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title = "Heat Transfer"
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author = ["Dehaeze Thomas"]
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Tags
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:
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## Electrical Analogy - Lumped Mass Modeling {#electrical-analogy-lumped-mass-modeling}
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The difference in temperature \\(\Delta T\\) is driving potential energy flow \\(Q\\) (in watts):
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\begin{equation}
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\Delta T = R\_{th} \cdot Q
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\end{equation}
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\\(R\_{th}\\) is the analogy of a "thermal resistance", and is expressed in K/W.
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## Conduction (diffusion) {#conduction--diffusion}
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The _conduction_ corresponds to the heat transfer \\(Q\\) (in watt) through molecular agitation within a material.
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\begin{equation}
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R\_{th} = \frac{d}{\lambda A}
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\end{equation}
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with:
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- \\(\lambda\\) the thermal conductivity in \\([W/m \cdot K]\\)
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- \\(A\\) the surface area in \\([m^2]\\)
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- \\(d\\) the length of the barrier in \\([m]\\)
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## Convection {#convection}
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The convection corresponds to the heat transfer \\(Q\\) through flow of a fluid.
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It can be either _natural_ or _forced_.
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\begin{equation}
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R\_{th} = \frac{1}{h A}
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\end{equation}
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with:
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- \\(h\\) the convection heat transfer coefficient in \\([W/m^2 \cdot K]\\).
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\\(h \approx 10.5 - v + 10\sqrt{v}\\) with \\(v\\) the velocity of the object through the fluid in \\([m/s]\\)
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Typically:
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- \\(h = 5 - 10\ W/m^2/K\\) for free convection with air
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- \\(h = 500 - 5000\ W/m^2/K\\) for forced water cooling in a tube of 5mm diameter
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- \\(A\\) the surface area in \\([m^2]\\)
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Note that clean-room air flow should be considered as forced convection, and \\(h \approx 10 W/m^2/K\\).
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## Radiation {#radiation}
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_Radiation_ corresponds to the heat transfer \\(Q\\) (in watt) through the emission of electromagnetic waves from the emitter to its surroundings.
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In the general case, we have:
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\\[ Q = \epsilon \cdot \sigma \cdot A \cdot (T\_r^4 - T\_s^4) \\]
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with:
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- \\(\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.
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It is between 0 (no emissivity) and 1 (maximum emissivity)
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- \\(\sigma\\) the Stefan-Boltzmann constant: \\(\sigma = 5.67 \cdot 10^{-8} \\, \frac{W}{m^2 K^4}\\)
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- \\(T\_r\\) the temperature of the emitter in \\([K]\\)
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- \\(T\_s\\) the temperature of the surrounding in \\([K]\\)
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In order to use the lumped mass approximation, the equations can be linearized to obtain:
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\begin{equation}
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R\_{th} = \frac{1}{h\_{rad} A}
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\end{equation}
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with:
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- \\(h\_{rad}\\) the effective heat transfer coefficient for radiation in \\(W/m^2 \cdot K\\)
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- \\(A\\) the surface in \\([m^2]\\)
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### Practical Cases {#practical-cases}
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Two parallel plates:
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\begin{equation}
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h\_{rad} = \frac{\sigma}{1/\epsilon\_1 + 1/\epsilon\_2 - 1} (T\_1^2 + T\_2^2)(T\_1 + T\_2)
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\end{equation}
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Two concentric cylinders:
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\begin{equation}
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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)
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\end{equation}
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A small object enclosed in a large volume:
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\begin{equation}
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h\_{rad} = \epsilon\_1 \sigma (T\_1^2 + T\_2^2)(T\_1 + T\_2)
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\end{equation}
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### Emissivity {#emissivity}
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The emissivity of materials highly depend on the surface finish (the more polished, the lower the emissivity).
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Some examples are given in <tab:emissivity_examples>.
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Gold coating gives also a very low emissivity and is typically used in cryogenic applications.
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<a id="table--tab:emissivity-examples"></a>
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<div class="table-caption">
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<span class="table-number"><a href="#table--tab:emissivity-examples">Table 1</a>:</span>
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Some examples of emissivity (specified at 25 degrees)
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</div>
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| Substance | Emissivity |
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|----------------------------|------------|
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| Silver (polished) | 0.005 |
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| Silver (oxidized) | 0.04 |
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| Stainless Steel (polished) | 0.02 |
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| Aluminium (polished) | 0.02 |
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| Aluminium (oxidized) | 0.2 |
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| Aluminium (anodized) | 0.9 |
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| Copper (polished) | 0.03 |
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| Copper (oxidized) | 0.87 |
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<div class="exampl">
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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):
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\\[ P = \epsilon \cdot \sigma \cdot A \cdot (T\_r^4 - T\_s^4) = 0.36\\, J \\]
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</div>
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## Heat {#heat}
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The _heat_ \\(Q\\) (in Joules) corresponds to the energy necessary to change the temperature of the mass with a certain material specific heat capacity:
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\\[ Q = m \cdot c \cdot \Delta T \\]
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with:
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- \\(m\\) the mass in \\([kg]\\)
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- \\(c\\) the specific heat capacity in \\([J/kg \cdot K]\\)
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- \\(\Delta T\\) the temperature different \\([K]\\)
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<div class="exampl">
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Let's compute the heat (i.e. energy) necessary to increase a 1kg granite by 1 degree.
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The specific heat capacity of granite is \\(c = 790\\,[J/kg\cdot K]\\).
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The required heat is then:
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\\[ Q = m\cdot c \cdot \Delta T = 790 \\,J \\]
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</div>
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<a id="table--tab:specific-heat-capacity"></a>
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<div class="table-caption">
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<span class="table-number"><a href="#table--tab:specific-heat-capacity">Table 2</a>:</span>
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Some examples of specific heat capacity
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</div>
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| Substance | Specific heat capacity [J/kg.K] |
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|---------------------|---------------------------------|
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| Air | 1012 |
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| Aluminium | 897 |
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| Copper | 385 |
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| Granite | 790 |
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| Steel | 466 |
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| Water at 25 degrees | 4182 |
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## Heat Transport (i.e. Water cooling) {#heat-transport--i-dot-e-dot-water-cooling}
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<a id="figure--fig:heat-transfer-fluid"></a>
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{{< figure src="/ox-hugo/heat_transfer_fluid.png" caption="<span class='figure-number'>Figure 1: </span>Heat transfered to the fluid" >}}
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\begin{equation}
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Q\_{in} = h \cdot A \cdot (T\_{wall} - T\_{mean})
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\end{equation}
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<a id="figure--fig:heat-transport"></a>
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{{< figure src="/ox-hugo/heat_transport.png" caption="<span class='figure-number'>Figure 2: </span>Heat Transport in the fluid" >}}
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\begin{equation}
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Q\_{out} = \phi \rho c\_p (T\_{mean,in} - T\_{mean,out})
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\end{equation}
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with:
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- \\(Q\_{out}\\) the transported heat in W
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- \\(\phi\\) the flow in \\(m^3/s\\)
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- \\(\rho\\) the fluid density in \\(kg/m^3\\)
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- \\(c\_p\\) the specific heat capacity of the fluid in \\(J/(kg \cdot K)\\)
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- \\(T\_{mean}\\) the mean incoming and outgoing fluid temperature
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Because of energy balance, we have in the stationary condition: \\(Q\_{in} = Q\_{out}\\)
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## Heat flow {#heat-flow}
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The heat flow \\(P\\) (in watt) is the derivative of the heat:
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\\[ P = \cdot{Q} = \frac{dQ}{dt} = \frac{dT}{R\_T} = C\_T \cdot dT \\]
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with:
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- \\(Q\\) the heat in [W]
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- \\(R\_T\\) the thermal resistance in \\([K/W]\\)
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- \\(C\_T\\) the thermal conductance in \\([W/K]\\)
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## Bibliography {#bibliography}
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<style>.csl-entry{text-indent: -1.5em; margin-left: 1.5em;}</style><div class="csl-bib-body">
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</div>
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