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- FLAIR Library case: Cleansing station

to the another Air enters each cleansing room near the floor and leaves near the ceiling The source of pollution is modelled with several pipe like exhausts of different flow rate and pollutant concentration Temperature is not solved Turbulence is modelled by means of the k e model Geometry 1 Top view showing the vent outlet and pollutant exhausts Geometry 2 Air communication between the rooms Results Velocity vectors on

Original URL path: http://www.cham.co.uk/phoenics/d_polis/d_spp/flair/d_applic/libc106.htm (2016-02-15)

Open archived version from archive - FLAIR Library case: Comfort in a supermarket

coefficients for the freezers and chillers are all set to 12W m 2 K The results show both air temperature distribution and dry resultant temperature which also takes into account radiation and air velocity In addition the ISO 7730 comfort indices Predicted Percent Dissatisfied PPD and Draught Rating PPDR are calculated PPD is an index that predicts the percentage of a large group of people who are likely to feel

Original URL path: http://www.cham.co.uk/phoenics/d_polis/d_spp/flair/d_applic/libc105.htm (2016-02-15)

Open archived version from archive - Section 2 of the PHOENICS Encyclopaedia article on multi-phase flow

nuclear steam generators and in pressurized water nuclear reactors during loss of coolant accidents combustion of pulverized coal clouds in furnaces or of oil droplet sprays in furnaces diesel engines and gas turbines rain and snow fall phenomena in the atmosphere the motion of sand carried by the wind of by river flows fluidized bed phenomena An example from the PHOENICS Input Library 3D Nuclear Power STEAM GENERATOR Library Case W802 DETAILS A cylindrical tube and baffle steam generator is simulated The heat flux distribution from hot water in the immersed tube bank is prescribed The flow is three dimensional and steady Steam and water enthalpies volume fractions and velocities have been calculated by use of the IPSA procedure A polar coordinate grid is used Steam generators of this kind are employed in nuclear power plant of the pressurized water reactor kind Contours of volume fraction of steam formed from water entering at the bottom of the cylindrical vessel The steam generation pattern is not symmetrical because the temperature of the water in the immersed U tubes falls between inlet and outlet The water enthalpy measured above a base value corresponding to the entry temperature just below boiling point The

Original URL path: http://www.cham.co.uk/phoenics/d_polis/d_enc/enc_ipsa.htm (2016-02-15)

Open archived version from archive - 274

0 RSET B DWSTR 0 0 20 5 6 0 5 0 RSET B ROAD 0 0 0 5 0 20 10 0 GROUP 6 Body fitted coordinates or grid distortion GROUP 7 Variables stored solved named SOLVE P1 U1 V1 W1 Select whole field solution procedure for pressure and point by point for velocities SOLUTN P1 Y Y Y N N N SOLUTN U1 Y Y N Y N Y SOLUTN V1 Y Y N Y N Y SOLUTN W1 Y Y N Y N Y GROUP 8 Terms in differential equations devices GROUP 9 Properties of the medium or media ENUL 1 E 5 ENUT 1 0E 3 GROUP 10 Inter phase transfer processes and properties GROUP 11 Initialization of variable or porosity fields if not conwiz then FIINIT P1 0 0 FIINIT W1 14 0 mult endif Vehicle body CONPOR BODY1 0 0 VOLUME 1 2 2 2 2 4 CONPOR BODY2 0 0 VOLUME 1 2 3 3 3 4 Front wheel CONPOR FWHL 0 0 VOLUME 2 2 1 1 2 2 Rear wheel CONPOR RWHL 0 0 VOLUME 2 2 1 1 4 4 GROUP 13 Boundary conditions and special sources Upstream boundary INLET UPSTR LOW 1 NREGX 1 NREGY 1 1 1 1 VALUE UPSTR P1 14 0 mult VALUE UPSTR W1 14 mult Downstream boundary PATCH DWSTR HIGH 1 NREGX 1 NREGY NREGZ NREGZ 1 1 COVAL DWSTR P1 FIXP 0 COVAL DWSTR U1 ONLYMS 0 0 COVAL DWSTR V1 ONLYMS 0 0 COVAL DWSTR W1 ONLYMS 0 0 Road surface WALL ROAD SOUTH 1 NREGX 1 1 1 NREGZ 1 1 COVAL ROAD W1 LOGLAW 14 0 mult Following Patches are only necessary if BODY definition above is set with positive limits GROUP 15 Termination of sweeps LSWEEP 200 GROUP 17

Original URL path: http://www.cham.co.uk/phoenics/d_polis/d_lecs/conwiz/cases/co274/q1.htm (2016-02-15)

Open archived version from archive - \phoenics\d_priv0\occlec\alferov\alferov.htm

picked up a suggestion made by L Prandtl in 1904 next 2 3 The advent of digital computers Numerical methods executed by humans were too expensive for widespread use but in the 1950s the digital computer changed everything Now such methods could be used and experience quickly revealed that Many different sets of equations could be derived from the same differential equations some less approximate than others Many different procedures could be devised for solving them and these differed in efficiency and success Which were best seemed often to depend on the problem in question Mathematicians provided no guidance as to where superiority was to be found It was left to the engineers who needed the solutions to find the best technique by trial and error I was one of those engineers next 2 4 Solid stress techniques The appendix to the 1970 edition of Timoshenko s book contains a 2 page section on solutions by digital computer and concludes with Similar methods are included in what is now known as the finite element method The stress analysis community had now a flag round which to rally The originator of the name appears to have been the American RW Clough who used it in 1956 However I believe that the method itself was invented in 1953 by my former colleague at Imperial College London John Argyris although his name for it was different the matrix displacement method Interestingly Argyris worked in 1944 with HL Cox who appears to have used a finite stringer method for stresses in aircraft wing panels in 1936 We all climb on the shoulders of our predecessors next 2 5 Fluid flow techniques Argyris was in the Aeronautical Department I in Mechanical Engineering so we knew nothing of each others work My own development of computational techniques had a different origin My first 1951 research was on the burning of liquid fuels Like Kruzhilin for heat transfer and Eckert and Lieblein for mass transfer I exploited themethod of von Karman and Pohlhausen This involved choosing a polynomial form of velocity profile and expressing its coefficients in terms of weighted integrals of the differential equation But even high order polynomials could not well express the complex shapes of profile which appeared in flames so why not I thought use the infinitely flexible step wise approximation instead I had thus stumbled upon the finite volume technique and my student Suhas Patankar expressed the idea in Fortran later drawing crucially on the independent 1965 work of Francis Harlow of Los Alamos next 3 Scientific appraisal 3 1 Essential similarities and differences The solid stress and fluid flow problems are similar in that The differential equations for displacements in solids and velocities in fluids are almost identical The equations for each of the three direction have terms in common dilatation for solids pressure for fluids so they must be solved simultaneously The solid stress and fluid flow problems are different in that The equations for displacement are linked by Poisson

Original URL path: http://www.cham.co.uk/phoenics/d_polis/d_lecs/alferov/alferov.htm (2016-02-15)

Open archived version from archive - Simultaneous Prediction of Solid stress, Heat Transfer and Fluid flow by a Single Algorithm

