Submitted by: Submitted by aksizi7
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Pages: 5
Category: Science and Technology
Date Submitted: 02/18/2013 09:15 AM
Introduction/Background
The purpose of this project was to perform a radiation analysis of a storage tank of liquefied methane and its distribution. The net radiative transfer to the liquid without a radiation shield was found and compared to the radiative transfer with a shield to understand the effects of the shield. The calculated flow rate for each case was then compared to the boil-off gas compressor requirement of .2% full tank capacity per day in order to ensure that the set-up was feasible and desirable from a safety and engineering standpoint. For the first case, three radiated surfaces were considered: a black liquid, a gray cylindrical sidewall (emissivity 0.2), and a black roof. The second case included a gray radiation shield (emissivity 0.25) between the sidewall and the roof.
Theoretical Development (see attached need circuits, diagrams, etc)
The following equations were used in the numerical evaluation:
The constants are shown in the table below and are the same for each case:
CONSTANTS
Property | Value | Units |
diam1 | 50 | m |
diam4 | 50 | m |
height | 40 | m |
Area1 | 1963.50 | m^2 |
Area2 | 1963.50 | m^2 |
ε3 | 0.2 | |
ε4 | 0.25 | |
T1 | 113.15 | K |
T2 | 353.15 | K |
T3 | 233.15 | K |
σ | 5.67E-08 | W/(m^2*K^4) |
Eb1 | 9.29 | W/m^2 |
Eb2 | 881.90 | W/m^2 |
Eb3 | 167.54 | W/m^2 |
In each case, the values of Ri and Rj and S were the same, and were found using the equations for coaxial parallel disks:
Ri= riL
Rj= rjL
S=1+ 1+Rj2Ri2
From this, the first value of F for use in the equivalent circuits could be found:
Fij= 12 { S-[S2-4(rjri)2]12 }
Then, for Case 1, using this first value of F, the remaining were found and allowed for the complete equivalent resistances to be calculated:
F13 =1- F12
F31= A1*F13/A3
F32=F31
R12=1A1*F12
R13=1A1*F13
R23=1A3*F32
R3=1- ∈3∈3*A3
In case 2, the process was similar, but because of the extra surface the process was altered somewhat:
F14= 12 {...