Fisica

PROBLEM 3.98 KNOWN: Radius, thickness, and incident flux for a radiation heat gauge. FIND: Expression relating incident flux to temperature difference between center and edge of gauge. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r (negligible temperature drop across foil thickness), (3) Constant properties, (4) Uniform incident flux, (5) Negligible heat loss from foil due to radiation exchange with enclosure wall, (6) Negligible contact resistance between foil and heat sink.

ANALYSIS: Applying energy conservation to a circular ring extending from r to r + dr, ′′ q r + qi ( 2π rdr ) = q r+dr , Rearranging, find that q′′ ( 2π rdr ) = i d  dT  ( −k2π rt ) dr  dr dr   q r = − k ( 2π rt ) dT , dr q r+dr = q r + dq r dr. dr

d  dT  q′′ i  r dr  = − kt r. dr   Integrating, r dT q′′r 2 = − i + C1 dr 2kt and q′′r 2 T ( r ) = − i + C1lnr+C2 . 4kt

With dT/dr|r=0 =0, C1 = 0 and with T(r = R) = T(R), q′′R 2 T (R ) = − i + C2 4kt or q′′R 2 C2 = T ( R ) + i . 4kt

Hence, the temperature distribution is T (r ) = q′′ i R 2 − r 2 + T ( R ). 4kt

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Applying this result at r = 0, it follows that q′′ = i 4kt 4kt  T ( 0 ) − T ( R ) = ∆T.   R2 R2

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COMMENTS: This technique allows for determination of a radiation flux from measurement of a temperature difference. It becomes inaccurate if emission from the foil becomes significant.

PROBLEM 3.98 KNOWN: Radius, thickness, and incident flux for a radiation heat gauge. FIND: Expression relating incident flux to temperature difference between center and edge of gauge. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r (negligible temperature drop across foil thickness), (3) Constant properties, (4) Uniform incident flux, (5) Negligible heat loss from foil due to radiation exchange with enclosure wall, (6) Negligible contact resistance between foil and heat sink.

ANALYSIS: Applying energy conservation to a