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氫燃料汽車加注模型優化

[英]Hydrogen Vehicle Refueling Model Optimization

 原文  2021-06-22 22:41:08  6   1    python/ optimization/ gekko

問題描述

我目前正在嘗試使用 GEKKO 確定氫氣 (H2) 車輛加油過程的最佳流入條件。 下面是耦合的常微分方程，它們控制着 H2 和燃料箱壁的溫度如何隨加油時間變化。

T.dt() = (1+alpha)*(T_star - T)/(t_star + t)
T_w.dt() = (T - T_w)/t_w_star

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)

其中， m_0為罐內 H2 的初始質量， m_dot_in為 H2 進入罐內的質量流量， gamma為 H2 的比熱比， T_inf為 H2 的流入溫度，其他變量為中間值變量/坦克參數。 通過加油過程， m_dot_in被認為是常數（但未知），因此罐中 H2 的質量隨時間的變化定義為：

m = m_0 + m_dot_in*t

此外，罐內 H2 的壓力可以用真實的氣體狀態方程計算（我在這個模型中使用 Peng-Robinson 狀態方程）。

我試圖用這個模型做的是確定最佳m_dot_in 、 T_inf和m_0以最小化總加油時間t_f 。 對變量的一些限制是在整個加油過程中T<=358.15 K （出於安全原因），並且罐內 H2 的最終壓力必須為 35 MPa。 對於此模型，我認為t_f 、 m_dot_in 、 m_0和T_inf是具有以下邊界的固定變量：

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

下面我使用 GEKKO 復制了我針對這個優化問題的代碼：

# 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]))

此代碼目前給我以下錯誤：

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...

我對優化很陌生，我嘗試包含幾個不同的等式和不等式約束，但似乎沒有任何效果。 我認為我根據示例問題和來自 APMonitor 網站的信息正確地做這件事，但顯然我的實現有些問題。 我想知道是否有人知道我應該更改/添加什么，或者我做錯了什么？ 任何幫助將非常感激！

感謝您的時間，

埃文

編輯：基於 Hedengren 博士的回答，我嘗試簡化模型，以便不包括mass和p變量，因為它們僅取決於t_f 、 m_dot_in和T的最終值，並且可能在解決方案有后計算已獲得。 下面是我編輯的代碼：

# 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]))

我仍然遇到不可行性（不像以前那么多），但我無法理解所說的不可行性意味着什么以及如何解決它們。 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
************************************************

此外，這需要大約 2 分鍾才能運行，所以如果有關於如何減少計算時間的任何建議，將不勝感激！

1 個解決方案

解決方案1
 0 已采納  2021-06-23 18:03:37

可以看到 Gekko 用m.open_folder()編寫的模型，然后用文本編輯器打開模型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

問題在於前兩個方程將m.time和tm作為 Numpy 數組，而不是使用 Gekko 變量或參數。 如果優化問題中需要 Numpy 數組或 Python 列表，則創建一個新的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)))

由於最終時間被最小化，這些方程中的時間可能需要為t*t_f 。

諸如m.Equation(T_w*final<=85+273.15)方程應重新m.Equation((T_w-(85+273.15))*final<=0)為m.Equation((T_w-(85+273.15))*final<=0)以便當final=0它是0<=0 。 在這種情況下，您的原始方程沒問題，但將所有項放在方程的一側是一種很好的做法。

即使進行了這些修改，仍然存在Exception: @error: Solution Not Found 。 問題可能出在終端約束上。 獲得可行解決方案的一種方法是通過將約束作為目標來“軟化”約束。

m.Minimize(final*(mass-m_f)**2)
m.Minimize(final*(p-35.0e6)**2)

還是有提示問題不可行。 您可能希望通過關閉自由度 (STATUS=0) 並刪除不等式約束來繼續簡化您的問題，以查看其中一個方程是否存在問題，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]))

回復編輯

不可行性表明 wall_temp 是第一個塊的罪魁禍首。 求解器正試圖將該值推低，但它處於一個界限。 等式中還有其他參數（變量 0），但 wall_temp 是唯一一個處於下限的變量。 我進行了修改以創建一個死區，以懲罰超出上限 (SPHI) 和下限 (SPLO) 的任何偏差。 這樣，如果不能滿足約束條件，該解決方案仍然可以保持可行。 如果需要，可以增加權重 (WSPHI / WSPLO)。 這里是關於調整的附加信息。

# 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]))

簡化問題的好工作。 如果需要，可以重新添加那些刪除的方程，以便於繪制解決方案。 下一步可能會繪制解決方案，看看它是否對權重和任何約束違規具有直觀意義。

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