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Unformatted text preview: 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, ( ) ( ) r r i r+dr r r+dr r dT dq q q 2 rdr q , q k 2 rt , q q dr. dr dr + = = = + Rearranging, find that ( ) ( ) i d dT q 2 rdr k2 rt dr dr dr = i d dT q r r . dr dr kt = Integrating, ( ) 2 2 i i 1 1 2 dT q r q r r C and T r C lnr+C . dr 2kt 4kt = + = + With dT/dr| r=0 =0, C 1 = 0 and with T(r = R) = T(R), ( ) ( ) 2 2 i i 2 2 q R q R T R C or C T R . 4kt 4kt = + = + Hence, the temperature distribution is ( ) ( ) ( ) 2 2 i q T r R r T R . 4kt = + Applying this result at r = 0, it follows that ( ) ( ) i 2 2 4kt 4kt q T T R T . R R = = < 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....
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This note was uploaded on 01/24/2012 for the course MECHE ALL taught by Professor Any during the Spring '11 term at Carnegie Mellon.
- Spring '11