Difference between revisions of "Supermarket refrigeration system"

Supermarket refrigeration system
State dimension:	1
Differential states:	9
Discrete control functions:	4
Interior point equalities:	9

Revision as of 17:08, 11 August 2009

The supermarket refrigeration system problem is based on a model describing a refrigeration system with 2 parallel connected compressors (called a compressor rack) which only can be controlled stepwise (each single compressor can be turned on or off) and 2 open refrigerated display cases containing goods needed to be refrigerated. Each display case is connected to the refrigeration circuit through an expansion valve which also can only be closed or opened. This valve controls the flow of refrigerant into the evaporator, where it absorbs heat from the surrounding air. The refrigerated air then creates the well-known air-curtain at the front of the display case.

The air temperatures surrounding the goods in each display case are modeled by one differential state each. These states have to be bounded, so that the goods are properly refrigerated.

The model was published by Larsen et. al. in 2007 <bibref>Larsen2007</bibref>. The main goal is to control the refirgeration system energy-optimal. The problem was set up as a benchmark problem for MIOCPs.

The mathematical equations form an ODE model. The initial values of the differential states are not fixed but periodicity of the whole process is required.

The optimal integer control function shows chattering behavior, making the supermarket refrigeration system problem a candidate for benchmarking of algorithms.

Mathematical formulation

For $t \in [t_0, t_f]$ almost everywhere the mixed-integer optimal control problem is given by

$\min_{x,u} \frac {1}{t_f - t_0}\int_{t_0}^{t_f} \left( u_2 \cdot 0.5 \cdot \eta_{vol} \cdot V_{sl} \cdot f \right) dt$

$\begin{array}{llcl} \displaystyle \mbox{s.t.} & \dot{x_0}(t) &=& \dfrac{1}{V_{suc} \cdot \frac{d\rho_{suc}}{dP_{suc}}(x_0)} \cdot \bigg[ \left(\dfrac{UA_{wall-ref, max}}{M_{ref, max} \cdot \Delta h_{lg}(x_0)}\right) \Big( x_4 \big( x_2 - T_e(x_0) \big)\\ & && + \, x_8 \big( x_6 - T_e(x_0) \big) \Big) + \, M_{ref,const} - \eta_{vol} \cdot V_{sl} \cdot 0.5 \, u_2 \rho_{suc}(x_0) \bigg] \\ &\dot{x_1}(t) &=& - \dfrac{ UA_{goods-air} \left( x_1 - x_3 \right) }{ M_{goods} \cdot C_{p,goods} } \\ &\dot{x_2}(t) &=& \dfrac{ UA_{air-wall} \left( x_3-x_2 \right) - \dfrac{UA_{wall-ref,max}}{M_{ref,max}} \, x_4 \big( x_2 - T_e(x_0) \big) }{ M_{wall} \cdot C_{p,wall} } \\ [2.5ex] &\dot{x_3}(t) &=& \dfrac{ UA_{goods-air} \left( x_1-x_3 \right) + \dot{Q}_{airload} - UA_{air-wall} \, (x_3-x_2) }{ M_{air} \cdot C_{p,air} } \\ [2.5ex] &\dot{x_4}(t) &=& \left(\dfrac{M_{ref,max} - x_4}{\tau_{fill}} \right) u_0 - \left( \dfrac{UA_{wall-ref,max}}{M_{ref,max} \cdot \Delta h_{lg}(x_0)} \, x_4 \big( x_2 - T_e(x_0) \big) \right) (1-u_0) \\ \\ &\dot{x_5}(t) &=& - \dfrac{ UA_{goods-air} \left( x_5 - x_7 \right) }{ M_{goods} \cdot C_{p,goods} } \\ &\dot{x_6}(t) &=& \dfrac{ UA_{air-wall} \left( x_7-x_6 \right) - \dfrac{UA_{wall-ref,max}}{M_{ref,max}} \, x_8 \big( x_6 - T_e(x_0) \big) }{ M_{wall} \cdot C_{p,wall} } \\ [2.5ex] &\dot{x_7}(t) &=& \dfrac{ UA_{goods-air} \left( x_5-x_7 \right) + \dot{Q}_{airload} - UA_{air-wall} \, (x_7-x_6) }{ M_{air} \cdot C_{p,air} } \\ [2.5ex] &\dot{x_8}(t) &=& \left(\dfrac{M_{ref,max} - x_8}{\tau_{fill}} \right) u_1 - \left( \dfrac{UA_{wall-ref,max}}{M_{ref,max} \cdot \Delta h_{lg}(x_0)} \, x_8 \big( x_6 - T_e(x_0) \big) \right) (1-u_1) \\ [4ex] & x_3(t) &\geq& 2.0 \quad \forall t \in [t_0, t_f],\\ & x_3(t) &\leq& 5.0 \quad \forall t \in [t_0, t_f],\\ & x_7(t) &\geq& 2.0 \quad \forall t \in [t_0, t_f],\\ & x_7(t) &\leq& 5.0 \quad \forall t \in [t_0, t_f],\\ & x_0(t) &\leq& 1.7 \quad \forall t \in [t_0, t_f], \\ & x_i(t_0) &=& free \quad \forall i \in \{0,\dots, 8\}, \\ & x_i(t_f) &=& x_i(t_0) \quad \forall i \in \{0,\dots, 8\}, \\ & u_i(t) &\in& \{0, 1\} \quad \forall i \in \{0,\dots, 2\}, \\ & t_f &\in& [ 650, 750 ]. \end{array}$

