氢燃料汽车加注模型优化
[英]Hydrogen Vehicle Refueling Model Optimization
问题描述
I am currently in the process of trying to determine the optimal inflow conditions for the refueling process of a hydrogen (H2) vehicle using GEKKO.
我目前正在尝试使用 GEKKO 确定氢气 (H2) 车辆加油过程的最佳流入条件。 Below are the coupled, ordinary differential equations which govern how the temperature of H2 and the fuel tank wall change over the fueling time.下面是耦合的常微分方程，它们控制着 H2 和燃料箱壁的温度如何随加油时间变化。
T.dt() = (1+alpha)*(T_star - T)/(t_star + t)
T_w.dt() = (T - T_w)/t_w_star
where在哪里
alpha = (a_in*A_in)/(c_v*m_dot_in), t_star = m_0/m_dot_in, t_w_star = (m_w*c_w)/(a_in*A_in)
T_star = gamma_p*T_inf + alpha_p*T_w, gamma_p = gamma/(1 + alpha), alpha_p = alpha/(1 + alpha)
Here, m_0 is the initial mass of H2 in the tank, m_dot_in is the mass flow rate of H2 into the tank, gamma is the ratio of the specific heats for H2, T_inf is the inflow temperature of H2, and the other variables are intermediate variables/tank parameters.其中， m_0为罐内 H2 的初始质量， m_dot_in为 H2 进入罐内的质量流量， gamma为 H2 的比热比， T_inf为 H2 的流入温度，其他变量为中间值变量/坦克参数。 Through the refueling process, m_dot_in is taken to be constant (but unknown), so the mass of H2 in the tank over time is defined as:通过加油过程， m_dot_in被认为是常数（但未知），因此罐中 H2 的质量随时间的变化定义为：
m = m_0 + m_dot_in*t
Additionally, the pressure of H2 within the tank may be calculated with a real gas equation of state (I use the Peng-Robinson equation of state for this model).此外，罐内 H2 的压力可以用真实的气体状态方程计算（我在这个模型中使用 Peng-Robinson 状态方程）。
What I am trying to do with this model is determine the optimal m_dot_in , T_inf , and m_0 to minimize the total fueling time, t_f .我试图用这个模型做的是确定最佳m_dot_in 、 T_inf和m_0以最小化总加油时间t_f 。 Some constraints on the variables are that T<=358.15 K throughout the whole refueling process (for safety reasons), and that the final pressure of H2 within the tank must be 35 MPa.对变量的一些限制是在整个加油过程中T<=358.15 K （出于安全原因），并且罐内 H2 的最终压力必须为 35 MPa。 For this model, I consider t_f , m_dot_in , m_0 , and T_inf to be fixed variables that have the following bounds:对于此模型，我认为t_f 、 m_dot_in 、 m_0和T_inf是具有以下边界的固定变量：
60 sec <= t_f <= 300 sec
0.0005 kg/sec <= m_dot_in <= 0.03 kg/sec
5% of m_f <= m_0 <= 90% of m_f, where m_f = 1.79 kg
288.15 K <= T_inf <= 303.15 K
Below I have copied my code for this optimization problem using GEKKO:下面我使用 GEKKO 复制了我针对这个优化问题的代码：
# HYDROGEN TANK REFUELING MODEL
# OPTIMIZE MODEL WITH GEKKO OPTIMIZATION SUITE
from gekko import GEKKO
m = GEKKO()
# CONSTANTS
## TANK PARAMETERS (ASSUME TYPE III, ALUMINUM, 74 l, RATED FOR 35 MPa)
V = 0.074 # m^3
a_in = 167/19e-3 # W/m^2/K
c_w = 2730 # J/kg/K
rho_w = 900 # kg/m^3
m_w = rho_w*(np.pi*(((0.358+19e-3)/2)**2)*(0.893+19e-3) - np.pi*((0.358/2)**2)*0.893)
A_in = 2*np.pi*(0.358/2)*((0.358/2) + 0.893) # m^2
T_w0 = 293.15 # K
m_f = 1.79 # final mass of hydrogen in tank, kg
## HYDROGEN PARAMETERS
c_p = 14.615e3 # specific heat at constant pressure, J/kg/K
c_v = 10.316e3 # specific heat at constant volume, J/kg/K
gamma = c_p/c_v
R = 8.314/M_H2 # gas constant for H2, J/kgK
T_c = -240 + 273.15 # critical temperature for H2, K
p_c = 1.3e6 # critical pressure for H2, Pa
w_H2 = -0.219 # acentric factor for H2
a = 0.45724*(R**2 * T_c**2)/(p_c**2)
b = 0.0778*(R*T_c)/p_c
kappa = 0.37464 + 1.54226*w_H2 - 0.26992*(w_H2**2)
# SET TIME ANALYSIS POINTS
nt = 101
tm = np.linspace(0, 1, nt)
m.time = tm
# options
m.options.NODES = 6
m.options.SOLVER = 3
m.options.IMODE = 6
m.options.MAX_ITER = 500
m.options.MV_TYPE = 0
m.options.DIAGLEVEL = 0
# FIXED VARIABLES
t_f = m.FV(value=60.0,lb=60.0,ub=300.0) # final fuel time, s
t_f.STATUS = 1
m_dot_in = m.FV(value=0.001,lb=0.0005,ub=0.03) # mass flow rate into tank, kg/s
m_dot_in.STATUS = 1
m_0 = m.FV(value=0.1*m_f,lb=0.05*m_f,ub=0.9*m_f) # initial mass of H2 in tank (as % of m_f), kg
m_0.STATUS = 1
T_inf = m.FV(value=20 + 273.15,lb=15 + 273.15,ub=30 + 273.15) # inflow temperature, K
