 #1
baseballfan_ny
 91
 23
 Homework Statement:

A sheet (“sheet 1”) of area A having emissivity ##e_1## is held at temperature ##T_1## so that it emits a total power ##P_1## to the right (i.e., energy per unit time). Let σB be the StefanBoltzmann constant and neglect all edge effects.
a. In terms of ##P_1##, calculate temperature ##T_1## of sheet 1.
Two additional black sheets, each of area A and having emissivities ##e_2## = 1 and ##e_3## = 1, are now added to the right of sheet 1. (See figure below) The spacings between the pairs of sheets is d, such that ##d^2## is much less than A . The temperature ##T_2## of the middle sheet (2) is allowed to vary and the righthand sheet (3) is held at a fixed temperature ##T_3## = 0.
b. Assume that sheet 2 reaches a steady state temperature ##T_2##. What is the power flow ##P_{23}## (total power, not power per unit area) emitted by sheet 2 to the righthand sheet, i.e., from sheet 2 to sheet 3? Your answer should be given in terms of in terms of ##T_2##, A, d, ##\sigma_B##, and ##e_1## only. [Note: Not all of these parameters should appear in your answer.]
c. Determine ##T_2## in terms of ##T_1##, A, d, ##\sigma_B## and ##e_1## only.
 Relevant Equations:

StefanBoltzmann
## J_u = e \sigma_B T^4 ##
I need someone to check my work, because I'm getting weird results that I'm not able to interpret physically for parts b and c. Thanks in advance.
For part a...
##J_u = e_1 \sigma_B T^4##
##P_1 = AJ_u = e_1 \sigma_B AT_1^4##
## T_1 = \left( \frac {P_1} {e_1 \sigma_B A} \right)^{\frac 1 4} ##
For part b...
I'm using the idea that energy emitted by sheet 2 = energy absorbed by sheet 2
## \sigma_B T_2^4 + \frac {P_{23}} {A} = e_1 \sigma_B T_1^4 + (1  e_1) \sigma_B T_2^4 ##
## \frac {P_{23}} {A} = e_1 \sigma_B \left( T_1^4  T_2^4 \right) ##
## P_{23} = A e_1 \sigma_B \left( T_1^4  T_2^4 \right) ##
For part c...
I'm using the idea that energy emitted by sheet 1 = energy absorbed by sheet 1
## e_1 \sigma_B T_1^4 + \left( 1  e_1 \right) \sigma_B T_2^4 = \sigma_B T_2^4 ##
## e_1 \sigma_B T_1^4 = e_1 \sigma_B T_2^4 ##
## T_1 = T_2##
Now this has been bothering me, because if ##T_1 = T_2##, then ##P_{23} = 0##? I can't seem to make sense of that.
For part a...
##J_u = e_1 \sigma_B T^4##
##P_1 = AJ_u = e_1 \sigma_B AT_1^4##
## T_1 = \left( \frac {P_1} {e_1 \sigma_B A} \right)^{\frac 1 4} ##
For part b...
I'm using the idea that energy emitted by sheet 2 = energy absorbed by sheet 2
## \sigma_B T_2^4 + \frac {P_{23}} {A} = e_1 \sigma_B T_1^4 + (1  e_1) \sigma_B T_2^4 ##
## \frac {P_{23}} {A} = e_1 \sigma_B \left( T_1^4  T_2^4 \right) ##
## P_{23} = A e_1 \sigma_B \left( T_1^4  T_2^4 \right) ##
For part c...
I'm using the idea that energy emitted by sheet 1 = energy absorbed by sheet 1
## e_1 \sigma_B T_1^4 + \left( 1  e_1 \right) \sigma_B T_2^4 = \sigma_B T_2^4 ##
## e_1 \sigma_B T_1^4 = e_1 \sigma_B T_2^4 ##
## T_1 = T_2##
Now this has been bothering me, because if ##T_1 = T_2##, then ##P_{23} = 0##? I can't seem to make sense of that.