Here the differential state $x_0$ describes the suction pressure in the suction manifold (in bar). The next three states model temperatures in the first display case (in °C). $x_1$ is the goods' temperature, $x_2$ the one of the evaporator wall and $x_3$ the air temperature surrounding the goods. $x_4$ then models the mass of the liquefied refrigerant in the evaporator (in kg).

$(x_5, x_6, x_7, x_8)$ describe the corresponding states in the second display case.

$u_0$ describes the inlet valve of the first display case, $u_1$ respectively the valve of the second display case. $u_2, u_3$ denote the activity of a single compressor.

The following polynomial functions are used in the model description originating from interpolations:

$\begin{array}{rcl} T_e(x_0) &=& -4.3544 x_0^2 + 29.224 x_0 - 51.2005,\\ \Delta h_{lg}(x_0) &=& (0.0217 x_0^2 - 0.1704 x_0 + 2.2988)\cdot 10^5, \\ \rho_{suc}(x_0) &=& 4.6073 x_0 + 0.3798, \\ \frac{d\rho_{suc}}{dP_{suc}}(x_0) &=& -0.0329 {x_0}^3 + 0.2161 {x_0}^2 - 0.4742 x_0 + 5.4817,\\ f(x_0) &=& (0.0265 x_0^3 - 0.4346 x_0^2 + 2.4923 x_0 + 1.2189) \cdot 10^5. \end{array}$

Parameters

These fixed values are used within the model for the day scenario. A night scenario is also available, see Variants.

Symbol	Value	Unit	Description
$\dot{Q}_{airload}$	3000.00	$\frac{J}{s}$	Disturbance, heat transfer from outside the display case
$\dot{m}_{ref,const}$	0.20	$\frac{kg}{s}$	Disturbance, constant mass flow of refrigerant from unmodeled entities
$M_{goods}$	200.00	$kg$	Mass of goods
$C_{p,goods}$	1000.00	$\frac{J}{kg \cdot K}$	Heat capacity of goods
$UA_{goods-air}$	300.00	$\frac{J}{s \cdot K}$	Heat transfer coefficient between goods and air
$M_{wall}$	260.00	$kg$	Mass of evaporator wall
$C_{p,wall}$	385.00	$\frac{J}{kg \cdot K}$	Heat capacity of evaporator wall
$UA_{air-wall}$	500.00	$\frac{J}{s \cdot K}$	Heat transfer coefficient between air and wall
$M_{air}$	50.00	$kg$	Mass of air in display case
$C_{p,air}$	1000.00	$\frac{J}{kg \cdot K}$	Heat capacity of air
$UA_{wall-ref,max}$	4000.00	$\frac{J}{s \cdot K}$	Maximum heat transfer coefficient between refrigerant and evaporator wall
$\tau_{fill}$	40.00	$s$	Parameter describing the filling time of the evaporator under opened valve
$T_{SH}$	10.00	$K$	Superheat in the suction manifold
$M_{ref,max}$	1.00	$kg$	Maximum mass of refrigerant in evaporator
$V_{suc}$	5.00	$m^3$	Total volume of suction manifold
$V_{sl}$	0.08	$\frac{m^3}{s}$	Total displacement volume
$\eta_{vol}$	0.81	$-$	Volumetric efficiency

Reference Solutions

For the relaxed problem (we only demand $u_i(t) \in [0,1]$ instead of $u_i(t) \in \{0,1\}$ the optimal solution is 12072.45. If we restrict ourselves to a solution with only integer controls we obtain the optimum objective value 12252.81.

