I am new to Python and trying to get this script to run, but it seems to be hanging in an infinite loop. When I use ctrl+c to stop it, it is always on line 103.
vs = 20.05 * np.sqrt(Tb + Lb * (y - y0)) # m/s speed of sound as a function of temperature
I am used to MatLab (from school) and the editor it has. I ran into issues earlier with the encoding for this code. Any suggestions on a (free) editor? I am currently using JEdit and/or Notepad.
Here is the full script:
#!/usr/bin/env python
# -*- coding: ANSI -*-
import numpy as np
from math import *
from astropy.table import Table
import matplotlib.pyplot as plt
from hanging_threads import start_monitoring#test for code hanging
start_monitoring(seconds_frozen=10, test_interval=100)
"""Initial Conditions and Inputs"""
d = 154.71/1000 # diameter of bullet (in meters)
m = 46.7 # mass of bullet ( in kg)
K3 = 0.87*0.3735 # drag coefficient at supersonic speed
Cd1 = 0.87*0.108 #drag coefficient at subsonic speed
v0 = 802 # muzzle velocity in m/sec
dt = 0.01 # timestep in seconds
"""coriolis inputs"""
L = 90*np.pi/180 # radians - latitude of firing site
AZ = 90*np.pi/180 # radians - azimuth angle of fire measured clockwise from North
omega = 0.0000727 #rad/s rotation of the earth
"""wind inputs"""
wx = 0 # m/s
wz = 0 # m/s
"""initializing variables"""
vx = 0 #initial x velocity
vy = 0 #initial y velocity
vy0 = 0
y_max = 0 #apogee
v = 0
t = 0
x = 0
"""Variable Atmospheric Pressure"""
rho0 = 1.2041 # density of air at sea-level (kg/m^3)
T = 20 #temperature at sea level in celcius
Tb = T + 273.15 # temperature at sea level in Kelvin
Lb = -2/304.8 # temperature lapse rate in K/m (-2degrees/1000ft)- not valid above 36000ft
y = 0 # current altitude
y0 = 0 # initial altitude
g = 9.81 # acceleration due to gravity in m/s/s
M = 0.0289644 #kg/mol # molar mass of air
R = 8.3144598 # J/molK - universal gas constant
# air density as a function of altitude and temperature
rho = rho0 * ((Tb/(Tb+Lb*(y-y0)))**(1+(g*M/(R*Lb))))
"""Variable Speed of Sound"""
vs = 20.05*np.sqrt(Tb +Lb*(y-y0)) # m/s speed of sound as a function of temperature
Area = pi*(d/2)**2 # computing the reference area
phi_incr = 5 #phi0 increment (degrees)
N = 12 # length of table
"""Range table"""
dtype = [('phi0', 'f8'), ('phi_impact', 'f8'), ('x', 'f8'), ('z', 'f8'),('y', 'f8'), ('vx', 'f8'), ('vz', 'f8'), ('vy', 'f8'), ('v', 'f8'),('M', 'f8'), ('t', 'f8')]
table = Table(data=np.zeros(N, dtype=dtype))
"""Calculates entire trajectory for each specified angle"""
for i in range(N):
phi0 = (i + 1) * phi_incr
"""list of initial variables used in while loop"""
t = 0
y = 0
y_max = y
x = 0
z = 0
vx = v0*np.cos(radians(phi0))
vy = v0*np.sin(radians(phi0))
vx_w = 0
vz_w = 0
vz = 0
v = v0
ay = 0
ax = 0
wx = wx
wz = wz
