在Python中求解x的高度非线性方程 [英] Solve highly non-linear equation for x in Python
问题描述
我正在尝试求解以下dB方程(为简单起见,我在问题标题中将dB表示为x):
I am trying to solve the following equation for dB (for simplicity, I stated dB as x in the question title):
等式中的所有其他术语都是已知的.我尝试使用SymPy来象征性地解决dB,但我不断遇到超时错误.我也尝试过使用scipy.optimize
中的fminbound
,但是dB的答案是错误的(请参见下面的使用fminbound
方法的Python代码).
All of the other terms in the equation are known. I tried using SymPy to symbolically solve for dB but I kept getting time out errors. I also tried using fminbound
from scipy.optimize
but the answer for dB is wrong (see below for Python code using the fminbound
approach).
有人知道使用Python解决dB方程的方法吗?
Does anyone know of a way to solve the equation for dB using Python?
import numpy as np
from scipy.optimize import fminbound
#------------------------------------------------------------------------------
# parameters
umf = 0.063 # minimum fluidization velocity, m/s
dbed = 0.055 # bed diameter, m
z0 = 0 # position bubbles are generated, m
z = 0.117 # bed vertical position, m
g = 9.81 # gravity, m/s^2
#------------------------------------------------------------------------------
# calculations
m = 3 # multiplier for Umf
u = m*umf # gas superficial velocity, m/s
abed = (np.pi*dbed**2)/4.0 # bed cross-sectional area, m^2
# calculate parameters used in equation
dbmax = 2.59*(g**-0.2)*(abed*(u-umf))**0.4
dbmin = 3.77*(u-umf)**2/g
c1 = 2.56*10**-2*((dbed / g)**0.5/umf)
c2 = (c1**2 + (4*dbmax)/dbed)**0.5
c3 = 0.25*dbed*(c1 + c2)**2
dbeq = 0.25*dbed*(-c1 + (c1**2 + 4*(dbmax/dbed))**0.5 )**2
# general form of equation ... (term1)^power1 * (term2)^power2 = term3
power1 = 1 - c1/c2
power2 = 1 + c1/c2
term3 = np.exp(-0.3*(z - z0)/dbed)
def dB(d):
term1 = (np.sqrt(d) - np.sqrt(dbeq)) / (np.sqrt(dbmin) - np.sqrt(dbeq))
term2 = (np.sqrt(d) + np.sqrt(c3)) / (np.sqrt(dbmin) + np.sqrt(c3))
return term1**power1 * term2**power2 - term3
# solve main equation for dB
dbub = fminbound(dB, 0.01, dbed)
print 'dbub = ', dbub
推荐答案
以下是四种单调根方法:
Here are the four single-dim root-methods:
from scipy.optimize import brentq, brenth, ridder, bisect
for rootMth in [brentq, brenth, ridder, bisect]:
dbub = rootMth(dB, 0.01, dbed)
print 'dbub = ', dbub, '; sanity check (is it a root?):', dB(dbub)
还有牛顿-拉普森(割线/海利)方法:
Also the newton-raphson (secant / haley) method:
from scipy.optimize import newton
dbub = newton(dB, dbed)
print 'dbub = ', dbub, '; sanity check (is it a root?):', dB(dbub)
如果有包围间隔,scipy文档建议使用brentq.
The scipy documentation recommends brentq if you have a bracketing interval.
这篇关于在Python中求解x的高度非线性方程的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!