Python中6个非线性方程组的疑难解答

2024-04-27 09:47:35 发布

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我正在尝试用Python解一个由6个非线性方程组成的系统。到目前为止,我已经:

from scipy.optimize import fsolve 
import math 
from numpy import log

ka1 = 1.045
ka2 = 0.1759
ka3 = 0.3159

def equations(p):
    yh2o, yco2, yh2, ych4, yco = p

    return (yh2o + yco2 + yh2 + ych4 + yco - 1,
        ka1 - (yh2o ** 2)/(yco * (yh2 ** 2),
        ka2 - (yco ** 2) / yco2,
        ka3 - ych4 / (yh2 ** 2),
        0.5 - (2.0 * yco2 + yco + yh2o) / (2 * yh2o + 2 * yh2 + 4 * ych4)))

yh2o, yco2, yh2, ych4, yco = fsolve(equations, [0.2, 0.2, 0.2, 0.2, 0.2])

print(f"yh2o = {yh2o}, yco2 = {yco2}, yh2 = {yh2}, ych4 = {ych4}, yco = {yco}")

当我试着运行这个时,我得到了

TypeError: fsolve: there is a mismatch between the input and output shape of the 'func' argument 'equations'.Shape should be (5,) but it is (2,).

我试图改变最初的猜测,但错误仍然是一样的

我也尝试过将方程转换成多项式,如下所示:

    return (yh2o + yco2 + yh2 + ych4 + yco - 1.0,
        ka1 * (yco * (yh2 ** 2.0)) - (yh2o * yh2o),
        ka2 * (yco2) - (yco ** 2.0),
        ka3 * (yh2 ** 2.0) - ych4,
        0.5 * (2.0 * yh2o + 2.0 * yh2 + 4 * ych4) - (2.0 * yco2 + yco + yh2o)) 

但是我有一个错误,说

TypeError: can't multiply sequence by non-int of type 'float'

Tags: fromimportreturnistypeerrorequationsfsolveyco
1条回答
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1楼 · 发布于 2024-04-27 09:47:35

我认为你只是在等式(返回的元组)中使用括号(即())有问题

以下是完整更正的工作代码:

Try it online!

from scipy.optimize import fsolve 
import math 
from numpy import log

ka1 = 1.045
ka2 = 0.1759
ka3 = 0.3159

def equations(p):
    yh2o, yco2, yh2, ych4, yco = p

    return (
        (yh2o + yco2 + yh2 + ych4 + yco - 1),
        ka1 - (yh2o ** 2) / (yco * (yh2 ** 2)),
        ka2 - (yco ** 2) / yco2,
        ka3 - ych4 / (yh2 ** 2),
        0.5 - (2.0 * yco2 + yco + yh2o) / (2 * yh2o + 2 * yh2 + 4 * ych4)
    )

yh2o, yco2, yh2, ych4, yco = fsolve(equations, [0.2, 0.2, 0.2, 0.2, 0.2])

print(f"yh2o = {yh2o}, yco2 = {yco2}, yh2 = {yh2}, ych4 = {ych4}, yco = {yco}")

输出:

yh2o = 0.17829101808889491, yco2 = 0.17513942402710805, yh2 = 0.4163023675952086, ych4 = 0.05474789024926531, yco = 0.1755193000395232

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