# General relation between power density of any engine and dissiapation rate and temperature

Many years ago ( before my university studies ) I read that renewable resources are fundamentally limited by laws of thermodynamics to produce energy very slowly (low specific power or power density) because they work at low temperature and with small change of entropy. Or, on the other hand, that any engine with high specific power (like jet engine or rocket engine ) must work at high temperature and/or dissipate lot of energy.

I don't remember the courses of thermodynamics properly to make it clear in my mind. The best I found on this topic is in this book Renewable Energy Conversion, Transmission and Storage. For example for system with linear thermal conductivity $\sigma$ using some Onsager reciprocal relations the power $P$ would be limited by:

$P < dG/dt = -T dS /dt = \sigma^{-1} (dQ/dt)^2$

where, $G$ is free energy, $t$ is time, $T$ is temperature, $S$ is entropy, $Q$ is heat.

Then transformation of heat $\Delta Q$ should fulfill equation $\Delta Q = \sqrt{ T \sigma \Delta S \Delta t }$ so the increase of heat flow (as well as power) can be achieved by increasing temperature or entropy change or thermal conductivity. ( please see the chapter 2.2 Irreversible thermodynamics in that book for more details )

• Is this really a general law? Could it be somehow generally qunantified irrespective of the particular system ( irrespective of particular realization of the heat engine).
• Is there any better, more general, more broadly applicable and more illustrative approach to demnostrate this effects? (e.g. I don't like that this is limited just to heat conduction with linear conduction coefficient)
Define the efficiency of a heat engine as \begin{equation} \eta = \frac{\text{Net work out}}{\text{Heat in}} \end{equation} The Second Law of Thermodynamics tells us that the efficiency can't be greater than the Carnot efficiency: \begin{equation} \eta \leq 1 - \frac{T_L}{T_H} \end{equation} where the $T$'s must be in absolute units (e.g. Kelvin) for the answer to be meaningful. Combining these \begin{equation} \text{Net work out} \leq {\text{Heat in}}\cdot\left( 1 - \frac{T_L}{T_H} \right) \end{equation} Thus to get a high rate of work output, you need a large heat flux, a large temperature ratio, or both. This is a general law.
• ok, this I know, my question was rather about the heat flow rate. At higher temperature or higher temperature gradient the heat flow is higher. ( eg. termal conduction scales as $dQ/dt \propto (T_2-T_1)$, thermal radiation scales as $dQ/dt \propto (T_2^4-T_1^4)$ – Prokop Hapala Apr 12 '15 at 9:05