T_inf.STATUS = 1
# PARAMETERS
f = np.zeros(nt)
f[-1] = 1.0
final = m.Param(value=f)
# VARIABLES
T = m.Var(value=15+273.15,lb=15+273.15,ub=85+273.15)
T_w = m.Var(value=T_w0,lb=T_w0,ub=85+273.15)
mass = m.Var(value=m_0,lb=m_0,ub=m_f)
p = m.Var(value=1.0e6,lb=0.0,ub=35.0e6)
# INTERMEDIATES
alpha = m.Intermediate((a_in*A_in)/c_v/m_dot_in)
gamma_p = m.Intermediate(gamma/(1 + alpha))
alpha_p = m.Intermediate(alpha/(1 + alpha))
t_star = m.Intermediate(m_0/m_dot_in)
t_w_star = m.Intermediate((m_w*c_w)/(a_in*A_in))
T_star = m.Intermediate(gamma_p*T_inf + alpha_p*T_w)
alpha_T = m.Intermediate(1 + kappa*(1 - (T/T_c)**0.5))
v = m.Intermediate(V/mass) # specific volume, m^3/kg
# EQUATIONS
m.Equation(mass==t_f*(m_0 + m_dot_in*m.time))
m.Equation(T.dt()==t_f*(1 + alpha)*((T_star-T)/(t_star+tm)))
m.Equation(T_w.dt()==t_f*((T-T_w)/t_w_star))
m.Equation(p*1.0e6==t_f*((R*T/(v-b)) - ((a*alpha_T**2)/(v*(v+b) + b*(v-b)))))
m.Equation(T*final<=85+273.15)
m.Equation(T_w*final<=85+273.15)
# SPECIFIY ENDPOINT CONDITIONS
m.fix(mass, pos=len(m.time)-1, val=m_f)
m.fix(p, pos=len(m.time)-1, val=35.0e6)
# MINIMIZE FINAL FUEL TIME
m.Obj(t_f)
# SOLVE
m.solve()
# RESULTS
print('Final Time: ' + str(t_f.value[0]))
This code currently gives me the following error:此代码目前给我以下错误：
apm 45.3.69.90_gk_model46 <br><pre> ----------------------------------------------------------------
APMonitor, Version 1.0.1
APMonitor Optimization Suite
----------------------------------------------------------------
@error: Equation Definition
Equation without an equality (=) or inequality (>,<)
0.140.150.160.170.180.190.20.210.220.230.240.250.260.27
STOPPING...
---------------------------------------------------------------------------
Exception Traceback (most recent call last)
<ipython-input-102-4d40bf2f7c9c> in <module>
87
88 # SOLVE
---> 89 m.solve()
90
91 # RESULTS
~\anaconda3\lib\site-packages\gekko\gekko.py in solve(self, disp, debug, GUI, **kwargs)
2172 #print APM error message and die
2173 if (debug >= 1) and ('@error' in response):
-> 2174 raise Exception(response)
2175
2176 #load results
Exception: @error: Equation Definition
Equation without an equality (=) or inequality (>,<)
0.140.150.160.170.180.190.20.210.220.230.240.250.260.27
STOPPING...
I am very new to optimization in general, and I tried including several different equality and inequality constraints, but nothing seems to work.我对优化很陌生，我尝试包含几个不同的等式和不等式约束，但似乎没有任何效果。 I thought I was doing it correctly based off example problems and information from the APMonitor website, but obviously something is off with my implementation.我认为我根据示例问题和来自 APMonitor 网站的信息正确地做这件事，但显然我的实现有些问题。 I was wondering if someone knew what I should change/add or if I am doing something completely wrong?我想知道是否有人知道我应该更改/添加什么，或者我做错了什么？ Any help would be much appreciated!任何帮助将非常感激！
Thank you for your time,感谢您的时间，
Evan埃文
EDIT : Based on Dr. Hedengren's answer, I tried to simplifying the model such that the variables of mass and p were not included, as they only depend on the final values for t_f , m_dot_in , and T and may be calculated after the solution has been obtained.编辑：基于 Hedengren 博士的回答，我尝试简化模型，以便不包括mass和p变量，因为它们仅取决于t_f 、 m_dot_in和T的最终值，并且可能在解决方案有后计算已获得。 Below is my edited code:下面是我编辑的代码：
# HYDROGEN TANK REFUELING MODEL
# OPTIMIZE MODEL WITH GEKKO OPTIMIZATION SUITE
from gekko import GEKKO
import numpy as np
m = GEKKO(remote=False)
# CONSTANTS
## TANK PARAMETERS (ASSUME TYPE III, ALUMINUM, 74 l, RATED FOR 35 MPa)
V = 0.074 # m^3
a_in = 167/19e-3 # W/m^2/K
c_w = 2730 # J/kg/K
rho_w = 900 # kg/m^3
m_w = rho_w*(np.pi*(((0.358+19e-3)/2)**2)*(0.893+19e-3) - np.pi*((0.358/2)**2)*0.893)
A_in = 2*np.pi*(0.358/2)*((0.358/2) + 0.893) # m^2
T_w0 = 293.15 # K
m_f = 1.79 # final mass of hydrogen in tank, kg
## HYDROGEN PARAMETERS
c_p = 14.615e3 # specific heat at constant pressure, J/kg/K
c_v = 10.316e3 # specific heat at constant volume, J/kg/K
