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+++ title = "Heat Transfer" author = ["Dehaeze Thomas"] draft = false +++
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Electrical Analogy - Lumped Mass Modeling
The difference in temperature \(\Delta T\) is driving potential energy flow \(Q\) (in watts):
\begin{equation} \Delta T = R_{th} \cdot Q \end{equation}
\(R_{th}\) is the analogy of a "thermal resistance", and is expressed in K/W.
Conduction (diffusion)
The conduction corresponds to the heat transfer \(Q\) (in watt) through molecular agitation within a material.
\begin{equation} R_{th} = \frac{d}{\lambda A} \end{equation}
with:
- \(\lambda\) the thermal conductivity in \([W/m \cdot K]\)
- \(A\) the surface area in \([m^2]\)
- \(d\) the length of the barrier in \([m]\)
Convection
The convection corresponds to the heat transfer \(Q\) through flow of a fluid. It can be either natural or forced.
\begin{equation} R_{th} = \frac{1}{h A} \end{equation}
with:
- \(h\) the convection heat transfer coefficient in \([W/m^2 \cdot K]\).
\(h \approx 10.5 - v + 10\sqrt{v}\) with \(v\) the velocity of the object through the fluid in \([m/s]\)
Typically:
- \(h = 5 - 10\ W/m^2/K\) for free convection with air
- \(h = 500 - 5000\ W/m^2/K\) for forced water cooling in a tube of 5mm diameter
- \(A\) the surface area in \([m^2]\)
Note that clean-room air flow should be considered as forced convection, and \(h \approx 10 W/m^2/K\).
Radiation
Radiation corresponds to the heat transfer \(Q\) (in watt) through the emission of electromagnetic waves from the emitter to its surroundings.
In the general case, we have: \[ Q = \epsilon \cdot \sigma \cdot A \cdot (T_r^4 - T_s^4) \] with:
- \(\epsilon\) the emissivity which corresponds to the ability of a surface to emit energy through radiation relative to a black body surface at equal temperature. It is between 0 (no emissivity) and 1 (maximum emissivity)
- \(\sigma\) the Stefan-Boltzmann constant: \(\sigma = 5.67 \cdot 10^{-8} \, \frac{W}{m^2 K^4}\)
- \(T_r\) the temperature of the emitter in \([K]\)
- \(T_s\) the temperature of the surrounding in \([K]\)
In order to use the lumped mass approximation, the equations can be linearized to obtain:
\begin{equation} R_{th} = \frac{1}{h_{rad} A} \end{equation}
with:
- \(h_{rad}\) the effective heat transfer coefficient for radiation in \(W/m^2 \cdot K\)
- \(A\) the surface in \([m^2]\)
Practical Cases
Two parallel plates:
\begin{equation} h_{rad} = \frac{\sigma}{1/\epsilon_1 + 1/\epsilon_2 - 1} (T_1^2 + T_2^2)(T_1 + T_2) \end{equation}
Two concentric cylinders:
\begin{equation} h_{rad} = \frac{\sigma}{1/\epsilon_1 + r_1/r_2 (1/\epsilon_2 - 1)} (T_1^2 + T_2^2)(T_1 + T_2) \end{equation}
A small object enclosed in a large volume:
\begin{equation} h_{rad} = \epsilon_1 \sigma (T_1^2 + T_2^2)(T_1 + T_2) \end{equation}
Emissivity
The emissivity of materials highly depend on the surface finish (the more polished, the lower the emissivity). Some examples are given in tab:emissivity_examples.
Gold coating gives also a very low emissivity and is typically used in cryogenic applications.
| Substance | Emissivity |
|---|---|
| Silver (polished) | 0.005 |
| Silver (oxidized) | 0.04 |
| Stainless Steel (polished) | 0.02 |
| Aluminium (polished) | 0.02 |
| Aluminium (oxidized) | 0.2 |
| Aluminium (anodized) | 0.9 |
| Copper (polished) | 0.03 |
| Copper (oxidized) | 0.87 |
Let's take a polished aluminum plate (20 by 20 cm) at 125K (temperature of zero thermal expansion coefficient of silicon) surrounded by elements are 25 degrees (300 K): \[ P = \epsilon \cdot \sigma \cdot A \cdot (T_r^4 - T_s^4) = 0.36\, J \]
Heat
The heat \(Q\) (in Joules) corresponds to the energy necessary to change the temperature of the mass with a certain material specific heat capacity: \[ Q = m \cdot c \cdot \Delta T \] with:
- \(m\) the mass in \([kg]\)
- \(c\) the specific heat capacity in \([J/kg \cdot K]\)
- \(\Delta T\) the temperature different \([K]\)
Let's compute the heat (i.e. energy) necessary to increase a 1kg granite by 1 degree. The specific heat capacity of granite is \(c = 790\,[J/kg\cdot K]\). The required heat is then: \[ Q = m\cdot c \cdot \Delta T = 790 \,J \]
| Substance | Specific heat capacity [J/kg.K] |
|---|---|
| Air | 1012 |
| Aluminium | 897 |
| Copper | 385 |
| Granite | 790 |
| Steel | 466 |
| Water at 25 degrees | 4182 |
Heat Transport (i.e. Water cooling)
{{< figure src="/ox-hugo/heat_transfer_fluid.png" caption="<span class="figure-number">Figure 1: Heat transfered to the fluid" >}}
\begin{equation} Q_{in} = h \cdot A \cdot (T_{wall} - T_{mean}) \end{equation}
{{< figure src="/ox-hugo/heat_transport.png" caption="<span class="figure-number">Figure 2: Heat Transport in the fluid" >}}
\begin{equation} Q_{out} = \phi \rho c_p (T_{mean,in} - T_{mean,out}) \end{equation}
with:
- \(Q_{out}\) the transported heat in W
- \(\phi\) the flow in \(m^3/s\)
- \(\rho\) the fluid density in \(kg/m^3\)
- \(c_p\) the specific heat capacity of the fluid in \(J/(kg \cdot K)\)
- \(T_{mean}\) the mean incoming and outgoing fluid temperature
Because of energy balance, we have in the stationary condition: \(Q_{in} = Q_{out}\)
Heat flow
The heat flow \(P\) (in watt) is the derivative of the heat: \[ P = \cdot{Q} = \frac{dQ}{dt} = \frac{dT}{R_T} = C_T \cdot dT \] with:
- \(Q\) the heat in [W]
- \(R_T\) the thermal resistance in \([K/W]\)
- \(C_T\) the thermal conductance in \([W/K]\)