate the crank in the same direction, now

the piston will come out of the cylinder edge and again it moves towards the top end.

The bottom edge of the piston will again coincide with the bottom edge of the cylinder.

Make a mark on flywheel rim against pointer. Find the center position of these two

points, which represents the bottom dead center (BDC) position of the piston.

Measure the circumference of the flywheel from BDC position. Mark out half

the circumference of the flywheel. This point represents Top Dead Center (TDC)

position of the piston.

Inlet valve opening and closing position

The flywheel is rotated slowly in the direction of rotation and a point is marked

on the flywheel, when the rocker arm of the inlet valve just touches the valve. The point

represents the inlet valve opening position. This can be determined by inserting a paper

and feeling grip. Another point is marked in the flywheel when the rocker arm of the

inlet valve becomes just free. This point represents the inlet valve closing position. This

can be determined by inserting a paper and its free movement.

Exhaust valve Opening and Closing

In the same manner two points are marked on the circumference of the flywheel

rim for exhaust valve opening and closing positions.

Exercise No: 01

VALVE TIMING DIAGRAM OF FOUR STROKE DIESEL ENGINE Date:



Measure the circumference of the flywheel marked on the rim of the flywheel

from the nearest dead centers and tabulate readings.

RESULT

Various points of four stroke diesel engine are studied and the valve-timing

diagram is drawn for the present set of values.



PORT TIMING DIAGRAM OF TWO STROKE PETROL ENGINE

S. No.

Description

Angle in degree

Events Before TDC/BDC

After TDC/BDC



AIM

To draw the port timing diagram for a two stroke petrol engine.

APPARATUS REQUIRED

1. The engine model

2. Tape

3. Chalk etc.

DESCRIPTION

In the case of two stroke cycle engines the inlet and exhaust valves are not

present. Instead, the slots are cut on the cylinder itself at different elevation and they

are called ports. There are three ports are present in the two stroke cycle engine.

1. Inlet port

2. Transfer port

3. Exhaust port

The diagrams which show the position of crank at which the above ports are

open and closed are called as port timing diagram.

The extreme position of the piston at the bottom of the cylinder is called

“Bottom Dead Center” [BDC]. The extreme position of the piston at the top of the

cylinder is called “Top Dead Center” [TDC].

In two stroke petrol engine the inlet port open when the piston moves from

BDC to TDC and is closed when the piston moves from TDC to BDC.

The transfer port is opened when the piston is moves from TDC to BDC and the

fuel enters into the cylinder through this transport from the crank case of the engine.

The transfer port is closed when piston moves from BDC to TDC. The transfer port

opening and closing are measured with respect to the BDC.

The exhaust port is opened, when the piston moves from TDC to BDC and is

closed when piston moves from BDC to TDC. The exhaust port opening and closing

are measured with respect to the BDC.

Exercise No: 02

PORT TIMING DIAGRAM OF TWO STROKE PETROL ENGINE Date:



PROCEDURE

Mark the direction of rotation of the flywheel (usually clockwise) looking from

the flywheel.

Mark the position of TDC and BDC on the flywheel rim.

Rotate the flywheel in clockwise direction and make on the flywheel rim when

the bottom position just uncover (open) the lowermost point of the suction port

during its upward movement. This mark is represented for suction port open.

Continue the rotation of the flywheel by hand, when piston’s bottom edge

reached the lowermost point of the suction port, during its downward movement

make a mark on the flywheel rim to represent for suction port close.

If we rotate the flywheel further, the piston will reach BDC and begin to move

upward. When the piston crown just covers the transfer port, make a mark on

the flywheel rim, which gives transfer port close. If we rotate further, the piston

crown just covers the exhaust port, make a mark on the flywheel rim, which

gives exhaust port close.

Now again rotate the flywheel in the same direction make a mark on the

flywheel rim when the top edge of piston (piston crown) just uncover the

uppermost point of the exhaust port during downward movement. This mark is

represented for Exhaust port open.

RESULT

Various points of two stroke petrol engine are studied and the port-timing

diagram is drawn for the present set of values.



TEST ON HEAT EXCHANGER-PARALLEL FLOW

Sl.no

Time for

Hot water

flow 4

litres (sec)

Temperature in ºC Heat

transfer

Q

Watts

LMTD

ºC

Overall heat

transfer

co-efficient U

W/ m2 K

Effectiveness

E T1

(Thi )

T2

(Tho )

T3

(Tci )

T4

(Tco )



AIM

To determine the effectiveness of the parallel flow heat exchanger.