gamma = c_p/c_v
M_H2 = 2.02
R = 8.314/M_H2 # gas constant for H2, J/kgK
T_c = -240 + 273.15 # critical temperature for H2, K
p_c = 1.3e6 # critical pressure for H2, Pa
w_H2 = -0.219 # acentric factor for H2
a = 0.45724*(R**2 * T_c**2)/(p_c**2)
b = 0.0778*(R*T_c)/p_c
kappa = 0.37464 + 1.54226*w_H2 - 0.26992*(w_H2**2)
## SET INFLOW TEMPERATURE AND INITIAL MASS IN TANK
m_0 = 0.1*m_f
T_inf = 20 + 273.15
# SET TIME ANALYSIS POINTS
nt = 101
tm = np.linspace(0, 1, nt)
m.time = tm
t = m.Param(tm, name='time')
# options
m.options.NODES = 6
m.options.SOLVER = 3
m.options.IMODE = 6
m.options.MAX_ITER = 500
m.options.MV_TYPE = 0
m.options.DIAGLEVEL = 0
# FIXED VARIABLES
t_f = m.FV(value=60.0,lb=60.0,ub=300.0) # final fuel time, s
t_f.STATUS = 0
m_dot_in = m.FV(value=0.001,lb=0.0005,ub=0.03) # mass flow rate into tank, kg/s
m_dot_in.STATUS = 0
# m_0 = m.FV(value=0.1*m_f,lb=0.05*m_f,ub=0.9*m_f) # initial mass of H2 in tank (as % of m_f), kg
# m_0.STATUS = 0
# T_inf = m.FV(value=20 + 273.15,lb=15 + 273.15,ub=30 + 273.15) # inflow temperature, K
# T_inf.STATUS = 0
# PARAMETERS
f = np.zeros(nt)
f[-1] = 1.0
final = m.Param(value=f, name='final')
# VARIABLES
T = m.Var(value=15+273.15,lb=15+273.15,ub=85+273.15, name='H2 Temp')
T_w = m.Var(value=T_w0,lb=T_w0,ub=85+273.15, name='Wall Temp')
# mass = m.Var(value=m_0,lb=m_0,ub=m_f, name='H2 Mass')
# p = m.Var(value=1.0e6,lb=0.0,ub=35.0e6, name='H2 Press')
# INTERMEDIATES
alpha = m.Intermediate((a_in*A_in)/c_v/m_dot_in, name='alpha')
gamma_p = m.Intermediate(gamma/(1 + alpha), name='gamma_p')
alpha_p = m.Intermediate(alpha/(1 + alpha), name='alpha_p')
t_star = m.Intermediate(m_0/m_dot_in, name='t_star')
t_w_star = m.Intermediate((m_w*c_w)/(a_in*A_in), name='t_w_star')
T_star = m.Intermediate(gamma_p*T_inf + alpha_p*T_w, name='Temp_star')
# alpha_T = m.Intermediate(1 + kappa*(1 - (T/T_c)**0.5))
# v = m.Intermediate(V/mass) # specific volume, m^3/kg
# EQUATIONS
# m.Equation(mass==t_f*(m_0 + m_dot_in*t*t_f))
m.Equation(T.dt()==t_f*(1 + alpha)*((T_star-T)/(t_star+t*t_f)))
m.Equation(T_w.dt()==t_f*((T-T_w)/t_w_star))
# m.Equation(p==t_f*((R*T/(v-b)) - ((a*alpha_T**2)/(v*(v+b) + b*(v-b)))))
# m.Equation((T-(85+273.15))*final<=0)
# m.Equation((T_w-(85+273.15))*final<=0)
# SPECIFIY ENDPOINT CONDITIONS
# m.Minimize(final*(mass-m_f)**2)
# m.Minimize(final*(p-35.0e6)**2)
m.Minimize(final*(T-351)**2)
#m.fix(mass, pos=len(m.time)-1, val=m_f)
#m.fix(p, pos=len(m.time)-1, val=35.0e6)
# MINIMIZE FINAL FUEL TIME
m.Minimize(t_f)
# SOLVE
m.options.SOLVER = 3
m.open_folder()
m.solve()
# RESULTS
print('Final Time: ' + str(t_f.value[0]))
I am still getting infeasibilities (not as many as before), but I am having trouble understanding what said infeasibilities mean and how to go about fixing them.我仍然遇到不可行性（不像以前那么多），但我无法理解所说的不可行性意味着什么以及如何解决它们。 Below are the infeasibilities I am getting:以下是我遇到的不可行性：
************************************************
***** POSSIBLE INFEASBILE EQUATIONS ************
************************************************
____________________________________________________________________________
EQ Number Lower Residual Upper Infeas. Name
1 0.0000E+00 -7.5600E-04 0.0000E+00 7.5600E-04 p(1).n(2).Eqn(1): 0 = $h2_temp-(((((p2)*((1+alpha))))*((((temp_star-h2_temp))/((t_star+((time)*(p2))))))))
Variable Lower Value Upper $Value Name
0 -1.2346E+20 1.0000E+00 1.2346E+20 0.0000E+00 p(1).n(2).time
0 6.0000E+01 6.0000E+01 3.0000E+02 0.0000E+00 p(1).n(1).p2
0 5.0000E-04 1.0000E-03 3.0000E-02 0.0000E+00 p(1).n(1).p3
1 2.8815E+02 2.8949E+02 3.5815E+02 9.7624E+02 p(1).n(2).h2_temp
2 2.9315E+02 2.9315E+02 3.5815E+02 -7.9554E+01 p(1).n(2).wall_temp
1 2.8815E+02 2.8949E+02 3.5815E+02 9.7624E+02 p(1).n(2).h2_temp
____________________________________________________________________________
EQ Number Lower Residual Upper Infeas. Name
