+++
title = "Heat Transfer"
author = ["Dehaeze Thomas"]
draft = false
category = "Mechanics and Physics"
+++
Tags
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## Electrical Analogy - Lumped Mass Modeling {#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) {#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 {#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}
_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 {#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 {#emissivity}
The emissivity of materials highly depend on the surface finish (the more polished, the lower the emissivity).
Some examples are given in [Table 1](#table--tab:emissivity-examples).
Gold coating gives also a very low emissivity and is typically used in cryogenic applications.
Table 1:
Some examples of emissivity (specified at 25 degrees)
| 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 {#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 \\]
Table 2:
Some examples of specific heat capacity
| 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) {#heat-transport--i-dot-e-dot-water-cooling}
{{< figure src="/ox-hugo/heat_transfer_fluid.png" caption="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="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 {#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]\\)