rho = rho0 * ((Tb / (Tb + Lb * (y - y0))) ** (1 + (g * M / (R * Lb))))
vs = 20.05 * np.sqrt(Tb + Lb * (y - y0)) # m/s speed of sound as a function of temperature
ax_c = -2 * omega * ((vz * sin(L)) + vy * cos(L) * sin(AZ))
ay_c = 2 * omega * ((vz * cos(L) * cos(AZ)) + vx_w * cos(L) * sin(AZ))
az_c = -2 * omega * ((vy * cos(L) * cos(AZ)) - vx_w * sin(L))
Mach = v/vs
""" initializing variables for plots"""
t_list = [t]
x_list = [x]
y_list = [y]
vy_list = [vy]
v_list = [v]
phi0_list = [phi0]
Mach_list = [Mach]
while y >= 0:
phi0 = phi0
"""drag calculation with variable density, Temp and sound speed"""
rho = rho0 * ((Tb / (Tb + Lb * (y - y0))) ** (1 + (g * M / (R *Lb))))
vs = 20.05 * np.sqrt(Tb + Lb * (y - y0)) # m/s speed of sound as a function of temperature
Cd3 = K3 / sqrt(v / vs)
Mach = v/vs
"""Determining drag regime"""
if v > 1.2 * vs: #supersonic
Cd = Cd3
elif v < 0.8 * vs: #subsonic
Cd = Cd1
else: #transonic
Cd = ((Cd3 - Cd1)*(v/vs - 0.8)/(0.4)) + Cd1
"""Acceleration due to Coriolis"""
ax_c = -2*omega*((vz_w*sin(L))+ vy*cos(L)*sin(AZ))
ay_c = 2*omega*((vz_w*cos(L)*cos(AZ))+ vx_w*cos(L)*sin(AZ))
az_c = -2*omega*((vy*cos(L)*cos(AZ))- vx_w*sin(L))
"""Total acceleration calcs"""
if vx > 0:
ax = -0.5*rho*((vx-wx)**2)*Cd*Area/m + ax_c
else:
ax = 0
""" Vy before and after peak"""
if vy > 0:
ay = (-0.5 * rho * (vy ** 2) * Cd * Area / m) - g + ay_c
else:
ay = (0.5 * rho * (vy ** 2) * Cd * Area / m) - g + ay_c
az = az_c
vx = vx + ax*dt # vx without wind
# vx_w = vx with drag and no wind + wind
vx_w = vx + 2*wx*(1-(vx/v0*np.cos(radians(phi0))))
vy = vy + ay*dt
vz = vz + az*dt
vz_w = vz + wz*(1-(vx/v0*np.cos(radians(phi0))))
"""projectile velocity"""
v = sqrt(vx_w**2 + vy**2 + vz**2)
"""new x, y, z positions"""
x = x + vx_w*dt
y = y + vy*dt
z = z + vz_w*dt
if y_max <= y:
y_max = y
phi_impact = degrees(atan(vy/vx)) #impact angle in degrees
""" appends selected data for ability to plot"""
t_list.append(t)
x_list.append(x)
y_list.append(y)
vy_list.append(vy)
v_list.append(v)
phi0_list.append(phi0)
Mach_list.append(Mach)
if y < 0:
break
t += dt
"""Range table output"""
table[i] = ('%.f' % phi0, '%.3f' % phi_impact, '%.1f' % x,'%.2f' % z, '%.1f' % y_max, '%.1f' % vx_w,'%.1f' % vz,'%.1f' % vy,'%.1f' % v,'%.2f' %Mach, '%.1f' % t)
""" Plot"""
plt.plot(x_list, y_list, label='%d°' % phi0)#plt.plot(x_list, y_list, label='%d°' % phi0)
plt.title('Altitude versus Range')
plt.ylabel('Altitude (m)')
plt.xlabel('Range (m)')
plt.axis([0, 30000, 0, 15000])
plt.grid(True)
print(table)
legend = plt.legend(title="Firing Angle",loc=0, fontsize='small', fancybox=True)
plt.show()
Thank you in advance
Which Editor Should I Use?
Personally, I prefer VSCode, but Sublime is also pretty popular. If you really want to go barebones, try Vim. All three are completely free.
