Exemplo n.º 1
0
from Tabular_values import Na
import matplotlib.pyplot as plt

ENERGY_ISOTHERM_RHO = [[], [], [], [], []]
RHO = [[], [], [], [], []]
k = 0

while k < 3:
    rho_is = 0.001
    for i in (0.001, 0.01, 0.1, 1., 10.):
        while rho_is < i * 10:
            RHO[k].append(rho_is)
            T_h = 3.98 / (10**3) * 10**k / 13**(4 / 3)
            rho_h = rho_is * Na * 11.19 * 1.4818 * 10**(-25) / 26 / 13
            ENERGY_ISOTHERM_RHO[k].append(
                2.626 * 10**3 / 26 * Energy(T_h, rho_h, 1, 1) * 13**(7 / 3))
            rho_is += i
    k += 1

plt.plot(RHO[0], ENERGY_ISOTHERM_RHO[0], label="3.98 эв")
plt.plot(RHO[1], ENERGY_ISOTHERM_RHO[1], label="39.8 эв")
plt.plot(RHO[2], ENERGY_ISOTHERM_RHO[2], label="398 эв")
plt.xscale('log')
plt.yscale('log')
plt.xlim(0.001, 100)
plt.ylim(1, 100000)
plt.xlabel('rho, г / см^3')
plt.ylabel('E, кДж / г')
plt.title('Энергия алюминия по различным моделям')
plt.grid('true')
plt.legend()
Exemplo n.º 2
0
for i in range(2, n + 1):
    TABLE_P[i][0] = TABLE_P[1][0] + (i - 1) * (-0.02)

for i in range(1, n + 2):
    for j in range(1, m + 1):
        #print(10 ** TABLE_E[i][0], 10 ** TABLE_E[0][j])
        found_rho = mixture_calculation(10**TABLE_E[i][0], 10**TABLE_E[0][j],
                                        mixture)
        for k in range(len(mixture)):
            T_h = 10**TABLE_E[i][0] / mixture[k]["Z"]**(4 / 3)
            rho_h = 10**TABLE_E[0][j] * Na * 11.19 * 1.4818 * 10**(
                -25) / mixture[k]["Atom_weight"] / mixture[k]["Z"]
            PHI = progonka(T_h, rho_h, 1, 1)
            TABLE_P[i][j] += P(T_h, rho_h, PHI) * mixture[k]["Z"]**(
                10 / 3) * mixture[k]["amount"] / nuclear_amount
            TABLE_E[i][j] += Energy(T_h, rho_h, 1, 1, PHI) * mixture[k]["Z"]**(
                7 / 3) * mixture[k]["amount"] / Weight
        print(i, j)

f = open('energy_sio2.txt', 'w')
for i in range(len(TABLE_E)):
    for j in range(len(TABLE_E[i])):
        f.write(str(TABLE_E[i][j]))
        f.write(" ")
    f.write("\n")

f = open('pressure_sio2.txt', 'w')
for i in range(len(TABLE_P)):
    for j in range(len(TABLE_P[i])):
        f.write(str(TABLE_P[i][j]))
        f.write(" ")
Exemplo n.º 3
0
#TA
TABLE_E[0][1] = -4.000
TABLE_E[1][0] = 3.000
for j in range(2, m + 1):
    TABLE_E[0][j] = TABLE_E[0][1] + (j - 1) / m * 13

for i in range(2, n + 1):
    TABLE_E[i][0] = TABLE_E[1][0] + (i - 1) / n * (-8)

for i in range(1, n + 1):
    for j in range(1, m + 1):
        T_now = 10**TABLE_E[i][0] / 36.7493224786
        rho_now = 1.0 / (8.923608963522 * 0.01 * 10**TABLE_E[0][j])
        # print(T_now, rho_now)
        # print(P(T_now, rho_now))
        TABLE_E[i][j] = log10(Energy(T_now, rho_now, 1, 1))
        print("T = ", TABLE_E[i][0], "rh0 = ", TABLE_E[0][j], "LG E = ",
              TABLE_E[i][j])
#T_now = 10 ** (-0.167) /36.749
#rho_now = 1.0 / (8.923608963522 * 0.01 * 10 ** 4)
#print(P(T_now, rho_now))
#print(log10(P(T_now,rho_now)))

for i in range(len(TABLE_E)):
    for j in range(len(TABLE_E[i])):
        print(TABLE_E[i][j], end=' ')
    print()
#
#
#        T_now = 10 ** LG_T[j] / 36.7493224786
#        rho_now = 1.0 / (8.923608963522 * 0.01 * 10 ** LG_V[i])
Exemplo n.º 4
0
def f2(T, rho):
    return Energy(
        T, rho, 13,
        26) - E_0 - 1 / 2 * (P(T, rho, 26, 13) + P_0) * (1 / rho_00 - 1 / rho)
Exemplo n.º 5
0
from Pressure import P
from Internal_energy import Energy
from numpy import dot
from numpy import transpose
from working_progonka import progonka

rho_00 = 2.712
E_0 = Energy(293 * 10**(-3) / (1.160 * 10**4), rho_00, 13, 26)
P_0 = P(293 * 10**3 / (1.160 * 10**4), rho_00, 26, 13)


def f1(T, rho):
    return P(T, rho, 26, 13) - P_0


def f2(T, rho):
    return Energy(
        T, rho, 13,
        26) - E_0 - 1 / 2 * (P(T, rho, 26, 13) + P_0) * (1 / rho_00 - 1 / rho)


def newton_solver(P, T_pribl, rho_pribl):
    delta = [[0 for i in range(1)] for j in range(2)]
    delta[0][0] = 10**(-8)
    delta[1][0] = 10**(-8)
    epsilon = 10**(-8)
    mod_delta = 100

    W = [[0 for i in range(2)] for j in range(2)]
    F = [[0 for i in range(1)] for j in range(2)]
    priblizhenie = [[0 for i in range(1)] for j in range(2)]
Exemplo n.º 6
0
TABLE_P[n + 1][0] = -9.000

for j in range(2, m + 1):
    TABLE_P[0][j] = TABLE_P[0][1] + (j - 1) * 0.01

for i in range(2, n + 1):
    TABLE_P[i][0] = TABLE_P[1][0] + (i - 1) * (-0.02)

for i in range(1, n + 2):
    for j in range(1, m + 1):
        #T_h = 10 ** TABLE_E[i][0] / z**(4 / 3)
        #rho_h = 10 ** TABLE_E[0][j] * Na * 11.19 * 1.4818 * 10**(-25) / Atom_weight / z
        T_h = 36.4 / z**(4 / 3)
        rho_h = 100 * Na * 11.19 * 1.4818 * 10**(-25) / Atom_weight / z
        PHI = progonka(T_h, rho_h, 1, 1)
        TABLE_E[i][j] = Energy(T_h, rho_h, 1, 1, PHI) * z ** (7 / 3)
        #TABLE_P[i][j] = P(T_h, rho_h, PHI) * z ** (10 / 3)
        TABLE_P[i][j] = delta_P(T_h, rho_h) * z ** (8 / 3)

        print("e =", TABLE_E[i][j], "p = ", TABLE_P[i][j])

f = open('energy_aluminium.txt', 'w')
for i in range(len(TABLE_E)):
    for j in range(len(TABLE_E[i])):
        f.write(str(TABLE_E[i][j]))
        f.write(" ")
    f.write("\n")

f = open('pressure_aluminium.txt', 'w')
for i in range(len(TABLE_P)):
    for j in range(len(TABLE_P[i])):
Exemplo n.º 7
0
def state_function_energy(T, rho, Atom_weight, z):

    #return 2.626 * 10**3 / Atom_weight * bilinear_interpolation_energy(T, rho) + ionic_contribution_energy(T, volume(rho, Atom_weight)) + cold_curve_energy(volume(rho, Atom_weight))
    return 2.626 * 10**3 / Atom_weight * Energy(T, rho, z, Atom_weight)