Heat Transfer Radiation Lab Report

3378 WordsMar 19, 201314 Pages
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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