Question

我有一个代码，以简化的方式计算高炉煤气168小时（例如一周）

它读入一些输入值和化学值，并计算系统中的摩尔质量。之后，求解器计算出气体以何种化学形式（主要是CO，CO2）排出。问题是它非常慢。如果我有这个excel工作簿打开它需要几分钟，我甚至不打算让它在有更多工作簿打开时运行到最后。

我对VBA很新，但是如果我能设置它来解决VBA中的方程而不是让值在excel工作表和之间来回“迭代”，我希望它会快一点。 VBA解算器，通过逐步求解单元格值 - 如果我只知道如何，如果可能或一个好主意。

其实体中的代码，首先是一般计算：

   Sub BFgas()

Datamatrix = Range(Cells.Find("Datamatrix").Offset(1, 0).Address, Cells.Find("Datamatrix").Offset(21, 0).Address)
ReDim BFoutput(1 To 168, 1 To 3) As Double



M_pigiron_Matrix = Range(Cells.Find("BF1 Pig iron production").Offset(1, 0).Address, Cells.Find("BF1 Pig iron production").Offset(168 + 1, 0).Address)
Bf_blast_Matrix = Range(Cells.Find("BF 1 - Blast").Offset(1, 0).Address, Cells.Find("BF 1 - Blast").Offset(168 + 1, 0).Address)
Bf_oxygen_Matrix = Range(Cells.Find("BF 1 - Oxygen").Offset(1, 0).Address, Cells.Find("BF 1 - Oxygen").Offset(168 + 1, 0).Address)

M = 1
Do

M_pigiron = M_pigiron_Matrix(M, 1) 'Tons of pig iron
Bf_blast = Bf_blast_Matrix(M, 1) 'Nm3
Bf_oxygen = Bf_oxygen_Matrix(M, 1) 'Nm3

If Bf_blast = 0 Or Bf_oxygen = 0 Then
Do
M_pigiron = M_pigiron_Matrix(M, 1) 'Tons of pig iron
Bf_blast = Bf_blast_Matrix(M, 1) 'Nm3
Bf_oxygen = Bf_oxygen_Matrix(M, 1) 'Nm3
M = M + 1
Loop While Bf_oxygen = 0 Or Bf_blast = 0
End If



n_N2_blast = Bf_blast * Datamatrix(19, 1) / Datamatrix(17, 1) 'kmol
n_O2_blast = Bf_blast * Datamatrix(18, 1) / Datamatrix(17, 1) 'kmol
n_O2_oxygenintake = Bf_oxygen / Datamatrix(17, 1) 'kmol
n_total_O_in = (n_O2_blast + n_O2_oxygenintake) * 2 'kmol




'Calculates the amounts of coke, briquettes and scrap

Cokeratio = Cells.Find("Input data").Offset(1, 1).Value2
Briqratio = Cells.Find("Input data").Offset(2, 1).Value2
Scrapratio = Cells.Find("Input data").Offset(3, 1).Value2
m_oil = Cells.Find("Input data").Offset(4, 1).Value2


m_coke = Cokeratio * M_pigiron * 1000 'kg
m_briq = Briqratio * M_pigiron 'kg
m_scrap = Scrapratio * M_pigiron 'kg


'Fe/Iron calculations
'Calculates the molar masses of iron and coal in pig iron, briqettes and scrap

n_Fe_pigiron = Datamatrix(3, 1) * M_pigiron * 1000 / Datamatrix(15, 1) 'kmol

n_Fe_briq = Datamatrix(12, 1) * m_briq / Datamatrix(15, 1) 'kmol

n_Fe_scrap = Datamatrix(13, 1) * m_scrap / Datamatrix(15, 1) 'kmol

'Calculates how many kmol is needed from pellets

n_Fe_pellets = n_Fe_pigiron - n_Fe_briq - n_Fe_scrap
m_pellets = n_Fe_pellets / Datamatrix(11, 1) * Datamatrix(15, 1) 'Divides by the iron content 0.72, to get the total mass



'O/Oxygen calculations
'Calculates the total incoming oxygen
'(m_pel*x_pellets,O + m_briq*x_O,briq)/M_O + n_blast,O2*2 + n_Oxygen,O2*2