APPARATUS REQUIRED

1. Heat exchanger apparatus

2. Stop watch

3. Measuring flask

SPECIFICATIONS

Length of the heat exchanger = 1800 mm

Inner copper tube ID = 12 mm

OD = 15 mm

Outer GI tube ID = 40 mm

Specific heat capacity of water Cp = 4.178 KJ/Kg K

Mass flow rate of hot water m = 0.041667 Kg/sec

THEORY

Heat exchangers are devices in which heat is transferred from one fluid to

another by conduction and convection. Common examples of the heat exchangers are

the radiator of a car, condenser at the back of domestic refrigerator etc. Heat

exchangers are classified mainly into three categories:

1. Transfer type

2. Storage type

3. Direct contact type

Transfer types of heat exchangers are most widely used. A transfer type of heat

exchanger is one in which both fluids pass simultaneously through the device and heat

is transferred through separating walls. Transfer type of exchangers are further

classified as

1. Parallel flow type

2. Counter flow type

3. Cross flow type

FORMULAE USED

Heat transfer rate Qh = m x Cp x ( Thi – Tho)

Logarithmic mean temperature difference

LMTD = (( Thi - Tci) - ( Tho - Tco))

Ln((Thi - Tci)/( Tho - Tco))

Area of inner copper tube A = x d x l m2

Exercise No: 03

TEST ON HEAT EXCHANGER-PARALLEL FLOW Date:



MODEL CALCULATION


=

= Watts


LMTD = (( Thi - Tci) - ( Tho - Tco))

Ln((Thi - Tci)/( Tho - Tco))

=

=

Area of inner copper tube A = x d x l

=

= m2

The overall heat transfer Co-efficient

U = Qh

( A x LMTD )

=

= W/ m2 K

Effectiveness E = Actual heat transfer

Maximum possible heat transfer

= (Thi – Tho)

( Thi - Tci)

=




U = Qh W/ m2 K

( A x LMTD )

Effectiveness E = Actual heat transfer __


= (Thi – Tho)

( Thi - Tci)

PROCEDURE

Connect water supply at the back of the unit. The inlet water flows through

geyser and inner pipe of the heat exchanger and flows out in only one direction.

Also the inlet water flows through the annulus gap of the heat exchanger and

flows out.

For parallel flow open valve V1,V3 and V5(Hot water)

Control the hot water flow approximately 2 lit. /min. and cold water flow

approximately 5 lit. /min.

Switch ON the geyser. Allow the apparatus to run for some times say

5-10 minutes to reach steady state.

Note temperature T1 and T2 (hot water inlet and outlet temperature

respectively).

Under parallel flow condition T3 is the cold water inlet temperature and T4 is the

cold water outlet temperature.

Note the temperature T3 and T4.

Note the time for 1 litre flow of hot and cold water. Calculate mass flow rate

Kg/sec.

Change the water flow rates and repeat the experiment.

RESULT

Thus, the heat transfer test in parallel flow heat exchanger was conducted and

the following values were determined.

LMTD =

Heat transfer =

Overall heat transfer =

Effectiveness =



TEST ON HEAT EXCHANGER-COUNTER FLOW

Sl.no

Time for

Hot water

flow 4

litres (sec)

Temperature in ºC Heat

transfer

Q

Watts

LMTD

ºC

Overall heat

transfer

co-efficient U

W/ m2 K

Effectiveness

E T1

(Thi )

T2

(Tho )

T3

(Tci )

T4

(Tco )



AIM

To determine the effectiveness of the counter flow heat exchanger.

APPARATUS REQUIRED

1. Heat exchanger apparatus

2. Stop watch

3. Measuring flask

SPECIFICATIONS

Length of the heat exchanger = 1800 mm

Inner copper tube ID = 12 mm

OD = 15 mm

Outer GI tube ID = 40 mm

Specific heat capacity of water Cp = 4.178 KJ/Kg K

Mass flow rate of hot water m = 0.041667 Kg/sec

DESCRIPTION

The apparatus consists of a concentric tube heat exchanger. The hot fluid i.e. hot

water is obtained from an electric geyser and flows through the inner tube. 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 apparatus or a counter flow apparatus. This can be done by

operating the different valves provided. Temperatures of the fluids can be measured

using thermometers. Flow rate may be measured using stop clock and measuring flask.

The outer tube is provided with adequate asbestos rope insulation to minimize the heat

loss to the surroundings.

