Lagged Pipe 6r3o2n

HEAT TRANSFER LAB Presentation

Jadavpur University Power Engineering Dept

Created by : ** PRASUN CHOWDHURY (000911501001) **KISHORE MANDI (000911501002) **SUBEDIT DAS (000911501005)

TOPIC : To determine the thermal conductivity of insulation for a LAGGED PIPE.

INTRODUCTION 

Lagging of pipes is required to prevent LEAKAGES of heat. The apparatus is designed to study the lagging phenomenon. In Lagged Pipe apparatus, three concentric pipes are arranged between two s.The gap between the pipes are filled compactly by two different insulating materials and heater is provided at the centre of inner pipe. Temperature at various points are measured with thermocouples. Heat input is measured by voltmeter and ammeter readngs.

APPARATUS USED voltmeter

ammeter

Temperature indicator

knob

SECTIONAL VIEW OF THE APPARATUS

ASSUMPTIONS

There is no heat loss from the two ends of the pipe.

Insulations No heat lossoo preventing the heat to come out axially

There is one dimensional radial heat conduction through the insulator.

The other assumptions are: We assume steady state heat conduction.  We have to assume the length of the pipe and the length of the heater to be the same. 

OBSERVATION TABLE Run No.

Voltage Current (I) (V)

1

40

2

50

3

70

Ta (K)

Tb (K)

Tc (K)

Td (K)

R1=25

R2=38

R3=50

R4=62

0.18 62. 41. 35. 26 7 3 7 0.22 74 46. 38. 28 7 7 0.31 10 59. 47. 31 2 3 7

CALCULATIONS(experimental) The rate of heat generation is given by q̇=Vi…….(1) Where V=applied voltage i=current through the heater At the steady state condition, the above heat is dissipated radially. From the governing equation of heat transfer by conduction we get q̇=-kA(dT/dr) Now A=2πrL, So q̇=-k(2πrL)dT/dr……………………(2) So equating (1) and (2) we get the following equation k=Vi/(2πrLdT/dr) This is the value of k obtained experimentally.

CALCULATIONS(theoretical) T1 -T12 are the temperatures of 12 thermocouples. Ta=(T1 + T2 + T3 )/3 Tb=(T4 + T5 + T6)/3 Tc=(T7 + T8 + T9 )/3 Td=(T10 + T11 + T12 )/3 The theoretical working equation is q̇=( Ta - Td)/[ln(R4/R1)/2πkL] where k=thermal conductivity L=length of the pipe k= q̇ ln(R4/R1)/ [2πL(Ta - Td)] This is the value of k obtained theoretically.

RESULTS RUN no.

Ktheo

Kexpt

(W/mK)

(W/(mK)

1

0.244

0.298

18

2

0.25

0.36

38

3

0.31

0.45

32

% Error

Temperature vs Radial length graph 300 250 200 t3 t2 t1

150 100 50 0 25

38

50

62

ACKNOWLEDGEMEN T We are very much grateful to the following teachers,without them this expt would not have been possible.

Prof. Amitava Dutta  Prof. A. K. Satra  Sri Bireshwar Paul  Sri Atish Nandi 

THANK YOU