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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) Negligibleheat 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

(

)

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 ofgauge. 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 acircular 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

(

)

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

<

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

PROBLEM 3.99 KNOWN: Net radiative flux to absorber plate. FIND: (a) Maximum absorber plate temperature, (b) Rate of energy collected per tube. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (x) conduction along absorber plate, (3) Uniform radiation absorption at plate surface, (4) Negligible losses by conduction throughinsulation, (5) Negligible losses by convection at absorber plate surface, (6) Temperature of absorber plate at x = 0 is approximately that of the water. PROPERTIES: Table A-1, Aluminum alloy (2024-T6): k ≈ 180 W/m⋅K. ANALYSIS: The absorber plate acts as an extended surface (a conduction-radiation system), and a differential equation which governs its temperature distribution may be obtained by applyingEq.1.11a to a differential control volume. For a unit length of tube q′x + q′′ ( dx ) − q′x+dx = 0. rad With and it follows that, q′′ − rad d 2T dx 2 d  dT   − kt dx  = 0 dx   q′x+dx = q′x + q′x = − kt dT dx dq′x dx dx

q′′ + rad = 0 kt

Integrating twice it follows that, the general solution for the temperature distribution has the form, q′′ T ( x ) = − rad x 2 + C1x+C2 . 2kt...
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