5 0.0000E+00 -7.5600E-04 0.0000E+00 7.5600E-04 p(1).n(3).Eqn(1): 0 = $h2_temp-(((((p2)*((1+alpha))))*((((temp_star-h2_temp))/((t_star+((time)*(p2))))))))
Variable Lower Value Upper $Value Name
0 -1.2346E+20 1.0000E+00 1.2346E+20 0.0000E+00 p(1).n(3).time
0 6.0000E+01 6.0000E+01 3.0000E+02 0.0000E+00 p(1).n(1).p2
0 5.0000E-04 1.0000E-03 3.0000E-02 0.0000E+00 p(1).n(1).p3
3 2.8815E+02 2.9123E+02 3.5815E+02 5.2535E+02 p(1).n(3).h2_temp
4 2.9315E+02 2.9315E+02 3.5815E+02 -4.1620E+01 p(1).n(3).wall_temp
3 2.8815E+02 2.9123E+02 3.5815E+02 5.2535E+02 p(1).n(3).h2_temp
____________________________________________________________________________
EQ Number Lower Residual Upper Infeas. Name
9 0.0000E+00 -7.5600E-04 0.0000E+00 7.5600E-04 p(1).n(4).Eqn(1): 0 = $h2_temp-(((((p2)*((1+alpha))))*((((temp_star-h2_temp))/((t_star+((time)*(p2))))))))
Variable Lower Value Upper $Value Name
0 -1.2346E+20 1.0000E+00 1.2346E+20 0.0000E+00 p(1).n(4).time
0 6.0000E+01 6.0000E+01 3.0000E+02 0.0000E+00 p(1).n(1).p2
0 5.0000E-04 1.0000E-03 3.0000E-02 0.0000E+00 p(1).n(1).p3
5 2.8815E+02 2.9229E+02 3.5815E+02 2.5164E+02 p(1).n(4).h2_temp
6 2.9315E+02 2.9315E+02 3.5815E+02 -1.8591E+01 p(1).n(4).wall_temp
5 2.8815E+02 2.9229E+02 3.5815E+02 2.5164E+02 p(1).n(4).h2_temp
____________________________________________________________________________
EQ Number Lower Residual Upper Infeas. Name
13 0.0000E+00 -7.5600E-04 0.0000E+00 7.5600E-04 p(1).n(5).Eqn(1): 0 = $h2_temp-(((((p2)*((1+alpha))))*((((temp_star-h2_temp))/((t_star+((time)*(p2))))))))
Variable Lower Value Upper $Value Name
0 -1.2346E+20 1.0000E+00 1.2346E+20 0.0000E+00 p(1).n(5).time
0 6.0000E+01 6.0000E+01 3.0000E+02 0.0000E+00 p(1).n(1).p2
0 5.0000E-04 1.0000E-03 3.0000E-02 0.0000E+00 p(1).n(1).p3
7 2.8815E+02 2.9274E+02 3.5815E+02 1.3550E+02 p(1).n(5).h2_temp
8 2.9315E+02 2.9315E+02 3.5815E+02 -8.8200E+00 p(1).n(5).wall_temp
7 2.8815E+02 2.9274E+02 3.5815E+02 1.3550E+02 p(1).n(5).h2_temp
____________________________________________________________________________
EQ Number Lower Residual Upper Infeas. Name
25 0.0000E+00 -7.5600E-04 0.0000E+00 7.5600E-04 p(2).n(3).Eqn(1): 0 = $h2_temp-(((((p2)*((1+alpha))))*((((temp_star-h2_temp))/((t_star+((time)*(p2))))))))
Variable Lower Value Upper $Value Name
0 -1.2346E+20 1.0000E+00 1.2346E+20 0.0000E+00 p(2).n(3).time
0 6.0000E+01 6.0000E+01 3.0000E+02 0.0000E+00 p(1).n(1).p2
0 5.0000E-04 1.0000E-03 3.0000E-02 0.0000E+00 p(1).n(1).p3
13 2.8815E+02 2.9312E+02 3.5815E+02 3.9285E+01 p(2).n(3).h2_temp
14 2.9315E+02 2.9315E+02 3.5815E+02 -7.2493E-01 p(2).n(3).wall_temp
13 2.8815E+02 2.9312E+02 3.5815E+02 3.9285E+01 p(2).n(3).h2_temp
____________________________________________________________________________
EQ Number Lower Residual Upper Infeas. Name
29 0.0000E+00 -7.5598E-04 0.0000E+00 7.5598E-04 p(2).n(4).Eqn(1): 0 = $h2_temp-(((((p2)*((1+alpha))))*((((temp_star-h2_temp))/((t_star+((time)*(p2))))))))
Variable Lower Value Upper $Value Name
0 -1.2346E+20 1.0000E+00 1.2346E+20 0.0000E+00 p(2).n(4).time
0 6.0000E+01 6.0000E+01 3.0000E+02 0.0000E+00 p(1).n(1).p2
0 5.0000E-04 1.0000E-03 3.0000E-02 0.0000E+00 p(1).n(1).p3
15 2.8815E+02 2.9320E+02 3.5815E+02 1.8882E+01 p(2).n(4).h2_temp
16 2.9315E+02 2.9315E+02 3.5815E+02 9.9172E-01 p(2).n(4).wall_temp
15 2.8815E+02 2.9320E+02 3.5815E+02 1.8882E+01 p(2).n(4).h2_temp
____________________________________________________________________________
EQ Number Lower Residual Upper Infeas. Name
2006 0.0000E+00 -1.0946E-01 0.0000E+00 1.0946E-01 p(1).c(2).t(2): not available
Variable Lower Value Upper $Value Name
2 2.9315E+02 2.9315E+02 3.5815E+02 -7.9554E+01 p(1).n(2).wall_temp
4 2.9315E+02 2.9315E+02 3.5815E+02 -4.1620E+01 p(1).n(3).wall_temp
6 2.9315E+02 2.9315E+02 3.5815E+02 -1.8591E+01 p(1).n(4).wall_temp
8 2.9315E+02 2.9315E+02 3.5815E+02 -8.8200E+00 p(1).n(5).wall_temp
10 2.9315E+02 2.9316E+02 3.5815E+02 -6.0293E+00 p(1).n(6).wall_temp