Code Errors
After scanning your code snippet, it appears that you are caught in an infinite loop, which you enter with the statement while y >= 0. The reason you always get line 103 when you hit Ctrl+C is likely because that takes the longest, making it more likely to land there at any given time.
Note that currently, you can only escape your while loop through this branch:
if y_max <= y:
y_max= y
phi_impact = degrees(atan(vy/vx)) #impact angle in degrees
""" appends selected data for ability to plot"""
t_list.append(t)
x_list.append(x)
y_list.append(y)
vy_list.append(vy)
v_list.append(v)
phi0_list.append(phi0)
Mach_list.append(Mach)
if y < 0:
break
t += dt
This means that if ymax never drops below y, or y never drops below zero, then you will infinitely loop. Granted, I haven't looked at your code in any great depth, but from the surface it appears that y_max is never decremented (meaning it will always be at least equal to y). Furthermore, y is only updated when you do y = y + vy*dt, which will only ever increase y if vy >= 0 (I assume dt is always positive).
Debugging
As #Giacomo Catenazzi suggested, try printing out y and y_max at the top of the while loop and see how they change as your code runs. I suspect they are not decrementing like you expected.
Related
I am tring to solve the equation of motion of charged particle in planetary magnetic field to see the path of the particle using Forward Euler's and RK5 method in python (as an excercise in learning Numerical methods) I encounter two problems:
The 'for loop' in the RK4 method does not update the new values. It give the values of the first iteration for all iteration.
With the change of the sing of 'β = charge/mass' the path of particle which is expected does not change. It seems the path is unaffected by the nature(sign) of the particle. What does this mean physically or mathematically?
The codes are adapted from :
python two coupled second order ODEs Runge Kutta 4th order
and
Applying Forward Euler Method to a Three-Box Model System of ODEs
I would be immensely grateful if anyone explain to me what is wrong in the code.
thank you.
The Code are as under:
import numpy as np
import matplotlib.pyplot as plt
from math import sin, cos
from scipy.integrate import odeint
scales = np.array([1e7, 0.1, 1, 1e-5, 10, 1e-5])
def LzForce(t,p):
# assigning each ODE to a vector element
r,x,θ,y,ϕ,z = p*scales
# constants
R = 60268e3 # metre
g_20 = 1583e-9
Ω = 9.74e-3 # degree/second
B_θ = (R/r)**4*g_20*cos(θ)*sin(θ)
B_r = 2*(R/r)**4*g_20*(0.5*(3*cos(θ)**2-1))
β = +9.36e10
# defining the ODEs
drdt = x
dxdt = r*(y**2 +(z+Ω)**2*sin(θ)**2-β*z*sin(θ)*B_θ)
dθdt = y
dydt = (-2*x*y +r*(z+Ω)**2*sin(θ)*cos(θ)+β*r*z*sin(θ)*B_r)/r
dϕdt = z
dzdt = (-2*x*(z+Ω)*sin(θ)-2*r*y*(z+Ω)*cos(θ)+β*(x*B_θ-r*y*B_r))/(r*sin(θ))
return np.array([drdt,dxdt,dθdt,dydt,dϕdt,dzdt])/scales
def ForwardEuler(fun,t0,p0,tf,dt):
r0 = 6.6e+07
x0 = 0.
θ0 = 88.
y0 = 0.
ϕ0 = 0.
z0 = 22e-3
p0 = np.array([r0,x0,θ0,y0,ϕ0,z0])
t = np.arange(t0,tf+dt,dt)
p = np.zeros([len(t), len(p0)])
p[0] = p0
for i in range(len(t)-1):
p[i+1,:] = p[i,:] + fun(t[i],p[i,:]) * dt
return t, p
def rk4(fun,t0,p0,tf,dt):
# initial conditions
r0 = 6.6e+07
x0 = 0.
θ0 = 88.
y0 = 0.