which makes use of the wall distance WDIS field This like WGAP is also derived from the LTLS distribution The contours of WDIS are displayed in Fig 13 which exhibits the expected maximum of 0 004 between the parallel horizontal walls and a somewhat greater value near the cavity where the true distance from the wall depends on the direction in which it is measured next back or contents LVEL like IMMERSOL is a heuristic model by which is meant that it is incapable of rigorous justification but is nonetheless useful WDIS is calculated once for all at the start of the computation From it and from the developing velocity distribution the evolving distribution of ENUT the effective turbulent viscosity is derived The resulting contours of ENUT are shown in Fig 14 Since the laminar viscosity is of the order of 1 e 5 m 2 s it is evident that turbulence raises the effective value far from the walls by an order of magnitude next back or contents c How the stresses were calculated As will be shown below the equations governing the displacements are very similar to those governing the velocities The CFD code PHOENICS like many others can calculate velocities in fluids but this ability is not needed in the solid region so such codes are usually idle there However PHOENICS can be tricked into calculating what it thinks are velocities everywhere whereas what it actually calculates in the solid regions are displacements The details of the trickery now follow next back or contents 3 The mathematics of the method a Similarities between the equations for displacement and velocity The similarities already referred to are here described for only one cartesian direction x but they prevail for all three directions next back or contents The x direction displacement U obeys the equation del 2 U d dx D C1 Te C3 Fx C2 0 where Te local temperature measured above that of the un stressed solid in the zero displacement condition multiplied by the thermal expansion coefficient D d dx U d dy V d dz W which is called the dilatation Fx external force per unit volume in x direction V and W displacements in y and z directions C1 C2 and C3 are functions of Young s modulus and Poisson s ratio next back or contents When the viscosity is uniform and the Reynolds number is low so that convection effects are negligible the x direction velocity u obeys the equation del 2 u d dx p c1 fx c2 0 where p pressure fx external force per unit volume in x direction c1 c2 the reciprocal of the viscosity next back or contents Notes The two equations are now set one below the other so that they can be easily compared del 2 U d dx D C1 Te C3 Fx C2 0 del 2 u d dx p c1 fx c2 0 The equations can thus be seen to become identical if p c1 D C1 Te C3 which implies D p c1 Te C3 C1 and fx c2 Fx C2 next back or contents The expressions for C1 C2 and C3 are C1 1 1 2 PR C2 2 1 PR YM where PR Poisson s Ratio and YM Young s Modulus and C3 2 1 PR 1 2 PR next back or contents A solution procedure designed for computing velocities will therefore in fact compute the displacements if the convection terms are set to zero within the solid region and a suitable linear relation between D ie d dx U and p is introduced by inclusion of a pressure and temperature dependent mass source term next back or contents b Deduction of the associated stresses and strains The strains ie extensions ex ey and ez are obtained from differentiation of the computed displacements Thus ex d dx U ey d dx V ez d dx W next back or contents Then the corresponding normal stresses sx sy sz and shear stresses tauxy tauyz tauzx are obtained from the strains by way of equations such as sx YM 1 PR 2 ex PR ey and tauxy YM 1 PR 2 0 5 1 PR gamxy where gamxy d dy U d dx V next back or contents c The SIMPLE algorithm for the computation of displacements PHOENICS employs a variant of the SIMPLE algorithm of Patankar Spalding 1972 for computing velocities from pressures under a mass conservation constraint Its essential features are All the velocity equations are solved first with the current values of p The consequent errors in the mass balance equations are computed These errors are used as sources in equations for corrections to p The corresponding corrections are applied and the process is repeated next back or contents All that it is necessary to do in order to solve for displacements simultaneously is in solid regions to treat the dilatation D as the mass source error and to ensure that p obeys the above linear relation to it Therefore a CFD code based on SIMPLE can be made to solve the displacement equations by eliminating the convection terms ie setting Re low and making D linearly dependent on p and temperature T The staggered grid used as the default in PHOENICS proves to be extremely convenient for solid displacement analysis for the velocities and displacements are stored at exactly the right places in relation to p next back or contents 4 Details of the auxiliary models a IMMERSOL summary The solved differential equation is div effective conductivity T3 source 0 effective conductivity 0 75 sigma T3 3 abso scat 1 WGAP source abso sigma T1 4 T3 4 in solids abso large so T3 T1 surface resistances account for non unity emissivities next back or contents Notes The main novelty is the inclusion of WGAP ie the distance between the walls in the formulae This enables a conduction type model to be used even with non participating media Of