2 2.9315E+02 2.9315E+02 3.5815E+02 -7.9554E+01 p(1).n(2).wall_temp
0 2.9315E+02 2.9315E+02 3.5815E+02 0.0000E+00 p(1).n(1).wall_temp
____________________________________________________________________________
EQ Number Lower Residual Upper Infeas. Name
2007 0.0000E+00 -2.5022E-01 0.0000E+00 2.5022E-01 p(1).c(2).t(3): not available
Variable Lower Value Upper $Value Name
2 2.9315E+02 2.9315E+02 3.5815E+02 -7.9554E+01 p(1).n(2).wall_temp
4 2.9315E+02 2.9315E+02 3.5815E+02 -4.1620E+01 p(1).n(3).wall_temp
6 2.9315E+02 2.9315E+02 3.5815E+02 -1.8591E+01 p(1).n(4).wall_temp
8 2.9315E+02 2.9315E+02 3.5815E+02 -8.8200E+00 p(1).n(5).wall_temp
10 2.9315E+02 2.9316E+02 3.5815E+02 -6.0293E+00 p(1).n(6).wall_temp
4 2.9315E+02 2.9315E+02 3.5815E+02 -4.1620E+01 p(1).n(3).wall_temp
0 2.9315E+02 2.9315E+02 3.5815E+02 0.0000E+00 p(1).n(1).wall_temp
____________________________________________________________________________
EQ Number Lower Residual Upper Infeas. Name
2008 0.0000E+00 -3.3207E-01 0.0000E+00 3.3207E-01 p(1).c(2).t(4): not available
Variable Lower Value Upper $Value Name
2 2.9315E+02 2.9315E+02 3.5815E+02 -7.9554E+01 p(1).n(2).wall_temp
4 2.9315E+02 2.9315E+02 3.5815E+02 -4.1620E+01 p(1).n(3).wall_temp
6 2.9315E+02 2.9315E+02 3.5815E+02 -1.8591E+01 p(1).n(4).wall_temp
8 2.9315E+02 2.9315E+02 3.5815E+02 -8.8200E+00 p(1).n(5).wall_temp
10 2.9315E+02 2.9316E+02 3.5815E+02 -6.0293E+00 p(1).n(6).wall_temp
6 2.9315E+02 2.9315E+02 3.5815E+02 -1.8591E+01 p(1).n(4).wall_temp
0 2.9315E+02 2.9315E+02 3.5815E+02 0.0000E+00 p(1).n(1).wall_temp
____________________________________________________________________________
EQ Number Lower Residual Upper Infeas. Name
2009 0.0000E+00 -3.6373E-01 0.0000E+00 3.6373E-01 p(1).c(2).t(5): not available
Variable Lower Value Upper $Value Name
2 2.9315E+02 2.9315E+02 3.5815E+02 -7.9554E+01 p(1).n(2).wall_temp
4 2.9315E+02 2.9315E+02 3.5815E+02 -4.1620E+01 p(1).n(3).wall_temp
6 2.9315E+02 2.9315E+02 3.5815E+02 -1.8591E+01 p(1).n(4).wall_temp
8 2.9315E+02 2.9315E+02 3.5815E+02 -8.8200E+00 p(1).n(5).wall_temp
10 2.9315E+02 2.9316E+02 3.5815E+02 -6.0293E+00 p(1).n(6).wall_temp
8 2.9315E+02 2.9315E+02 3.5815E+02 -8.8200E+00 p(1).n(5).wall_temp
0 2.9315E+02 2.9315E+02 3.5815E+02 0.0000E+00 p(1).n(1).wall_temp
____________________________________________________________________________
EQ Number Lower Residual Upper Infeas. Name
2010 0.0000E+00 -3.8212E-01 0.0000E+00 3.8212E-01 p(1).c(2).t(6): not available
Variable Lower Value Upper $Value Name
2 2.9315E+02 2.9315E+02 3.5815E+02 -7.9554E+01 p(1).n(2).wall_temp
4 2.9315E+02 2.9315E+02 3.5815E+02 -4.1620E+01 p(1).n(3).wall_temp
6 2.9315E+02 2.9315E+02 3.5815E+02 -1.8591E+01 p(1).n(4).wall_temp
8 2.9315E+02 2.9315E+02 3.5815E+02 -8.8200E+00 p(1).n(5).wall_temp
10 2.9315E+02 2.9316E+02 3.5815E+02 -6.0293E+00 p(1).n(6).wall_temp
10 2.9315E+02 2.9316E+02 3.5815E+02 -6.0293E+00 p(1).n(6).wall_temp
0 2.9315E+02 2.9315E+02 3.5815E+02 0.0000E+00 p(1).n(1).wall_temp
____________________________________________________________________________
EQ Number Lower Residual Upper Infeas. Name
2017 0.0000E+00 -7.1275E-04 0.0000E+00 7.1275E-04 p(2).c(2).t(3): not available
Variable Lower Value Upper $Value Name
12 2.9315E+02 2.9315E+02 3.5815E+02 -3.6833E+00 p(2).n(2).wall_temp
14 2.9315E+02 2.9315E+02 3.5815E+02 -7.2493E-01 p(2).n(3).wall_temp
16 2.9315E+02 2.9315E+02 3.5815E+02 9.9172E-01 p(2).n(4).wall_temp
18 2.9315E+02 2.9315E+02 3.5815E+02 1.6707E+00 p(2).n(5).wall_temp
20 2.9315E+02 2.9316E+02 3.5815E+02 1.8694E+00 p(2).n(6).wall_temp
14 2.9315E+02 2.9315E+02 3.5815E+02 -7.2493E-01 p(2).n(3).wall_temp
10 2.9315E+02 2.9316E+02 3.5815E+02 -6.0293E+00 p(1).n(6).wall_temp
************************************************
Also, this took around 2 minutes to run, so if there is any advice on how to decrease the computation time, that would be much appreciated!此外，这需要大约 2 分钟才能运行，所以如果有关于如何减少计算时间的任何建议，将不胜感激！
