Modelado Matematico De Un Cautin

Project Number 2

Group Number 5

Hugo Alfonso de Jesus Camacho Cervantes
MN: 06350811

FENG Yanwen
Bibl-Nr:86336488019
HAN Yiqi
MN:06207911

Objective:
The general objective of the project is to provide a simulation for the soldering iron presented in the system, using the data established and creating a model that can be run in Simulink. The model should be run in 2 differentmaterials, and we should be able to determine the output temperature at the end of the tip in the soldering iron.

The data varies from group to group, and there are general considerations to be made for the creation of the mathematical models. The model will be tested in two different varieties to detect mistakes and to be sure the answer is correctly represented. The Simulink model will be basedin the model based in "disks" or sections, while the other simulation will be represented in excel trough numerical methods and this one it's done trough node implementation. Both models are done using the finite differences method for analysis.
There is an small gap of 0.1 mm between the heating element and the exterior part of the material.
General Data provided:
Lsp=0.02 m
LHK=0.2 m∝L=260 W/m2K
∝SU=10 W/m2K
Pel=10 W
r1=0.005 m
r2=0.007 m
Data for copper:
λcu=399 W/mK
ccu=382 J/kgK
ρcu=8960 kg/m3
Data of Aluminum:
λal=220 W/mK
cal=896 J/kgK
ρal=2702 W/m3

TASK NUMBER 1
For the first task we have to make a model from the cylinder where the heating element rest, taking into account the considerations showed in the document showed before and the initial values specifiedfor this system.
Graphical representation:
This is the graphical representation provided, the system has a resistive element represented by the alfa between each section.
Then it is provided a resistive relation as follows:

This relationship is provided in as a sum of resistances in the system:
Rtotal= RL2RSU2RT
RL2=1αL(2π*r1*∆X)
RT=ln⁡(r2r1)2π(r2-r1)
RSU2=1αSU(2π*r2*∆X)

This willbe useful to make the model for both simulations and to understanding the behavior of the system.

The flow of energy trough the first part of the soldering iron is represented in the next manner:

From this we can make the energy balance to this part of the system, where the input is kPel, and the axial energy in the left side of the cut Pleit, with a constant area in the whole body for acylindrical body. The outputs are the radial eat released to the environment and the heat transferred to the right side of the disk. For the kind of system, the balance is equal to the heat in the central part, considered as the medium heat for the disk section.

Then, the thermal balance goes as showed:
PLeitk-10+Pelk-PLeitk-PHKk=PTHk

The system starts at 293 K and there will be a differencebetween the exterior temperature and the interior temperature. For that we need to calculate the heat differential for the center cut of the disc section. Following the thermodynamics conventions the heat released is:
q=Aλ∆Tdr
The area is the whole area of the cross section, therefore:
∆T=qAλdr
∆T=qλ2πlr1r21rdr=(Tinternal-Texternal)
∆T=qλ2πl(lnr1-lnr2=qλ2πl(ln(r2r1))
Therefore with theresistive sum we get
∆X=l
Tinternal-Texternal=q*12πλl(lnr2r1+RL2+RSU2
(Tinternal-Texternal)=q1πl*12λl(lnr2r1+1αL(2*r1)+1αSU(2*r2)

Therefore:
FIRST PART OF THE MODEL
q=PTH=πl(Tinternal-Texternal)12λl(lnr2r1+1αL(2*r1)+1αSU(2*r2)
For the input in the left side of the section, the area of the central element is not taken into account because the temperature there does not change, we know for thedifference of temperature will be between the left and central temperature:
Tk-1-Tk=PLeitk-1lAλ
SECOND PART OF THE MODEL
PLeitk-1=AλTk-1-Tkl
For the middle flow of energy we know:
Q=cm∆T
THIRD PART OF THE MODEL
PTHk=cVρTHKk

For the input, we already know it is constant along the axys, therefore:
FOURTH PART OF THE MODEL
Pelk=PelG*lLHK=Pel
And last, the right side of the disk...