ϕ0 = 0.
z0 = 22e-3
p0 = np.array([r0,x0,θ0,y0,ϕ0,z0])
t = np.arange(t0,tf+dt,dt)
p = np.zeros([len(t), len(p0)])
p[0] = p0
for i in range(len(t)-1):
k1 = dt * fun(t[i], p[i])
k2 = dt * fun(t[i] + 0.5*dt, p[i] + 0.5 * k1)
k3 = dt * fun(t[i] + 0.5*dt, p[i] + 0.5 * k2)
k4 = dt * fun(t[i] + dt, p[i] + k3)
p[i+1] = p[i] + (k1 + 2*(k2 + k3) + k4)/6
return t,p
dt = 0.5
tf = 1000
p0 = [6.6e+07,0.0,88.0,0.0,0.0,22e-3]
t0 = 0
#Solution with Forward Euler
t,p_Euler = ForwardEuler(LzForce,t0,p0,tf,dt)
#Solution with RK4
t ,p_RK4 = rk4(LzForce,t0, p0 ,tf,dt)
print(t,p_Euler)
print(t,p_RK4)
# Plot Solutions
r,x,θ,y,ϕ,z = p_Euler.T
fig,ax=plt.subplots(2,3,figsize=(8,4))
plt.xlabel('time in sec')
plt.ylabel('parameters')
for a,s in zip(ax.flatten(),[r,x,θ,y,ϕ,z]):
a.plot(t,s); a.grid()
plt.title("Forward Euler", loc='left')
plt.tight_layout(); plt.show()
r,x,θ,y,ϕ,z = p_RK4.T
fig,ax=plt.subplots(2,3,figsize=(8,4))
plt.xlabel('time in sec')
plt.ylabel('parameters')
for a,q in zip(ax.flatten(),[r,x,θ,y,ϕ,z]):
a.plot(t,q); a.grid()
plt.title("RK4", loc='left')
plt.tight_layout(); plt.show()
[RK4 solution plot][1]
[Euler's solution methods][2]
''''RK4 does not give iterated values.
The path is unaffected by the change of sign which is expected as it is under Lorentz force''''
[1]: https://i.stack.imgur.com/bZdIw.png
[2]: https://i.stack.imgur.com/tuNDp.png
You are not iterating more than once inside the for loop in rk4 because it returns after the first iteration.
for i in range(len(t)-1):
k1 = dt * fun(t[i], p[i])
k2 = dt * fun(t[i] + 0.5*dt, p[i] + 0.5 * k1)
k3 = dt * fun(t[i] + 0.5*dt, p[i] + 0.5 * k2)
k4 = dt * fun(t[i] + dt, p[i] + k3)
p[i+1] = p[i] + (k1 + 2*(k2 + k3) + k4)/6
# This is the problem line, the return was tabbed in, to be inside the for block, so the block executed once and returned.
return t,p
For physics questions please try a different forum.
I have two points in 2D space as we can see in the figure to move from the blue point to the red points we can use the equation (1). Where b is a constant used to limit the shape of the logarithmic spiral, l is a random number in [−1,1], which is used to
control the indentation effect of the movement, D indicates the distance between blue points and the current point
I need another movement that can move from blue points to the red points like in the figure
You can use sinusoidal model.
For start point (X0, Y0) and end point (X1,Y1) we have vector end-start, determine it's length - distance between points L, and angle of vector Fi (using atan2).
Then generate sinusoidal curve for some standard situation - for example, along OX axis, with magnitude A, N periods for distance 2 * Pi * N:
Scaled sinusoid in intermediate point with parameter t, where t is in range 0..1 (t=0 corresponds to start point (X0,Y0))
X(t) = t * L
Y(t) = A * Sin(2 * N * Pi * t)
Then shift and rotate sinusoid using X and Y calculated above
X_result = X0 + X * Cos(Fi) - Y * Sin(Fi)
Y_result = Y0 + X * Sin(Fi) + Y * Cos(Fi)
Example Python code:
import math
x0 = 100
y0 = 100
x1 = 400
y1 = 200
nperiods = 4
amp = 120
nsteps = 20
leng = math.hypot(x1 - x0, y1 - y0)
an = math.atan2(y1 - y0, x1 - x0)
arg = 2 * nperiods* math.pi
points = []
for t in range(1, nsteps + 1):
r = t / nsteps
xx = r * leng
yy = amp * math.sin(arg * r)
rx = x0 + xx * math.cos(an) - yy * math.sin(an)
ry = y0 + xx * math.sin(an) + yy * math.cos(an)
points.append([rx, ry])
print(points)
Draw points:
The following code generates numpy 2D lists of r and E values for the specified intervals.