FORMULAE USED



LMTD = (( Thi - Tco) - ( Tho - Tci))

Ln((Thi - Tco)/( Tho - Tci))

Area of inner copper tube A = x d x l m2

Exercise No: 04

TEST ON HEAT EXCHANGER-COUNTER FLOW Date:



MODEL CALCULATION


=

= Watts


LMTD = (( Thi - Tco) - ( Tho - Tci))

Ln((Thi - Tco)/( Tho - Tci))

=

=

Area of inner copper tube A = x d x l

=

= m2


U = Qh

( A x LMTD )

=

= W/ m2 K

Effectiveness E = Actual heat transfer


= (Thi – Tho)

( Thi - Tci)

=




U = Qh W/ m2 K

( A x LMTD )

Effectiveness E = Actual heat transfer __


= (Thi – Tho)

( Thi - Tci)

PROCEDURE

Connect water supply at the back of the unit. The inlet water flows through

geyser and inner pipe of the heat exchanger and flows out in only one direction.

Also the inlet water flows through the annulus gap of the heat exchanger and

flows out.

For counter flow open valve V2,V4 and V5(Hot water)

Control the hot water flow approximately 2lit. /min. and cold water flow

approximately 5 lit. /min.

Switch ON the geyser. Allow the apparatus to run for some times say

5-10 minutes to reach steady state.

Note temperature T1 and T2 (hot water inlet and outlet temperature

respectively).

Under counter flow condition T4 is the cold water inlet temperature and T3 is the

cold water outlet temperature.

Note the temperature T3 and T4.

Note the time for 1 litre flow of hot and cold water. Calculate mass flow rate

Kg/sec.

Change the water flow rates and repeat the experiment.

RESULT

Thus, the heat transfer test in counter flow heat exchanger was conducted and

the following values were determined.

LMTD =

Heat transfer =

Overall heat transfer =

Effectiveness =



LAGGED PIPE

Sl. No. Voltage Current Heater temperature Average

temperature

Asbestos

temperature Average

temperature

Sawdust

temperature Average

temperature T1 T2 T3 T4 T5 T6 T7 T8

Volts Amps (ºC) (ºC) (ºC) (ºC) (ºC) (ºC) (ºC) (ºC) (ºC) (ºC) (ºC)



AIM

To determine the heat transfer and thermal conductivity at the lagged pipe

apparatus.

DESCRIPTION

The insulation is defined as a material, which retards the heat flow with

reasonable effectiveness. Heat is transferred through insulation by conduction,

convection and radiation or by the combination of these three. There is no insulation

that is 100% effective to prevent the flow of heat under temperature gradient.

The apparatus consists of a rod heater with asbestos lagging. The assembly is

covered by MS pipe. Saw dust is filled between the asbestos lagging and MS pipe.

The set-up is shown in the figure. Let r1 be the radius of the heater, r2 be the

radius of the heater with asbestos lagging and r3 be the inner radius of the outer MS

pipe.

Now the heat flow through lagging materials is given by

Q = K12L (dt) or K22L (dt)

ln (r2/r1) ln (r3/r2)

Where,

dt → the temperature across the lagging

K1 → the thermal conductivity of asbestos lagging material

K2 → the thermal conductivity of saw dust

L → the length of the cylinder

Knowing the thermal conductivity of one lagging material the thermal

conductivity of the other insulating material can be found.

PROCEDURE

Switch on the unit and check if all channels of temperature indicator showing

proper temperature.

Switch on the heater using the regulator and keep the power input at some

particular value.

Allow the unit to stabilize for about 20 to 30 minutes.

Exercise No: 07

HEAT FLOW THROUGH LAGGED PIPE Date:



MODEL CALCULATION

Average Temp. of heater = T1+T2+T3

3

=

Average Temp. of asbestos = T4+T5+T6

3

=

Average Temp. of Sawdust lagging = T7+T8

2

=

The heat flow from the heater to

outer surface of asbestos lagging Q = (K12l (dt) )

( ln (r2/r1) )

where,

K1→ Thermal conductivity of asbestos Lagging = 0.1105 W/mºC

r1→Radius of the heater = 10 mm

r2→Radius of the asbestos lagging = 20 mm

r3→radius of sawdust = 40 mm

l→Length of the heater = 500 mm

Q = Watts

The thermal conductivity of saw dust lagging can be determined by

Q = (K22l (dt) )

( ln (r3/r2) )

=

=

K2 = W/mºC



Now note down the ammeter, voltmeter reading which gives the heat input.

Temperatures 1, 2 and 3 are the temperatures of heater rod, 4, 5 and 6 are the

temperatures on the asbestos layer; 7 and 8 are temperatures on the saw dust

lagging. The average temperature of each cylinder is taken for calculation.

The temperatures are measured by thermocouple (Fe/Ko) with multipoint digital

temperature indicator.

