a 7.298 g sample of copper reacts completely with oxygen, forming a copper oxide. the final mass of the copper oxide is 8.217 g.

Lammps ERROR on proc 0: Out of range atoms - cannot compute PPPM

Alexeev081 [22]

Using the knowledge in computational language in C++ it is possible to write a code that Out of range atoms:

<h3>Writting the code in C++:</h3>

#include <cmath>

#include "pppm_tip4p.h"

#include "atom.h"

#include "domain.h"

#include "force.h"

#include "memory.h"

#include "error.h"

#include "math_const.h"

using namespace LAMMPS_NS;

using namespace MathConst;

#define OFFSET 16384

#ifdef FFT_SINGLE

#define ZEROF 0.0f

#define ONEF 1.0f

#else

#define ZEROF 0.0

#define ONEF 1.0

#endif

void PPPMTIP4P::init()

{

// TIP4P PPPM requires newton on, b/c it computes forces on ghost atoms

if (force->newton == 0)

error->all(FLERR,"Kspace style pppm/tip4p requires newton on");

PPPM::init();

}

void PPPMTIP4P::particle_map()

{

int nx,ny,nz,iH1,iH2;

double *xi,xM[3];

int *type = atom->type;

double **x = atom->x;

int nlocal = atom->nlocal;

if (!std::isfinite(boxlo[0]) || !std::isfinite(boxlo[1]) || !std::isfinite(boxlo[2]))

error->one(FLERR,"Non-numeric box dimensions - simulation unstable");

int flag = 0;

for (int i = 0; i < nlocal; i++) {

if (type[i] == typeO) {

find_M(i,iH1,iH2,xM);

xi = xM;

} else xi = x[i];

nx = static_cast<int> ((xi[0]-boxlo[0])*delxinv+shift) - OFFSET;

ny = static_cast<int> ((xi[1]-boxlo[1])*delyinv+shift) - OFFSET;

nz = static_cast<int> ((xi[2]-boxlo[2])*delzinv+shift) - OFFSET;

part2grid[i][0] = nx;

part2grid[i][1] = ny;

part2grid[i][2] = nz;

if (nx+nlower < nxlo_out || nx+nupper > nxhi_out ||

ny+nlower < nylo_out || ny+nupper > nyhi_out ||

nz+nlower < nzlo_out || nz+nupper > nzhi_out) flag++;

}

int flag_all;

MPI_Allreduce(&flag,&flag_all,1,MPI_INT,MPI_SUM,world);

if (flag_all) error->all(FLERR,"Out of range atoms - cannot compute PPPM");

}

void PPPMTIP4P::make_rho()

{

int i,l,m,n,nx,ny,nz,mx,my,mz,iH1,iH2;

FFT_SCALAR dx,dy,dz,x0,y0,z0;

double *xi,xM[3];

FFT_SCALAR *vec = &density_brick[nzlo_out][nylo_out][nxlo_out];

for (i = 0; i < ngrid; i++) vec[i] = ZEROF;

int *type = atom->type;

double *q = atom->q;

double **x = atom->x;

int nlocal = atom->nlocal;

for (int i = 0; i < nlocal; i++) {

if (type[i] == typeO) {

find_M(i,iH1,iH2,xM);

xi = xM;

} else xi = x[i];

nx = part2grid[i][0];

ny = part2grid[i][1];

nz = part2grid[i][2];

dx = nx+shiftone - (xi[0]-boxlo[0])*delxinv;

dy = ny+shiftone - (xi[1]-boxlo[1])*delyinv;

dz = nz+shiftone - (xi[2]-boxlo[2])*delzinv;

compute_rho1d(dx,dy,dz);

z0 = delvolinv * q[i];

for (n = nlower; n <= nupper; n++) {

mz = n+nz;

y0 = z0*rho1d[2][n];

for (m = nlower; m <= nupper; m++) {

my = m+ny;

x0 = y0*rho1d[1][m];

for (l = nlower; l <= nupper; l++) {

mx = l+nx;

density_brick[mz][my][mx] += x0*rho1d[0][l];

}

}

See more about C++ at brainly.com/question/19705654

#SPJ1

6 0

2 years ago

The population of two different villages are modeled by the equations shown below. The population (in thousands) is represented

PilotLPTM [1.2K]

Answer:

1990 and 2010

$y_l_w(15)=315\hspace{3}thousands\\y_l_c(15)=315\hspace{3}thousands\\y_l_w(35)=715\hspace{3}thousands\\y_l_c(35)=715\hspace{3}thousands$

Explanation:

Let:

$y_l_w=Population\hspace{3}of\hspace{3}Lewiston\\y_l_c=Population\hspace{3}of\hspace{3}Lockport$

We need to know, in what year(s) the villages had the same population, mathematically this is:

$y_l_w=y_l_c$

So:

$x^2-30x+540=20x+15\\\\Subtract\hspace{3}20x\hspace{3}from\hspace{3}both\hspace{3}sides\\\\x^2-50x+540=15\\\\Subtract\hspace{3}15\hspace{3}from\hspace{3}both\hspace{3}sides\\\\x^2-50x+525=0$

Solving for x:

Factoring

$(x-15)(x-35)=0$

Hence:

$x=15\\\\or\\\\x=35$

Therefore the year(s) which the village had the same population are:

$1975+15=1990\\\\and\\\\1975+35=2010$

In order to find the population of both cities during the year(s) of equal population, just evalue the equations at x=15 and x=35:

$y_l_w(15)=315\hspace{3}thousands\\y_l_c(15)=315\hspace{3}thousands\\y_l_w(35)=715\hspace{3}thousands\\y_l_c(35)=715\hspace{3}thousands$

7 0

3 years ago

Which traditional model of nursing care in which one nurse provides total care to a patient is still used today?

natima [27]

Answer:

Total patient care

Total patient care is a nursing model where one nurse provides total care to a single patient or a group of patients during his/her shift.

Explanation:

6 0

2 years ago

Why grade 12 learners should write their nbts as early as possible

Vinvika [58]

Answer: help to interpret your NSC results and can be used by universities in different ways. so these are two reasond.

Explanation:

4 0

2 years ago

4x - 5y = 15

Tresset [83]

Solution is (5,1)
3(5) - 3(1)
15 - 3 = 12
the answer is B

5 0

3 years ago