r = np.linspace(3, 14, 10)
E = np.linspace(0.05, 0.75, 10)
r, E = np.meshgrid(r, E)
I am then using the following nested loop to generate output from the function ionisationGamma for each r and E interval value.
for ridx in trange(len(r)):
z = []
for cidx in range(len(r[ridx])):
z.append(ionisationGamma(r[ridx][cidx], E[ridx][cidx]))
Z.append(z)
Z = np.array(Z)
This loop gives me a 2D numpy array Z, which is my output and I am using it for a 3D graph. The problem with it is: it is taking ~6 hours to generate the output for all these intervals as there are so many values due to np.meshgrid. I have just discovered multi-threading in Python and wanted to know how I can implement this by using it. Any help is appreciated.
See below code for ionisationGamma
def ionisationGamma(r, E):
I = complex(0.1, 1.0)
a_soft = 1.0
omega = 0.057
beta = 0.0
dt = 0.1
steps = 10000
Nintervals = 60
N = 3000
xmin = float(-300)
xmax = -xmin
x = [0.0]*N
dx = (xmax - xmin) / (N - 1)
L = dx * N
dk = 2 * M_PI / L
propagator = None
in_, out_, psi0 = None, None, None
in_ = [complex(0.,0.)] * N
psi0 = [complex(0.,0.)] * N
out_ = [[complex(0.,0.)]*N for i in range(steps+1)]
overlap = exp(-r) * (1 + r + (1 / 3) * pow(r, 2))
normC = 1 / (sqrt(2 * (1 + overlap)))
gammai = 0.5
qi = 0.0 + (r / 2)
pi = 0.0
gammai1 = 0.5
gammai2 = 0.5
qi1 = 0.0 - (r / 2)
qi2 = 0.0 + (r / 2)
pi1 = 0.0
pi2 = 0.0
# split initial wavepacket
for i in range(N):
x[i] = xmin + i * dx
out_[0][i] = (normC) * ((pow(gammai1 / M_PI, 1. / 4.) * exp(complex(-(gammai1 / 2.) * pow(x[i] - qi1, 2.), pi1 * (x[i] - qi1)))) + (pow(gammai2 / M_PI, 1. / 4.) * exp(complex(-(gammai2 / 2.) * pow(x[i] - qi2, 2.), pi2 * (x[i] - qi2)))))
in_[i] = (normC) * ((pow(gammai1 / M_PI, 1. / 4.) * exp(complex(-(gammai1 / 2.) * pow(x[i] - qi1, 2.), pi1 * (x[i] - qi1)))) + (pow(gammai2 / M_PI, 1. / 4.) * exp(complex(-(gammai2 / 2.) * pow(x[i] - qi2, 2.), pi2 * (x[i] - qi2)))))
psi0[i] = in_[i]
for l in range(1, steps+1):
for i in range(N):
propagator = exp(complex(0, -potential(x[i], omega, beta, a_soft, r, E, dt, l) * dt / 2.))
in_[i] = propagator * in_[i];
in_ = np.fft.fft(in_, N)
for i in range(N):
k = dk * float(i if i < N / 2 else i - N)
propagator = exp(complex(0, -dt * pow(k, 2) / (2.)))