The experiment may be repeated for different heat inputs. RESULT

Thus, the heat transfer and thermal conductivity of the lagged pipe apparatus

were determined

Q = Watts

K2 = W/m ºC



HEAT CONDUCTION THROUGH COMPOSITE WALLS

The temperatures are measured from bottommost plate to top plate in the order of T1 to T8 as per diagram.

Sl.No. Voltage Current Wood

Temp.

Asbestos

Temp.

Mild steel

Temp.

Heater

Temp.

Heater

Temp.

Mild steel

Temp.

Asbestos

Temp.

Wood

Temp.

V (volts) A (Amps) T1 (ºC) T2 (ºC) T3 (ºC) T4 (ºC) T5 (ºC) T6 (ºC) T7 (ºC) T8 (ºC)



AIM

To determine the heat transfers through various materials.

INTRODUCTION

When heat conduction takes place through two or more solid materials of

different thermal conductivity, the temperature drop across each material depends on

the resistance offered to heat conduction path and the thermal conductivity of each

material.

DESCRIPTION

The experimental set-up consists of test specimens made of different materials

aligned together on both sides of the heater unit. The first test disc is next to a

controlled heater. The temperatures at the interface between the heater and the disc is

measured by a thermocouple, similarly temperatures at the interface between discs are

measured. Similar arrangement is made to measure temperatures on the other side of

the heater. The whole set-up is kept in a convection free environment. The temperatures

are measured using thermocouples (Fe/ ko) with multi point digital temperature

indicator. A channel frame with a screw rod arrangement is provided for proper

aligning of the plates without air gap.

The apparatus uses known insulating of large area of heat transfer to enable

unidirectional heat flow. The apparatus is used mainly to study the resistance offered by

different slab materials and to establish the heat flow is similar to that of current flow in

an electrical circuit. However due to certain limitations of the experimental set-up of

this nature, such as air gap between the plates and heat flow through the sides of the

plates it is difficult to get theoretical results.

PROCEDURE

Turn the screw rod handle clockwise to tighten the plates.

Switch on the unit and then the regulator clockwise to provide any desired heat

input.

Note the ammeter and volt meter readings.

Wait still steady state temperatures are reached.

(The steady state condition is defined as when the temperatures gradient across the

plates does not change with time)

When steady state is reached, note temperatures and find the temperature

gradient across each slab.

The average temperature gradient between top and bottom slaps from the heater

are considered for calculations.

Exercise No: 08

HEAT CONDUCTION THROUGH COMPOSITE WALLS Date:



MODEL CALCULATION

(T4+T5) - (T1+T8)

Total heat flow through plates Q = 2 2

R

=

Q = Watts

(T4+T5) - (T3+T6)

Heat flow through MS plate Q1 = 2 2

R1

=

Q1 = Watts

(T3+T6) - (T2+T7)

Heat flow through Asbestos plate Q2 = 2 2

R2

=

Q2 = Watts

(T2+T7) - (T1+T8)

Heat flow through Wooden plate Q3 = 2 2

R3

=

Q3 = Watts



FORMULAE USED

(T4+T5) - (T1+T8)

Total heat flow through plates Q = 2 2

R

(T4+T5) - (T3+T6)

Heat flow through MS plate Q1 = 2 2

R1

(T3+T6) - (T2+T7)

Heat flow through Asbestos plate Q2 = 2 2

R2

(T2+T7) - (T1+T8)

Heat flow through Wooden plate Q3 = 2 2

R3

Where,

R1 → Resistance of mild steel = L1/AK1

R2 → Resistance of asbestos = L2/AK2

R3 → Resistance of wood = L3/AK3

Where,

A → Area of the plate

K → Thermal conductivity of the material

L → Thickness of the plate

Note:

Thermal conductivity of mild steel = 72.7 w/mk

Thickness of mild steel = 10 mm

Thermal conductivity of asbestos sheet = 0.069 w/mk

Thickness of asbestos sheet = 5 mm

Thermal conductivity of wood = 0.052 w/mk

Thickness of wood = 8 mm

Diameter of plates = 300 mm

Area of plates (/4) x D2 = 0.070686 m2

Resistance of mild steel R1 = 0.001946

Resistance of asbestos R2 = 1.025153

Resistance of wood R3 = 2.176478

Total resistance offered by the plates (R) = R1+R2+R3

R = 3.2035764



RESULT

Thus, the heat transfers through various materials were determined.

Heat flow through MS plate Q1 = Watts

Heat flow through Asbestos plate Q2 = Watts

Heat flow through Wooden plate Q3 = Watts

Total heat flow through plates Q = Watts

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