Its a summer day and the air in my house has been heated up. I could switch on my air conditioning, but then I'd be using energy from the grid in order to reduce the amount of energy in my house.
What I'd much rather do is capture the heat energy from the air, so cooling it down, and turn that heat energy into electricity. (So I could store it in a battery or sell it to the grid or use some other means of disposing of it.)
Is such an apperatus possible within our present understand of nature? Why or why not?
(I presume not, as people would be selling kits.)
Just to pre-empt a possible answer: I've seen similar conversations in the past and "inefficient" has been given as the answer as to why not. I don't understand this as an objection, because the heat energy is effectively free. The sun rudely went and gave me an excessive amount of heat energy without being asked to. Is converting heat to electricity really so ineffiecient that I'd be better off running air conditioning?
Answer
In order to build any thermal engine as envisioned by you, you need both a cold and a hot reservoir, such that heat can flow from the hot part to the cold part and the entropy doesn’t decrease while you’re making energy.
The efficiency of such a machine has an upper limit of $(T_{\textrm{hot}} - T_{\textrm{cold}})/T_{\textrm{cold}}$ (as given by the perfect Carnot engine).
Given that you are usually well off when you get a cold reservoir of $T_{\textrm{cold}} = 290\textrm{ K}$ on a hot summer day ($T_{\textrm{hot}} = 320\textrm{ K}$, the efficiency of your machine has an upper bound of 10%, which does not necessarily include loss due to friction, electric resistance, escaping air etc. If you include these, you get (probably) well below 1%. But, for the sake of argument, let us continue with an assumed efficiency of 10%.
What you want to know next is the maximum heat you can transfer out of your hot reservoir into your cold reservoir. For simplicity, we will assume that the cold reservoir is very large and stays at a constant temperature, heat then flows from the hot to the cold reservoir as long as $T_{\textrm{hot}} > T_{\textrm{cold}}$. The thermal energy hence available to you is $W = c_V \times \delta T \times N \times 10\%$, where $\delta T = 30 \textrm{ K}$ is the temperature difference, $N$ is the mass/number of air (particles) and $c_V$ is the heat capacity.
For air at sea-level, Wikipedia gives me $c_V \approx 29 \textrm{ J}\textrm{K}^{-1}\textrm{mol}^{-1}$ (I am using the constant-pressure one, as we cannot compress air without putting more work into it). I shall then assume that you have a really large house of $10 \times 10 \times 10 \textrm{m}^3 = 10^6 \textrm{ L } \hat{=} 4.27 \times10^4 \textrm{ mol} = N$ where the last but one equality stems from the fact that there are about $6.02 \times 10^{23}$ particles in about $23 \textrm{ L}$. Great, we can now calculate W!
$$ W = 1.03 \textrm{ kWh } \hat{=} 21 \textrm{ cent}$$
where the last equality is a top-of-my-head number I have floating around for electricity prices in Germany during the summer of 2011.
To conclude: Even assuming that you somehow manage to build a perfect Carnot engine using the heat in the air in your house and find a magical reservoir of constant temperature (some part of the earth, possibly), you would get about 20 € per year out of it (four months of high temperature and one "charge" per day).
Really, just put some solar cells on your roof and appropriate insulation on/in your walls. ☺
No comments:
Post a Comment