Original URL path: http://www.cham.co.uk/phoenics/d_polis/d_lecs/solstr/vancouvr.htm (2016-02-15)

Open archived version from archive - x-direction tensile stress in a cube; s001.htm

Variables STOREd SOLVEd NAMEd ONEPHS T SOLVE P1 U1 STORE PRPS DEN1 ENUL DVO1 DRH1 STORE PTH TH signifies theoretical i e textbook STORE STRX SXTH STORE STRY SYTH STORE STRZ SZTH STORE EPSX EXTH STORE EPSY EYTH STORE EPSZ EZTH Group 8 iteration numbers etc RESREF U1 0 0 GROUP 9 PROPERTIES CSG10 Q1 signal use of the following properties line which correspond to steel MATFLG T NMAT 1 160 7800 0 0 3 473 0 43 0 0 37e 5 0 5E 11 REAL YOUNG POISSON POISSON 0 3 must conform with matflg value YOUNG 1 0 5E 11 must conform with matflg value GROUP 11 INITIAL VALUES FIINIT PRPS 0 FIINIT P1 0 0 FIINIT U1 0 0 FIINIT V1 0 0 FIINIT W1 0 0 FIINIT EXTH 1 234E 11 to print as none FIINIT EYTH 1 234E 11 to print as none FIINIT EZTH 1 234E 11 to print as none FIINIT SXTH 1 234E 11 to print as none FIINIT SYTH 1 234E 11 to print as none FIINIT SZTH 1 234E 11 to print as none PATCH BLOCK INIVAL 2 NX 1 1 1 1 1 1 1 INIT BLOCK PRPS FIXVAL 160 GROUP 13 BOUNDARY SPECIAL SOURCES IF DIRECT THEN PATCH WES WEST 1 1 1 1 1 1 1 1 COVAL WES U1 FIXVAL 0 0 PATCH EFORCE EAST NX 1 NX 1 1 1 1 1 1 1 COVAL EFORCE U1 FIXFLU APPSTR ELSE PATCH EAS EAST NX 1 NX 1 1 1 1 1 1 1 COVAL EAS U1 FIXVAL 0 0 PATCH EFORCE EAST 1 1 1 1 1 1 1 1 COVAL EFORCE U1 FIXFLU APPSTR ENDIF GROUP 15 TERMINATE SWEEPS LSWEEP 100 ISG21 20 to ensure at least isg21 sweeps GROUP 17 RELAXATION CONPROM GROUP 19

Original URL path: http://www.cham.co.uk/phoenics/d_polis/d_lecs/fluisol/q1.htm (2016-02-15)

