This sample problem demonstrates the use of FLUID220 to predict the sound wave reflection due to the existence of an unsymmetric 2×2 admittance matrix. The transfer admittance matrix is defined as:
Both ports of the 2 x 2 network are connected to the fluid.
The sound wave characteristic impedance is Z0 = 417.45 N⋅s/m3.
In terms of the network theory, the reflection coefficients are 0.15 and 0.20 at port 1 and port 2, respectively.
/batch,list /title, Transfer Admittance Matrix in the Fluid /nopr /prep7 pi=acos(-1) k=8*pi ! wave number rho0 = 1.21 ! mass density c0 = 345 ! sound speed freq = k*344/(2*pi) ! working frequency wavelen = 2*pi/k ! wavelength ! structure dimensions h = wavelen/12 ! mesh size d = 2*h l = 0.75*wavelen z1 = l/2 z2 = z1+0.05*h z3 = z2+l/2 ! define elements and material et,1,220,,1 mp,dens,1,rho0 mp,sonc,1,c0 mp,dens,2,rho0 mp,sonc,2,c0 ! set transfer admittance matrix parameters y11r=0.634852E-02 ! Re(Y11) y11i=0. ! Im(Y11) y12r=-0.437959E-02 ! Re(Y12) y12i=0.437959E-02 ! Im(Y12) y21r=-0.437959E-02 ! Re(Y21) y21i=-0.437959E-02 ! Im(Y21) y22r=0.598419E-02 ! Re(Y22) y22i=0. ! Im(Y22) ! define transfer admittance matrix tbdel,perf,2 tb,perf,2,,,YMAT tbdata,1,y11r,y11i,y12r,y12i,y21r,y21i tbdata,7,y22r,y22i tblist,perf,2 ! define normal velocity pa=1 z0=rho0*c0 un=-pa/z0 ! create model block,-d/2,0,-d/2,d/2,0,z1 block,-d/2,0,-d/2,d/2,z1,z2 block,-d/2,0,-d/2,d/2,z2,z3 vglue,all mshape,0,3d esize,h type,1 mat,1 vsel,s,loc,z,0,z1 vsel,a,loc,z,z2,z3 vmesh,all ! mesh with normal material type,1 mat,2 vsel,s,loc,z,z1,z2 vmesh,all ! mesh with admittance matrix ! define ports for admittance matrix esel,s,mat,,2 nsle,s nsel,s,loc,z,z1 sf,all,port,1 ! port 1 of 2-port network nsel,s,loc,z,z2 sf,all,port,2 ! port 2 of 2-port network alls ! define boundary condition nsel,s,loc,z,z3 sf,all,impd,z0 ! matched outlet ! excitation at transparent port nsel,s,loc,z,0 sf,all,shld,un ! normal velocity excitation sf,all,impd,z0 ! matched inlet sf,all,port,10 ! transparent port alls fini ! perform solution /solu antype,harmic hropt,auto harfrq,freq solve fini ! post-processing /post1 nsel,s,loc,z,0 nsel,r,loc,x,0 nsel,r,loc,y,-d/2 nod=0 node=ndnext(nod) set,,,,,,,1 ! real solution *get,frq,active,0,set,freq pr=pres(node) set,,,,,,,2 ! imaginary solution pi=pres(node) ref=sqrt((pa-pr)**2+pi**2)/pa /com, /com,***** Computed Reflection Coefficient ***** /com, *vwrite, (' Frequency Computed |R|') *vwrite,frq,ref (1x,e14.7,6x,e14.7) fini