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Heat Transfer Radiation Lab Report

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Module : Heat Transfer – Free Convection and Radiation Laboratory

Date : 22nd March 2012

CONTENTS

INTRODUCTION 3

AIMS & OBJECTIVES 3

Objectives 3 To investigate Free Convection and Radiation 3 Theory 3

EXPERIMENT 3

Apparatus Used 3 Procedure 4

RESULTS, CALCULATIONS, OBSERVATIONS & CONCLUSIONS 5

Observations During Tests 5 Table 1 5 Table 2 5

Calculations 6

Calculating Power (Watts) 6 Calculating Heat Transfer Emissivity (Ɛ) 6 Emisssivity of a black body 6 Calculating Q rad 6 Calculating Q rad 6 Calculating Q conv 7 Equation for Free Convection 7 Percentage values calculation 7 Absolute Pressure calculation 7 Graph of Pressure Against Temp Difference 8

Conclusions 8

Conclusion …show more content…

|
|TEL –TV (K) |(Mb)^1/4 |W |W |% |% | WM^-2K^-1 |
|144 |2^1/4 = 1.19 |4.87 |1.14 |81 |19 |2.57 WM^-2K^-1 |
|133 |16^1/4 = 2 |4.31 |1.66 |72 |28 |4.06 WM^-2K^-1 |
|123 |61^1/4 = 2.79 |3.81 |2.13 |64 |36 |5.64 WM^-2K^-1 |
|111 |203^1/4 = 3.77 |3.25 |2.71 |55 |45 |7.95 WM^-2K^-1 |
|97 |503^1/4 = 4.73 |2.68 |3.24 |45 |55 |10.88 WM^-2K^-1 |
|87 |1018^1/4 = 3.22 |2.27 |3.65 |38 |62 |13.66 WM^-2K^-1 |

Table 2

Calculations

Heat losses in the connecting leads Q = (0.94 x Volts x Amperes) in watts

Calculating Power (Watts)

Power = Volts x Amperes (Watts)

Power = 8.21volts x 0.779 amps = 6.39 (W) x Heat loses

Power = 6.39 (W) x 0.94 = 6.01 Watts

Heat Transfer = 0.94 x 8.21 x 0.779 = 6.01 watts

Calculating Heat Transfer Emissivity (Ɛ)

Emisssivity of a black body ( copper ) = 1

If Ɛ = >1 Use Ɛ = 0.97 to calculate Q rad

Ɛ = Q rad Joules or Watts A x σ x

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