Open archived version from archive - stress & strain in solids

for example are zero everywhere the equations for nx ny and nz become nx R1 ex R3 et 3 3 1 ny R2 ex R3 et 3 3 2 nz R2 ex R3 et 3 3 3 If however there are no constraints in the y and z direction ey and ez may be non zero while ny and nz obey ny 0 and nz 0 then ey and ez may be eliminated from the equations of section 3 2 with the result nx R8 ex R9 et 3 3 4 The derivation is as follows Equations 3 1 1 3 1 2 and 3 1 3 become nx R1 ex R2 ey ez R3 et 3 3 5 0 R1 ey R2 ex ez R3 et 3 3 6 0 R1 ez R2 ex ey R3 et 3 3 7 Since there is nothing to distinguish ey from ez they may be taken as equal with the results nx R1 ex 2 R2 ey R3 et 3 3 5 0 R1 R2 ey R2 ex R3 et 3 3 6 Substitution for ey from 3 3 6 in 3 3 5 yields nx R1 ex 2 R2 R1 R2 R2 ex R3 et R3 et 3 3 7 ie nx R1 2 R2 R2 R1 R2 ex R3 1 2 R2 R2 R1 R2 et 3 3 8 In terms of R this is 4 Balance of forces The balances of the forces acting on unit volume in the three coordinate directions lead to ddx nx ddy sxy ddz sxz fx 0 4 1 ddy ny ddx sxy ddz syz fy 0 4 2 ddz nz ddx sxz ddy syz fz 0 4 3 5 The differential equations for the displacements 3D 5 1 The x direction displacement u Differentiation of the equation 4 1 leads after substitution from the equations of section 3 1 to ddx R1 ex R2 ey ez R3 et ddy R4 ddy u ddx v ddz R4 ddz u ddx w fx 0 0 Since by definition ex ddx u ey ddy v ez ddz w the strains ex ey ez can be eliminated from the above differential equation with the result ddx R1 ddx u R2 ddy v ddz w R3 et ddy R4 ddy u ddx v ddz R4 ddz u ddx w fx 0 0 If Poisson s ratio R is taken as constant implying that all Rs are also constant this equation can be written in terms of second order operators as follows ddx R1 ddx u R2 ddy v ddz w R3 et R4 d2dyy u R4 ddx ddy v R4 d2dzz u R4 ddx ddz w fx 0 0 with further rearrangement to ddx R1 R4 ddx u R2 R4 ddy v ddz w ddx R3 et R4 d2dxx u d2dyy u d2dzz u fx 0 0 But as a consequence of the above definitions R1 R4 R2 R4 1 2 1 R 1 2 R R10 Therefore the equation can be reduced to R10 ddx dil R3 ddx et R4 div grad u fx 0 where dil ddx u ddy v ddz w the dilatation Re arrangement so as to give prominence to the div grad term leads to div grad u ddx R10 R4 dil R3 R4 et fx R4 0 0 5 1 1 This should be compared with equation C 15 of H T In that equation the multiplier of dil is 1 1 2R R10 R4 2 1 R 2 1 R 1 2R i e 1 1 2R So they equations agree The multiplier of fx in H T is 2 1 R E which is again in agreement However H T are silent about et 5 2 The y and z direction displacements v and w Replacing x successively by y and z and u by v and w leads to the corresponding equations div grad v ddy R10 R4 dil R3 R4 et fy R4 0 0 div grad w ddz R10 R4 dil R3 R4 et fz R4 0 0 Note that as a consequence of the definitions R10 R4 1 1 0 2 R R3 and R3 R4 2 1 R 1 2 R The three equations can thus be written as u x x div grad v R3 dd y dil R11 et R11 f y 0 0 w z z 5 2 1 where it is to be noted that R3 1 1 2 R R11 1 R4 2 1 R 6 The differential equations for the displacements 2D 6 1 When displacements in the third direction are zero Let the zero displacement direction be z Then the equation of section 5 2 still holds for directions x and y with the proviso that ddz terms are absent from both the div grad and dil definitions 6 2 When stressess in the third direction are zero When nz say is zero the analysis must start from nx R5 ex R6 ey R7 et rather than from nx R1 ex R2 ey R3 et Thereafter apart from the substitutions of Rn 4 for Rn and from the absence of z direction derivatives the analysis proceeds as in section 6 The resulting equations are u x x div grad v R3 dd y dil R11 et R11 f y 0 0 6 2 1 whereby it should be noted that R11 remains in place but R13 has replaced R12 7 The differential equations for the displacements 1D 7 1 When displacements in the second and third directions are zero Let the zero displacement directions be y and z Then the equation of section 5 2 still holds for directions x and y with the proviso that ddz terms are absent from both the div grad and dil definitions 7 2 When stressess in the second and third directions are zero When ny and nz say are zero the analysis must start from nx R8 ex R9 et Thereafter

Original URL path: http://www.cham.co.uk/phoenics/d_polis/d_lecs/stress/solstreq.htm (2016-02-15)

Open archived version from archive

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