in_[i] = propagator * in_[i]
in_ = np.fft.ifft(in_, N)
for i in range(N):
propagator = exp(complex(0, -potential(x[i], omega, beta, a_soft, r, E, dt, l) * dt / 2.))
in_[i] = propagator * in_[i]
out_[l][i] = in_[i]
initialGammaCentre = 0.0
finalGammaCentre = 0.0
for i in range(500, 2500 +1):
initialGammaCentre += pow(abs(out_[0][i]), 2) * dx
finalGammaCentre += pow(abs(out_[steps][i]), 2) * dx
ionisationGamma = finalGammaCentre / initialGammaCentre
return ionisationGamma
def potential(x, omega, beta, a_soft, r, E, dt, l):
V = (-1. / sqrt((x - (r / 2)) * (x - (r / 2)) + a_soft * a_soft)) + ((-1. / sqrt((x + (r / 2)) * (x + (r / 2)) + a_soft * a_soft))) + E * x
return V
Since the question is about how to use multiprocessing, the following code will work:
import multiprocessing as mp
if __name__ == '__main__':
with mp.Pool(processes=16) as pool:
Z = pool.starmap(ionisationGamma, arguments)
Z = np.array(Z)
Where the arguments are:
arguments = list()
for ridx in range(len(r)):
for cidx in range(len(r[ridx])):
arguments.append((r[ridx][cidx], E[ridx][cidx]))
I am using starmap instead of map, since you have multiple arguments that you want to unpack. This will divide the arguments iterable over multiple cores, using the ionisationGamma function and the final result will be ordered.
However, I do feel the need to say that the main solution is not really the multiprocessing but the original function code. In ionisationGamma you are using several times the slow python for loops. And it would benefit your code a lot if you could vectorize those operations.
A second observation is that you are using many of those loops separately and it would be nice if you could separate that one big function into multiple smaller functions. Then you can time every function individually and speed up those that are too slow.
I want to perform Monte Carlo simulation to the particles which are interacting via Lennard-Jones potential + FENE potential. I'm getting negative values in the FENE potential which have the log value in it. The error is "RuntimeWarning: invalid value encountered in log return (-0.5 * K * R**2 * np.log(1-((np.sqrt(rij2) - r0) / R)**2))" The FENE potential is given by:
import numpy as np
def gen_chain(N, R0):
x = np.linspace(1, (N-1)*0.8*R0, num=N)
y = np.zeros(N)
z = np.zeros(N)
return np.column_stack((x, y, z))
def lj(rij2):
sig_by_r6 = np.power(sigma/rij2, 3)
sig_by_r12 = np.power(sig_by_r6, 2)
lje = 4.0 * epsilon * (sig_by_r12 - sig_by_r6)
return lje
def fene(rij2):
return (-0.5 * K * R**2 * np.log(1-((np.sqrt(rij2) - r0) / R)**2))
def total_energy(coord):
# Non-bonded
e_nb = 0
for i in range(N):
for j in range(i-1):
ri = coord[i]
rj = coord[j]
rij = ri - rj
rij2 = np.dot(rij, rij)
if (np.sqrt(rij2) < rcutoff):
e_nb += lj(rij2)
# Bonded
e_bond = 0
for i in range(1, N):
ri = coord[i]
rj = coord[i-1]
rij = ri - rj
rij2 = np.dot(rij, rij)
e_bond += fene(rij2)
return e_nb + e_bond
def move(coord):
trial = np.ndarray.copy(coord)
for i in range(N):
delta = (2.0 * np.random.rand(3) - 1) * max_delta
trial[i] += delta
return trial
def accept(delta_e):
beta = 1.0/T
if delta_e <= 0.0:
return True
random_number = np.random.rand(1)
p_acc = np.exp(-beta*delta_e)
if random_number < p_acc:
return True
return False
if __name__ == "__main__":
# FENE parameters
K = 40
R = 0.3
r0 = 0.7
# LJ parameters
sigma = r0/0.33
epsilon = 1.0
# MC parameters
N = 50 # number of particles
rcutoff = 2.5*sigma
max_delta = 0.01
n_steps = 10000000
T = 0.5
coord = gen_chain(N, R)
energy_current = total_energy(coord)
traj = open('traj.xyz', 'w')
for step in range(n_steps):
if step % 1000 == 0:
traj.write(str(N) + '\n\n')
for i in range(N):
traj.write("C %10.5f %10.5f %10.5f\n" % (coord[i][0], coord[i][1], coord[i][2]))
print(step, energy_current)
coord_trial = move(coord)
energy_trial = total_energy(coord_trial)
delta_e = energy_trial - energy_current
if accept(delta_e):
coord = coord_trial
energy_current = energy_trial
traj.close()
The problem is that calculating rij2 = np.dot(rij, rij) in total energy with the constant values you use is always a very small number. Looking at the expression inside the log used to calculate FENE, np.log(1-((np.sqrt(rij2) - r0) / R)**2), I first noticed that you're taking the square root of rij2 which is not consistent with the formula you provided.