1 个解决方案
解决方案1
0 已采纳 2021-06-23 18:03:37
You can see the model that Gekko writes with m.open_folder() and then open the model gk_model0.apm with a text editor:可以看到 Gekko 用m.open_folder()编写的模型，然后用文本编辑器打开模型gk_model0.apm ：
Model
Parameters
p1 = 60.0, <= 300.0, >= 60.0
p2 = 0.001, <= 0.03, >= 0.0005
p3 = 0.17900000000000002, <= 1.611, >= 0.08950000000000001
p4 = 293.15, <= 303.15, >= 288.15
p5
End Parameters
Variables
v1 = 288.15, <= 358.15, >= 288.15
v2 = 293.15, <= 358.15, >= 293.15
v3 = 0.17900000000000002, <= 1.79, >= p3
v4 = 1000000.0, <= 35000000.0, >= 0.0
End Variables
Intermediates
i0=(1.0272572651140832/p2)
i1=(1.416731291198139/(1+i0))
i2=((i0)/((1+i0)))
i3=((p3)/(p2))
i4=2.762640041099948
i5=(((i1)*(p4))+((i2)*(v2)))
i6=(1+((0.02393942687999997)*((1-((((v1)/(33.14999999999998)))^(0.5))))))
i7=(0.074/v3)
End Intermediates
Equations
v3=((p1)*((p3+((p2)*([0. 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.13
0.14 0.15 0.16 0.17 0.18 0.19 0.2 0.21 0.22 0.23 0.24 0.25 0.26 0.27
0.28 0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36 0.37 0.38 0.39 0.4 0.41
0.42 0.43 0.44 0.45 0.46 0.47 0.48 0.49 0.5 0.51 0.52 0.53 0.54 0.55
0.56 0.57 0.58 0.59 0.6 0.61 0.62 0.63 0.64 0.65 0.66 0.67 0.68 0.69
0.7 0.71 0.72 0.73 0.74 0.75 0.76 0.77 0.78 0.79 0.8 0.81 0.82 0.83
0.84 0.85 0.86 0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97
0.98 0.99 1. ])))))
$v1=((((p1)*((1+i0))))*((((i5-v1))/((i3+[0. 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.13
0.14 0.15 0.16 0.17 0.18 0.19 0.2 0.21 0.22 0.23 0.24 0.25 0.26 0.27
0.28 0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36 0.37 0.38 0.39 0.4 0.41
0.42 0.43 0.44 0.45 0.46 0.47 0.48 0.49 0.5 0.51 0.52 0.53 0.54 0.55
0.56 0.57 0.58 0.59 0.6 0.61 0.62 0.63 0.64 0.65 0.66 0.67 0.68 0.69
0.7 0.71 0.72 0.73 0.74 0.75 0.76 0.77 0.78 0.79 0.8 0.81 0.82 0.83
0.84 0.85 0.86 0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97
0.98 0.99 1. ])))))
$v2=((p1)*((((v1-v2))/(i4))))
((v4)*(1000000.0))=((p1)*((((((4.1158415841584155)*(v1)))/((i7-8.165418118811874e-06)))-((((5.036651227998414e-09)*(((i6)^(2)))))/((((i7)*((i7+8.165418118811874e-06)))+((8.165418118811874e-06)*((i7-8.165418118811874e-06)))))))))
((v1)*(p5))<=358.15
((v2)*(p5))<=358.15
minimize p1
End Equations
Connections
p(100).n(end).v3=1.79
p(100).n(end).v3=fixed
p(100).n(end).v4=35000000.0
p(100).n(end).v4=fixed
End Connections
End Model
The problem is with the first two equation that have m.time and tm as Numpy arrays instead of using a Gekko variable or parameter.问题在于前两个方程将m.time和tm作为 Numpy 数组，而不是使用 Gekko 变量或参数。 If a Numpy array or Python list is is needed in the optimization problem then create a new m.Param() such as:如果优化问题中需要 Numpy 数组或 Python 列表，则创建一个新的m.Param()例如：
t = m.Param(tm)
m.Equation(mass==t_f*(m_0 + m_dot_in*t))
m.Equation(T.dt()==t_f*(1 + alpha)*((T_star-T)/(t_star+t)))
Because the final time is minimized, the time in those equations may need to be t*t_f .由于最终时间被最小化，这些方程中的时间可能需要为t*t_f 。
Equations such as m.Equation(T_w*final<=85+273.15) should be reformulated as m.Equation((T_w-(85+273.15))*final<=0) so that when final=0 then it is 0<=0 .诸如m.Equation(T_w*final<=85+273.15)方程应重新m.Equation((T_w-(85+273.15))*final<=0)为m.Equation((T_w-(85+273.15))*final<=0)以便当final=0它是0<=0 。 In this case, your original equations are okay but it is good practice to put all terms on one side of the equation.在这种情况下，您的原始方程没问题，但将所有项放在方程的一侧是一种很好的做法。
Even with these modifications there is still Exception: @error: Solution Not Found .即使进行了这些修改，仍然存在Exception: @error: Solution Not Found 。 The problem may be with the terminal constraint.问题可能出在终端约束上。 One way to get a feasible solution is to "soften" the constraint by including it as an objective.获得可行解决方案的一种方法是通过将约束作为目标来“软化”约束。
m.Minimize(final*(mass-m_f)**2)
