PLAN DE MEJORA
Fase 1 del plan de acción
˙v = −b mv − k
mx + A mPa
˙x = v 7. Combine relations:
k1�
Ps(t) − Pa− k2(θv)√Pa=ρAv +ρV0
β ˙Pa
˙Pa= β ρV0
�k1�
Ps(t) − Pa− k2(θv)√Pa− ρAv�
8.2 Thermal Systems
We will model thermal systems using continuity principles, just as we have each of the other sys-tems we have studied. The main difference is that there is no analog to inertia or flow storage components in thermal systems. Instead, we consider just thermal resistance and thermal capaci-tance.
8.2.1 Thermal Resistance
Thermal resistances are characterized by a relationship between temperature and heat flux, the effort and flow variables for thermal systems. Note that, unlike each of the other types of systems, thermal resistances do not dissipate energy; they merely limit the flow of heat from one part of a system to another. There are three types of thermal resistances: conduction, convection, and radiation.
Conduction
Thermal conduction is characterized by heat flow through a material. The temperature difference is given by
∆T = Rqh
where R is the thermal resistance and qh is the heat flux. For a rectangular solid of length L, cross-sectional area A, and conductivityκ, the resistance is
R = L κA
Fluid and Thermal Systems 137
Convection
Convection occurs when heat is carried by a moving fluid. The heat flux from a solid to a fluid is often written
qh= hA∆T
where h is the convection coefficient and A is the area over which convective heat transfer occurs.
h is large for forced convection and for liquids, and small for free convection and for gases.
Radiation
Radiative heat transfer occurs when heated bodies radiate energy in the form of electromagnetic radiation. Unlike conduction and convection, it is a nonlinear phenomenon.
qh=β(T14− T24)
whereβ is the radiative heat transfer coefficient, which depends on the materials involved, geom-etry, and sometimes even on the temperatures T1 and T2. The temperatures must be given on an absolute scale (Rankine or Kelvin), and they represent the temperatures of the two bodies exchang-ing heat due to radiation.
8.2.2 Thermal Capacitance
All materials store heat, and their temperature is a measure of how much heat they store. The thermal capacitance of a body results in the relation
qh= CdT
dt =ρVcpdT
dt = mcpdT dt where cpis the heat capacity of the material.
8.2.3 Thermal Sources
Thermal sources are normally of two types: a prescribed temperature or a prescribed heat flux.
A temperature source represents the case of a thermal reservoir. A heat flux source represents a heater, such as an electrical heater, in which case
qh,s= i2R
where i and R are the electrical current and resistance of the heater.
138 8.2 Thermal Systems
8.2.4 Biot Number
The Biot number is the ratio of the convective heat transfer to the conductive heat transfer.
NB=hA∆T
κA∆TL
= hL κ
L is a representative length of the system — often the volume to surface ratio.
If NB is small (less than 0.1 or so), the body can be treated as having a single temperature (a lumped model). Otherwise, it will need to be split into multiple “lumps,” or a distributed model must be used. In addition, a Biot number analysis can be used to determine the relative importance of convective and conductive heat transfer.
8.2.5 Conservation of Energy
The rate of change of the energy stored in a system is governed by the principle of conservation of energy, which is also known as the first law of thermodynamics. In words, this law states that
Rate at which the energy of the system changes
=
Net rate at which heat is transferred
Nate rate at which work is done by the system
Assuming there are no work interactions and the system is stationary, the first law simplifies to mcpdT
dt = qh,in− qh,out
8.2.6 Analyzing Thermal Systems
To analyze thermal systems,
1. Define temperature nodes in the system. If necessary, use a Biot number analysis to deter-mine important temperature nodes.
2. Write the conservation of energy equation for each temperature node.
3. Use thermal resistance relations to define qh,inand qh,out for each node. Again, a Biot number analysis may be required to determine the important heat transfer modes.
4. Combine relations to give a set of state equations. State variables are simply the temperatures at each node.
Fluid and Thermal Systems 139
Example
The figure below shows a micromachined switch that carries current. The current flowing through the electrical contact resistance at the contact spot causes heating of the switch. This heat flows through the switch beam, as well as the thermal contact resistance, to the environment at a temper-ature of T0. The dimensions of the gold beam and other parameters are:
L = 300µm w = 100µm t = 3µm
h = 1 W/m2·K κ = 317 W/m·K Rt= 1000 K/W ρ = 19,000 kg/m3 cp= 129 J/kg·K
t L width = w
electrical contact resistance Re
thermal contact resistance Rt heat generated in contact
Determine the equations of motion, and use them to determine the thermal time constant of the beam.
Solution: First, we will calculate the Biot number for the beam. The effective length for calculating the Biot number is
Le=V and the Biot number is
NB= hLe κ = ht
2κ = 2.37 × 10−9
Since the Biot number is so small, we can treat the beam as having a single temperature node, and we can ignore convective heat transfer. Hence, we will define a single temperature node, at the center of the beam. The heat flux into this node is the electrical heating qh,e(t). There are three paths for heat to leave this node: conduction through both sides of the beam, and conduction through the thermal contact resistance.
The conservation of energy equation is
qh,in− qh,out = mcpdT dt
140 8.2 Thermal Systems The heat flux into the node is
qh,in= qh,e(t)
where qh,e(t) is the heat generated in the node. The heat conduction through one half of the beam is
qh,L/2=κwt
L/2(T − T0) and the heat conduction through the thermal contact resistance is
qh,R= 1
Rt(T − T0).
Since the total heat leaving the node is the sum of the heat traveling through each half of the beam and the contact, So the characteristic equation is
mcps +�4κwt
L + 1
Rt
�
= 0 and the root is
s = −
�4κwt L +R1t�
mcp Therefore, the time constant is
τ = −1
s = mcp
�4κwt
L +R1t� = 97 µs