Reference solution plots
Optimal relaxed control.
Optimal integer control.

Source Code

Model descriptions are available in

Variants

Since the compressors are parallel connected one can introduce a single control $u_2 \in \{0,1,2\}$ instead of two equivalent controls. The same holds for scenarions with $n$ parallel connected compressors.

In the paper <bibref>Larsen2007</bibref> mentioned above, the problem was stated slightly different:

The temperature constraints weren't hard bounds but there was a penalization term added to the objective function to minimize the violation of these constraints.
The differential equation for the mass of the refrigerant had another switch, if the valve (e.g. $u_0$ ) is closed. It was formulated this way:

$\dot{x_4} = \begin{cases} \dfrac{M_{ref,max} - x_4}{\tau_{fill}} & \qquad \text{if} \quad u_0 = 1 \\ \\ - \dfrac{UA_{wall-ref,max}}{M_{ref,max} \cdot \Delta h_{lg}(x_0)} x_4 \big( x_2 - T_e(x_0) \big) & \qquad \text{if} \quad u_0 = 0 \quad \text{and}\quad x_4 > 0 \\ \\ 0 & \qquad \text{if} \quad u_0 = 0 \quad \text{and} \quad x_4 = 0 \end{cases}$

This additional switch is redundant because the mass itself is a factor on the right hand side and so the complete right hand side is 0 if $x_4 = 0$ .

A night scenario with two different parameters was given. At night the following parameters change their value:

$\begin{array}{lcrr} \dot{Q}_{airload} &=& 1800.00 & \frac{J}{s}, \\ \dot{m}_{ref,const} &=& 0.00 & \frac{kg}{s}, \\ \end{array}$

Additionally the constraint on the suction pressure $x_0(t)$ is softened to $x_0(t) \leq 1.9$ .

No periodicity was required but the solution on a fixed time horizon 4 hours - 2 in day scenario and 2 in night scenario - with $t_f = 14400$ was asked.
The number of compressors and display cases is not fixed. Larsen also proposed the problem with 3 compressors and 3 display cases. This leads to a change in the compressor rack's preformance to $V_{sl} = 0.095 \frac{m^3}{s}$ . Unfortunately this constant is only given for these two cases although Larsen proposed scenarios with more compressors and display cases.

References

@@ Line 22: / Line 22: @@
 <p>
 <math>
-\min_{x,u} \frac {1}{t_f - t_0}\int_{t_0}^{t_f} \left( (u_2)\cdot 0.5 \cdot \eta_{vol} \cdot V_{sl} \cdot f \right) dt
+\min_{x,u} \frac {1}{t_f - t_0}\int_{t_0}^{t_f} \left( u_2 \cdot 0.5 \cdot \eta_{vol} \cdot V_{sl} \cdot f \right) dt
 </math>
 </p>
@@ Line 34: / Line 34: @@
                      \left(\dfrac{UA_{wall-ref, max}}{M_{ref, max} \cdot
                      \Delta h_{lg}(x_0)}\right) \Big( x_4 \big( x_2 - T_e(x_0) \big)\\
-&           &&  + \, x_8 \big( x_6 - T_e(x_0) \big) \Big) + \, M_{ref,const}   - \eta_{vol} \cdot V_{sl} \cdot 0.5 \, \left(u_2\right) \rho_{suc}(x_0)
+&           &&  + \, x_8 \big( x_6 - T_e(x_0) \big) \Big) + \, M_{ref,const}   - \eta_{vol} \cdot V_{sl} \cdot 0.5 \, u_2 \rho_{suc}(x_0)
                  \bigg] \\
   &\dot{x_1}(t) &=&  - \dfrac{
@@ Line 179: / Line 179: @@
 == Source Code ==
-=== C ===
+Model descriptions are available in
-The differential equations in C code:
+* [[:Category:C | C code]] at [[Supermarket refrigeration system (C)]]
-<source lang="cpp">
+* [[:Category:optimica | optimica]] at [[Supermarket refrigeration system (optimica)]]
-// number of compressors
-#define NVALVES 2
-// constants
-#define M_GOODS 200.0
-#define C_P_GOODS 1000.0
-#define UA_GOODS_AIR 300.0
-#define M_WALL 260.0
-#define C_P_WALL 385.0
-#define UA_AIR_WALL 500.0
-#define M_AIR 50.0
-#define C_P_AIR 1000.0
-#define UA_WALL_REF_MAX 4000.0
-#define M_REF_MAX 1.0
-#define TAU_FILL 40.0
-#define T_SH 10.0
-#define V_SUC 5.0
-#define V_SL 0.08 // 2 display cases - 2 compressors
-// #define V_SL 0.095 // 3 display cases - 3 compressors
-#define ETA_VOL 0.81
-// disturbances - day scenario
-#define Q_AIRLOAD 3000.0
-#define M_REF_CONST 0.2
-// disturbances - night scenario
-// #define Q_AIRLOAD 1800.0
-// #define M_REF_CONST 0.0
-double delta_h = (0.0217*xd[0]*xd[0] - 0.1704*xd[0] + 2.2988)*1e5;
-double T_e = -4.3544*xd[0]*xd[0] + 29.224*xd[0] - 51.2005;
-double rho_suc = 4.6073*xd[0] + 0.3798;
-double rho_suc__P_suc = -0.0329*xd[0]*xd[0]*xd[0] + 0.2161*xd[0]*xd[0] - 0.4742*xd[0] + 5.4817;
-double f = (0.0265*xd[0]*xd[0]*xd[0] -0.4346*xd[0]*xd[0] + 2.4923*xd[0] + 1.2189)*1e5;
-double Q_goods_air[NVALVES];
-double Q_air_wall[NVALVES];
-double UA_wall_ref[NVALVES];
-double Q_e[NVALVES];
-double m[NVALVES];
-double m_in_suc = 0.0;
-int i;
-for (i=0; i<NVALVES; i++){
-    Q_goods_air[i] = UA_GOODS_AIR*(xd[1 + i*4] - xd[3 + i*4]);
-    Q_air_wall[i] = UA_AIR_WALL*(xd[3 + i*4] - xd[2 + i*4]);
-    UA_wall_ref[i] = UA_WALL_REF_MAX * xd[4 + 4*i]/M_REF_MAX;
-    Q_e[i] = UA_wall_ref[i]*(xd[2 + 4*i] - T_e);
-    m[i] = Q_e[i]/delta_h;
-    m_in_suc += m[i];
-}
-double V_comp = 0.0;
-double comp_scale = (double) 1.0/NCOMPS;
-V_comp = comp_scale*u[NVALVES]*ETA_VOL*V_SL;
-// suction pressure
-rhs[0] = (m_in_suc + M_REF_CONST - V_comp*rho_suc) / (V_SUC*rho_suc__P_suc);
-// for each display/valve
-for (i=0; i<NVALVES; i++){
-    // temperatures:
-    // goods
-    rhs[1 + i*4] = - Q_goods_air[i]/(M_GOODS*C_P_GOODS);
-    // wall
-    rhs[2 + i*4] = (Q_air_wall[i] - Q_e[i])/(M_WALL*C_P_WALL);
-    // air
-    rhs[3 + i*4] = ((Q_goods_air[i] + Q_AIRLOAD - Q_air_wall[i]) /(M_AIR*C_P_AIR));
-    // mass of liquefied refrigerant:
-    rhs[4 + i*4] = ((M_REF_MAX - xd[4 + 4*i])/TAU_FILL * u[i] - m[i] * (1 - u[i]));
-}
-</source>
-=== Optimica ===
-<source lang="optimica">
-package Supermarket_pack
-  optimization Supermarket_Opt (objective = cost(tf),
-                         startTime = 0,
-                         finalTime=1)
-    // Parameters
-    parameter Real Q_airload = 3000.00;
-    parameter Real m_ref = 0.20;
-    parameter Real M_goods=200.00;
-    parameter Real Cp_goods=1000.00;
-    parameter Real UA_goods=300.00;
-    parameter Real M_wall=260.00;
-    parameter Real Cp_wall=385.00;
-    parameter Real UA_air=500.00;
-    parameter Real M_air=50.00;
-    parameter Real Cp_air=1000.00;
-    parameter Real UA_wall=4000.00;
-    parameter Real t_fill=40.00;
-    parameter Real T_sh=10.00;
-    parameter Real M_ref=1.00;
-    parameter Real V_suc=5.00;
-    parameter Real V_sl=0.08;
-    parameter Real eta_vol=0.81;
-    parameter Real tf(free=true,min=650,max=750)=650;
-   // The states
-    Real x0;
-    Real x1;
-    Real x2;
-    Real x3;
-    Real x4;
-    Real x5;
-    Real x6;
-    Real x7;
-    Real x8;
-    Real cost(start=0);
-    // The control signal
-    input Real u0(free=true,min=0,max=1);
-    input Real u1(free=true,min=0,max=1);
-    input Real u2(free=true,min=0,max=1);
-//Te=(-4.3544*x0*x0+29.224*x0-51.2005)
-//delta_h=((0.0217*x0*x0-0.1704*x0+2.2988)*10^5)
-//rho_suc=(4.6073*x0+0.3798)
-//drho=(-0.0329*x0*x0*x0+0.2161*x0*x0-0.4742*x0+5.4817)
-//f=((0.0265*x0*x0*x0-0.4346*x0*x0+2.4923*x0+1.2189)*10^5)
-  equation
-    der(x0) = ((1.0/(V_suc*(-0.0329*x0*x0*x0+0.2161*x0*x0-0.4742*x0+5.4817)))*((UA_wall/(M_ref*((0.0217*x0*x0-0.1704*x0+2.2988)*10^5)))*(x4*(x2-(-4.3544*x0*x0+29.224*x0-51.2005))+x8*(x6-(-4.3544*x0*x0+29.224*x0-51.2005)))+M_ref-eta_vol*V_sl*0.5*(u2)*(4.6073*x0+0.3798)))*tf;
-    der(x1) =(-UA_goods*(x1-x3)/M_goods*Cp_goods)*tf;
-    der(x2)=((UA_air*(x3-x2)-(UA_wall/M_ref)*x4*(x2-(-4.3544*x0*x0+29.224*x0-51.2005)))/(M_wall*Cp_wall))*tf;
-    der(x3)=((UA_goods*(x1-x3)+Q_airload-UA_air*(x3-x2))/(M_air*Cp_air))*tf;
-    der(x4)=(((M_ref-x4)/t_fill)*u0-((UA_wall/(M_ref*((0.0217*x0*x0-0.1704*x0+2.2988)*10^5)))*x4*(x2-(-4.3544*x0*x0+29.224*x0-51.2005)))*(1-u0))*tf;
-    der(x5)=(-UA_goods*(x5-x7)/(M_goods*Cp_goods))*tf;
-    der(x6)=((UA_air*(x7-x6)-(UA_wall/M_ref)*x8*(x6-(-4.3544*x0*x0+29.224*x0-51.2005)))/(M_wall*Cp_wall))*tf;
-    der(x7)=((UA_goods*(x5-x7)+Q_airload-UA_air*(x7-x6))/(M_air*Cp_air))*tf;
-    der(x8)=(((M_ref-x8)/t_fill)*u1-((UA_wall/(M_ref*((0.0217*x0*x0-0.1704*x0+2.2988)*10^5)))*x8*(x6-(-4.3544*x0*x0+29.224*x0-51.2005)))*(1-u1))*tf;
-    der(cost) = ((-1.0/((1/*time*/)))*cost+(1.0/((1/*time*/)))*((u2)*0.5*eta_vol*V_sl*((0.0265*x0*x0*x0-0.4346*x0*x0+2.4923*x0+1.2189)*10^5)))*tf;
-   // der(cost) = 1.0 / 650 * (((u2)*0.5*eta_vol*V_sl*((0.0265*x0*x0*x0-0.4346*x0*x0+2.4923*x0+1.2189)*10^5)))*tf;
-  constraint
-    x3>=2.0;
-    x3<=5.0;
-    x7>=2.0;
-    x7<=5.0;
-    x0<=1.7;
-    x0(tf)=x0(startTime);
-    x1(tf)=x1(startTime);
-    x2(tf)=x2(startTime);
-    x3(tf)=x3(startTime);
-    x4(tf)=x4(startTime);
-    x5(tf)=x5(startTime);
-    x6(tf)=x6(startTime);
-    x7(tf)=x7(startTime);
-    x8(tf)=x8(startTime);
-  end Supermarket_Opt;
-end Supermarket_pack;
-</source>
 == Variants ==

Difference between revisions of "Supermarket refrigeration system"

Revision as of 17:08, 11 August 2009

Contents

Mathematical formulation

Parameters

Reference Solutions

Source Code

Variants

References

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