m.Minimize(final*(p-35.0e6)**2)
There is still a message that the problem is not feasible.还是有提示问题不可行。 You may want to continue to simplify your problem by turning off degrees of freedom (STATUS=0) and remove inequality constraints just to see if there is a problem with one of the equations such as divide-by-zero.您可能希望通过关闭自由度 (STATUS=0) 并删除不等式约束来继续简化您的问题，以查看其中一个方程是否存在问题，例如被零除。
# HYDROGEN TANK REFUELING MODEL
# OPTIMIZE MODEL WITH GEKKO OPTIMIZATION SUITE
from gekko import GEKKO
import numpy as np
m = GEKKO(remote=False)
M_H2 = 2.02
# CONSTANTS
## TANK PARAMETERS (ASSUME TYPE III, ALUMINUM, 74 l, RATED FOR 35 MPa)
V = 0.074 # m^3
a_in = 167/19e-3 # W/m^2/K
c_w = 2730 # J/kg/K
rho_w = 900 # kg/m^3
m_w = rho_w*(np.pi*(((0.358+19e-3)/2)**2)*(0.893+19e-3) - np.pi*((0.358/2)**2)*0.893)
A_in = 2*np.pi*(0.358/2)*((0.358/2) + 0.893) # m^2
T_w0 = 293.15 # K
m_f = 1.79 # final mass of hydrogen in tank, kg
## HYDROGEN PARAMETERS
c_p = 14.615e3 # specific heat at constant pressure, J/kg/K
c_v = 10.316e3 # specific heat at constant volume, J/kg/K
gamma = c_p/c_v
R = 8.314/M_H2 # gas constant for H2, J/kgK
T_c = -240 + 273.15 # critical temperature for H2, K
p_c = 1.3e6 # critical pressure for H2, Pa
w_H2 = -0.219 # acentric factor for H2
a = 0.45724*(R**2 * T_c**2)/(p_c**2)
b = 0.0778*(R*T_c)/p_c
kappa = 0.37464 + 1.54226*w_H2 - 0.26992*(w_H2**2)
# SET TIME ANALYSIS POINTS
nt = 101
tm = np.linspace(0, 1, nt)
m.time = tm
t = m.Param(tm)
# options
m.options.NODES = 6
m.options.SOLVER = 3
m.options.IMODE = 6
m.options.MAX_ITER = 500
m.options.MV_TYPE = 0
m.options.DIAGLEVEL = 0
# FIXED VARIABLES
t_f = m.FV(value=60.0,lb=60.0,ub=300.0) # final fuel time, s
t_f.STATUS = 0
m_dot_in = m.FV(value=0.001,lb=0.0005,ub=0.03) # mass flow rate into tank, kg/s
m_dot_in.STATUS = 1
m_0 = m.FV(value=0.1*m_f,lb=0.05*m_f,ub=0.9*m_f) # initial mass of H2 in tank (as % of m_f), kg
m_0.STATUS = 1
T_inf = m.FV(value=20 + 273.15,lb=15 + 273.15,ub=30 + 273.15) # inflow temperature, K
T_inf.STATUS = 1
# PARAMETERS
f = np.zeros(nt)
f[-1] = 1.0
final = m.Param(value=f)
# VARIABLES
T = m.Var(value=15+273.15,lb=15+273.15,ub=85+273.15)
T_w = m.Var(value=T_w0,lb=T_w0,ub=85+273.15)
mass = m.Var(value=m_0,lb=m_0,ub=m_f)
p = m.Var(value=1.0e6,lb=0.0,ub=35.0e6)
# INTERMEDIATES
alpha = m.Intermediate((a_in*A_in)/c_v/m_dot_in)
gamma_p = m.Intermediate(gamma/(1 + alpha))
alpha_p = m.Intermediate(alpha/(1 + alpha))
t_star = m.Intermediate(m_0/m_dot_in)
t_w_star = m.Intermediate((m_w*c_w)/(a_in*A_in))
T_star = m.Intermediate(gamma_p*T_inf + alpha_p*T_w)
alpha_T = m.Intermediate(1 + kappa*(1 - (T/T_c)**0.5))
v = m.Intermediate(V/mass) # specific volume, m^3/kg
# EQUATIONS
m.Equation(mass==t_f*(m_0 + m_dot_in*t))
m.Equation(T.dt()==t_f*(1 + alpha)*((T_star-T)/(t_star+t)))
m.Equation(T_w.dt()==t_f*((T-T_w)/t_w_star))
m.Equation(p*1.0e6==t_f*((R*T/(v-b)) - ((a*alpha_T**2)/(v*(v+b) + b*(v-b)))))
m.Equation((T-(85+273.15))*final<=0)
m.Equation((T_w-(85+273.15))*final<=0)
# SPECIFIY ENDPOINT CONDITIONS
m.Minimize(final*(mass-m_f)**2)
m.Minimize(final*(p-35.0e6)**2)
#m.fix(mass, pos=len(m.time)-1, val=m_f)
#m.fix(p, pos=len(m.time)-1, val=35.0e6)
# MINIMIZE FINAL FUEL TIME
m.Minimize(t_f)
# SOLVE
m.options.SOLVER = 3
m.solve()
# RESULTS
print('Final Time: ' + str(t_f.value[0]))
Response to edit回复编辑
The infeasibility reveals that wall_temp is a culprit from the first block.不可行性表明 wall_temp 是第一个块的罪魁祸首。 The solver is trying to push that value lower but it is at a bound.求解器正试图将该值推低，但它处于一个界限。 There are other parameters (variable 0) in the equation but wall_temp is the only one that is a variable that is at the lower bound.等式中还有其他参数（变量 0），但 wall_temp 是唯一一个处于下限的变量。 I made a modification to create a deadband to penalize any deviation outside of the upper (SPHI) and lower (SPLO) bounds.我进行了修改以创建一个死区，以惩罚超出上限 (SPHI) 和下限 (SPLO) 的任何偏差。 This way, the solution can still remain feasible if the constraints cannot be satisfied.这样，如果不能满足约束条件，该解决方案仍然可以保持可行。 The weights (WSPHI / WSPLO) can be increased if needed.如果需要，可以增加权重 (WSPHI / WSPLO)。 Here is additional information on tuning .这里是关于调整的附加信息。
# HYDROGEN TANK REFUELING MODEL
# OPTIMIZE MODEL WITH GEKKO OPTIMIZATION SUITE
from gekko import GEKKO
import numpy as np
m = GEKKO(remote=False)
# CONSTANTS
## TANK PARAMETERS (ASSUME TYPE III, ALUMINUM, 74 l, RATED FOR 35 MPa)
V = 0.074 # m^3
a_in = 167/19e-3 # W/m^2/K
c_w = 2730 # J/kg/K
rho_w = 900 # kg/m^3
m_w = rho_w*(np.pi*(((0.358+19e-3)/2)**2)*(0.893+19e-3) - np.pi*((0.358/2)**2)*0.893)
A_in = 2*np.pi*(0.358/2)*((0.358/2) + 0.893) # m^2
T_w0 = 293.15 # K
m_f = 1.79 # final mass of hydrogen in tank, kg
## HYDROGEN PARAMETERS
c_p = 14.615e3 # specific heat at constant pressure, J/kg/K
c_v = 10.316e3 # specific heat at constant volume, J/kg/K
gamma = c_p/c_v
M_H2 = 2.02
R = 8.314/M_H2 # gas constant for H2, J/kgK
T_c = -240 + 273.15 # critical temperature for H2, K
p_c = 1.3e6 # critical pressure for H2, Pa
w_H2 = -0.219 # acentric factor for H2
a = 0.45724*(R**2 * T_c**2)/(p_c**2)
b = 0.0778*(R*T_c)/p_c
kappa = 0.37464 + 1.54226*w_H2 - 0.26992*(w_H2**2)
## SET INFLOW TEMPERATURE AND INITIAL MASS IN TANK
m_0 = 0.1*m_f
T_inf = 20 + 273.15
# SET TIME ANALYSIS POINTS
nt = 101
tm = np.linspace(0, 1, nt)
m.time = tm
t = m.Param(tm, name='time')
# options
m.options.NODES = 6
m.options.SOLVER = 3
m.options.IMODE = 6
m.options.MAX_ITER = 500
m.options.MV_TYPE = 0
m.options.DIAGLEVEL = 0
# FIXED VARIABLES
t_f = m.FV(value=60.0,lb=60.0,ub=300.0) # final fuel time, s
t_f.STATUS = 1
m_dot_in = m.FV(value=0.001,lb=0.0005,ub=0.03) # mass flow rate into tank, kg/s
m_dot_in.STATUS = 1
# m_0 = m.FV(value=0.1*m_f,lb=0.05*m_f,ub=0.9*m_f) # initial mass of H2 in tank (as % of m_f), kg
# m_0.STATUS = 0
# T_inf = m.FV(value=20 + 273.15,lb=15 + 273.15,ub=30 + 273.15) # inflow temperature, K
# T_inf.STATUS = 0
# PARAMETERS
f = np.zeros(nt)
f[-1] = 1.0
final = m.Param(value=f, name='final')
# VARIABLES
T = m.CV(value=15+273.15,name='H2 Temp')
T.SPLO=15+273.15
T.SPHI=85+273.15
T.WSPLO = 100
T.WSPHI = 100
T.TR_INIT = 0
T.STATUS = 1
T_w = m.CV(value=T_w0,name='Wall Temp')
T_w.SPLO=T_w0
T_w.SPHI=85+273.15
T_w.WSPLO = 100
T_w.WSPHI = 100
T_w.TR_INIT = 0
T_w.STATUS = 1
# INTERMEDIATES
alpha = m.Intermediate((a_in*A_in)/c_v/m_dot_in, name='alpha')
gamma_p = m.Intermediate(gamma/(1 + alpha), name='gamma_p')
alpha_p = m.Intermediate(alpha/(1 + alpha), name='alpha_p')
t_star = m.Intermediate(m_0/m_dot_in, name='t_star')
t_w_star = m.Intermediate((m_w*c_w)/(a_in*A_in), name='t_w_star')
T_star = m.Intermediate(gamma_p*T_inf + alpha_p*T_w, name='Temp_star')
# alpha_T = m.Intermediate(1 + kappa*(1 - (T/T_c)**0.5))
# v = m.Intermediate(V/mass) # specific volume, m^3/kg
# EQUATIONS
# m.Equation(mass==t_f*(m_0 + m_dot_in*t*t_f))
m.Equation(T.dt()==t_f*(1 + alpha)*((T_star-T)/(t_star+t*t_f)))
m.Equation(T_w.dt()==t_f*((T-T_w)/t_w_star))
# m.Equation(p==t_f*((R*T/(v-b)) - ((a*alpha_T**2)/(v*(v+b) + b*(v-b)))))
# m.Equation((T-(85+273.15))*final<=0)
# m.Equation((T_w-(85+273.15))*final<=0)
# SPECIFIY ENDPOINT CONDITIONS
# m.Minimize(final*(mass-m_f)**2)
# m.Minimize(final*(p-35.0e6)**2)
m.Minimize(final*(T-351)**2)
#m.fix(mass, pos=len(m.time)-1, val=m_f)
#m.fix(p, pos=len(m.time)-1, val=35.0e6)
# MINIMIZE FINAL FUEL TIME
m.Minimize(t_f)
# SOLVE
m.options.SOLVER = 3
m.solve()
# RESULTS
print('Final Time: ' + str(t_f.value[0]))
Nice work simplifying the problem.简化问题的好工作。 Those deleted equations can likely be added back in, if desired, for convenience in plotting the solution.如果需要，可以重新添加那些删除的方程，以便于绘制解决方案。 The next step is likely to plot the solution and see if it make intuitive sense with the weights and any constraint violations.下一步可能会绘制解决方案，看看它是否对权重和任何约束违规具有直观意义。
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