Secondly, notice that ((rij2 - r0) / R)**2 is the same as ((r0 - rij2) / R)**2, since the sign gets lost when squaring. Because rij2 is very small (already in the first iteration -- I checked by printing the values), this will be more or less equal to ((r0 - 0.05)/R)**2 which will be a number bigger than 1. Once you subtract this value from 1 in the log expression, 1-((np.sqrt(rij2) - r0) / R)**2 will be equal to np.nan (standing for "Not A Number"). This will propagate through all the function calls (for example, calling energy_trial = total_energy(coord_trial) will effectively set energy_trial to np.nan), until an error will be raised by some function.
Maybe you could do something with np.isnan() call, documented here. Moreover, you should check how you iterate through the coord (there's some inconsistencies throughout the code) -- I suggest you check the code review community as well.
I converted a Matlab code into python by manually typing it out. However i keep getting an error message which i still have not been able to fix. what am i doing wrong and how do i get the plot as that in Matlab? Just is little information about the code; this is a Explicit finite difference method for solving pressure distribution in an oil reservoir with production from the middle block only. Its similar to the heat equation, Ut=Uxx. I was told to add more text because my question is mostly code so had to add all these details. I think that notification has vanished now.
[P_new[N] = 4000 #last blocks at all time levels equals 4000
IndexError: index 9 is out of bounds for axis 0 with size 9]
The Matlab code which runs ok is below: The python code follows.
clear
clc
% Solution of P_t = P_{xx}
L = 1000 ; %ft length of reservoir
W = 100 ; %ft reservoir width
h = 50 ;%ft pay thickness
poro = 0.25; % rock porosity
k_o = 5; %md effective perm to oil
P_i = 4000; %psia initial pressure
B_o = 1.25; %oil formation vol fact
mu = 5; %cp oil visc
c_t = 0.0000125; %1/atm total compressibility
Q_o = 10;%stb/day production rate from central well
alpha = c_t*mu*poro/k_o;
T = 1;
N_time = 50;
dt = T/N_time;
% % Number of grid cells
N =9; %number of grid cells
%N =11;%number of grid cells
dx = (L/(N-1)); %distance between grid blocks
x = 0+dx*0.5:dx:L+dx; %points in space
for i=1:N
P_old(i)=P_i;
FPT(i)=0;
end
FPT((N+1)/2)=-Q_o*B_o*mu/1.127/W/dx/h/k_o; %source term at the center block of grid cell
P_new = P_old;
for j = 1:N_time
for k = 1: N
if k<2
P_new(k)=4000;%P_old(k)+dt/alpha*((P_old(k+1)-2*P_old(k)+P_old(k))/dx^2+FPT(k));
elseif k > N-1
P_new(k) = 4000;%P_old(k)+dt/alpha*((P_old(k)-2*P_old(k)+P_old(k-1))/dx^2+FPT(k));
else
P_new(k) = P_old(k)+dt/alpha*((P_old(k+1)-2*P_old(k)+P_old(k-1))/dx^2+FPT(k));
end
end
plot(x,P_new, '-x')
xlabel('X')
ylabel('P(X)')
hold on
grid on
%%update "u_old" before you move forward to the next time level