Oxygen_in = m_pellets * Datamatrix(10, 1) / Datamatrix(16, 1) + m_briq * Datamatrix(9, 1) / Datamatrix(16, 1) + n_total_O_in 'kmol
Cells.Find("Solutions").Offset(0, 1).Value = Oxygen_in

'C/Coal calculations
'Calculates the incoming coal minus what comes out with the pig iron, leaving what comes out with the bf-gases
'm_coke,*x_C,coke + m_oil*x_C,oil + m_br*x_br,C = m_rj*x_C,rj + V_tg*(y,co + y,co2)

Coal_for_bf_gas = (m_coke * Datamatrix(4, 1) / Datamatrix(14, 1) + m_oil * Datamatrix(5, 1) / Datamatrix(14, 1) + m_briq * Datamatrix(6, 1) / Datamatrix(14, 1)) - M_pigiron * 1000 * Datamatrix(1, 1) / Datamatrix(14, 1)
Cells.Find("Solutions").Offset(0, 2).Value = Coal_for_bf_gas

'N/Nitrogen
'Nitrogen is mainly what comes in with the blast

N2_for_bf_gas = n_N2_blast
Cells.Find("Solutions").Offset(0, 3).Value = N2_for_bf_gas



'Sets in the hydrogen just in case
'H/hydrogen
'H_for_bf_gas = m_coke * Datamatrix(21, 1) / Datamatrix(20, 1) + m_oil * Datamatrix(7, 1) / Datamatrix(20, 1)

求解器部分：

SolverReset 'Code solves the problem for a specific set of lines, in this case meaning hours
SolverOptions Precision:=1, Iterations:=100, AssumeNonNeg:=True
SolverOk setCell:=Cells.Find("Differences").Offset(1, 0).Address, MaxMinVal:=3, ValueOf:="0", ByChange:=Range(Cells.Find("Testing here").Offset(0, 1).Address, Cells.Find("Testing here").Offset(0, 3).Address)

SolverAdd cellRef:=Range(Cells.Find("Testing here").Offset(0, 2).Address, Cells.Find("Testing here").Offset(0, 3).Address), _
relation:=3, _
formulaText:=0.1

SolverAdd cellRef:=Range(Cells.Find("Testing here").Offset(0, 2).Address, Cells.Find("Testing here").Offset(0, 3).Address), _
relation:=1, _
formulaText:=0.4

SolverAdd cellRef:=Cells.Find("Testing here").Offset(0, 1).Address, _
relation:=3, _
formulaText:=(Bf_blast + Bf_oxygen) * 1.2

SolverAdd cellRef:=Cells.Find("Testing here").Offset(0, 1).Address, _
relation:=1, _
formulaText:=(Bf_blast + Bf_oxygen) * 2

SolverSolve userFinish:=True

BFoutput(M, 1) = Cells.Find("Testing here").Offset(0, 1).Value
BFoutput(M, 2) = Cells.Find("Testing here").Offset(0, 2).Value
BFoutput(M, 3) = Cells.Find("Testing here").Offset(0, 3).Value



M = M + 1
Loop While M < 169

Cells.Find("BF1 - Output data").Offset(2, 0).Resize(UBound(BFoutput, 1), 3).Value = BFoutput

Answer 1

我不是化学工程师，所以我不知道你想要解决的方程式。

我猜它们是非线性的，瞬态的和迭代的。 168 * 3 = 504自由度对我来说似乎不大，但如果你有很多小的时间步骤并且每个步骤都有迭代，那么它可能会很多。

我无法判断你发布的代码是否存在暂时或稳定状态问题。

我更熟悉的数值问题（固体力学和热传递）对算法非常敏感。出于稳定性原因，方程式可能会受到时间步长的限制，具体取决于所选择的积分方案。

如果您正在解决非线性稳态问题，则会应用相同的注释，但迭代步长而不是时间步长除外。

我无法从您的VB代码中收集到很多相关信息，但我会提供以下建议：

写出你的方程式，进行傅里叶变换，看看是否有任何稳定性限制。
想想像Matlab这样的工具包。他们开箱即用的实现可能比您的自定义代码更加优化。
我不知道VB或Excel的任何分析功能，但是没有数据就无法解决问题。在假设解决方案之前，我会看看是否可以获得有关时间花费的信息。

VBA：慢速求解器循环

1 个答案: