>[!Tip]- Where's Convection? >It's fluid mechanics. I am NOT learning multivariable calc for it - so it can go stay in a little corner of any underworld for what it matters. >No more Laplacians!!! # Emissivity - An Intro # The Conduction Equation We're mostly able to generalise the process of conduction - it's really useful when you've got a bunch of layers of a material you're hoping is heat resistant. No more shall you shiver in the cold of the damp with this mathematical model! To find the total rate of conduction we basically transmute it as a rate of heat transfer: $r_{cond} = \frac{dQ}{dt}$ We can expand this into: $\frac{dQ}{dt} = kA\frac{{ T_{R}-T_{L}}}{L}$ That's why hotter objects seem to conduct faster. This is an example of **Fourier's Law**, given by: $\frac{dQ}{dt} \propto -k \frac{dT}{dx}$ For more read on the partial derivation breakdown of this system - namely: [[Fourier's Law (Thermo)]]. This equation is dependent on the incident area of the blocks! Each object has a resistance to this conductivity - the *insulative strength* of the object. This is given by: $R = \frac{L}{G}$ Where: - $L$ is the length of the conductive rod (we don't really use slabs for their conductive properties) - $G$ is the conductivity constant of the material As you might expect, this is higher for substances like plastic or air and lower for metals. Amazing! >[!Example]- Problem Taster - TBA ## Case Study - Composite Slab Conduction A building isn't all concrete! Though some on the more brutalist side of things may choose to use more of it, in essence making a building wholly out of concrete isn't the best idea - there are certain issues with concrete's tensile strength.[^1] Most of the time, construction requires loads and loads of materials - from insulators to rebar, it's a pretty complex logistical nightmare to be frank! ![[Pasted image 20231208213204.png]] $k$ is the thermal conductivity constant of the material. The conductivity of the combined slabs is equal as the potential energy differentials across heat transfer slabs *must* be equal. Using the [[The First Law of Thermodynamics|First Law of thermodynamics]] we can get: (TBA. will add.) [1 ]:This means that it has a low Young's modulus - concrete's compressive strength comes from its shear strength, which is pretty high! # Radiative Processes - The Stefan Boltzmann Constant ![[Pasted image 20230911142002.png]] *BEHOLD! IT RADIATES!* Here's a big, massive (>8 solar mass) star, straight from the [[An Astrophysics Landing Page|astrophysics section]]. However, it's a little too eager to greet you and is quite close to us at this point. The question it poses: *how far must it be from our solar system in order to not fry us?* We can take the radiative power over an area $A$ as the following equation: $P = \sigma \epsilon AT^4$ Where: - $\sigma$ is the **Stefan-Boltzmann Constant** - very useful in [[Everything Astronomy|astrophysics!]] - $\epsilon$ is the emissivity constant of the blackbody - empirical! We tend to use this equation for approximated [[Blackbody Radiation (Thermodynamics)#Defining the Black Body Radiator|black body radiators]], so an example of when it's really useful is when we want to know the luminosity of a big star! *This is an excuse to add an astrophysics insert. While you're here, check it out!* [[Stars (Astro)]] For binary star systems, one star is typically larger than the other, meaning that there's an extra heat transfer incident on the second, cooler object. Let's start by thinking of the logic - the additional energy transferred is going to be dependent on the area incident of the object facing it. This is known as the **view factor**. $P = \sigma \epsilon F(T_{1}^4 - T_{2}^4)$ >[!Success]- Solving the Problem >Let's lay the groundwork for the problem first! > >The temperature at which fat begins to fry (approximated to be on a pan) is 393 Kelvins. > >Our high-mass sample star is [Zeta Ophiuchi](https://www.nasa.gov/image-article/embracing-rejected-star/), a late O-Type star more commonly known for being a runaway star, perhaps the remains of a supernova. Since it's a single star, it's a perfect candidate for us!With this, let's list out our information: > >- $\zeta$ Oph Absolute Magnitude: -3.07 >We need find the absolute bolometric luminosity of Zeta Ophiuchi. So: >$-3.18 - 0 = -2.5 \log_{10} \left( \frac{L_{\zeta}}{} \right)$ >We need to find the temperature. To do this, let's employ an expression for the total flux over a given area! >$L = \frac{4\pi \sigma T^4}{1}$ > >- Target Temperature: 393 K > >- Surface Area: >[!Example]- Sun Example Question >The sun imparts a power of $P$ on the earth and has a temperature of $5800 K$. Find an expression for the temperature of the Earth and solve for it! >%%diagram of the system!%% >**Solution:** >The sun's emissions can be interpreted as proportional to the inverse of the surface area of any arbitrary sphere orbital distance $D$ $1 AU$ away from it. This means that for the Earth that occupies a cross-sectional area of $\pi r_E^2$ we can (have to) make the equation as shown: >$\sigma (4\pi r_{S}^2)T_{S}^4 \cdot \frac{\pi r_{E}^2}{4\pi D^2}$ >This is as the sun emits across it's entire surface - though if you really want to be pedantic about it try to incorporate sunspots into your model - I dare you. >Back to the question - this gives us the total power from the sun going to the Earth from the sun's perspective, so let's write it as the equation: >$\sigma 4\pi r_{S}^2 T_{S}^4 \cdot \frac{\pi r_{E}^2}{4\pi D^2} = \sigma 4\pi r_{E}^2 T_{E}^4$ >I'll derive the rest of the expression for you this time, but you could probably just 'insight gremlin' the question. >Cancel the $\sigma$ and $4\pi$ constants from both sides. The emissivity constant is already gone - as negligent physics students, we shouldn't worry about it: >$r_{S}^2 T_{S}^4 \cdot \frac{\pi r_{E}^2}{4\pi D^2} = r_{E}^2 T_{E}^4 $ >Now we cancel the $r_E$ and $\pi$ terms on both sides. Is there no one else! >$r_{S}^2 T_{S}^4 \cdot \frac{ 1}{4 D^2} = T_{E}^4$ >Get rid of the fourth power on both sides and you're done! >$T_{E} = T_{S} \sqrt{ \frac{R_{S}}{2D} }$ >1 AU is $1.5 \times 10^{11}$ metres. Substitute this value into the equation along with the the temperature of the sun to get our value for the Earth's temperature! >$$ > >[!Example]- Manipulating Emissivity # What's Next? If you've read this article carefully I'd like you to take notice of something - the convection part almost has no maths in it. This is because it is glorified fluid dynamics and looking at all the laplacians makes me want to cry. Self-defeating attitudes otherwise, feel free to check out this skeleton if you want to learn about it. I did warn you... 1. [[Convection - Down the metaphorical Drain]] (unmade for math skill issue reasons) Want to learn more about radioactivity? Join us in learning about Planck's Law! 1. [[Blackbody Radiation (Thermodynamics)]]