P_old = P_new;
end
hold off
Python Code:
import numpy as np
import matplotlib.pyplot as plt
# Solution of P_t = P_{xx}
L = 1000 #ft length of reservoir
W = 100 #ft reservoir width
h = 50 #ft pay thickness
poro = 0.25 # rock porosity
k_o = 5 #md effective perm to oil
P_i = 4000 #psia initial pressure
B_o = 1.25 #oil formation vol fact
mu = 5 #cp oil visc
c_t = 0.0000125 #1/atm total compressibility
Q_o = 10 #stb/day production rate from central well
alpha = c_t * mu * poro / k_o
T = 1
N_time = 20
dt = T / N_time
# % Number of grid cells
N = 9 #number of grid cells
dx = (L / (N - 1)) #distance between grid blocks
x= np.arange(0.0,L+dx,dx)
P_old = np.zeros_like(x) #pressure at previous time level
P_new = np.zeros_like(x) #pressure at previous time level
FPT = np.zeros_like(x)
for i in range(0,N):
P_old[i]= P_i
FPT[int((N + 1) / 2)]= -Q_o * B_o * mu / (1.127 * W * dx * h * k_o) # source term at the center block of grid cell
P_new = P_old
d=np.arange(0,N)
for j in range(0,N_time):
for k in range(0,N):
P_new[0] = 4000 #pressure at first block for all time levels equals 4000
P_new[N] = 4000 #pressure at last block for all time levels equals 4000
P_new[k]= P_old[k] + dt / alpha * ((P_old[k+1] - 2 * P_old[k] + P_old[k - 1]) / dx ** 2 + FPT[k])
plt.plot(x, P_new)
plt.xlabel('X')
plt.ylabel('P(X)')
P_old = P_new
Matlab uses 1 based indexing , Python arrays use "0" based indexing. If you define an array of length N in python, the indices are from 0 to N-1.
So just replace the index N to index N-1 in your code as below and it works.
import numpy as np
import matplotlib.pyplot as plt
# Solution of P_t = P_{xx}
L = 1000 #ft length of reservoir
W = 100 #ft reservoir width
h = 50 #ft pay thickness
poro = 0.25 # rock porosity
k_o = 5 #md effective perm to oil
P_i = 4000 #psia initial pressure
B_o = 1.25 #oil formation vol fact
mu = 5 #cp oil visc
c_t = 0.0000125 #1/atm total compressibility
Q_o = 10 #stb/day production rate from central well
alpha = c_t * mu * poro / k_o
T = 1
N_time = 20
dt = T / N_time
# % Number of grid cells
N = 9 #number of grid cells
dx = (L / (N - 1)) #distance between grid blocks
x= np.arange(0.0,L+dx,dx)
P_old = np.zeros_like(x) #pressure at previous time level
P_new = np.zeros_like(x) #pressure at previous time level
FPT = np.zeros_like(x)
for i in range(0,N):
P_old[i]= P_i
FPT[int((N + 1) / 2)]= -Q_o * B_o * mu / (1.127 * W * dx * h * k_o) # source term at the center block of grid cell
P_new = P_old
d=np.arange(0,N)
for j in range(0,N_time):
for k in range(0,N-1):
P_new[0] = 4000 #pressure at first block for all time levels equals 4000
P_new[N-1] = 4000 #pressure at last block for all time levels equals 4000
P_new[k]= P_old[k] + dt / alpha * ((P_old[k+1] - 2 * P_old[k] + P_old[k - 1]) / dx ** 2 + FPT[k])
plt.plot(x, P_new)
plt.xlabel('X')
plt.ylabel('P(X)')
P_old = P_new
output: