me555 design optimization project final report...
TRANSCRIPT
ME555 Design Optimization
Term Project Final Report
Design optimization for Building Entryway
Submitted by, YoungJae Kim Sehyun Chang Byungsik Lee
Il Yeo
04/19/2005
CONTENTS i. List of contents ii. Abstract 1. Problem Statement 2. Subsystem 1 : Heat Pump and Heat Exchanger
2.1 Problem statement 2.2. Nomenclature 2.3 Mathematical Models
2.3.1 Heat pump model 2.3.2. Heat exchanger model
2.4 Model Analysis 2.5 Numerical Result
2.5.1. DOE result – full factorial for finding global minimum location
2.5.1.1. Parameter information 2.5.1.2 DOE scheme 2.5.1.3 DOE result 2.5.2 Optimization result 2.5.2.1 Optimization setup 2.5.2.2 Execution Results 2.5.2.3 Best design parameter values 2.5.2.4 History Plots (improvements only)
3. Subsystem 2 : Heat Curtain
3.1 Problem statement 3.2 Nomenclature 3.3 Mathematical Model
3.3.1 Objective function 3.3.2 Constraints 3.3.3 Design Variables and Parameters 3.3.4 Summary Model
3.4 Model Analysis 3.5 Optimization Study
3.5.1 Procedure 3.5.2 Model Construction
ii
3.5.3 Results 3.6 Parametric Study 3.7 Discussion of Result
3.7.1 Variables and Constraints 3.7.2 Physical Interpretation 3.7.3 Model Construction and Implementation
4.Subsystem 3: Motor driven sliding door system
4.1 Problem statement 4.2. Nomenclature 4.3. Mathematical Model
4.3.1 Objective function 4.3.2 Constraints 4.3.3 Design Variables and Parameters 4.3.4 Summary Model
4.4. Model Analysis 4.5 Numerical Results and Discussion
5. Subsystem 4: Structure of The Sliding Door
5.1 Problem statement 5.2 Nomenclature 5.3 Mathematical Model
5.3.1Objective function 5.3.2 Constraint 5.3.3 Design Variables and Parameters 5.3.4Summary Model
5.4 Model Analysis 5.5 Discussion of Results
6. System Integration Study
6.1 Objective 6.2 Methodology: All-In-One Approach 6.3 Results and Discussion
Reference APPENDICES
A.1. Heat Pump and Heat Exchanger Matlab M files and Simulink Models A.2 iSight description file of the integration system A.3 GAMBIT journal file for the heat curtain
iii
A.4 FLUENT journal file for the heat curtain A.5 Monotoncity Analysis of the motor system
iv
ABSTRACT In this project, the minimization of maintenance cost of an entryway to the building is optimized. The entryway consists of sliding door structure, motor system, heat pump and heat exchanger system and heat curtain. The objective of this system is to maintain the inner building temperature reducing chimney effect with the lowest cost while people are coming in and out of the entryway.
The subsystems and objectives are related to various engineering fields like thermodynamics, heat transfer, solid mechanics and fluid mechanics and system control. Each of 4 subsystems has the different objectives as follows. 1) Heat pump and heat exchanger design: minimizing the compressor and fan work; 2) Heat Curtain: minimizing the heat transfer between the inside of the building and the buffer area; 3) Motor driven sliding door system: minimizing of the annual maintenance cost; 4) Structure of the sliding door: minimizing of stress of door.
The objective of the integrated system is to minimize the total cost for maintaining the entryway. The total cost is the sum of the cost of each subsystem – construction cost of the doors, motor price, motor maintenance cost, energy cost for maintaining the building temperature. The cost is calculated for one year. This integrated problem has been simulated using an optimization tool, iSIGHT and subsystem are solved by Matlab Simulink, Fluent, and MS Excel program.
v
CHAPTER 1
Problem Statement In this project, an entryway configuration is optimized. The entryway consists of sliding
doors, heat exchanger system and heat curtain. The objective of this system is to maintain the inner building temperature reducing chimney effect with the lowest cost while people are coming in and out of the gateway.
The subsystems and objectives are given as below:
1) Heat pump and heat exchanger design
The objective function can be minimizing the compressor work and fan work.
2) Heat Transfer in the buffer area
The objective for this subsystem is to minimize the heat transfer between the
inside of the building and the buffer area.
3) Motor driven sliding door system
The objective function of this sub-system is the minimization of the annual
maintenance cost. This cost consists of the annual consumed power cost of a
motor and the annual mechanical cost combined with the initial cost of motor
size plus the maintenance cost of motor.
4) The optimal structure of the sliding door
Objective function for an automatic sliding door is to calculate the minimum
value of stress.
The objective of the integrated system is: To minimize the total cost for maintaining the entryway. The total cost is the sum
of the cost of each subsystem – construction cost of the doors, motor price, motor maintenance cost, energy cost for maintaining the building temperature. The cost is calculated for one year.
1
CHAPTER 2
Subsystem 1 : Heat Pump and Heat Exchanger 2.1 Problem statement
As a tool of heating air in the air-room, the heat pump was selected. Usually a heat pump may
use only one-third as much energy as electric resistance heat (electric furnace and baseboards,
for example) during mild winter weather (outdoor temperature about 45 degrees F). In the heat
pump industry, this is described as a COP (Coefficient of Performance) of 3. COP is the ratio of
heat output, to electrical energy input.
In this modeling, Heat pump works through a thermodynamic cycle of reverse refrigerator and
heat exchanger model is applied to calculate the heat transfer phenomenon between the
condenser and fan. To make a air-room temperature stable and constant, the heat pump absorbs
the heat from outside atmosphere through evaporator and the fan attached to condenser blows the
cold air through the compact heat exchanger module which is the condenser and the heated air
flows into the air-room.
The objective function for minimization is the power consumption of the fan attach at
condenser and the compressor in the heat pump cycle. The number of variables is 2 and they are
mass flow rate of air and R12 as a refrigerant. The number of constraint is 7. The constraint is
reasonable COP range and condenser width. Thermodynamic 1st law also provides the lower
limit of refrigerant’s condenser outlet temperature which is higher than that of cold air flowing
into the condenser by the fan.
2.2. Nomenclature
Parameter
Alk = 237 Aluminum conduction coefficient [W/m]
oD = 16.4*10-3 Tube outside diameter [m]
2
iD =
es in radiator
Pitc
10 lic diameter [m]
13.8*10-3 Tube inside diameter [m]
tubeN = 1 Number of cooling tub
h_fin = 275 Fin pitch [per meter]
Dh = 6.68* -3 Flow passage hydrau
t = 0.254 Fin thickness [mm]
σ = 0.449 Free flow area/frontal area
α = 2 tal volume [m2/m3]
= 2]
[K]
Air heat transfer surface [m2]
Interm
,12 e [K]
f
G
PrR
69 Heat transfer area/to
A/ = 0.830 Fin area/total area Af
frA 0.2 Condenser frontal area [m
inairT , = 273 Air inlet temperature [K]
outairT , = 303 Air outlet temperature [K]
in,12 = 323.8 RRT 12 condenser inlet temperature
hA
ediate variable
outRT R12 outlet temperatur
friction factor of air
Hj Colburn j factor of air
Mass velocity of air [ 2/ mskg ⋅ ]
airRe , 12 Reynolds number of air and R12ReR
airPr , 2 Prandtle number of air and R12 1
airµ 12Rµ, Kinetic viscosity of air [Ns/m^2]
airk , 12Rk Thermal conductivity of air and R12[ W/mK]
gK] aircp , 12 Specific heat of air and R12 [J/kRcp
airρ , 12Rρ kg/m3]
qmax ]
Density of air and R12[
q Heat transfer rate [W]
Maximum possible heat transfer rate [W
3
ε Heat exchanger effectiveness (q/qmax)
unit
Nu
R12 [J/kgK]
Cair,
t, average [m3/kg] 3]
Width_con
ube of air and R12 [N/m2]
NTU Number of heat transfer
Nusselt number of R12
airh , 12Rh Convective heat transfer coefficient of air and
U Overall convection coefficient [J/kgK]
12RC Heat capacity rate of air and R12 [W/K]
moi vvv ,, Specific volume of air inlet, outle
V finned-tube total volume [m
Condenser width [m]
airp∆ , 12Rp∆ Pressure drop in finned-t
ai Fan power [kW] rfanP ,
fanη Fan efficiency coefficient
airQ Volume flow rate of air [m3/s]
Variable
m& R12 mass flow rate [kg/s]
.3 Mathematical Models
.3.1 Heat pump model
om the
cle
ent on QH. QH is decided by the heat exchanger
odel. The diagram is shown in Fig. 2.1.
airm& Air mass flow rate [kg/s]
12R
2
2
The heat pump cycle is reverse of the refrigeration cycle. Heat pump absorbs the heat fr
cold outside and transfer the heat into the air-room through the cycle. In this heat pump
modeling, the ideal vapor-compression refrigeration cycle was used which means that no entropy
is generated and state 1 and state 3 are saturated gas and liquid. To evade complicated all cy
modeling, the condenser and evaporator out condition were fixed and only the condenser in
temperature was unknown which is depend
m
4
Figure 2.1 The ideal vapor-compression heat pump cycle
bles. Each process
steady state, steady flow with no changes in kinetic or potential energy.
up
ble from saturation pressure. T2 is fixed for simplifying problem and the value is 323K.
First law:
For each control volume analyzed, the thermodynamic model is the R-12 ta
is
Control volume: compressor
Inlet state: T1 is known as -253K, saturated vapor, state fixed
Exit state: P2 is known as same to the saturation pressure at T3, The T3 calculated from look-
ta
Analysis
5
wc = h2 - h1
Second law:
s2 = s1
6, s1=0.7082
s described earlier, T2=323K and h2=211.38
c=h2-h1=211.38-178.61=32.77kJ/kg
At calculated T3, h3 can be given by R-12 table.
Then COP(heat pump) =
From R-12 table, h1=178.
Therefore s2=s1=0.7082
A
W
Control volume: Condenser
Qh=h2-h1
c
H
wq =
12
32
hhhh
−−
The pump work =
.3.2. Heat exchanger model
the condenser is formulated as following.
The pressure drop depend
cR wm ×12&
2
When the refrigerant flows in the tube, refrigerant loses its pressure due to friction loss and
minor losses. A pressure drop due to friction loss in
s on velocity as following,
][10002
2 kpaUfpp inout ∞−= ρ 11
Friction coefficient, f, is depended on Reynolds number,
Re64
=f , RE<2000
s linear function. Heat exchanging process
rom coolant to air leads to heat exchanger analysis.
2)64.1Reln79.0( −−=hLf , 6105Re3000 ⋅≤≤
Between 3000Re2000 <≤ , friction factor is assumed a
f
6
However, in this modeling the loss in the refrigerant is not considered. The heat exchanger
model is made for calculation of heat transfer between hot refrigerant and air. Compact heat
exchangers are typically used when a large heat transfer surface area per unit volume is desired
and at leat one of the fluid is a gas.
In heat exchanger modeling the configuration of Fig 2.2 of circular tube-circular fin heat
exchanger is used.
Figure 2.2 Heat transfer and friction factor for a circular tube-circular fin heat exchanger, surface
CF-7.0-5/8J from Kays and London
For fixed configuration and dimension of Fig2, the heat transfer coefficient can be given as
follows depending on the flow property like turbulent or laminar.
From the Nusselt numbers calculated as follows the heat transfer coefficients of air and R12 can
be calculated.
12RNu =4.36 Laminar, fully developed, uniform q”, Pr>0.6
)1(Pr)8/(7.121Pr)1000)(Re8/(
3/22/112
12 −+−
=f
fNu RR Turbulent, fully developed, 3000<Re<5x106 , 0.5<Pr<2000
7
Neglecting fouling effects and tube inner surface is not finned, the overall heat transfer
coefficient based on the gas side surface area is given by
( ) airairowair
RairR hRA
AAhU ,1212
1/
11η
++=
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
AA
DD
AA f
o
i
R
air 112
)/(2)/ln(
12 airRAl
iowh AAk
DDRA =
The gas side convection coefficient may be obtained by first and mass velocity is evaluated like
GfrA
mVσ
ρ&
=≡ max
Then Re is calculated and hair can be calculated as follows
hair 3/2Pr0096.0 pGc
≈
Then )1(1, ff
ho AA
ηη −−=
With Cc= cpccm ,&
q=Cc(Tc,o-Tc,i)
qmax=Cmin(Th,i-Tc,i)
To fine out refrigerant outlet temperature, use NTU−ε method,
minCAU
NTU h⋅=
V=α
hA
Width_con =frA
V
8
In cross flow (single pass), both fluid unmixed
⎥⎦
⎤⎢⎣
⎡−−⎟⎟
⎠
⎞⎜⎜⎝
⎛−= }1])({exp[)(1exp1 78.022.0 NTUCNTU
C rr
ε
A coolant flow rate is determined by pressure increase between pump inlet and pump outlet
Pressure drop across finned-tube bank
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎟⎟
⎠
⎞⎜⎜⎝
⎛−+=∆
i
m
fri
oi
vv
AVf
vvvG
pσασ 1)1(
22
2
airfrair PAconstm ∆⋅=& = airairQρ
To calculate power for fan, filter power calculation formula (http://ateam.lbl.gov/Design-
Guide/DGHtm/filterpowercalculation.htm) was adapted
PQP airfanairfan ∆= η,
Conventionally, 7.0=fanη is used.
2.4 Model Analysis
Physical constraint
Physical constrain in this subsystem is thermodynamic 1st law which is energy conservation and
mass conservation. The heat output of heat pump needs to be enough to maintain the pathway
temperature at desired value. The air velocity is large enough to function as an air curtain in the
given condenser frontal area.
Assumption:
• The heat pump is fixed at evaporator and pump inlet and outlet. Outlet of R12
temperature is variable.
• The cycle is ideal, so pump works ideally and no entropy is generated.
• Negligible heat loss to the surroundings
9
• Negligible kinetic and potential energy changes
• Negligible fouling factors
• Fully developed conditions for the air and R12
To simulate real world, the condenser can be modeled as a finned-tube, compact heat exchanger
and the shape is given. The overall convection coefficient should be calculated to given shape
and air and R12 material property is calculated from the temperature.
Objective function
The objective function can be minimizing the fan and pump work.
Min f = 12,, Rpumpairfan PowerPower +
Constraint
g1=- 0≤− airm&
g2= 012 ≤− Rm&
g3=1.5-COP 0≤
g4=COP-5.5 0≤
g5=0.0385-Width_con 0≤
g6=Width_con-0.5 0≤
g7=273-TR12out 0≤
10
2.5 Numerical Result
2.5.1. DOE result – full factorial for finding global minimum location
2.5.1.1. Parameter information
Inputs Type Current Value mdotR12 REAL 0.005 mdotair REAL 0.001 Outputs Type Current Value fanpower REAL 0 pumppower REAL 0.17255177 cop REAL 3.89410894 widthcon REAL 0.0875637 Tr12out REAL 317.09012396 G5 REAL -0.0490637 G1 REAL -0.005 G6 REAL -0.4124363 G2 REAL -0.001 G7 REAL -44.09012396 G3 REAL -2.39410894 F REAL 0.17255177 G4 REAL -1.60589106 Objective REAL 0.17255177 Feasibility INTEGER 9 TaskProcessStatus REAL -1.0 2.5.1.2 DOE scheme Technique: Full-Factorial Design Matrix: Full-(9:2) 81 experiments Name Levels mdotR12 0.001 0.005 0.01 0.05 0.1 0.5 1.0 5.0 10.0 mdotair 0.001 0.005 0.01 0.05 0.1 0.5 1.0 5.0 10.0
11
2.5.1.3 DOE result Item Best Point Best Levels DOE Objective 0.17255177 184.155185 Experiment number 1,10 FACTORS mdotR12 0.005 5 mdotair 0.001 0.001 RESPONSES ObjectiveAndPenalty 0.17255177 184.155185 fanpower 0 0 pumppower 0.17255177 184.155185 cop 3.89410894 3.54584515 widthcon 0.0875637 0.06828142 Tr12out 317.090124 322.9940901 G5 -0.0490637 -0.02978142 G1 -0.005 -5 G6 -0.4124363 -0.43171858 G2 -0.001 -0.001 G7 -44.09012396 -49.99409012 G3 -2.39410894 -2.04584515 F 0.17255177 184.155185 G4 -1.60589106 -1.95415485
12
Effect of R12 Mass Flow Rate
-1
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5
log Mass Flow Rate (kg/s)
log
Pow
er (k
W)
-3-2.301029996-2-1.301029996-1-0.30102999600.698970004
Effect of Air Mass Flow Rate
-1
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1
log Mass Flow Rate (kg/s)
log
Pow
er (k
W)
-2.301029996-2-1.301029996-1-0.30102999600.6989700041
2.5.2 Optimization result
2.5.2.1 Optimization setup Optimization Technique: Sequential Quadratic Programming - NLPQL Design Variables
Type Lower Bound Current Value Upper Bound
mdotR12 REAL 0.0001 0.0027040609 mdotair REAL 0.0001 0.00176585046861812
13
Output Constraints
Type Lower Bound Current Value Upper Bound
cop REAL 1.5 5.00481621 5.5 widthcon REAL 0.0385 0.13310511 0.5 Tr12out REAL 273.0 303.70317616 G5 REAL -0.09460511 0.0 G1 REAL -0.0027040609 0.0 G6 REAL -0.36689489 0.0 G2 REAL 0.00176585046861812 0.0 G7 REAL -30.70317616 0.0 G3 REAL -3.50481621 0.0 G4 REAL -0.49518379 0.0 Objectives Type Direction Current Value F REAL minimize 0.07657256 2.5.2.2 Execution Results Task Task1 Total runs 45 Feasible runs 16 Infeasible runs 29 Failed runs 0 Database file Task1.db 2.5.2.3 Best design parameter values mdotR12 0.0027040609 Mdotair 0.00176585046861812 Fanpower 2e-008 Pumppower 0.07657254 Cop 5.00481621 Widthcon 0.13310511 Tr12out 303.70317616 G5 -0.09460511 G1 -0.0027040609 G6 -0.36689489 G2 -0.00176585046861812 G7 -30.70317616 G3 -3.50481621 F 0.07657256 G4 -0.49518379
14
2.5.2.4 History Plots (improvements only)
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0 5 10 15 20 25 30
RunCounter
f
15
CHAPTER 3
Subsystem 2 : Heat Curtain 3.1 Problem statement
This subsystem is about the heat transfer between the air-room and the outside through the
outer door. To minimize the heat transfer, some geometric variables will be changed. Firstly,
there are some variables about duct. Duct will be positioned somewhere between two doors and
the position will be optimized. Also the duct size can be changed. Width and length of the duct
become the other design variables. Volumetric flow rate from the heater is fixed in this
subsystem, so the size of the duct output decides the flow rate at the output.
Figure 3.1 Top view of the subsystem
Other design variables are the size of the room. The area cannot be infinitely large due to the
space limit so the size of the area is limited a specific value . With this upper limit of the area, bA
Outer Door
LdW Inside of Outside
Duct Output
Building
Inner Door
SW
Airroom
Heat Transfer
hP
dL
16
the length and width of the area can be adjusted. If the length is too short, there would be longer
time in which both the outer and inner door are open and the heat transfer would rapidly
increase. In contrast, if the length is too long, the heat loss through the side wall would increase,
which in turn decrease the temperature of the air-room. Therefore, optimal dimension should be
decided through optimization process.
The heat transfer involved in this problem is quite complex and it cannot be solved without
numerical analysis tool with appropriate assumptions. For the steady state flow, it is assumed
that the two doors are opened simultaneously and calculates the heat transfer during this opening
period. For the analysis, FLUENT was used and GAMBIT for mesh generation.
Figure 3.2 Side view of the subsystem
Outer Door
Flow from Duct output
Inner Door
Boundary condition: Temperature To
dW
Boundary condition: Temperature Ti
H Heat Transfer
Boundary condition: Temperature Td
17
3.2 Nomenclature
Symbol Unit Description
To oC Outside temperature
Ti oC Inside temperature
Td oC Temperature at the duct output boundary
oP Pa Outside pressure
iP Pa Inside pressure
H m Height of the air-room
L m Length of the air-room
W m Width of the air-room
dV m/s Flow velocity at the duct output
dL m Duct output length
dW m Duct output width
hP m Distance of the duct from the outer door
dP m Distance of the duct from the origin in y-direction
sP m Distance of the door from the origin in y-direction
S m Width of the door
rT sec Duration in which the door is opening or closing
rD m Distance from which the door is start to open or
close
wS m/s Walking speed of average person
bA 2m Maximum area of the air-room
Q& sm /3 Volume flow rate from the duct
Table 3.1 Nomenclature
18
3.3 Mathematical Model
3.3.1 Objective function
The objective for this subsystem is to minimize the heat transfer between the air-room and the
outside of the building. The objective function is given as:
min f = f(To, Ti, Td, , , , , , , H, L, W, S) oP iP dV hP dW dL
Firstly, the geometry variables are passed into GAMBIT and the mesh model will be constructed
and then, the mesh model with some property variables like temperature, pressure and velocity
would be passed into FLUENT and the velocity vectors of the air at the outer door will be
returned as output file. And then, a parsing program is used to calculate the average of these
velocities and returns the value to the iSight.
The process mentioned above need parameters as well as variables to get the objective, but in
this subsystem, parameters are not changed and they are fixed value. To enhance the readability,
the objective function can be expressed only with independent design variables, rewritten as:
min f = f( , , , L, W) hP dW dL
19
3.3.2 Constraints
There are some constraints for this subsystem. Constraints are categorized to two groups –
physical and practical constraints.
1) Physical Constraints
Firstly, there are some constraints with the duct. It should be located with a margin of 0.1m
from the outer door introducing a constraint:
hP ≥ 0.1
Also, the duct should be located in the air-room. Width of the duct and the margin is
considered.
hP + + 0.1 ≤ L dW
A flow rate is given as parameter in this subsystem. When the flow rate and the duct width
are given, the velocity should be given as:
ddd WL
QV×
=&
The width of air-room should be larger than the length of the duct with some margin. 0.1m
are assigned for both sides,
W ≥ + 0.2 dL
Some intermediate variables are defined for the modeling convenience.
)(21
dd LWP −×=
20
)(21 SWPs −×=
2) Practical Constraints
There are some practical constraints which express limitations of this subsystem.
It is assumed that the area of the air-room has to be less than some value, say Ab,
L * W ≤ bA
To generate a mesh model, the length of the duct should either always bigger than the door
width or always smaller than the door width. In this problem, it is assumed that the duct
length is always bigger than the door width. Actually, it is the way the real system is.
S + 0.2 ≤ dL
Also, for some manufacturing constraints, the duct should have some minimum size. It is
assumed that the width of the duct is bigger than 0.1m
dW ≥ 0.1
It is not desirable if both doors are opened simultaneously and it increases flow
tremendously. Therefore the length of the air-room should be set in a way that at least one
door is closed in any moment. In other words, before a person reaches the point that the inner
door respond to, the outer door should be closed.
w
rr S
DLT )2( ×−≤
21
3.3.3 Design Variables and Parameters
Design variables of this subsystem are listed below:
Symbol Unit Description
L m Length of the air-room
W m Width of the air-room
dL m Duct output length
dW m Duct output width
hP m Distance of the duct from the outer door
dP m Distance of the duct from the origin in y-direction
sP m Distance of the door from the origin in y-direction
dV m/s Flow velocity at the duct output
Table 3.2 Design variables
Table 3.2 shows design variables. Not all variables are independent. Some variables are
dependent on others according to the equality constraints. It will turn out that there are five
independent variables and they will be the input variable for this optimization study.
22
Parameters are listed below:
Symbol Unit Description
To oC Outside temperature
Ti oC Inside temperature
Td oC Temperature at the duct output boundary
oP Pa Outside pressure
iP Pa Inside pressure
H m Height of the air-room
S m Width of the door
rT sec Duration in which the door is opening or closing
rD m Distance from which the door is start to open or
close
wS m/s Walking speed of average person
bA 2m Maximum area of the air-room
Q& sm /3 Volume flow rate from the duct
Table 3.3 Parameters
Table 3.3 shows parameters for this subsystem. Some of these parameters are from the other
subsystem and it will be changed in the integration study.
23
3.3.4 Summary Model
In summary, the problem posed for optimization of this subsystem can be written as follows:
For variables given as:
L, W, , , > 0 dL dW hP
min f = f( , , , L, W) hP dW dL
Subject to:
g1 = 0.1 – ≤ 0 hP
g2 = + + 0.1 – L ≤ 0 hP dW
g3 = + 0.2 – W ≤ 0 dL
g4 = L * W – ≤ 0 bA
g5 = S + 0.2 – ≤ 0 dL
g6 = 0.1 – ≤ 0 dW
g7 = 0)2( ≤×+−× rwr DLST
h1 = 0)( =−×× QWLV ddd&
h2 = 0)(21
=−×− dd LWP
h3 = 0)(21
=−×− SWPs
24
3.4 Model Analysis
One of the most frequently used model analysis methods is monotonicity analysis, but in this
problem, the objective function is not analytical and it is gotten from the analysis software.
Therefore, monotonicity analysis cannot be applicable in this problem.
In this subsystem, there are three equality constraints. Therefore three variables can be
eliminated from the model through direct elimination. From this condition, variable , and
are removed from the model and there remains 5 variables.
dV sP
dP
Also, there are two inequalities g2 and g6 that is dominated by the other inequalities and they
need not to be constrained.
In short, the model was reduced to have only 5 inequalities and no equality constraint.
25
3.5 Optimization Study
FLUENT is used for the numerical analysis to solve the heat exchange between the outside and
the air-room. For the calculation, a mesh model should be constructed first in GAMBIT. Also,
iSight is to be used for optimization.
3.5.1 Procedure
iSight
Mesh Model Batch Process Geometric design variables (GAMBIT journal file)
Analysis Model Batch Process Other design variables (FLUENT journal file) for boundary conditions
Generate Objective from output (Custom-built Parsing Program)
Algorithm (SQP) (Optimization Process)
Figure 3.3 Optimization procedure
As Figure 3.3 shows, overall process is managed by iSight (version 9.0). From the initial
variables set by the user, it generates a mesh model using GAMBIT and analyzes the model
using FLUENT. FLUENT solves the mesh with the given boundary conditions, and then exports
the velocities at the door at the outside. A parsing program was programmed to calculate the
average velocity at the door. The output of the parsing program becomes the objective of this
system. iSight, accepting this value, regenerates the variables using SQP algorithm and iterates
again until it meets a termination criteria.
26
3.5.2 Model Construction
• Mesh model from GAMBIT
Figure 3.4 Mesh model generated in GAMBIT
• Analysis result from FLUENT
Figure 3.5 Analysis result from FLUENT
27
3.5.3 Results
Optimized values for the independent variables are shown in Table 3.4 below:
Inputs Type Current Value
Ph REAL 0.1
Wd REAL 0.1
Ld REAL 1.2
W REAL 1.5
L REAL 10.0
Table 3.4 Independent input variables found through optimization
We also have some depende om the input variables.
able 3.5 shows the optimized values. These three dependent variables are calculated from the
ree equality constraints.
nt variables and the objective calculated fr
T
th
Outputs Type Current Value
Vd REAL 16.667
Pd REAL 0.15
Ps REAL 0.25
Table 3.5 Dependent variables
Table 3.6 shows the value of the objective function in the optimized system. In this optimized
configuration, the objective, average velocity at the outer door, is calculated as 1.00 m/s.
Outputs Type Current Value
ObjF REAL 1.00
Table 3.6 Objective of the system
28
Outputs Type Current Value
g1 REAL 0.0
g2 REAL -9.7
g3 REAL -0.1
g4 REAL 0.0
g5 REAL 0.0
g6 REAL 0.0
g7 REAL -6.5
Table 3. ty constraints
Table 3.7 show nstraints were satisfied
eaning that the there is a solution in this problem. There are four active constraints - g1, g4,
7 Inequali
s the constraints value after the optimization. All the co
m
g5, and g6. The meaning of this result will be discussed in the discussion section.
Outputs Type Current Value
Ti REAL 300.0
To REAL 270.0
Td REAL 310.0
Pi REAL 0.0
Po REAL 5.0
Sw REAL 0.5
Tr REAL 3.0
Dr REAL 1.0
Qdot REAL 2.0
Ab REAL 15.0
Tab meters
Table 3.8 shows parameters u
le 3.8 Para
sed in this optimization process.
29
3.6 Parametric Study
Parametric Study
0
2
4
6
8
10
12
14
16
18
2.00000.0.01
Vd (m/s)
1.0.5010
Q (Volumetric Flow rate m^3/s )
Objective (m/s)
. In this
subsystem is parameter and it was fixed to 2.0 when the optimization was performed. In this
section, was va he chang output was ause this parameter is a
design variable in the other subsyste tly speakin flow rate is the design variable of
Heat Exchanger subsystem, but volu low rate ha elation with mass flow rate).
Table 3.9 shows the objective change with different flow rates. If is increased, the
velocity at the duct output is increas increased velocity improves the heat curtain effect
and it lessens the transfer decre objective so shows that the lower limit of the
influx Vd is around 2 m/s and that if low 0.10 t he heat curtain cannot function at all.
Table 3.9 Inequality constraints
Parametric study was undertaken with respect to the volumetric flow rate Q&
Q&
Q& ried and t e of the investigated bec
m (Stric g, mass
metric f s a linear r
Q&
ed. The
heat asing the . It al
Q& is be hen t
30
3.7 Discussion of Result
3.7.1 Variables and Constraints
As shown in Table 3.7, there are four active constraints. From these constraints, some
design rules would be retrieved. Firstly, constraint g1 is active. g1 is a constraint which
sets a lower limit for the distance between the duct and the outer door. When this
constraint is active, the duct is positioned right next to the outer door with the smallest
allowable margin 0.1m. Constraint g4 is active and it means that the optimized design has
the maximum room size. Constraints g5 and g6 are about the size of the duct. Both g5
and g6 are active, and it mean e smallest as possible. We have a
onstant flow rate, therefore the smaller the duct becomes, the higher the velocity is. In
th oor
elocity at the duct which is installed in ceiling would be maximized, and it minimizes
eat transfer between the air-room and the outside. The only difference is the room shape.
this optimization study, it turns out that a thin and long room is the optimized shape,
nd it seems unrealistic. I suppose that the reason is the simplified model. During the
odeling, heat loss through heat conduction was ignored for simplification, so there is no
eat loss through the wall. This assumption might eliminate some existing constraints on
the room size.
s that the duct should be th
c
short, we have, in this optimized setup, the smallest duct right next to e outer d
in the largest room.
3.7.2 Physical Interpretation
This result is almost the same as the real air-room configuration. In the real air-room
configuration, we can see that the width of the room is the smallest possible (A little
larger that the door width), and the duct is located right next to the outer door with
maximum length, and the duct size is short and wide. In this configuration, the air
v
h
In
a
m
h
31
3.7.3 Model Construction and Implementation
difficult parts was to generate the proper model. The
oncept was quite straightforward. However, the objective function cannot be calculated
In this subsystem, one of the most
c
from an analytical equation, and the function is an output from the numerical analysis
software. The function was critically dependent on the mesh size. It was another
optimization problem that the accuracy of the result with variable of mesh size and
constraint of given time.
32
CHAPTER 4
4.1 Pr
In an a
concer ssed by
the mo
Theref m of this sub-system is focused on the minimization of
consuming motor energy with a constraint of motor size during the door opening operation
because the motion of closing door is slower than that of opening door and the other continuous
operation can be combined using these open and close door motions.
Figure 4.1 is showing the schematic motor driven sliding door system.
Figure 4.1 Schematic Motor driven Sliding Door System
This system is assembled and operated as follows.
The door is built in the lead screw through the nut. The motor generates driving torque and the
gear box reduces the speed and increases the torque from motor to the lead screw along with the
gear ratio. This rotational hardware affects directly the motor size and energy consumption. In
addition, the mass of door and the desired moving velocity versus time infect the strength of a
lead screw and the required motor power.
We set the following assumption to simplify this optimization problem.
Subsystem 3: Motor driven sliding door system
oblem statement
utomatic sliding door system, the cost of initial installation and maintenance is the main
n of the selection of motor and mechanical systems. This cost can be simply expre
tor size and the consumption energy during operation.
ore the optimization proble
DOOR(CLOSED)
GEARBOX
MOTOR
DOOR(OPEN)
LEAD SCREW
Ls
S
33
The motor power is equivalent with th sed on the desired door velocity.
Moreover, we can exclude the control problem ectly
achieved u
d as ideal. No disadvantage of gear box such as backlash, additional
of the
n be derived from
Lagrange equation.
Figure 4.2 Free body diagram of sliding door System
or is
at of door operation ba
by assuming that the desired velocity is perf
sing the proper tracking controller.
The gear box is considere
friction and the gear compliance is considered and the gear size is assumed small enough to
ignore the moment inertia of gear because the length and weight of lead screw is one of main
inertia factors of this door operation system.
The lead screw is assumed as a rigid with a constant diameter. Thus the diameter and pitch
lead screw are considered as the only design variables.
In addition, we assumed that there is no packaging problem with respect to the assembly of these
components such as a motor, a gear box, a lead screw and a door.
The mathematical equations of this motor driven sliding door system ca
GEARBOX MOTOR
Y
X
)(BL tV
fLFmT
v
Js Jmsθ& mθ&
≡mL
GEARBOX MOTOR
Y
X
The kinematical relation between lead screw and do
sss PVxPx
Px θ
πθ
ππθ &&
2,
22===→= (4-1)
In addition, the kinematical relation between motor and lead screw is
sm nθθ = (4-2)
Then, the energy of this system is as follows.
Kinetic energy:
34
2222
222
21
21
221
21
21
21
smsssLmmssLT nJJpmJJxmE θθθπ
θθ &&&&&& ++⎟⎠⎞
⎜⎝⎛=++= (4-3)
Potential energy: 0=VE (4-4)
Damping energy: 2
2
221
21
⎟⎠⎞
⎜⎝⎛== sLL
PBxBD θπ&& (4-5)
Generalized force term: smsfLmmfLPTxFQ θ⋅+⋅−= nTF θθπ
⋅⋅+⋅−=2
(4-6)
Apply Lagrange equation using equations of (4-3)-(4-6).
smsssLTs
nJJmEdt
θθθπθ
&&&&&&&
2
2++⎟
⎠⎜⎝
=⎟⎟⎠
⎜⎜⎝ ∂
, Pd 2⎞⎛⎞⎛ ∂ 0=
∂∂
Vs
Eθ
, xBBD LsLs
&&& ⎟
⎠⎜⎝
=⎟⎠
⎜⎝
=∂ π
θπθ 22
PP ⎞⎛⎞⎛∂ 2
nTPFQ mfLs
⋅+⋅−=∂ πθ 2
∂
→ nTPFxPBnJJPm mfLLsmsssL ⋅+⋅−=⎟⎠⎞
⎜⎝⎛+++⎟
⎠⎞
⎜⎝⎛
ππθθθ
π 2222
2
&&&&&&& (4-7)
Then motor torque can is
nPFx
nPBx
PnJ
nJ
nPm
nPFx
nPBnJ
nJ
nPmT
fLLms
L
fLLsms
Lm
πππ
π
ππθ
π
222
2
222
2
2
2
2
⋅+⎟⎟⎠
⎞⎜⎜⎝
⎛+
⎥⎥⎦
⎤
⎢⎢⎣
⎡++⎟⎟
⎠
⎞⎜⎜⎝
⎛=
⋅+⎟⎟⎠
⎞⎜⎜⎝
⎛+
⎥⎥⎦
⎤
⎢⎢⎣
⎡++⎟⎟
⎠
⎞⎜⎜⎝
⎛=
&&&
&&&
(4-8)
As shown in equation (4-8), motor torque can be calculated if the desired door trajectory is
known. To simplify the optim
with a rising time of ta and a stop time of tb in figure 4.3.
Figure 4.3 Desired door velocity vs. time
ization problem of this door system, we will use the desired door
velocity profile which is triangular
Vm
ta tb
V(t)
TIME
35
For each time, the desired door velocity is
)(:0
aaa
aa
ttatatVtt
−=<<
(4-9)
Then, load force due to the door motion is
tVBFMatFttt
tVBFMa
LfLLbba
LfLLa
++−=<<
++
for maximum values,
()(: bba tatVttt −=<< )
)()(:
)()(:0 tFtt a =<<
Rewrite kinematical relation equations of (4-1) and (4-2)
PV
PV msmmsm
πθπθ 2,2 &&&& == , (4-10)
From the desired door velocity, the maximum acceleration of the lead screw is
smm nθθ &&&& =
a
sm
a
msm tPt
V θπθ&
&& =⋅=2 (4-11)
g equations of (4-8) and (4-10), the peak motor torque is Usin
mLfLms
Lm
mLfLsms
LmpmPmJT +⎟⎟
⎞⎜⎜⎛
+= (
VPBn
PFtaV
pn
nJ
nPmJ
Vn
PBn
PFnnJ
n
⎟⎞
⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛+=
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+
⎠⎝
ππ
π
ππθ
π
222)
2(
22)
2
2
2
2
2
&&
(4-12)
With respect to the desired door velocity, the motor peak torque is only a function of door
velocity because the door acceleration is a constant.
n ⎟⎠π
Then, motor power is
mpmpspmpmpmpm VP
nTnTTP πθθ 2⋅=⋅=⋅= && (4-13)
Moreover, the energy consumption during the operation time is
∫ ∫∫∫ ⋅=⋅=⋅=⋅= bt
mmsmmmm dttVP
nTdxP
nTdnTdTE0
)(22 ππθθ (4-14)
The consumed power is defined as follows
bm t
EmP = (4-15)
This equation can be calculated using equations of (4-8) ,(4-9) and (4-14).
36
4.2. Nomenclature
S Width of door (m)
ta Rise time (sec)
tb Stop time (sec)
Maximum door velocity (m/sec)
Deceleration of door (m/sec2)
µfL Coefficient of friction of door motion (0.01)
BL Damping coe
Ls Screw length (m)
Ds Screw diameter (m)
µ Coefficient of friction of screw
Ix Moment of Inertia of screw
I Polar moment of inertia of screw
een the motor and the screw
Tm Motor torque (N-m)
Tpm_spec Peak Motor Torque Spec (N-m)
Peak Motor power (W)
C
Maximum screw angular velocity (rad/sec)
ML Mass of door (Kg)
Vm
aa Acceleration of door (m/sec2)
ba
fLF Friction force against door motion (N)
fficient against door motion (N-sec /m)
P Screw pitch (m)
z
Jm Motor inertia (N-m- sec2)
J Screw inertia (N-m- sec2) s
n Gear ratio betw
Tpm Peak Motor torque (N-m)
pmP
mE Energy consumption of motor (J)
mP onsumed Motor power (W)
sθ Screw angular velocity (rad/sec) &
smθ&
37
Maximum screw angular acceleration (rad/sec2) smθ&&
mθ& r acceleration (rad/sec2) & Motor angula
mmθ Maximum motor angular acceleration (rad/sec& 2) &
allσ Maximum allowable bending stress (N/m2)
allτ Maximum allowable shear stress(N/m2)
K
electricity cost of a motor
al cost for the initial cost of motor size and motor
el
b sub-system is the minimization of the annual door maintenance
sumed power cost of a motor and the annual mechanical
f motor size plus the maintenance cost of motor.
(4-16)
Stress concentration factor
1motor Weight factor of the consumedW
2motor Weight factor of the mechanicW
maintenance cost
N Number of the door operation cycle per hour
4.3. Mathematical Mod
4.3.1 Objective function
The o jective function of this
cost. This cost consists of the annual con
cost combined with the initial cost o
mpmotormmotor TWPWf ⋅+⋅= 21min
where
bm t
P = Em
∫∫ ⋅=⋅=⋅ msmmm dxp
nTdnTdE ∫∫ ⋅== bt
mm dttVp
nTT0
)(22 ππ θθ
mLfLms
Lm tapnn ⎟⎠
⎜⎝ π2 2pm V
nPB
nPFVnmJT ⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛++⎟⎜+=
πππ
22)(
Th e ws.
Th of annual electricity cost consumed by a motor and
is the weight factor of the mechanical cost of motor due to the initial cost of motor size and
motor m
JP ⎞⎛ 22
e w ight factors are defined as follo
e term 1W represents the weight factor 2W
aintenance cost.
38
The rocedure. The annual electricity cost is
ob multiplying the electricity cost per Kw. The
number of the door operation cycle per hour (N), which corresponds to one cycle when the door
is c tinuous door operation time is
calc la the same as that of the door opening time
and this operation continues every day for a year.
co
The m mation of the initial cost for
motor st of
motor a cost based on the common sense.
Table 4.1 Example of weight factor calculation with N=60
4.3.2 Constraints
Physical constraints
The lead screw is assumed as a rigid with a constant diameter. Moreover, the moment of inertia
of screw affects directly to the motor size as shown in motor torque equation. Thus, we consider
bles.
c
following table shows the weight factor calculation p
tained from the annual consumed power by
losed after open, is defined as parameter. Then the con
u ted by assuming that the door closing time is
NtHo boperationdoorntinuous ⋅⋅= 2 ur
echanical cost of motor per torque is calculated from the sum
torque size and the motor maintenance cost. We used the assumed values for the co
nd the maintenance
weight 1: Annual Electricity Cost / Kw Hour $76.51
Kilowatt Hours X Cost per KW Hour
kilowatt hour $0.08 /Kw Hour
0.109 24 365 956.338 hr
Standard cost with respect to motor torque Motor Peak Torque Cost
2 [N-m] $100.00
ii. Motor maintenance co
=Annual
electricity cost per
Annual Kilow att Hours
(continueous / 1 hr) hour day
weight 2: Mechanical cost of motor / torque $44.49 /N-m
i. Initial cost for motor torque size $44.44 /N-m
0.2 [N-m] $20.00
s $0.05 /annual $0.00005 /cycle/torque
the screw dimensions as optimization varia
We set the onstraints of the lead screw strength as follows.
Constraint against bending in normal stress:
39
allxIDsM σ≤/)2/( where /2LgMM sL ⋅⋅= , )64/( 4DsIx π= (4-17)
Constraint against fatigue failure in shear
allzps IDsTK τ≤⋅⋅ /)2/( pmps TnTwhere = , K is a stress concentration factor (4-18)
No slip constraint during the operation of a lead screw:
specpmsnTss
L nTTNDDm _1
1 cos=≤⎥
⎤⎢⎡ +
⎟⎞
⎜⎛
−
− απµ Tns ND cos2 ⎦⎣ −⎠⎝ µαπ
(4-19)
Where PNT /1= ; Number of threads per inch
⎟⎟⎠
⎞⎜⎜⎝
⎛= −
sn D
Pπ
α 1tan ; Thread angle measured in the normal plane
Practical constraints
The distance of door moved during the full open is limited as follows because of the door
package. The value of 0.05 means the spare space, 0.05m.
05.021
+≤≤ StVS bm (4-20)
To avoid inconvenient using of a door when it opens too slowly, we impose the rise time
constraint by assuming that the door should be open at least 0.5m within 1 second. This situation
subscribed 1 can occur before or after ta and the constraint equation for each case is as follows.
Case 1 5.021
≥⋅= maa Vtd where da : moving distance at ta
111 12111 ≤→≤⇒=→=
m
a
a
m
a
m
Vt
ttt
tV
21
21
11 =⋅⋅tVVandtV
01 ≤−⇒m
a
V t
Case 2 5.02
<⋅= maa Vtd 1
→−−==⋅⋅ )(21
21
1111 abaa ttataVandtV
(4-21)
→≤−+−−= 1//21 baabbb ttVmtVmtttt
0)//()1( 22 ≤−+−−− baabbb ttVmtVmttt
40
41
Fortunately, we fou s occurs before t o minimize objective function. Thus we
nuity due to the case switch is avoided.
vs. time
f a lead screw is limited as follows.
n (4-12). To minimize the motor
peak torque, the equality constraint is used.
a b
nd that t1 alway a t
considered only the case 1 and the disconti
da
ta tb
d(t)
TIME
d(t)
da
ta tb TIME
case 1
Figure 4.3 Plot of moving distance of door
The length o
sLS (4-22)
The motor peak torque has to be greater than that in equatio
≤
specpmpm TT _= (4-23)
From the definition of time constant, the stop time is greater than the rise time and the
acceleration aa is greater than the deceleration ab.
t < t and baab tt −
mm ttVVaa ≤⇒≥→≥ 2 (4-24) a
ba t
There is also upper bound on the screw pitch by production manufacture.
(4-25)
sign Variables and rameters
Variables
ta Rise time (sec)
tb Stop time (sec)
Maximum Load velocity (m/sec)
Screw pitch (m)
2.0≤P
4.3.3 De Pa
mV
P
d db
0.5
t1
b
0.5
t1
case 2
Ds Screw diameter (m)
n Gear ratio between the motor and the screw
Ls Screw length (m)
Parameters
S Width of Door (1 m)
ML Mass of Load (Door) (1 Kg)
µfL Coefficient of friction of door motion (0.01)
BL Damping coefficient against door motion (0.01 N-sec /m)
µ Coefficient of friction of screw (0.01)
allσ Maximum allowable bending stress (200 MPa)
Maximum allowable shear stress (100 MPa) allτ
K Stress concentration factor (3)
sρ Density of steel screw (7850 kg/m3)
Jm Motor - sec2)
cost of a motor
anical cost
maintenance cost
termediate variables
Friction force against door motion (N)
inertia (N-m
specpmT _ Motor peak torque spec (0.2 N-m)
1motorW Weight factor of the consumed electricity
2motorW Weight factor of the mech for the initial cost of motor size and motor
N Number of the door operation cycle per hour
In
gMF LfLfL µ= fLF
)64/( 4DsIx π= Ix Moment of Inertia of screw
r moment of inertia of screw
-m- sec2)
)32/( 4DsI z π= Iz Pola
IzLsJ ss ⋅⋅= ρ Js Screw inertia (N
a
ma t
Va = 2 aa Acceleration of door (m/sec )
42
ab
mb tt
Va−
= m/sec2) ba Deceleration of door (
4.3.4 Summary Model
⋅+ 2 (4-26) Wf ⋅= 1min mpm TWP
where
bm t
P = Em
∫∫∫ =⋅=⋅=⋅= msmmmm dxp
nTdnTdTE ∫ ⋅m dttVp
nT0
)(bt 22 ππ θθ
mLfLLm nF
tapn
nnmJ ⎟
⎠⎜⎝
++⎟⎠
⎜⎝
+=ππ 2
)2
( 2ms
pm Vn
PBT ⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎜⎟⎜
π2
Su ct (4-27)
PVJP ⎞⎛⎞⎛ π22
bje to
0/)2/(:1 ≤− allxIDsMg σ
0/)2/( ≤−⋅⋅ allzpm IDsTn:2 Kg τ
0cos2 ⎦⎣ −⎠ ns PD µαπ
cos≤−
⎤⎡ +⎞ nss nTPDD απµ :3 ⎥⎢⎟⎜⎝⎛
pmLmg
01≤bmtV :4 −Sg
2
005.021:5 ≤−− StVg bm
01: ≤−mV
6 ag
S
at
t
07 ≤− sLg :
08 ≤− btg 2:
02.0:9 ≤Pg −
43
4.4. Model Analysis
e can derive the analytical solution of the objective function using Maple software with
and to simplify the monotonicity analysis because
is a function of m
.03333333333 Vm2 ta 3.141592654
W
symbolic weight factors 1motorW 2motorW
2motorW otor torque which is dependent all variables.
f W1 06.283185307 ⎛
⎝⎜⎜⎜ +
0.02533029591 P2
n2245.
P
⎛
⎝
⎜⎜⎜⎜⎜ +
⎛
⎝
⎜⎜⎜⎜⎜ :=
0.01561309992 Pn +
⎞
⎠
⎟⎟⎟⎟⎟ n Vm ta P/ 0.03333333333 Vm2 ( ) − 3tb 1. ta3
( ) − tb 1. ta 2 + +
0.5000000000 6.2831853076.283185307 ⎛
⎝⎜⎜⎜ +
0.02533029591 P2
n2245.3125000 Ls π Ds
n2
P ( ) − tb 1. ta
⎛
⎝
⎜⎜⎜⎜⎜−
⎛
⎝
⎜⎜⎜⎜⎜
0.01591549431 P ⎛⎝⎜⎜
⎞⎠⎟⎟ + Vm Vm ta
− tb 1. tan
0.01561309992 Pn +
⎞
⎠
⎟⎟⎟⎟⎟ m P − tb 1/(n V ( ). ta ) +
0.1000000000 Vm ⎛⎝⎜⎜
⎞⎠⎟⎟ + Vm Vm ta
− tb 1. ta − tb 1. ta −
⎞
⎠
⎟⎟⎟⎟⎟ ( ) − tb2 1. ta2 6.2831853076.283185307 ⎛
⎝⎜⎜⎜
0.0⎛
⎝
⎜⎜⎜⎜⎜ +
0.01591549431 P ⎛⎝⎜⎜
⎞⎟⎟⎠ + Vm Vm ta
− tb 1. tan
0.01561309992 Pn + +
⎞
⎠
⎟⎟⎟⎟⎟ n ⎛⎝⎜⎜
⎞⎠⎟⎟ + Vm Vm ta
− tb 1. ta
( ) − tb 1. ta P/
⎞
⎠
⎟⎟⎟⎟⎟ tb/ W22 ⎛
⎝⎜⎜⎜
⎞
⎠⎟⎟⎟ + +
P2
4 π2 n23925 Ls π Ds4
16 n2 0.005036 n π Vm
P ta
⎛
⎝
⎜⎜⎜⎜⎜ +
0.05000000000 P Vmπ n
0.04905000000 Pπ n + +
⎞
⎠
⎟⎟⎟⎟⎟
In addition, the inequality constraints are in equation 4-27.
The monotonicity of the objective function can be investigated using the derivatives with respect
to the variables and the detail derivative equations are in Appendix 5.
44
Table 4.2 Monotonicity table for the specific parameters in Appendix 1
Using MP1 with respect to Ls, g7 is active. However, for the other variables as shown in the
objective equation and the other constraint equations, the equations are too complicate to inspect
the monotonicity with respect to variables.
4.5 Numerical Results and Discussion
The Excel solver is used to solve the optimization problem of this subsystem. The objective
function and constraints equation are in equations 4-26 and 4-27 except the constraint equation
g1 and g2. These g1 and g2 are scaled by multiplying the scale factor 1/10^8 to achieve better
optimization. Moreover, we have to guarantee the positive values for 7 variables by adding extra
constraints to prevent numerical error like divided zero because Excel solver tried to use the
automatic scaling is
ance. In this
project, Excel solver solved this optimization problem very fast, actually less than 3 seconds.
Variablesta tb Vm n P Ls Ds
+ U
g3 - U U
g4 - -g5 + +
g6 + -g7 -g8 + -g9 +
Functions
f U U U U U + +
g1 + -g2 - + U U
equal lower bound value, zero. As shown in figure 4.5 Excel solver option,
used and other default options are modified slightly to achieve a better perform
45
Figure 4.5 Option values of Excel Solver
To investigate the optimization trend of this subsystem, we performed two parametric studies:
the different door weight (ML) and door width (S). In particular, in the integration system these
two parameters are used as the elp to
. The other parameters are in Table 4.3.
Table 4.3 Parameter list
Table 4.4 Optimal variables and objective function for different door weights (ML)
coupling variables so that the parametric studies can h
understand the roll of this subsystem
46
Table 4.4 shows the results of parametric study for the door weight with a fixed door width and
Table 4.5 provides the constraint values and constraint activity. The Figure 4.6 and 4.7 provide
graphical expression of Table 4.4.
Gearratio
S ML ta tb Vm P Ds Ls nm kg sec sec m/sec m m m -
case 1 1 1 0.806 1.613 1.240 0.046 0.006 1.000 0.078 1.63 37.67 1.633 44.47 0.251 $54.71
Variables
DoorMotor
Pm(W) w1 Torque
(Nm)Power(W)w2
Screw
case 2 1 10 0.866 2.308 0.866 0.068 0.014 1.000 0.325 5.27 53.92 5.275 44.48 0.526 $96.29
case 3 1 20 0.724 2.764 0.724 0.087 0.017 1.000 0.612 6.45 64.57 6.454 44.48 0.728 $120.20
case 4 1 30 0.651 3.074 0.651 0.101 0.02 1.000 0.890 7.29 71.82 7.292 44.49 0.883 $137.33
ConsumedPower
Objective Function
f(x)=w1*P m+w2*Tpm
S ML µfL BL µm kg - N-sec /m -
Door Screw Motorρall τall K ρsMPa MPa - kg/m3
S ML 0.01 0.1 0.1 200 100 3 7850
JmN-m-sec2
0.005
Annual Maintenance Cost vs Door weight (S=1m)
$72.91$87.80
$32.39
$137.33
$54.71
$96.29
$120.20
$98.07
$43.55$23.38
$39.26$11.16$0.00
$50.00
$100.00
$150.00
0 10 20 30
D o o r weight (kg)
Dol
lar Total Cost($)
Electricity Cost
Initial Motor Cost
Figure 4.6 Optimal objective functions for different door weights (ML)
Design variables vs Door weight (S=1m)
3.500
0.000
0.500
1.000
1.500
2.000
2.500
3.000
0 10 20 30
D o o r weight (kg)
Des
ign
varia
bles
ta (sec)
tb (sec)
Vm (m/s)
P(x10)
Ds(cm)
Ls(m)
n
Figure 4.7 Optimal variables L)
l concept. Figure 4.7 provides the variables
ange of the door weight.
We can also find the competition between variables as follows.
for different door weights (M
Figure 4.6 shows the objective values along to the door weight change and we can notify that the
consumed electricity cost is dominant and covers almost 70% of total cost. This is attributed to
the method of cost calculation based on the annua
trend with respect to the ch
( ) ( ) SLtaVmnDsPtbVariablesandCostObjectiveMAs L =↓↑↑↑ ,,,,,,::,
In other words, as the door weight increases, four variables ( )nDsPtb ,,, increase and two
variables decrease. This competition can be explained using the objective function. In
particular, the motor torque is proportional to the maximum door velocity (Vm) and the cost also
increases as the motor torque increases. Therefore, the optimal process tends to reduce this door
( )taVm,
47
velocit
terpret the physical meaning directly because their contributions to the operation time or motor
torque are very complicated. For example, the stop time (tb) increases the total energy
consumption time, however this contributes to reduce the motor operation torque.
From Table 4.5, the number of active constraints is five for all cases. In particular, the door
length constraint g7 is always active as shown in this table. This result is consistent with the
previous Monotonicity analysis. Moreover, based on this table we can conclude that the normal
stress constraint of a screw g1, the no slip condition of a lead screw gear g3 and the lower bound
of the moving distance of a door g4 are always active. The minimum opening time constraint g6
is always active when the door size is large enough to be considered as a real system and this can
be confirmed ired door
velocity time constraint g8 is only active when the door size is small like the case 1 in Table 4.5.
(a) Constraint values
L)
The results of parametric study for the door width are shown in Table 4.6-7 and Figure 4.8-9.
ws the cons
y so that the objective decreases. With respect to other terms it may be difficult to
in
from the results of parametric studies, Table 4.5 and Table 4.7. The des
(b) Constraint activity
Table 4.5 Constraint values and activity for different door weights (M
S ML g1 g2 g3 g4 g5 g6 g7 g8 g9case 1 1 1 0 -1 -3E-14 4E-14 -0.05 -0.35 0 -2E-14 -0.154
case 2 1 10 0 -1 5E-12 3E-11 -0.05 -5E-12 0 -0.575 -0.132
case 3 1 20 0 -1 -4E-14 -2E-12 -0.05 0 0 -0.832 -0.113
case 4 1 30 0 -1 2E-12 2E-11 -0.05 -5E-13 0 -1.08 -0.096
S M g1 g2 g3 g4 g5 g6 g7 g8 g9case 1 -
case 2 1 10 Active - Active Active - Active Active - -
L
1 1 Active - Active Active - - Active Active
case 3 1 20 Active - Active Active - Active Active - -
case 4 1 30 Active - Active Active - Active Active - -
Table 4.6 provides the numerical results and the Table 7 sho traint values and
constraint activity.
The overall results are very similar to those of the parametric study of a door weight.
48
The electricity cost dominant the total cost and the competition between variables is
( ) ( ) SLVmandnDsPtbtaVariablesCostObjectiveSAs =↓↑↑↑ ,,,,,:,:,
Table 4.6 Optimal variables and objective function for different door width(S)
Design variables vs Door width (ML=20kg)
0.000
0.500
1.000
1.500
2.000
2.500
3.000
3.500
1 1.2 1.4 1.6
D o o r Width(m)
Des
ign
varia
bles
ta (sec)
tb (sec)
Vm (m/s)
P(x10)
Ds(cm)
Ls(m)
n
Figure 4.8 Optim nt door width(S)
al variables for differe
Annual Maintenance Cost vs Door width (ML=20kg)
$96.09 $103.77
$32.98$32.70 $3
$144.19
$120.20 $128.79$136.74
$110.96$87.80
3.24$32.39
$50.00
$100.00
$150.00
D o o r weight (kg)
Dol
lar
$0.001 1.2 1.4 1.6
Total Cost($)
Electricity Cost
Initial Motor Cost
Figure 4.9 Optimal objective functions for different door width(S)
Gearratio
S ML ta tb Vm P Ds Ls nm kg sec sec m/sec m m m -
case 1 1 20 0.724 2.764 0.724 0.087 0.017 1.000 0.612 6.45 64.57 6.454 44.48 0.728 $120.20
case 2 1.2 20 0.793 3.025 0.793 0.092 0.018 1.200 0.639 7.06 70.66 7.064 44.49 0.735 $128.79
case 3 1.4 20 0.857 3.266 0.857 0.096 0.019 1.400 0.665 7.63 76.3 7.629 44.49 0.741 $136.74
case 4 1.6 20 0.916 3.493 0.916 0.101 0.02 1.600 0.692 8.16 81.59 8.158 44.5 0.747 $144.19
Pm(W) w1 Torque
(Nm)Power(W)w2
Door
VariablesMotor
ScrewConsumed
PowerObjective Function
f(x)=w1*P m+w2*Tpm
49
As shown in the following tables, there are five active constraints, g1, g3, g4, g6 and g7.
(a) Constraint values
(b) Constraint activity
Table 4.7 Constraint values and activity for different door width(S)
From these parametric studies for a door weight and a door width we can conclude that the
consumed electricity cost is dominant in the objective function and five constraints are always
active. The competition between variables during optimization process can be explained as
follows: when the door weight or width increases, the maximum door velocity decreases to
reduce the motor tor e the increase of an
operation time and the decrease of a motor torque.
que and the other variables are slightly adjusted to balanc
S ML g1 g2 g3 g4 g5 g6 g7 g8 g9case 1 1 20 Active - Active Active - Active Active - -
case 2 1.2 20 Active - Active Active - Active Active - -
case 4 1.6 20 Active - Active Active - Active Active - -
case 3 1.4 20 Active - Active Active - Active Active - -
S ML g1 g2 g3 g4 g5 g6 g7 8 g9case 1 1 20 0 -1 -3E-13 2E-14 -0.05 -1E-14 0 -1.317 -0.113
case 2 1.2 20 0 -1 2E-11 1E-11 -0.05 3E-12 0 -1.438 -0.108
case 3 1.4 20 0 -1 8E-12 7E-12 -0.05 2E-12 0 -1.552 -0.104
case 4 1.6 20 0 -1 3E-12 5E-12 -0.05 2E-12 0 -1.66 -0.099
g
50
CHAPTER 5
Subsystem 4: Structure of The Sliding Door 5.1 Problem statement
When it comes to the structure of an automatic sliding door, the ultimate object is to find out the
minimum weight of the door after satisfying conditions with respect to shear stress and normal
stress caused by the force originated building. Door design is concerned
with finding the optimal weight for an automatic sliding door in our project during this semester,
and is on of the most challenging problems in architecture as well as mechanical engineering.
5.2 Nomenclature
W: total weight of the door (Kg)
H: height of th
S: Width of the door (m)
UB1: upper boundary value of the upper width the aluminum frame of
a door (m)
WUB2: upper boundary value of the lateral width of the aluminum frame of
a door (m)
F : Force caused by acceleration of a motor (N)
Tg : thickness of glass (m)
Ta : thickness of aluminum frame (m)
a: acceleration of a motor (m/s2)
Ce : combined height, exposure and gust factor coefficient
from environment of a
e door (m)
σ: normal stress on the door (Pa)
σ1 : normal stress on the upside of a door (Pa)
σ2 : normal stress on the side of a door (Pa)
τ: shear stress on the door (Pa)
ε: Poisson’s ratio
w: width of aluminum frame of a door (m)
W
51
Cq : pressure coefficient for the structure or portion of structure under consideration
Iw : importance factor
P :design wind pressure (N/m2)
qs ; wind stag
)
a : volume of a aluminum frame (m3)
ress of plate glass (Pa)
inum (Pa)
inum (Pa)
pact loads (Pa)
a e from human impact loads (Pa)
inum from wind pressure (Pa)
inum frame from human impact loads (Pa)
wind pressure (Pa)
: shear s
lateral aluminum frame from human impact loads (Pa)
lateral aluminum frame from wind pressure (Pa)
inum frame from human impact loads (Pa)
nation pressure at the standard height of 33feet (N/m2) 2q : load per unit area on glass (N/m
m1: mass of an impactor (kg)
v : velocity of an impactor (m/s)
g : gravity (9.81m/s2)
ρg: density of glass (kg/m3)
Vg : volume of a glass (m3)
ρ : density of aluminum (kg/m3) a
V
σmax1 : yield normal stress of plate glass (Pa)
τ : yield shear stmax1
σ : yield normal stress of alummax2
τmax2 : yield shear stress of alum
σ g1 : normal stress of glass from human im
σ : normal stress of glass from wind pressure (Pa) g2
σ 1 : normal stress of lateral aluminum fram
σ a2 : normal stress of lateral alum
σ : normal stress of upper alum a3
σ a4 : normal stress of upper aluminum from
τg1 tress of glass from human impact loads (Pa)
τ : shear stress of glass from wind pressure (Pa) g2
τ : shear stress ofa1
τa2 : shear stress of
τ : shear stress of upper aluma3
τ : shear stress of upper aluminum frame from wind pressure (Pa) a4
Sg : width of glass (m)
Hg : height of glass (m)
A : area of glass (m2) g
52
Af : area of frame (m2)
Lf : length of frame(m)
yg : distance to centroid with respect to y axis in glass(m)
o x axis in glass(m)
art of the aluminum frame(m4)
ct to y axis in in lateral part of the aluminum frame(m)
f upper part of the aluminum frame(m4)
respect to y axis in upper part of the aluminum frame(m)
an impact loads on the glass(m3)
t loads on the lateral part of the frame(N)
ads on the lateral part of the frame(m3)
the upper frame(N)
on the upper part of the frame(m3)
(N)
glass(m3)
ame(N)
of the frame(m3)
(m3)
part of the
on the upper part of the
m)
ame(Nm)
Ig :moment of inertia of glass(m4)
bg : distance to centroid with respect t
Ih; moment of inertia of later p
yh; distance to centroid with respe
Is ; moment of inertia o
ys : distance to centroid with
Qg1; first moment from hum
Vh1;shear force from human impac
Qh1; first moment from human impact lo
Vs1; shear force from human impact loads on
Qs1; first moment from human impact loads
Vg2; shear force from wind loads on the glass
Qg2; ;first moment from wind loads on the
Vh2;shear force from wind loads on the lateral part of the fr
Qh2;first moment from wind loads on the lateral part
Vs2;shear force from wind loads on the upper part of the frame(N)
Qs2; first moment from wind loads on the upper part of the frame
Mmaxg1;maximum moment by human impact loads on the glass(Nm)
Mmaxh1; maximum moment by human impact loads on the lateral
frame (Nm)
Mmaxs1; maximum moment by human impact loads
frame(Nm)
Mmaxg2; maximum moment by wind loads on the glass(Nm)
Mmaxh2; maximum moment by wind loads on the lateral part of the frame (N
Mmaxs2; maximum moment by wind loads on the upper part of the fr
53
5.3 Mathematical Model
ubtopic is to calculate the minimum value of weight of the
cing injuries when
ecognized safety
rchitectural Glazing Materials
glazing material
shall be identified by a permanent label that
taller, and state that safety glazing material
onsidered specific hazardous
s and panels in swinging
all unframed swing doors.
aunas, steam rooms,
and showers. Glazing in any portion of a building wall enclosing these
f the glazing is less than 60 inches
rest exposed edge
of the glazing is within a 24 inch(610mm) arc of either vertical edge of the door in a
closed position and where the bottom exposed edge of the glazing is less than 60
inches(1525mm) above the walking surface.
5.3.1Objective function
Objective function for this s
sliding door.
min f = ρg*Vg+ ρa*Va
where Vg : volume of a glass = Hg *Ag
Va : volume of a aluminum frame= Af *Lf
5.3.2 Constraint
We adopt a safety glass designed to minimize cutting and pier
impacted by people; fully tempered glass, laminated glass and wired glass are r
glazing materials. Safety glazing should meet Safety Standard of A
of the United Sates Consumer Product Safety Commission. Each light of safety
installed in hazardous locations as defined below
specifies the labeler, whether the manufacturer or ins
has been utilized in such installation. The following shall be c
locations for the purposes of glazing.
1. Glazing in ingress and egress doors except jalousies.
2. Glazing in fixed and sliding panels of sliding door assemblie
doors other that wardrobe doors.
3. Glazing in storm doors.
4. Glazing in
5. Glazing in doors and enclosures for hot tubs, whirlpools, s
bathtubs
compartments where the bottom exposed edge o
(1525mm) above a standing surface and drain inlet.
6. Glazing in fixed or operable panels adjacent to a door where the nea
54
7. Glazing in an individual fixed or operable panel, other than those locations described
in Item 5,6 that meets all of the following conditions.
Exposed bottom edge less than 18 inches(457mm) above the floor.
top edge greater than 36 inches(914mm)
aces within 36 inches(914mm) horizontally of the plane
es to the human impact loads, individual glazed areas in hazardous locations
such a
er wind loads, human impact loads and dead loads for my
subtopi
n
pressure shall be taken for the shielding effect of adjacent structures. Basic wind speed is the
fastest- e with an annual probability of 0.02 measured at a point 33 ft
(10,000mm g exposure category C. Exposure B has terrain
with bu i
area extend
generally o
D represen speeds of 80miles per hour (mph)
(129km
over 1 mile
extends inl
greater. Fa ined from wind velocity maps prepared
7.1 Exposed area of an individual pane greater than 9 square feet(0.84㎡).
7.2
7.3 Exposed
7.4 One or more walking surf
of the glazing.
When it com
s those indicated above, including glazing used in fire assemblies shall pass the test
requirements of UBC (Uniform Building Codes) Standard 24-2, Part 1. This fact reflects that we
can apply the conditions to a constraint for our topic. When we design a door, we should
consider allowable stress design, dead loads, live loads. Dead loads consist of the weight of all
materials and fixed equipment incorporated into the door. Live loads are those loads produced by
the use and occupancy of the building or other structures and do not include dead load,
construction load, or environmental loads such as wind load, snow load, rain load, earthquake
load or flood load. I decided to consid
c.
Wi d shall be assumed to come from any horizontal direction. No reduction in wind
mil wind speed associated
) above the ground for an area havin
ild ngs, forest or surface irregularities, covering at least 20 percent of the ground level
ing 1 mile (1.61km) or more from the site. Exposure C has terrain that is flat and
pen, extending 1/2mile (0.81km) or more from the site in any full quadrant. Exposure
ts the most severe exposure in areas with basic wind
/h) or greater and has terrain that is flast and unobstructed facing large bodies of water
(1.61km) or more in width relative to any quadrant of the building site. Exposure D
and from the shoreline 1/4mile (0.40km) or 10 times the building height, whichever is
stest-mile wind speed is the wind speed obta
55
by the t
average w sample of air to pass a fixed
oint. Openings are apertures or holes in the exterior wall boundary of the structure. All
windows o enings and
their fram elements and
component Partially enclosed structure or
story is a s
and the area of opening on all other projected areas is less than half of that on the windward
rojection. Special wind region is an area where local record and terrain features indicate 50-year
fastest-
minimum basic wind speed at any site shall not be less than that shown in Figure 1.
For tho
Na ional Oceanographic and Atmospheric Administration and is the highest sustained
ind speed based on the time required for a mile-long
p
r doors or other openings shall be considered as opening unless such op
es are specifically detailed and designed to resist the loads on
s in accordance with the provisions of this section.
tructure or story that has more than 15 percent of any windward projected area open
p
mile basic wind speed is higher than shown in Fig. 1. Unenclosed structure or story is a
structure that has 85 percent or more opening on all sides.
Ce = combined height, exposure and gust factor coefficient
Cq= pressure coefficient for the structure or portion of structure under consideration
Iw = importance factor
P = design wind pressure
qs= wind stagnation pressure at the standard height of 33feet (10,000mm)
The
se areas designated in Figure 16-1 as special wind regions and other areas where local
records or terrain indicate higher 50-year (mean recurrence interval) fastest-mile wind speeds,
these higher values shall be the minimum basic wind speed. So the basic wind speed is 70mph in
Michigan. Design wind pressures for buildings and structures and elements therein shall be
determined for any height in accordance with the following formula:
P = Ce CqqsIw
Ce , Cq, qs, Iw are determined by UBC.
Ce = 1.39*1.06*0.62=0.91
Cq = 1.2
qs = 0.60354kN/㎡,
Iw = 1
P = 0.91*1.2*0.60354*1= 0.069kN/㎡ (load per unit area)
56
This pressure P is defined as load per unit area q in this subsystem.
<Fig 1>
When it comes to human impact on a door, the impact test based on the Safety Standard
Materials of the United States Consumer Product Safety Commission
e of impact test to
our constraints. The test frame is like Fig 2.
for Architectural Glazing
can be applied to a real human impact, so we made a decision to apply the cas
<Fig 2>
57
m1v=F∆t ; linear momentum equation.
here, m1 = mass of an impactor = 45.4 kg
v = velocity of an impactor = (2gh)1/2m/s = 3m/s
g = gravity = 9.81m/s2
h = height of an impactor before dropping = 460mm
There is an assumption that the value of F∆t can be replaced by mg×3 .
Physical constraints
Normal and shear stress of the glass and frame caused by human impact loads and wind
pressure should be less than their yield stress respectively. We can express these constraints as
several mathematical formulas as below.
⎟⎟⎠
⎞⎜⎜⎝
⎛
+×
××××
)(
32. 222
21
ggg
staticggg HST
PHS≤σ= 1σ max1
gg
ggg bI
QV×
×= 11
1τ ≤ τmax1
h
hha I
yM ×= 1max
1σ ≤σmax2 hh
hh QVa bI ×
×= 11τ ≤ τ 2 1 max
≤σmax2 ss
ssa bI
QV××
= 112τ ≤ τmax2
.
Iyss
a×
= 1max2σ M
⎟⎟⎠
⎞⎜⎜⎝
⎛×+×
××=
6
6
2
2
623.01
5.0
g
gg
gg
H
ST
qSσ ≤σmax1
gg
ggg bI
QV×
×= 22
2τ ≤ τmax1
h
hha I
yM ×= 2max
3σ ≤ σmax2 hh
hha bI
QV××
= 223τ ≤τmax
2
s
ssa I
yM ×= 2max
4σ ≤σmax2 ss
ssa bI
QV××
= 224τ ≤τmax
2
58
Practical constraints
I would like to determine thickness of the glass to be less than 80 percent of that of the
rpose like this,
Tg ≤0.8(Ta)
e at the same
eaning.
w ≥2*(Ta)
f width of the door as well as that of thickness of the aluminum frame
e
0, T 0, T 0
5.3.3 Design Variables and Para
First of all, design variables for this door are dimensions of each part. In other words, the
width of the late m are design variables
respectively. Both the height and the width of glass are also design variables in this subtopic.
a f glass and alu as parameters.
Moreover, maximum value of shear and normal stress is also given by a parameter.
List of Design Variables
w: width of aluminum frame(m)
Tg : thickness of glass(m)
Ta : thickness of aluminum frame(m)
List of parameters
: height of a door(2m)
aluminum frame for the esthetic pu
In addition, width of the frame is not less than twice of the thickness of the fram
m
Of course the value o
should be always positive in order to calculate it with excel solver. The value of thickness of th
glass should also be positive.
w ≥ g ≥ a ≥
meters
ral edge and the upside edge of aluminum fra e
nd aluminum frame, density o minum are givenThickness of glass
H
59
S: width of a door(1m)
0kg/m3)
ρa : den
τmax1 : yield shear stress of plate glass(70MPa)
σmax2 : yield normal stress of aluminum(40
impactor (45.4kg)
g: gravity (9.81m/s2)
Vg : volume of a glass = Hg *Ag
a : volume of a aluminum frame = Ta*HS- Vg
3 : width of glass = H-2*W1
g : he
f : total length of aluminum frame = 2*(S+H)
ass
ρg: density of glass(250
sity of aluminum(2700kg/m3)
σmax1 : yield normal stress of plate glass(70MPa)
0MPa)
τmax2 : yield shear stress of aluminum(400MPa)
m: mass of
q: load per unit area caused by wind (69N/m2)
Intermediate variables
V
W
W4 : width of glass = H-2*W2
H ight of glass = H-2w
Sg : width of glass = S-2w
Ag : area of glass = Sg* Tg
Af : area of frame = w * Ta
Va : volume of aluminum frame = Af* Lf
L
Ig :moment of inertia of gl
12
3gg
g
TbI
×=
bg : distance to centroid with respect to x axis in glass
wSbg 2−=
60
yg ; distance to centroid with respect to y axis in glass
2g
g
Ty =
m frame Ih; moment of inertia of later part of the aluminu
3
121
ah TwI ××=
d with respect to y axis in in lateral part of the aluminum frame yh; distance to centroi
2a
hT
y =
Is ; moment of inertia of upper part of the aluminum frame
12
3a
sI =Tw×
respect to y axis in upper part of the aluminum frame ys : distance to centroid with
2a
sT
y =
V ; shear force from humag1 n impact loads on the glass
staticg PV ×=23
1
Qg1; first moment from human impact loads on the glass
42
2gg
g
TbQ ×=
part of the frame Vh1;shear force from human impact loads on the lateral
statich PV ××=43
21
1
61
Qh1; first moment from human impact loads on the lateral part of the frame
42
2
1a
hTwQ ×=
me Vs1; shear force from human impact loads on the upper fra
statics PV ××=43
21
1
Qs1; first moment from human impact loads on the upper p rt of the frame
a
42
2Tw1
asQ ×=
Vg2; shear force from wind loads on the glass
22 )2(
21 wHwqVg −×××=
Qg2; ;first moment from wind loads on the glass
22 4
12 gg
g Tb
Q ××=
Vh2;shear force from wind loads on the lateral part of the frame
HwqVh ×××=21
2
Qh2;first moment from wind loads on the lateral part of the frame
22 4
12 ah
h Tb
Q ××=
Vs2;shear force from wind loads on the upper part of the frame
22 )2(
21 wHwqVg −×××=
62
Qs2; first moment from wind loads on the upper part of the frame
22 4
12 a
ss T
bQ ××=
Mmaxg1;maximum moment by human impact loads on the glass
)2(43
1max wHPM statixg −××=
Mmaxh1; maximum moment by human impact loads on the lateral part of the
e fram
HPM statich ×××=43
41
1max
Mmaxs1; maximum moment by human impact loads on the upper part of the
frame
SPM statics ×××=43
41
1max
Mmaxg2; maximum moment by wind loads on the glass
22max )2(
81 wHwqM g ×−×××=
Mmaxh2; maximum moment by wind loads on the lateral part of the frame
22max 8
1 HwqM h ×××=
Mmaxs2; maximum moment by wind loads on the upper part of the frame
22max 8
SwqM s ×××= 1
5.3.4Summary Model
in f: ρg*Vg+ ρa*Va
Subject to
g2:τg1 -τmax1≤0
Objective function
M
g1: σg1 - σmax1≤0
63
g3:σa1 -σmax2≤0
a1 - max2
g5: σa2 -σmax2≤0
g7: σ g2 - σmax1≤0
0
g9: σ a2 -σmax2≤0
g10; τa3 -τmax2≤0
g12: τa4 -τmax2≤0
g13: Tg ≤ Ta
≤0
a
g4: τ τ ≤0
g6: τa2 -τmax2≤0
g8: τg2 - τmax1≤
g11: σ a4 -σmax2≤0
×8.0
g14: -w≤0
g15: -T ≤0 g
g16: -Ta
g17: 2*(T )-w≤0
5.4 Model Analysis
64
Design Specification
e door
ssumption:
of the door
Assumption:
id, not hollow
the frame are same.
1. Feature of th
A
w1 = w2 =w
2. Cross section
Tg
w
Ta
w 1
w 2
Pla ste Glas
Aluminum F
S
H
-The Frame is sol
-All cross sections of
65
Case 1) Under the human impact condition
a. glass
H-2w
P=3*Pstatic
⎟⎟⎠
⎞⎜⎜⎝
⎛
+×
××××=
)(
32.1 222
21
ggg
staticggg HST
PHSσ
12
3gg
g
TbI
×=
wSbg 2−=
)2(43
1max wHPM statixg −××=
gmPstatic ×=
gg
ggg bI
QV×
×= 11
1τ
staticg PV ×=23
1
42
4gg
g
TbQ ×=
66
b. lateral frame H
c. upper frame S
S
P=(3*
H
=(3*Pstatic)/4
P
Htatic ×
12
yM s ×1max s
hh
hha bI
QV××
= 111τ
statich PV ××=43
21
1
42
2
1a
hTwQ ×=
h
hha I
yM ×= 1max
1σ
PM sh ××=43
41
1max
2a
hT
y =
31 wI ×= ah T×
Pstatic)/4
ss
ssa bI
QV××
= 112τ
statics PV ××=43
21
1
42
3
1a
sTwQ ×=
Ia =2σ
SPM ××=31
statics ×441max
12
3a
sTw
I×
=
2a
sT
y =
67
Case 2) Under the wind loads condition
a. Glass
lat
H-2w
q
b.
×× 25.0 g qS
ggV Q× 22
⎞⎛g =62
Sσ
eral frame H
c. upper
qf =q/4
H
⎟⎟⎠
⎜⎝⎜ ×+× 6623.01
g
gg H
T
2)2w−
h
hha I
yM ×= 2max
3σ
2
2max 81 HwqM h ×××=
τ
Q
ggg bI ×
=2τ
2 (21 HwqVg ×××=
21g T
bQ ××=
fra e S m
2 42 gg
hh
hha bI
QV××
=322
HwqVh =2 ×××21
22 4
12 ah
h Tb
××=
68
5.5 Di
Resul
First
w
ta
tg
qf =q/4
S
ss
ssa bI
QV××
= 224τ
SwqVs ×××=21
2
22 4
12 a
ss T
bQ ××=
ss yM ×2max
sa I
=4σ
22max 8
1 SwqM s ×××=
scussion of Results
ts from excel solver
iteration
Variables Constratins
-1.48132E+15 m g1 -70000000
10138053.13 m g2 -70000000
-3.80652E+14 m g3 -400000000
g4 -400000000
g5 -400000000
g6 -400000000
g7 -70001567.41
g8 -69999798.61
g9 -400000000
g10 -400000000
000
g12 -400000000
g11 -400000
69
objective function
f -8.35268E+48
ediate variables
axg1 9.89612E+17
g -1.36 7E+58
g 2.96264E+15
6E+14
3E+43
167.01525
Vh1 167.01525
1
-
maxs1 167.01525
s 5069026.566
s1 167.01525
s1 -1.90313E+28
maxg2 -1.12141E+47
g2 -1.51407E+32
g2 5.36593E+43
maxh2 -5.11055E+16
h2 -1.02211E+17
Qh2 -1.90313E+28
Interm
M m
I 1
bh -1.48132E+15
b
yg -1.9032
Vg1 668.061
Qg1 5.3659
Ih -1.28627E+35
Mmaxh1
yh 5069026.566
Qh -1.90313E+28
bs 1.48132E+15
Is -1.28627E+35
M
y
V
Q
M
V
Q
M
V
70
Mmaxs2 -1.27764E+16
s2 -5.11055E+16
s2 -1.90313E+28
g 2.96264E+15
g 2.96264E+15
g -1.12774E+30
g -3.34107E+45
6
-
-
eter
g/m3
g/m3
g
/s2
tic
maga 70000000 /m2
/m2
aaa 400000000 /m2
/m2
/m2
T sult is not av imization of the door. I need to modify my constraints and
m
V
Q
S
H
A
V
Lf
Af 1.50177E+22
Vf 9.01062E+22
param
S 1 m
H 2 m
rhog 2500 k
rhoa 2700 k
m 45.4 k
g 9.81 m
Psta 445.374 N
Sig N
tawga 70000000 N
sigm N
tawaa 400000000 N
q 69 N
his re ailable for the opt
odeling.
71
Final iteration
Variables dim nsion e
w 0.021559357 m
ta 0.010779678 m
tg 0.003293428 m
0
Para 0
meter
S 1 m 0
2 m 0
2 00 kg/m3 0
hoa 2 00 kg/m3 0
.4 kg 0
81 m/s2
445 74 N
ga 7. N/m
ga 7. N/m
/m
/m
N/m
ickness %
o
frame
ediate
g1 0
2.84 E 4
h 0.021559 57 m2
bg 0.956881287 m
668.061 N
H
rhog 5
r 7
m 45
g 9.
Pstatic .3
σ_ 00E+07 2
taw 00E+07 2
σ_aa 4.00E+08 N 2
tawaa 4.00E+08 N 2
q 69 2
glass th 80
f
Interm variables
M max 653.658 347 Nm
Ig 853 -09 m
b 3
yg 0.001646714 m
Vg1
72
Qg1 1.29737E-06 m3
2.25046E-09 m4
Mmaxh1 167.01525 Nm
yh 0.005389839
1
1
2.25046E-09 4
Mmaxs1 167.01525 m
0.00538983
1 167.0152
3.13154
xg2 0.71207
2 1.45552
2 1.29737
2 0.74
1.4
3.1
s2 0.
0.7
2 3.13154E-
Sg 0.956881287 m
1.9568812
g 0.00315142 2
Ih
m
Vh 167.01525 N
Qh 3.13154E-07
bs 0.021559357 m3
Is m
N
ys 9 m
Vs 5 N
Qs1 E-07 m3
M ma 1919 Nm
Vg 3999 N
Qg E-06 m3
Mmaxh 3797801 Nm
Vh2 87595603 N
Qh2 3154E-07 m3
Mmax 18594945 Nm
Vs2 43797801 N
Qs 07 m3
Hg 87 m
A m
Vg 0.006166954 3
Aa 0.000232403 2
m
m
La 6 m
Va 0.001394418 m3
73
f 19.1823123 kg
ht_g 15.4173
ight_a 3.76492
ratins
Weig 8487
We 7428
Const
g1 0.0000 <= 0
g2 -0.3500 <= 0
g3 -4.0000 <= 0
g4 -3.9892 <= 0
g5 0.0000 <= 0
g6 -3.9892 <= 0
g7 -0.6711 <= 0
g8 -0.6856 <= 0
g9 -3.9822 <= 0
g10 -3.9999 <= 0
g11 -3.9955 <= 0
g12 -4.0000 <= 0
g13 -0.0053 <= 0
g14 -2.06E-02 <= 0
g15 -9.78E-03 <= 0
g16 -2.29E-03 <= 0
g17 0 <= 0
ated zero 0.001
Approxim
74
σ_g1 7.00E+07 g1
tau_g1 3.50E+07 g2
σ_h1 2.90E-01 3
1.08E+06 4
1 0E+08
s1 8E+06
g
tau_h1 g
σ_s 4.0 g5
tau_ 1.0 g6
σ_g2 2.89E+06 g7
tau_g2 1.44E+06 g8
σ_h2 8E+06
h2 0E+03
2 5E+05
s2 0E+03
scale factor
1.00E-08 1.30E-04
1.00E-08 -3.50E+07
1.00E-08 -4.00E+08
1.00E-08 -3.99E+08
-6.70E+00
1.7 g9
tau_ 9.6 g10
σ_s 4.4 g11
tau_ 4.8 g12
1.00E-08
75
1.00E-08 -3.99E+08
1.00E-08 -6.71E+07
1.00E-08 -6.86E+07
1.00E-08 -3.98E+08
1.00E-08 -4.00E+08
0E-08 -4.
-08 -4.
jective J31
1.0 00E+08
1.00E 00E+08
Ob
Cost of the
door $57.65 $/kg
Glass $50.88 $3.30
Aluminum $6.78 $1.80
Answer report
from excel solver
Microsoft Excel 11.0 Answer Report
Worksheet: [il.x lver
Report Created: 4/17/2005 5:53:52 PM
Ta
(M
ls]so
rget Cell
in)
76
me
O
Final Valu Cell Na
riginal
Value e
19.1823123 19.1823123 $F$3 f
Ad lls
me
O
Final Valu
justable Ce
Cell Na
riginal
Value e
$B$2 w 0.021559357 0.021559357
$B$3 ta 0.010779678 0.010779678
$B$4 tg 0.003293428 0.003293428
Constraints
Cell Name Cell Value Formula Status Slack
$F$10 g1 0.0000 $F$10<=$H$10 Binding 0
$F$11 -0.3500 $F$11<=$H$11
Not
Binding 0.35 g2
$F$12 g3 $F$12<=$H$12
-4.0000
Not
Binding 3.999999997
$F$13 g4 -3.9892 $F$13<=$H$13
Binding 3.989220322
Not
$F$14 g5 0.0000 $F$14<=$H$14 Binding 0
$F$15 g6 -3.9892 $F$15<=$H$15
Not
Binding 3.989220322
$F$16 g7 -0.6711 $F$16<=$H$16
Not
Binding 0.67112274
$F$17 g8 -0.6856
$F$17<=$H$17
Not
Binding 0.68556137
$F$18 g9 -3.9822 F$18<=$H$18 ot .982186111 $ N 3
77
Binding
$ 99 $F$19<=$H$19
Not
inding .999903986 F$19 g10 -3.99 B 3
$F$20 11 -3.9955 F$20<=$H$20
ot
inding .995546528 g $
N
B 3
$F$21 g12 -4.0000 F$21<=$H$21
ot
inding .999951993 $
N
B 3
$ 53 $F$22<=$H$22
Not
inding .005330315 F$22 g13 -0.00 B 0
$F$23 g14 $F
ot
inding .020559357 -2.06E-02 $23<=$H$23
N
B 0
$F$24 15 -9.78E-03 F$24<=$H$24
ot
inding .009779678 g $
N
B 0
$F$25 g16 -2.29E-03 F$25<=$H$25
ot
inding .002293428 $
N
B 0
$ $F$26<=$H$26 Binding 0 F$26 g17 0
I used Excel solver to find out the optimal weight of the door used for automatic sliding
d o e results sh a s , we can design the most suitable
door for our project. In addition, width of the aluminum e, thickness of the aluminum frame
and thickness of glass is ab m el raints are g1,g5
and g17. g1 is the constraint that normal stress on the glass should be less than the yield stress on
i g ssociation u in d g7 means the
c n ickness of t a of the value o e width of door
for the aesthetic purpose. In this condition as I said above, it just cost $57.65 to make the
a t l g door for o
o r system. Th ow us th t when the door i 19.18 kg
fram
respectivout 21.5cm, 1cm and 0.3c y. Active const
t. 5 has also a with normal stress on the pper alum um frame. An
o dition that th he frame should be less th n the half f th
u omatic s idin our air-ro m.
78
CHAPTER 6
System Integration Study
6.1 Objective
tive of th t f the entryway
system for one year. This cost consists of the m nd sliding door price and the cost
e city to run r m
6.2 Methodology: All-In-One Approach
For the integration, iSight v9.0 was used. All th iables were put together and the
m ion was ra m as used for the
optimization algorithm.
The objec e overall system is to minimize the cos or maintaining
otor a
. for lectri the cont ol and heat syste
e var
tic Prograopti izat performed. Sequential Quad ming (SQP) w
Fig 6.1 Integrated systems
As shown in Fig 6.1, four subsystems – Door system, motor system, heat curtain, and
heat exchanger – were integrated using iSight.
79
Table 6.1 shows the descriptio
n of each step.
Index Name Job
1 DoorSystem Running Excel solver to find a optimum door
structure
2 MotorSystem Running Excel solver to find a optimum motor
control system
3 HCConstraints Setup constraints equations for the heat curtain
subsystem
4 HCGAMBIT Running GAMBIT to generate mesh for heat
curtain
5 HCFLUENT Running FLUENT to analyze the heat transfer
6 HCParser Running Parsing program for the post-processing
7 HeatExchanger Running MATLAB to find an optimum heat
exchanger system
8 HEConstraints Generating constraints equations for the heat
exchanger
9 CalcObjective Calculating the overall objective summing
individual objectives Table 6.1 Step description in the integrated systems
80
6.3 Result and Discussion
9350 9145
4716
28 5 58 58 119 1190
1000
3000
4000
6000
7000
9000
10000D
)
2000
5000
olla
r($
8000
Total
Objective function (C
HC HE Door Motor
ost)
Integration case
Independent subsystem
Fig 6.2 shows the objective f r this system. As can be seen, the total
cos $93 o . Heat curtain
system spent $4716 when it worked alone, but in the overall system, mass flow rate was
con ined ha at curtain was
reduced almost the half, and dividual cost. Also, the cost for the
heat exchanger system i eason for these
hanges is the change of mass flow rate. In the individual subsystem, heat curtain could
draw more mass flow rate and it could increases the performance. Also the heat
exchanger system didn’t have to supply large mass flow and it didn’t have to overburden
the motor pump and fan pump.
If the heat exchanger had a slacker constraint on the relevant variables, the cost of the
heat curtain and the heat transfer would have been similar. It can be concluded that the
mass flow rate constrained by the heat exchanger is the active constraint that has the most
dominant effect on this system, and that the cost for maintaining the heat curtain system
is dominant over other systems.
Fig 6.2 Objective function: Annual cost
unction, the cost, fo
t is 50. Most of the c st comes from the heat curtain system (97.8%)
stra by the heat exc nger system, and the performance of the he
spent 193.9% of the in
ncreased to 560% ($5 to $28). The major r
c
81
9
1
4
7
0 0
7
3 3
17
5 5
9
0
2
4
6
8
10
12
14
16
18
Total HC HE Door Motor
Number of active constraints
Active for the integrated system
Active for ind. Subsystem
Total constraints for Ind. Subsystem
Fig 6.3 Objective function: Annual cost
Fig 6.3 shows the number constraints and the number of active constraints. As can be
en, only heat curtain subsystem has a difference (4 to 1). For individual subsystem, it
.
se
has four active constraints, but in the integrated system, it only has one active constraint
It seems that the interaction between the other subsystem – heat exchanger – affect the
subsystem and the boundary optimum was moved.
82
Reference [1] David P. Dewitt, Frank P. Incropera , Fundamentals of Heat and Mass Transfer, 6th Edition
[2] Richard E. Sonntag, Claus Borgnakke, Gordon J. Van Wylen , Fundamentals of
Thermodynamics, 6th Edition
[3] Fussell, B.K. and Taft, C.K., Brushless DC motor selection, Electrical Manufacturing & Coil
[4] Winding Conference. Proceedings , 18-21 Sept. 1995 Pages:345 – 353
[5] Mechanics of Materials : Gere & Timoshenko 3rd edition.
[6] ASTM standards on materials
[7] Uniform Building Code : 1997 yr version
[8] Machine Design Databook: Lingaiah, McGraw-Hill
83
APPENDICES
% Ideal vapor-compression refrigeration cycle example
%==================================================================
T1=-20; %Evaporator out temperature [oC]
h1=178.7; %Evaporator out enthalpy [kJ/kg]
s1=0.7087; %Evaporator out entropy [kJ/kg K]
%==================================================================
% Heat exchanger spec
%==================================================================
cond_al=237; %Al conduction coefficient [W/m]
Do=16.4*10^-3; %Tube outside diameter [m]
Di=13.8*10^-3; %Tube inside diameter [m]
ntube=1; %number of cooling channel in radiator
Pitch_fin=275; %Fin pitch
Dh=6.68*10^-3; %Flow passage hydraulic diameter [m]
t=0.254*10^-3; %Fin thickness [m]
Sigma=0.449; %Free flow area/frontal area
alpha=269; %Heat transfer area/total volume [m2/m3]
Afr=0.2; %Condensor Frontal Area [m2]
A.1. Heat Pump and Heat Exchanger Matlab M files and Simulink Models
Condenser.m
clear all;
close all;
time_step=0.1;
%==================================================================
84
AfperA=0.830; %Fin area/to
Ah=9.3; %Air heat transfer surface [m2]
ta_pump=0.8;
==================================================================
xchanger input
============================================================
_air_i=273; %Temperature air inlet
%Temperature air outlet
dot_R12=0.00220928932732473; %mass flow rate R12
ad power_pump.mat;
);
_r_out=%15.8f\n',...
tal area
eta_fan=0.7; %fan power conversion factor
e
%
% Heat e
%======
T
T_air_o=303;
m
mdot_air=0.00218517527639549; %mass flow rate air
lookup_input;
sim('condensor_r12');
load power_fan.mat;
lo
load cop.mat;
load width.mat;
load T_R_o.mat;
fid=fopen('p_fan.txt','w');
power_f=power_fan(2,length(power_fan));
power_p=power_pump(2,length(power_pump));
cop_p=cop(2,length(cop));
width_con=width(2,length(width)
T_r_out=T_R_o(2,length(T_R_o));
fprintf(fid,'fan power=%15.8f\npump
power=%15.8f\ncop=%15.8f\nwidth_con%15.8f\nT
power_f,power_p,cop_p,width_con,T_r_out);
fclose(fid);
85
exit;
lookup_input.m
280 290 300 310 320 330 340 350 360 370 380 385]; %[K]
mp_R12=[230 240 250 260 270 280 290 300 310 320]; %[K]
9 420 365 324 289 260 248]; %[N*s/m^2]
0 668 674 679 683 685]; %[W/mK]
4.195 ...
ho_w=1000.*[1.0 1./1. 1./1.001 1./1.003 1./1.007 1./1.011 1./1.016 ...
27 1./1.034 1./1.041 1./1.049 1./1.053]; % [kg/m^3]
==================================================================
===================================================
0.0385 0.0354 0.0322 0.0304 0.0283 0.0265 0.0254 ...
3]; %[N*s/m^2]
*[68 69 70 73 73 73 73 72 69 68]; %[W/mK]
8816 0.8923 0.9037 0.9163 0.9301 0.9450 0.9609 ...
[J/kgK]
072 1.3744 1.3405 1.3058 ...
===============================================
========================================
900]; %[K]
^2]
62.0]; %W/mK
7 1009 1014 1030 1051 1075 1099 1121]; % [J/kgK]
temp_w=[273.15
te
%==================================================================
% Properties of water
%==================================================================
mu_w=10^-6*[1750 1422 1080 855 695 577 48
cond_w=10^-3*[569 582 598 613 628 640 650 66
cp_w=1000.*[4.217 4.198 4.184 4.179 4.178 4.180 4.184 4.188
4.203 4.214 4.226 4.232]; %[J/kgK]
r
1./1.021 1./1.0
%
% Properties of R12
%===============
mu_R12=10^-2*[0.0457
0.0244 0.023
cond_R12=10^-3
cp_R12=1000.*[0.
0.9781 0.9963 1.0155]; %
rho_R12=1000.*[1.528 1.498 1.4695 1.439 1.4
1.2689 1.2286]; % [kg/m^3]
%===================
% Properties of air
%==========================
temp_air=[300 350 400 500 600 700 800
mu_air=10^-7*[184.6 208.2 230.1 270.1 305.8 338.8 369.8 398.1]; % [Ns/m
cond_air=10^-3*[26.3 30. 33.8 40.7 46.9 52.4 57.8
cp_air=[100
86
rho_air=[1.1614 0.9950 0.8711 0.6964 0.5804 0.4975 0.4354 0.3868]; % [kg/m^3]
%==================================================================
=====================
4.9 960.7 1219.3 1525.9];
=================
eat exchanger
7 8];
=====================
========================================
(:,2:8);
============================
==================
% R12 saturated liquid Enthalpy [kJ/kg]
%=============================================
T3=[0 10 20 30 40 50 60]+273;
P3=[308.6 423.3 567.3 74
hf=[36.05 45.37 54.87 64.59 74.59 84.94 95.74];
P2=[300 400 500 750 1000 1500 2000];
hg=[190.3 195.4 199.4 206.8 212.2 219.8 225.0];
T2=[3.99 14.7 23.4 39.8 52.4 71.9 86.5]+273;
%=================================================
% Heat transfer and friction factor for circular tube-circular fin h
%==================================================================
Re=10^4*[1.5 2 3 4 5 6
f=[0.041 0.038 0.035 0.032 0.03 0.029 0.028 0.027];
jH=[0.0125 0.011 0.0095 0.0089 0.008 0.0075 0.0071 0.0069];
%=============================================
% Pump Efficiency
%==========================
rpm_col=[1500 2000 2500 3000 3500 4000 4449];
%load('eta_lookup.txt');
%vol_row=eta_lookup(:,1)';
%eta_out=eta_lookup
%======================================
% NTU calculation (Cross flow, non-mixing)
%================================================
e=[0 0.2 0.4 0.6 0.8 0.9];
87
Cr=[0 0.25 0.5 0.75 1];
NTU=[0 0 0 0 0;
sor_compressor.mdl
0.21 0.22 0.23 0.24 0.25;
0.4 0.47 0.55 0.64 0.75;
0.8 1. 1.23 1.5 1.8;
1.7 2. 2.7 4.2 6.5;
2.3 3.1 5.4 8 12];
conden
0.09554
width
0.9821
pump power
mdot_R12
mdot_c[kg/s]
mdot_air
mdot_air[kg/s]
0.03
mdotR12
74.75
h3
211.4
h2
2.113e-006
fan power
hc and hh Out1
U
width.mat
To File4
cop.mat
To File3
power_pump.mat
To File2
T_R_o.mat
To File1
power_fan.mat
To File
T_air_i
Tari,i [K]
313.2
T_c,o
323
T_c,i1
323.6
T_c,i
T_air_o
T ar,o[K]
sf_fcn2
S-Function
Lookup Table3
Lookup Table2
Lookup Table1
Lookup Table
f(u)
Fcn
Divide
V
mdot_c
h_cool
T_c,o
T_ci h_air
T_air,i NTU
T_air,o Cmin
mdot_air Power_f an
Convective Heat Trans. Coeff.
Afr
Constant1
alpha
Constant
4.176
COP
5.14
Ah
88
6Power_fan
5T_c,o
4Cmin
3NTU
2h_air
1h_cool
u(1)^2
u^2
specific heat
rho*l_air/mu
rho*Di/mu
jH
hconv_cool
f(u)
h_ari
Re f /2.
f/2
f*rho*u^2/2
f
density of air1
density of air
f(u)
del_p (Pa)
alpha
alpha
V_cool
Di
Tube dia1
Di
Tube dia
Terminator
Specific heat cpR12
eta_pump
Sigma3
eta_fan
Sigma2
Sigma
Sigma1
Sigma
Sigma
Saturation
sf_fcn
S-Function
Re_air
Di^2*pi/4
R12_Tube_area
R12 Viscosity Mu
R12 Conductivity Pr_cool
Pr_air
ntub
Number of
e
Tube
f /2
Re
Pr
Nu
Nu_cool
Memory
LookupTable (2-D)
u(1)*0.47/Di
Lh/D*conversion pa
G2
G
f(u)
Fcn
0.07446
Display
Density of R12
DhChar. Length of Air
Calc. Re_cool1
Calc. Re_cool
C_air_mult
Afr
Air front. Area
Air conductivity
Air Viscosity Mu
f(u)
A/Aff6
mdot_air
5mdot_c
4T_air,o
3T_air,i
2T_ci
1V
A.2 iSight description file of the integration system ---------------------------------------------------------------------------------------------------- MDOLVersion: 9.0 CompilerOptions: warn Task Task1 TaskHeader Task1 Version: 1.0 Evaluation: optimize SQLNL ControlMode: user RunCounter: 14 BoundsPolicy: adjustvalue
89
CheckPoint: unknown End TaskHeader Task1 Inputs Task1 Parameter: Ph Type: real InitialValue: 0.1 Description: "Distance of the duct from the outer door" Parameter: Wd Type: real InitialValue: 0.1 Description: "Duct output width" Parameter: Ld Type: real InitialValue: 1.2 Description: "Duct output length" Parameter: W Type: real InitialValue: 1.5 Description: "Width of the air-room" Parameter: L Type: real InitialValue: 10.0 Description: "Length of the air-room" Parameter: ma Type: real InitialValue: 0.1 Description: "Mass influx through the duct" Parameter: mr12 Type: real InitialValue: 0.1 Description: "Mass influx through the pump" Parameter: S Type: real InitialValue: 1.0 Description: "Width of the door" Parameter: fanfactor Type: real InitialValue: 0.7 End Inputs Task1 Outputs Task1 Parameter: tb Type: real Description: "door opening time" Parameter: ML Type: real Description: "Weight of the door" Parameter: Afr Type: real Description: "Heat exchanger frontal area" Parameter: Vd Type: real Description: "Air velocity at the duct output" Parameter: Pd Type: real Description: "Distance between the duct and origin(Y-direction)" Parameter: Ps Type: real Description: "Distance between the door and origin(Y-direction)" Parameter: Ti Type: real Description: "Inside temperature" Parameter: To Type: real Description: "Outside temperature" Parameter: Td Type: real Description: "Duct output temperature" Parameter: Pi Type: real Description: "Inside gauge pressure" Parameter: Po Type: real Description: "Outside gauge pressure" Parameter: Sw Type: real Description: "Walking speed of average person" Parameter: Tr Type: real Description: "Duration in which the door is opened" Parameter: Dr Type: real Description: "Distance from which the door starts to open or lose"
area of the air-room" Parameter: g2 Type: real Description: "Constraint Eq. 2"
ype: real Description: "Constraint Eq. 4" : real Description: "Constraint Eq. 5"
Parameter: g7 Type: real Description: "Constraint Eq. 7" real Description: "Objective function"
Parameter: TotalObj Type: real Description: "Overall objective" r Type: real Description: "objective of motor subsystem"
otor[15] Type: real Description: "output constraints of motor subsystem" al Description: "Density of air"
or Type: real Type: real Description: "Average open time"
c Parameter: Ab Type: real Description: "Maximum Parameter: g3 Type: real Description: "Constraint Eq. 3" Parameter: g4 T Parameter: g5 Type Parameter: ObjF Type: Parameter: fmoto Parameter: gm Parameter: airDensity Type: re Parameter: Nopendo Parameter: avgTO Parameter: w1 Type: real
90
Parameter: fDoor Type: real Type: real
Parameter: fanpower Type: real Description: "power of fan" pumppower Type: real Description: "power of pump"
rameter: cop Type: real
ype: real Parameter: gh7 Type: real
sk1
/3600*0.08
ppower))+fmotor+fDoor
Parameters
Parameter: gDoor[17] Parameter: Parameter: tr12out Type: real Description: "R12 out temperature" Pa Parameter: g1 Type: real Parameter: g6 Type: real Parameter: gh1 Type: real Parameter: gh2 Type: real Parameter: gh3 Type: real Parameter: gh4 Type: real Parameter: gh5 Type: real Parameter: widthcon Type: real Parameter: gh6 T End Outputs Ta Calculations Task1 Calculation CalcObjective Parameters Nopendoor airDensity ObjF avgTO fDoor fanpower fmotor pumppower w1 TotalObj Statements w1 = 2*airDensity*30.12*Nopendoor*avgTO*24*365 TotalObj = w1*ObjF+(Nopendoor*avgTO*24*365/3600*0.08*(fanpower+pum End Statements End Calculation CalcObjective Calculation HCConstraints Ld Wd S ma tb W Ph L g1 g2 avgTO g3 Sw Ps g4 To Pd g5 airDensity g6 g7 Dr Tr Td Afr Pi Vd Ab Ti Po Statements Ti=300 To=270 Td=310 Pi=0 Po=5 Sw=0.5 Dr=1 Tr=3 Ab=15 airDensity = 1.169
91
avgTO = Tr+tb g1 = 0.1-Ph g2 = Ph+Wd+0.1-L g3 = Ld+0.2-W g4 = L*W - Ab g5 = S+0.2-Ld g6 = 0.1-Wd g7 = (Tr*Sw)-L+(2*Dr)
d*Wd)
widthcon mr12 ma tr12out cop gh5 gh1 gh6 gh2 gh7
op
h5=270-tr12out idthcon
End Calculations Task1
BIT
File: "Geo_0_1.jou.template" : "Geo_0_1.jou" scription Geo01jou HCGAMBIT
HCGAMBIT tion geo01msh
pe: fastparse ateFile: "geo_0_1.msh.1.template"
ile: "geo_0_1.msh" eo01msh
Vd = ma/airDensity/(L Pd = (W-Ld)/2 Ps = (W-S)/2 Afr = Wd*Ld End Statements End Calculation HCConstraints Calculation HEConstraints Parameters gh3 gh4 Statements gh1=-ma gh2=-mr12 gh3=1.5-c gh4=cop-5.5 g gh6=0.038-w gh7=widthcon-0.5 End Statements End Calculation HEConstraints SimCode HCGAM InputFiles HCGAMBIT FileDescription Geo01jou FileType: fastparse Template InputFile End FileDe End InputFiles OutputFiles FileDescrip FileTy Templ OutputF End FileDescription g
92
End OutputFiles HCGAMBIT
BIT /FLUENT6122/ntbin/ntx86/gambit.exe"
ient stored
ecs: geo01msh ram -inp geo_0_1.jou"
ss HCGAMBIT
End SimCode HCGAMBIT
ENT
ion FluentSlv1jou fastparse ile: "Fluent_Slv1.jou.template"
luent_Slv1.jou" tion FluentSlv1jou FLUENT
NT Batch
T6122/ntbin/ntx86/fluent.exe 3d -g -i Fluent_Slv1.jou del *.lok
nous
Prologue ntSlv1jou
ENT
SimCodeProcess HCGAM Program: "S:/CAEN ProcessType: trans Environment: unre ElapseTime: 10m Prologue WriteInputSpecs: Geo01jou Epilogue ReadOutputSp Execution: "$Prog End SimCodeProce SimCode HCFLU InputFiles HCFLUENT FileDescript FileType: TemplateF InputFile: "F End FileDescrip End InputFiles HC SimCodeProcess HCFLUE ScriptLanguage: DOS Script del result.dat S:/CAEN/FLUEN End Script ProcessType: asynchro Environment: unrestored ElapseTime: 16m 40s AsynchronousRun AsyncFile: "result.dat" AsyncDelay: 0s End AsynchronousRun WriteInputSpecs: Flue End SimCodeProcess HCFLU End SimCode HCFLUENT SimCode HCParser
93
OutputFiles HCParser FileDescription averagedat
dat" gedat
./Parser.exe"
AsynchronousRun at"
AsyncDelay: 0s Run
t Parser
InputFiles HeatExchanger
plateFile: "C:/temp/do/exchanger/condensor.m.template" temp/do/exchanger/condensor.m"
HeatExchanger
xchangerpfantxt
:/temp/do/exchanger/p_fan.txt.template" /exchanger/p_fan.txt"
Ctempdoexchangerpfantxt anger
Script ger
start matlab -nosplash -nodesktop -minimize -r condensor
FileType: fastparse TemplateFile: "average.dat.1.template" OutputFile: "average. End FileDescription avera End OutputFiles HCParser SimCodeProcess HCParser Program: " ProcessType: asynchronous Environment: unrestored ElapseTime: 5m AsyncFile: "average.d End Asynchronous Epilogue ReadOutputSpecs: averageda End SimCodeProcess HC End SimCode HCParser SimCode HeatExchanger FileDescription Ctempdoexchangercondensorm FileType: fastparse Tem InputFile: "C:/ End FileDescription Ctempdoexchangercondensorm End InputFiles OutputFiles HeatExchanger FileDescription Ctempdoe FileType: fastparse TemplateFile: "C OutputFile: "C:/temp/do End FileDescription End OutputFiles HeatExch SimCodeProcess HeatExchanger ScriptLanguage: DOSBatch cd c:\temp\do\exchan End Script
94
ProcessType: asynchronous
changer/p_fan.txt"
Prologue pdoexchangercondensorm
oexchangerpfantxt xchanger
nger
File: "C:/temp/do/motor/Excel_solver_cost_macro.xls"
ElapseTime: 1m
Visible: no
"Excel_solver_cost_macro.xls" Sheet: "Excel_Solver" Range:
ook: "Excel_solver_cost_macro.xls" Sheet: "Excel_Solver"
cel_solver_cost_macro.xls" Sheet: "Excel_Solver" Range: 6"
ok: "Excel_solver_cost_macro.xls" Sheet: "Excel_Solver"
] WorkBook: "Excel_solver_cost_macro.xls" Sheet:
_cost_macro.xls" Sheet:
End Excel MotorSystem
ks /door/Door_cost_Excel_solver_macro.xls"
ks
Environment: unrestored ElapseTime: 5m AsynchronousRun AsyncFile: "C:/temp/do/ex AsyncDelay: 0s End AsynchronousRun WriteInputSpecs: Ctem Epilogue ReadOutputSpecs: Ctempd End SimCodeProcess HeatE End SimCode HeatExcha Excel MotorSystem WorkBooks WorkBook MacroCalls: opt1 End WorkBooks Persistent: yes InputMapping Parameter: S WorkBook:"D16" Parameter: ML WorkBRange: "D17" End InputMapping OutputMapping Parameter: tb WorkBook: "Ex"D Parameter: fmotor WorkBoRange: "J4" Parameter: gmotor[0:14"Excel_Solver" Range: "J10:J24" Parameter: Nopendoor WorkBook: "Excel_solver"Excel_Solver" Range: "D27" End OutputMapping Excel DoorSystem WorkBoo WorkBookFile: "C:/temp/do MacroCalls: door End WorkBoo
95
ElapseTime: 1m Persistent: yes Visible: no InputMapping Parameter: S WorkBook: "Door_cost_Excel_solver_macro.xls" Sheet: "solver" Range:
ing r_macro.xls" Sheet: "solver" Range:
solver_macro.xls" Sheet: "solver"
Parameter: gDoor[0:16] WorkBook: "Door_cost_Excel_solver_macro.xls" Sheet:
End OutputMapping m
ts IT
HeatExchanger
jective
sk1
ization Task1
d Ld W L ma mr12 S fanfactor
tor: 1000.0 leFactor: 1000.0
actor: 1000.0 Parameter: W ScaleFactor: 1000.0
caleFactor: 1000.0 ma ScaleFactor: 1000.0
Factor: 1.0 actor ScaleFactor: 1.0
"B7" End InputMapping OutputMapp Parameter: ML WorkBook: "Door_cost_Excel_solve"F3" Parameter: fDoor WorkBook: "Door_cost_Excel_Range: "J31" "solver" Range: "F10:F26" End Excel DoorSyste TaskProcess Task1 Control: [ DoorSystem MotorSystem HCConstrain HCGAMB HCFLUENT HCParser HEConstraints CalcOb ] End TaskProcess Ta Optim PotentialVariables: Ph W Variables: ma mr12 L W Ld Ph Wd VariableScaling Parameter: Ph ScaleFac Parameter: Wd Sca Parameter: Ld ScaleF Parameter: L S Parameter: Parameter: mr12 ScaleFactor: 1000.0 Parameter: S Scale Parameter: fanf
96
InputConstraints Parameter: Ph LowerBound: 0.1 UpperBound: 10.0
: Wd LowerBound: 0.1 UpperBound: 3.0 d LowerBound: 0.1 UpperBound: 3.0
Parameter: L LowerBound: 0.1 UpperBound: 10.0 werBound: 0.001
tb ML Afr Vd Pd Ps Ti To Td Pi
otalObj fmotor gmotor airDensity Nopendoor avgTO w1 fDoor gDoor
7 Ph Wd Ld W L ma
Parameter: TotalObj Direction: minimize Weight: 1.0 ScaleFactor: 1.0
er: tb LowerBound: 0.0001 Weight: 1.0 ScaleFactor: 1.0 LowerBound: 0.1 Weight: 1.0 ScaleFactor: 1.0
owerBound: 0.1 Weight: 1.0 ScaleFactor: 1.0 pperBound: 0.0 Weight: 1.0 ScaleFactor: 1.0
UpperBound: 0.0 Weight: 1.0 ScaleFactor: 1.0 UpperBound: 0.0 Weight: 1.0 ScaleFactor: 1.0
g5 UpperBound: 0.0 Weight: 1.0 ScaleFactor: 1.0 perBound: 0.0 Weight: 1.0 ScaleFactor: 1.0
ut LowerBound: 270.0 Weight: 1.0 ScaleFactor: 1.0 LowerBound: 1.5 Weight: 1.0 ScaleFactor: 1.0 UpperBound: 5.5 Weight:
leFactor: 1.0 LowerBound: 0.038 Weight: 1.0 ScaleFactor: 1.0 UpperBound: 0.5
eight: 1.0 ScaleFactor: 1.0
L
ng: yes
Sequential Quadratic Programming - NLPQL"
Parameter Parameter: L Parameter: W LowerBound: 0.1 UpperBound: 10.0 Parameter: ma Lo Parameter: mr12 LowerBound: 0.001 PotentialObjectives: Po Sw Tr Dr Ab g2 g3 g4 g5 g7 ObjF T fanpower pumppower tr12out cop g1 g6 gh1 gh2 gh3 gh4 gh5 widthcon gh6 gh mr12 S fanfactor Objectives OutputConstraints Paramet Parameter: Pd Parameter: Ps L Parameter: g2 U Parameter: g3 Parameter: g4 Parameter: Parameter: g7 Up Parameter: tr12o Parameter: cop 1.0 Sca Parameter: widthconW OptimizePlan SQLN DefaultUpperBound: 1.0E15 UseScali OptimizeStep Step1 Technique: " Prologue RestoreBestSolution: no RerunTask: no Epilogue RestoreBestSolution: yes RerunTask: no Options MaxIterations: 4000 Control: [
97
Step1 ] End Optimization Task1 TaskPlan Task1 StopTaskPlanOnError: no Control: [ SQLNL ] End TaskPlan Task1 DataStorage Task1 Restore: no DataLog: "Task1.db" Mode: overwrite DataLookUp: "Task1.db"
: Exact
----------------
MatchMode Levels: all StoreGradRuns: yes StoreApproxRuns: yes End DataStorage Task1 End Task Task1 ------------------------------------------------------------------------------------
98
A.3 GAMBIT journal file for the heat curtain
------------------------------------------------------------------ Journal File for GAMBIT 2.1.6
ite Thu Mar 24 23:23:18 2005. saveprevious
ordinates 0 0 0 rtex.1" multiple 1 offset 6.90000000000003 0 0
cmove "vertex.2" multiple 1 offset 3 0 0 multiple 1 offset 10 0 0
ertex cmove "vertex.1" "vertex.2" "vertex.3" "vertex.4" multiple 1 offset 0 \
rtex.1" "vertex.2" "vertex.3" "vertex.4" multiple 1 offset 0 \
x.9" "vertex.10" "vertex.11" multiple 1 offset \
rtex.8" "vertex.5" "vertex.6" "vertex.7" multiple 1 offset 0 \
tex.2" "vertex.3" "vertex.4" multiple 1 offset 0 \
ertex cmove "vertex.1" "vertex.2" "vertex.3" "vertex.4" "vertex.5" \ x.7" "vertex.8" "vertex.9" "vertex.10" "vertex.11" \
"vertex.18" "vertex.19" "vertex.20" "vertex.21" "vertex.22" "vertex.23" \ "vertex.24" multiple 1 offset 0 0 2 edge create straight "vertex.1" "vertex.2" "vertex.3" "vertex.4" edge create straight "vertex.5" "vertex.6" "vertex.7" "vertex.8" edge create straight "vertex.9" "vertex.10" "vertex.11" "vertex.12" edge create straight "vertex.14" "vertex.15" "vertex.16" "vertex.13" edge create straight "vertex.18" "vertex.19" "vertex.20" "vertex.17" edge create straight "vertex.21" "vertex.22" "vertex.23" "vertex.24" edge create straight "vertex.1" "vertex.5" "vertex.9" "vertex.14" "vertex.18" \ "vertex.21" edge create straight "vertex.2" "vertex.6" "vertex.10" "vertex.15" \ "vertex.19" "vertex.22" edge create straight "vertex.3" "vertex.7" "vertex.11" "vertex.16" \ "vertex.20" "vertex.23" edge create straight "vertex.4" "vertex.8" "vertex.12" "vertex.13" \ "vertex.17" "vertex.24" edge create straight "vertex.25" "vertex.26" "vertex.27" "vertex.28" edge create straight "vertex.29" "vertex.30" "vertex.31" "vertex.32" edge create straight "vertex.33" "vertex.34" "vertex.35" "vertex.36" edge create straight "vertex.38" "vertex.39" "vertex.40" "vertex.37" edge create straight "vertex.42" "vertex.43" "vertex.44" "vertex.41" edge create straight "vertex.45" "vertex.46" "vertex.47" "vertex.48" edge create straight "vertex.25" "vertex.29" "vertex.33" "vertex.38" \
----------------------------------/ / File opened for wridentifier name "Geometry" newvertex create covertex cmove "vevertex vertex cmove "vertex.1"v 0.1 0 vertex cmove "ve 0.2 0 vertex cmove "vertex.12" "verte 0 1 0 vertex cmove "ve 1.2 0 vertex cmove "vertex.1" "ver 1.4 0 v "vertex.6" "verte "vertex.12" "vertex.13" "vertex.14" "vertex.15" "vertex.16" "vertex.17" \
99
"vertex.42" "vertex.45" edge create straight "vertex.26" "vertex.30" "vertex.34" "vertex.39" \
ertex.31" "vertex.35" "vertex.40" \
rtex.36" "vertex.37" \
ate straight "vertex.5" "vertex.29"
ate straight "vertex.7" "vertex.31"
ate straight "vertex.9" "vertex.33"
ate straight "vertex.11" "vertex.35"
ate straight "vertex.14" "vertex.38"
ex.43"
reframe "edge.24" "edge.62" "edge.78" "edge.82" real l
ge.34" "edge.72" "edge.80" "edge.84" real l
ge.25" "edge.63" "edge.82" "edge.86" real l
ge.35" "edge.73" "edge.84" "edge.88" real
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100
face create wireframe "edge.36" "edge.74" "edge.88" "edge.92" real
ge.28" "edge.66" "edge.94" "edge.98" real
ge.38" "edge.76" "edge.96" "edge.100" real
ge.2" "edge.40" "edge.78" "edge.79" real dge.79" "edge.80" real dge.81" "edge.82" real dge.82" "edge.83" real dge.83" "edge.84" real dge.85" "edge.86" real dge.86" "edge.87" real dge.87" "edge.88" real
"edge.89" "edge.90" real "edge.90" "edge.91" real
dge.91" "edge.92" real dge.93" "edge.94" real dge.94" "edge.95" real dge.95" "edge.96" real dge.97" "edge.98" real dge.98" "edge.99" real dge.99" "edge.100" real dge.61" "edge.66" real dge.60" "edge.65" real dge.59" "edge.64" real dge.58" "edge.63" real dge.57" "edge.62" real dge.66" "edge.71" real dge.65" "edge.70" real dge.64" "edge.69" real
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101
face create wireframe "edge.5" "edge.8" "edge.25" "edge.30" real face create wireframe "edge.2" "edge.5" "edge.24" "edge.29" real face create wireframe "edge.15" "edge.18" "edge.33" "edge.38" real face create wireframe "edge.12" "edge.15" "edge.32" "edge.37" real face create wireframe "edge.9" "edge.12" "edge.31" "edge.36" real face create wireframe "edge.6" "edge.9" "edge.30" "edge.35" real face create wireframe "edge.3" "edge.6" "edge.29" "edge.34" real volume create stitch "face.1" "face.2" "face.21" "face.24" "face.43" \
\
\
\
\
\
\
\
\
\
\
\
\
edge.95" \ e.88" \
ge.81" \ dge.74" \ dge.67" \
e.60" \ e.53" \
ge.46" \
"face.58" real volume create stitch "face.2" "face.3" "face.22" "face.25" "face.48" "face.63" real volume create stitch "face.3" "face.4" "face.23" "face.26" "face.53" "face.68" real volume create stitch "face.5" "face.6" "face.24" "face.27" "face.42" "face.57" real volume create stitch "face.6" "face.7" "face.25" "face.28" "face.47" \ "face.62" real volume create stitch "face.7" "face.8" "face.26" "face.29" "face.52" \ "face.67" real volume create stitch "face.13" "face.14" "face.27" "face.30" "face.41" "face.56" real volume create stitch "face.14" "face.15" "face.28" "face.31" "face.46" "face.61" real volume create stitch "face.15" "face.16" "face.29" "face.32" "face.51" "face.66" real volume create stitch "face.9" "face.10" "face.30" "face.33" "face.40" "face.55" real volume create stitch "face.10" "face.11" "face.31" "face.34" "face.45" "face.60" real volume create stitch "face.11" "face.12" "face.32" "face.35" "face.50" "face.65" real volume create stitch "face.17" "face.18" "face.33" "face.36" "face.39" "face.54" real volume create stitch "face.18" "face.19" "face.34" "face.37" "face.44" "face.59" real volume create stitch "face.19" "face.20" "face.35" "face.38" "face.49" "face.64" real undo begingroup edge picklink "edge.100" "edge.99" "edge.98" "edge.97" "edge.96" " "edge.94" "edge.93" "edge.92" "edge.91" "edge.90" "edge.89" "edg "edge.87" "edge.86" "edge.85" "edge.84" "edge.83" "edge.82" "ed "edge.80" "edge.79" "edge.78" "edge.77" "edge.76" "edge.75" "e "edge.73" "edge.72" "edge.71" "edge.70" "edge.69" "edge.68" "e "edge.66" "edge.65" "edge.64" "edge.63" "edge.62" "edge.61" "edg "edge.59" "edge.58" "edge.57" "edge.56" "edge.55" "edge.54" "edg "edge.52" "edge.51" "edge.50" "edge.49" "edge.48" "edge.47" "ed
102
"edge.45" "edge.44" "edge.43" "edge.42" "edge.41" "edge.40" "edge.39" \ dge.32" \
e.25" \ e.18" \
ge.11" \ "edge.3" \
ge.7" \ .9" "edge.10" "edge.11" "edge.12" "edge.13" "edge.14" \
.21" \ e.23" "edge.24" "edge.25" "edge.26" "edge.27" "edge.28" \
.35" \ e.37" "edge.38" "edge.39" "edge.40" "edge.41" "edge.42" \
.49" \ e.51" "edge.52" "edge.53" "edge.54" "edge.55" "edge.56" \
.63" \ e.65" "edge.66" "edge.67" "edge.68" "edge.69" "edge.70" \
.77" \ e.79" "edge.80" "edge.81" "edge.82" "edge.83" "edge.84" \
" \ e.93" "edge.94" "edge.95" "edge.96" "edge.97" "edge.98" \
" "face.10" "face.11" "face.12" "face.13" "face.14" \
\ .23" "face.24" "face.25" "face.26" "face.27" "face.28" \
.37" "face.38" "face.39" "face.40" "face.41" "face.42" \
.51" "face.52" "face.53" "face.54" "face.55" "face.56" \
.65" "face.66" "face.67" "face.68" map size 1 " "volume.6" \
lume.8" "volume.9" "volume.10" "volume.11" "volume.12" \
UENT 5/6" visible mesh
.16"
4" \
sh"
"edge.38" "edge.37" "edge.36" "edge.35" "edge.34" "edge.33" "e "edge.31" "edge.30" "edge.29" "edge.28" "edge.27" "edge.26" "edg "edge.24" "edge.23" "edge.22" "edge.21" "edge.20" "edge.19" "edg "edge.17" "edge.16" "edge.15" "edge.14" "edge.13" "edge.12" "ed "edge.10" "edge.9" "edge.8" "edge.7" "edge.6" "edge.5" "edge.4" "edge.2" "edge.1" edge mesh "edge.1" "edge.2" "edge.3" "edge.4" "edge.5" "edge.6" "ed "edge.8" "edge "edge.15" "edge.16" "edge.17" "edge.18" "edge.19" "edge.20" "edge "edge.22" "edg "edge.29" "edge.30" "edge.31" "edge.32" "edge.33" "edge.34" "edge "edge.36" "edg "edge.43" "edge.44" "edge.45" "edge.46" "edge.47" "edge.48" "edge "edge.50" "edg "edge.57" "edge.58" "edge.59" "edge.60" "edge.61" "edge.62" "edge "edge.64" "edg "edge.71" "edge.72" "edge.73" "edge.74" "edge.75" "edge.76" "edge "edge.78" "edg "edge.85" "edge.86" "edge.87" "edge.88" "edge.89" "edge.90" "edge.91 "edge.92" "edg "edge.99" "edge.100" successive ratio1 1 size 0.1 undo endgroup face mesh "face.1" "face.2" "face.3" "face.4" "face.5" "face.6" "face.7" \ "face.8" "face.9 "face.15" "face.16" "face.17" "face.18" "face.19" "face.20" "face.21" "face.22" "face "face.29" "face.30" "face.31" "face.32" "face.33" "face.34" "face.35" \ "face.36" "face "face.43" "face.44" "face.45" "face.46" "face.47" "face.48" "face.49" \ "face.50" "face "face.57" "face.58" "face.59" "face.60" "face.61" "face.62" "face.63" \ "face.64" "facevolume mesh "volume.1" "volume.2" "volume.3" "volume.4" "volume.5 "volume.7" "vo "volume.13" "volume.14" "volume.15" map size 1 solver select "FLwindow modify inphysics create "pressure_inlet.1" btype "PRESSURE_INLET" face "face.13" physics create "pressure_outlet.2" btype "PRESSURE_OUTLET" face "facephysics create "velocity_inlet.3" btype "VELOCITY_INLET" face "face.46"physics create "fluid.4" ctype "FLUID" volume "volume.1" "volume.2" \ "volume.3" "volume.4" "volume.5" "volume.6" "volume.7" "volume.8" \ "volume.9" "volume.10" "volume.11" "volume.12" "volume.13" "volume.1 "volume.15" export fluent5 "n:\Private\Class\ME555\Project\Fluent\geometry\geo_0_1.m
103
/ File closed at Thu Mar 24 23:24:10 2005, 20.13 cpu second(s), 45576456 maximum memory.
te-item "MenuBar*ReadSubMenu*Case...")
")
Cancel)")
ton1(Energy
*ToggleBox1*CheckButton1(Energy
" #f) ggleBox1(Model)*k-
(OK)")
hButton2(Cancel)")
Operating Conditions...") n "Operating
vity)*Table2(Gravity)*Frame1*ToggleBox1*CheckButton1(Gr
CheckButton1(Gr
Table2(Gravitational Acceleration)*RealEntry3(Z)" '( -9.81)) K)")
A.4 FLUENT journal file for the heat curtain (cx-gui-do cx-activa(cx-gui-do cx-set-text-entry "Select File*Text" "N:\Private\Class\ME555\Project\Fluent\geometry\geo_0_1.msh") (cx-gui-do cx-activate-item "Select File*OK") (cx-gui-do cx-activate-item "MenuBar*GridMenu*Scale...") (cx-gui-do cx-activate-item "Scale Grid*PanelButtons*PushButton2(Cancel)(cx-gui-do cx-activate-item "MenuBar*ModelsSubMenu*Solver...") (cx-gui-do cx-activate-item "Solver*PanelButtons*PushButton1(OK)") (cx-gui-do cx-activate-item "MenuBar*ModelsSubMenu*Multiphase...") (cx-gui-do cx-activate-item "Multiphase Model*PanelButtons*PushButton2((cx-gui-do cx-activate-item "MenuBar*ModelsSubMenu*Energy...") (cx-gui-do cx-set-toggle-button "Energy*Frame1(Energy)*Table1(Energy)*Frame1*ToggleBox1*CheckButEquation)" #f) (cx-gui-do cx-activate-item "Energy*Frame1(Energy)*Table1(Energy)*Frame1Equation)") (cx-gui-do cx-activate-item "Energy*PanelButtons*PushButton1(OK)") (cx-gui-do cx-activate-item "MenuBar*ModelsSubMenu*Viscous...")(cx-gui-do cx-set-toggle-button "Viscous Model*Table1*Frame1(Model)*ToggleBox1(Model)*k-epsilon (2 eqn)(cx-gui-do cx-activate-item "Viscous Model*Table1*Frame1(Model)*Toepsilon (2 eqn)") (cx-gui-do cx-set-position "Viscous Model" '(x 365 y 426)) (cx-gui-do cx-activate-item "Viscous Model*PanelButtons*PushButton1(cx-gui-do cx-activate-item "MenuBar*ModelsSubMenu*Species...") (cx-gui-do cx-activate-item "Species Model*PanelButtons*Pus(cx-gui-do cx-activate-item "MenuBar*DefineMenu*Materials...") (cx-gui-do cx-activate-item "Materials*PanelButtons*PushButton1(Close)") (cx-gui-do cx-activate-item "MenuBar*DefineMenu*(cx-gui-do cx-set-toggle-buttoConditions*Table1*Frame2(Graavity)" #f) (cx-gui-do cx-activate-item "Operating Conditions*Table1*Frame2(Gravity)*Table2(Gravity)*Frame1*ToggleBox1*avity)") (cx-gui-do cx-set-real-entry-list "Operating Conditions*Table1*Frame2(Gravity)*Table2(Gravity)*Frame2(Gravitational Acceleration)*(cx-gui-do cx-activate-item "Operating Conditions*PanelButtons*PushButton1(O
104
(cx-gui-do cx-activate-item "MenuBar*DefineMenu*Boundary Conditions...") (cx-gui-do cx-set-list-selections "Boundary Conditions*Table1*Frame1*List1(Zone)" '( 2))
x-gui-do cx-activate-item "Boundary Conditions*Table1*Frame1*List1(Zone)") x-gui-do cx-activate-item "Boundary Conditions*PanelButtons*PushButton1(OK)")
1*Table3*RealEntry5(Total Temperature)" '( 70))
ntry1(Gauge Total Pressure)"
Button1(OK)") ditions*Table1*Frame1*List1(Zone)" '( 3))
ame1*List1(Zone)") OK)")
ry4(Backflow Total
auge Pressure)" '( 0))
1*List1(Zone)" '( 4)) ndary Conditions*Table1*Frame1*List1(Zone)")
t-real-entry-list "velocity-inlet-4-1*Table3*RealEntry3(Velocity Magnitude)" '(
310)) -activate-item "velocity-inlet-4-1*PanelButtons*PushButton1(OK)")
tton2(Cancel)")
trols*PanelButtons*PushButton2(Cancel)")
7(Temperature)" '( 290)) ttons*PushButton2(Cancel)")
Values)*Table3(Initial
ushButton1(OK)") 2(Cancel)")
x-activate-item "Residual gleBox1(Options)*CheckButton1(Print)")
o cx-activate-item "MenuBar*InitializeSubMenu*Initialize...") itialization*Frame3(Initial Values)*Table3(Initial
1(OK)")
(c(c(cx-gui-do cx-set-real-entry-list "pressure-inlet-6-2(cx-gui-do cx-set-real-entry-list "pressure-inlet-6-1*Table3*RealE'( 5)) (cx-gui-do cx-activate-item "pressure-inlet-6-1*PanelButtons*Push(cx-gui-do cx-set-list-selections "Boundary Con(cx-gui-do cx-activate-item "Boundary Conditions*Table1*Fr(cx-gui-do cx-activate-item "Boundary Conditions*PanelButtons*PushButton1((cx-gui-do cx-set-real-entry-list "pressure-outlet-5-1*Table3*RealEntTemperature)" '( 300)) (cx-gui-do cx-set-real-entry-list "pressure-outlet-5-1*Table3*RealEntry1(G(cx-gui-do cx-activate-item "pressure-outlet-5-1*PanelButtons*PushButton1(OK)") (cx-gui-do cx-set-list-selections "Boundary Conditions*Table1*Frame(cx-gui-do cx-activate-item "Bou(cx-gui-do cx-activate-item "Boundary Conditions*PanelButtons*PushButton1(OK)") (cx-gui-do cx-se0.00277777777777778)) (cx-gui-do cx-set-real-entry-list "velocity-inlet-4-1*Table3*RealEntry25(Temperature)" '( (cx-gui-do cx(cx-gui-do cx-activate-item "Boundary Conditions*PanelButtons*PushBu(cx-gui-do cx-activate-item "MenuBar*ControlsSubMenu*Solution...")(cx-gui-do cx-activate-item "Solution Con(cx-gui-do cx-activate-item "MenuBar*InitializeSubMenu*Initialize...") (cx-gui-do cx-set-real-entry-list "Solution Initialization*Frame3(Initial Values)*Table3(Initial Values)*RealEntry(cx-gui-do cx-activate-item "Solution Initialization*PanelBu(cx-gui-do cx-activate-item "MenuBar*InitializeSubMenu*Initialize...") (cx-gui-do cx-set-real-entry-list "Solution Initialization*Frame3(InitialValues)*RealEntry7(Temperature)" '( 290)) (cx-gui-do cx-activate-item "Solution Initialization*PanelButtons*P(cx-gui-do cx-activate-item "Solution Initialization*PanelButtons*PushButton(cx-gui-do cx-activate-item "MenuBar*MonitorsSubMenu*Residual...") (cx-gui-do cx-set-toggle-button "Residual Monitors*Table1*Frame1(Options)*ToggleBox1(Options)*CheckButton1(Print)" #t) (cx-gui-do cMonitors*Table1*Frame1(Options)*Tog(cx-gui-do cx-activate-item "Residual Monitors*PanelButtons*PushButton1(OK)") (cx-gui-d(cx-gui-do cx-set-real-entry-list "Solution InValues)*RealEntry2(X Velocity)" '( 1.3)) (cx-gui-do cx-activate-item "Solution Initialization*PanelButtons*PushButton(cx-gui-do cx-activate-item "Warning*OK")
105
(cx-gui-do cx-activate-item "Solution Initialization*PanelButtons*PushButton2(Cancel)")
sure)" '(
i-do cx-activate-item "Solution
i-do cx-activate-item "Solution Controls*PanelButtons*PushButton1(OK)")
ations)" 20)
lections "Write Profile*Frame2*List2(Surfaces)" '( 1))
"Write Profile*PanelButtons*PushButton2(Cancel)")
(cx-gui-do cx-activate-item "MenuBar*ControlsSubMenu*Solution...") (cx-gui-do cx-set-list-selections "Solution Controls*Table2*Frame1(Discretization)*Table1(Discretization)*DropDownList1(Pres4)) (cx-guControls*Table2*Frame1(Discretization)*Table1(Discretization)*DropDownList1(Pressure)") (cx-gu(cx-gui-do cx-activate-item "MenuBar*SolveMenu*Iterate...") (cx-gui-do cx-set-integer-entry "Iterate*Table1*Frame2(Iteration)*Table2(Iteration)*IntegerEntry1(Number of Iter(cx-gui-do cx-activate-item "Iterate*PanelButtons*PushButton1(OK)") (cx-gui-do cx-activate-item "MenuBar*WriteSubMenu*Profile...") (cx-gui-do cx-set-list-se(cx-gui-do cx-activate-item "Write Profile*Frame2*List2(Surfaces)") (cx-gui-do cx-set-list-selections "Write Profile*Frame3*List3(Values)" '( 9)) (cx-gui-do cx-activate-item "Write Profile*Frame3*List3(Values)") (cx-gui-do cx-activate-item "Write Profile*PanelButtons*PushButton1(OK)") (cx-gui-do cx-set-text-entry "Select File*Text" "result.dat") (cx-gui-do cx-activate-item "Select File*OK") (cx-gui-do cx-activate-item(cx-gui-do cx-activate-item "MenuBar*FileMenu*Exit") (cx-gui-do cx-activate-item "Warning*OK") ----------------------------------------------------------------------------------------------------
106
A.5 Monotoncity Analysis of the motor system
The following table shows the parameter list used for the monotonicity analysis.
The derivatives of objective function with respect to each variable are as follows.
dentafta __∂numtaff __
=∂
f_ta_num n Vm 0.002350 W1 n3 ta4 Vm 0.000800 W1 n ta5 P2 + (− = 0.4650 W2 P3 tb3 ta 0.4650 W2 P3 tb2 ta2 0.1550 W2 P3 tb ta3 + − + 0.03100 W2 P tb4 n2 0.1550 W2 P3 tb4 0.0117500 W1 n ta4 Vm P2 − − + 0.001000 W1 n ta2 Vm P2 tb3 0.00100 W1 n ta3 Vm P2 tb2 − + 0.0352500 W1 n ta2 Vm P2 tb2 0.035250 W1 n ta3 Vm P2 tb + − 0.00705000 W1 n3 ta2 Vm tb2 0.0070500 W1 n3 ta3 Vm tb + − 0.01175 W1 n ta Vm P2 tb3 117.5 W1 n ta Vm Ls π Ds4 tb3 − − 352.500 W1 n ta2 Vm Ls π Ds4 tb2 0.002350 W1 n3 ta Vm tb3 + − 0.000800 W1 n ta2 P2 tb3 0.002400 W1 n ta3 P2 tb2 0.002400 W1 n ta4 P2 tb − + − 352.50 W1 n ta3 Vm Ls π Ds4 tb 117.500 W1 n ta4 Vm Ls π Ds − + 1550. W2 P tb4 Ls π Ds4 4650. W2 P tb3 Ls π Ds4 ta − + 4650. W2 P tb2 Ls π Ds4 ta2 1550. W2 P tb Ls π Ds4 ta3 − + + 0.09300 W2 P tb2 n2 ta2 0.03100 W2 P tb n2 ta3 − + )
4
0.09300 W2 P tb3 n2 ta
= f_ta_den n2 P2 ta2 ( )− + tb 1. ta 3 tb
System Parameter Value Unit RemarkS m Width of door
friction
ress
tion factor
ur
1ML 1 kg Mass of door
Doorµf L 0.01 - Coefficient of friction
eff icient BL 0.1 N-sec /m Damping co
.1 - Coefficient ofµ 0ρall 200 MPa Max allow able bending stress
τall 100 MPa Max allow able shear stScrewK 3 - Stress concentra
ρ 7850 kg/m3 Density of steel s2Motor Jm 0.005036 N-m-sec Motor inertia
vityetc g 9.81 m/sec2 acceleration of gra
Operation N 60 cycle/hour # of door operation cycle per ho
107
dentbftb __∂numtbff __
=∂
f_tb_num W1 Vm n2 ta 0.19687500 Vm n2 ta3 0.9843750 Vm P2 ta3 + (− = 0.001000 Vm P2 ta tb3 0.001000 Vm P2 ta2 tb2 2.9531250 Vm P2 tb2 ta + − + 2.9531250 Vm P2 tb ta2 0.59062500 Vm n2 tb2 ta 0.59062500 Vm n2 tb ta2 − + − 0.9843750 Vm P2 tb3 0.000800 P2 ta4 9843.750 Vm Ls π Ds4 tb3 − − − 29531.250 Vm Ls π Ds4 tb2 ta 0.19687500 Vm n2 tb3 0.000800 P2 ta tb3 + − + 0.002400 P2 ta2 tb2 0.002400 P2 ta3 tb 29531.250 Vm Ls π Ds4 tb ta2 − + − 9843.750 Vm Ls π Ds4 ta3 + )
= f_tb_den n2 P2 ( )− + tb 1. ta 3 tb2 ta
denVmfnumVmf
Vf
m ____
=∂
∂
f_Vm_num n 0.1550 W2 P3 tb3 0.016 W2 P3 tb3 ta 0.032 W2 P3 tb2 ta2 + − ( =
0.016 W2 P3 tb ta3 0.1356 W1 n ta2 Vm P2 tb2 0.068800 W1 n ta3 Vm P2 tb + − + 0.066800 W1 n ta Vm P2 tb3 0.000800 W1 n ta4 P2 0.39610000 W1 n3 ta3 Vm + − + 0.3100 W2 P3 tb2 ta 0.1550 W2 P3 tb ta2 0.03100 W2 P tb3 n2 − + + 0.101600 W1 n ta2 P2 tb2 0.052000 W1 n ta3 P2 tb 0.050400 W1 n ta P2 tb3 − + + 1.9805000 W1 n ta Vm P2 tb2 1.9805000 W1 n ta3 Vm P2 + + 0.396100 W1 n3 ta Vm tb2 19805.000 W1 n ta Vm Ls π Ds4 tb2 + + 19805.000 W1 n ta3 Vm Ls π Ds4 3.96100 W1 n ta2 P2 Vm tb + − 0.792200 W1 n3 ta2 tb Vm 39610.0 W1 n ta2 tb Vm Ls π Ds4 − − 1550. W2 P tb3 Ls π Ds4 3100. W2 P tb2 Ls π Ds4 ta + − 1550. W2 P tb Ls π Ds4 ta2 0.06200 W2 P tb2 n2 ta 0.03100 W2 P tb n2 ta2 + − + )
= f_Vm_den n2 P2 ta ( )− + tb 1. ta 2 tb
dennfnumnf
nf
____
=∂∂
108
f_n_num n 0.39375000 W1 Vm2 n3 ta tb2 0.39375000 W1 Vm2 n3 ta3− − (− = 0.787500 W1 Vm2 n3 ta2 tb 0.1550 W2 P3 tb3 Vm 0.3100 W2 P3 tb2 Vm ta + + − 0.1550 W2 P3 tb Vm ta2 1488. W2 P tb3 Vm Ls π Ds4 + + 2976. W2 P tb2 Vm Ls π Ds4 ta 1488. W2 P tb Vm Ls π Ds4 ta2 − + 0.03100 W2 P tb3 Vm n2 0.06200 W2 P tb2 Vm n2 ta − + 0.03100 W2 P tb Vm n2 ta2 0.016 W2 P3 tb3 Vm ta 0.032 W2 P3 tb2 Vm ta2 − + − 0.016 W2 P3 tb Vm ta3 0.016 W2 P3 tb3 ta 0.032 W2 P3 tb2 ta2 + + − 0.016 W2 P3 tb ta3 + )
= f_n_den n3 P2 ta ( )− + tb 1. ta 2 tb
denPfnumPf
Pf
____
=∂∂
f_P_num n 0.0312500 W1 Vm2 n ta P2 tb2 0.0312500 W1 Vm2 n ta3 P2− − (− = 0.39375000 W1 Vm2 n3 ta tb2 0.39375000 W1 Vm2 n3 ta3 + + 19687.500 W1 Vm2 n ta Ls π Ds4 tb2 19687.500 W1 Vm2 n ta3 Ls π Ds4 + + 0.06250 W1 Vm2 n ta2 P2 tb 0.787500 W1 Vm2 n3 ta2 tb + − 39375.0 W1 Vm2 n ta2 Ls π Ds4 tb 0.1650 W2 P3 tb3 Vm − − 0.3300 W2 P3 tb2 Vm ta 0.1650 W2 P3 tb Vm ta2 1550. W2 P tb3 Vm Ls π Ds4 + − + 3100. W2 P tb2 Vm Ls π Ds4 ta 1550. W2 P tb Vm Ls π Ds4 ta2 − + 0.03100 W2 P tb3 Vm n2 0.06200 W2 P tb2 Vm n2 ta + − 0.03100 W2 P tb Vm n2 ta2 0.016 W2 P3 tb3 Vm ta 0.032 W2 P3 tb2 Vm ta2 + − + 0.016 W2 P3 tb Vm ta3 0.016 W2 P3 tb3 ta 0.032 W2 P3 tb2 ta2 − − + 0.016 W2 P3 tb ta3 − )
= f_P_den n2 P3 ta ( )− + tb 1. ta 2 tb
denLsfnumLsf
Lf
s ____
=∂∂
f_Ls_num π Ds4 Vm 10000. W1 n ta Vm tb2 20000. W1 n ta2 tb Vm − ( = 10000. W1 n ta3 Vm 1500. W2 P tb3 3000. W2 P tb2 ta 1500. W2 P tb ta2 + + − + )
= f_Ls_den P2 ( )− + tb 1. ta 2 tb n ta
109
denDsfnumDsf
Df
s ____
=∂∂
f_Ds_num Ls π Ds3 Vm 40000. W1 n ta Vm tb2 80000. W1 n ta2 tb Vm − ( = 40000. W1 n ta3 Vm 6200. W2 P tb3 12400. W2 P tb2 ta 6200. W2 P tb ta2 + + − + )
= f_Ds_den P2 ( )− + tb 1. ta 2 tb n ta
The constraint equations are
:= g1 − 156.9600000 Ls
Ds3 π200000000
g2 48 n2 ⎛
⎝⎜⎜⎜
⎞
⎠⎟⎟⎟ + +
P2
4 π2 n23925 Ls π Ds4
16 n2 0.005036 n π Vm
P ta0.05000000000 P Vm
π n +
⎛
⎝
⎜⎜⎜⎜⎜ :=
0.04905000000 Pπ n +
⎞
⎠
⎟⎟⎟⎟⎟ Ds3 π( ) 100000000 −
:= g3 −
Ds ⎛
⎝
⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟
+ 0.1 π Ds P
+ 1 P2
π2 Ds2
2 ⎛
⎝
⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟
− π Ds
+ 1P2
2 Ds2
0.1 P0.2 n
π
:= g4 − 1 Vm tb2
:= g5 − Vm tb
2 1.05
:= g6 − ta
Vm 1
110
:= g7 − 1 Ls := g8 − ta 2 tb := g9 − P 0.2
Then, the derivatives with respect to variables ar ollows.
G1:
e as f
:= g1_ta 0 := g1_tb 0
:= g1_Vm 0 := g1_n 0 := g1_P 0
:= g1_Ls156.9600000
Ds3 π
:= g1_Ds −470.8800000 Ls
Ds4 π G2:
:= g2_ta −96 n2 ⎛
⎝⎜⎜⎜
⎞
⎠⎟⎟⎟ + +
P2
4 π2 n23925 Ls π Ds4
16 n2 0.005036 π Vm
P ta2 Ds3 π
:= g2_tb 0
:= g2_Vm48 n
⎛
⎝
⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟ + 2 ⎛
⎝⎜⎜⎜
⎞
⎠⎟⎟⎟ + +
P2
4 π2 n23925 Ls π Ds4
16 n2 n π
P ta
0.0050360.05000000000 P
π nDs3 π
g2_n 482 ⎛
⎝⎜⎜⎜
⎞
⎠⎟⎟⎟ + +
P2
4 π2 n23925 Ls π Ds4
16 n2 0.005036 n π Vm
P ta0.05000000000 P Vm
π n +
⎛
⎝
⎜⎜⎜⎜⎜ :=
0.04905000000 Pπ n +
⎞
⎠
⎟⎟⎟⎟⎟ Ds3 π( ) 48 n2 ⎛
⎝⎜⎜⎜
⎞
⎠⎟⎟⎟− −
P2
2 π2 n33925 Ls π Ds4
8 n3 n π Vm
P ta
⎛
⎝
⎜⎜⎜⎜⎜⎜ +
111
2 ⎛
⎝⎜⎜⎜
⎞
⎠⎟⎟⎟ + +
P2
4 π2 n23925 Ls π Ds4
16 n2 0.005036 π Vm
P ta0.05000000000 P Vm
π n2 + −
0.04905000000 Pπ n2 −
⎞
⎠
⎟⎟⎟⎟⎟⎟ Ds3 π( )
2 ⎛
⎝⎜⎜⎜
⎞
⎠⎟⎟⎟ + +
P2
4 π2 n23925 Ls π Ds4
16 n2 0.005036 n π VmVmg2_P 48 n
π n ta P2 ta −
⎛
⎝
⎜⎜⎜⎜⎜⎜ :=
0.05000000000 Vmπ n
0.04905000000π n + +
⎞
⎠
⎟⎟⎟⎟⎟⎟ Ds3 π( )
:= g2_Ls 23550 Ds π VmP ta
g2_Ds 94200 Ls π VmP ta 144 n
2 ⎛
⎝⎜⎜⎜
⎞
⎠⎟⎟⎟ + +
P2
4 π2 n23925 Ls π Ds4
16 n2 0.005036 n π Vm
P ta
⎛
⎝
⎜⎜⎜⎜⎜ − :=
0.05000000000 P Vmπ n
0.04905000000 Pπ n + +
⎞
⎠
⎟⎟⎟⎟⎟ Ds4 π( )
G3: := g3_ta 0 := g3_tb 0
:= g3_Vm 0 := g3_n -0.2
112
g3_P
Ds ⎛
⎝
⎜⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟⎟
− 1
+ 1 P2
π2 Ds2
P2
⎛
⎝⎜⎜⎜
⎞
⎠⎟⎟⎟ + 1
P2
π2 Ds2
( )/3 2
π2 Ds2
2 ⎛
⎝
⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟
− π Ds
+ 1P2
π2 Ds2
0.1 P :=
Ds ⎛
⎝
⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟
+ 0.1 π Ds P
+ 1 P2
π2 Ds2
⎛
⎝
⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟
− − P
π Ds ⎛
⎝⎜⎜⎜
⎞
⎠⎟⎟⎟ + 1 P2
π2 Ds2
( )/3 2 0.1
2 ⎛
⎝
⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟
− π Ds
+ 1 P2
π2 Ds2
0.1 P2 −
:= g3_Ls 0
g3_Ds
+ 0.1 π Ds P
+ 1 P2
π2 Ds2
2 ⎛
⎝
⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟
− π Ds
+ 1P2
π2 Ds2
0.1 P
Ds ⎛
⎝
⎜⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟⎟
+ 0.1 πP3
⎛
⎝⎜⎜⎜
⎞
⎠⎟⎟⎟ + 1 P2
π2 Ds2
( )/3 2
π2 Ds3
2 ⎛
⎝
⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟
− π Ds
+ 1P2
π2 Ds2
0.1 P + :=
Ds ⎛
⎝
⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟
+ 0.1 π Ds P
+ 1 P2
π2 Ds2
⎛
⎝
⎜⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟⎟
+ π
+ 1 P2
π2 Ds2
P2
π Ds2 ⎛
⎝⎜⎜⎜
⎞
⎠⎟⎟⎟ + 1 P2
π2 Ds2
( )/3 2
2 ⎛
⎝
⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟
− π Ds
+ 1 P2
π2 Ds2
0.1 P2 −
G4: := g4_ta 0
:= g4_tb −Vm2
:= g4_Vm −tb2
:= g4_n 0
113
:= g4_P 0 := g4_Ls 0 := g4_Ds 0
G5: := g5_ta 0
:= g5_tbVm2
:= g5_Vm tb2
:= g5_n 0 := g5_P 0 := g5_Ls 0 := g5_Ds 0
G6:
:= g6_ta 1Vm
:= g6_tb 0
:= g6_Vm −ta
Vm2
:= g6_n 0 := g6_P 0 := g6_Ls 0 := g6_Ds 0
G7: := g7_ta 0 := g7_tb 0
:= g7_Vm 0 := g7_n 0 := g7_P 0
114
:= g7_Ls -1
a 0
:= g7_Ds 0
G8: := g8_ta 1 := g8_tb -2
:= g8_Vm 0 := g8_n 0 := g8_P 0 := g8_Ls 0 := g8_Ds 0
G9: g9_t :=
:= g9_tb 0 := g9_Vm 0
:= g9_n 0 := g9_P 1 := g9_Ls 0 := g9_Ds 0
115