A User's Manual for DELSOL3 - prod.sandia.gov - Sandia National ...

Sample Problem 3b - Generating a Flux MaD for a Cavity Roof
Problem Statement
For the system of Problem 3a, determine the incident flux on the roof of the
cavity receiver.
Input Cards
SAMPLE PROBLEM 3b
$BASIC IPROB=O. ITAPE=3. TDESP=125.0 $
$FIELD $
BHSTAT $
OREC B
BNLFLUX IFLX=1, IFXOUT(B.I)-I. IFI.AUT=3.
XFC-0.0. YFC=O.O. ZFC=16.10, POLF=180.0. AZMF=180.0.
NXFLX=5. NYFLX=5. FAZMIN=-22.5. FAZMAX=+22.5.
FZMIN=O.O. FZMAX=22.5 $
BNLEFF B
BREC W=-100. $
Analysis of Input
This is a performance rerun of a previously optimized system, as indicated by
the specified values in Namelist $BASIC$. The system description is read from
the file on Unit 30. The flux points are specified to be on a plane (IFLAUT=3)
whose center is at the location XFC, YFC, ZFC. The values of FAZMIN, FAZ-
MAX, FZMIN, and FZMAX are measured from that location. The plane is
downward facing (POLF=180.0). Because the center of the flux plane is also the
center of the tower, and because the cavity design (RWCAV and W) specify this
design’s back wall to be at the center of the tower, FZMIN is set to zero, and FZ-
MAX is set to the depth of the cavity (W/2). This means that the point (0,O) on
the plane is the same point as the top point on the centerline of the cavity wall
used in Sample Problem 3a.
Comments on OutDut
All of the output except for flux points and the flux map will be exactly the
same as for the final performance run of Problem 3a. The smart user might have
specified IPROB=2 on Namelist $BASIC$ as a means of saving computer time,
since the purpose of this run wils only to generate a flux map.
The flux map is generated by DELSOL assuming that the flux surface is the
only surface within the aperture. Thus, for points on the flux surface which are
outside the semi-circular area of the cavity, the flux levels would in real life be
nonsense, since the flux would strike the wall of the cavity instead. At the flux
point common with Problem 3a, the flux is calculated to be 56% of that calcu-
lated in 3a. The difference is caused by the difference in the incident angle of flux
on the two different flux surfaces. The cavity wall is more nearly normal to the
flux, and so has a higher normal flux level than the roof, which is closer to being
parallel to the incident radiation.
217