# Intro - Fridges and Freezers # The Thermodynamic Cycle If we harken back to our p-V diagram, drawing a cycle typically gives us: ![[isothermalcycle.png]] *Fig 1: An impression of a generic thermodynamic cycle.* So what happens here is that the gas in the engine compresses from A to B through some external piston exerting a force - doing work - on the gas, expanding after it reaches B through the gas itself doing work against the piston, then it loses heat energy to its surroundings through C to A. In other words, A-B is an [[Adiabatic Gas Expansion (Thermodynamics)|adiabatic compression]], where there is no heat transferred to or from the gas, B-C is an *isothermal expansion*, where the gas expands at a constant temperature (along what is known as an *isotherm*, the reciprocal function gained from the [[Kinetic Theory & Ideal Gases (Thermodynamics)#The Ideal Gas Law|ideal gas law]]), and C-A is an *isovolumetric expansion*, where heat energy is lost in order to reduce the pressure (since $W = p\Delta V$ and there is no change in the volume during this process). # The Heat Engine Let's consider the nuclear power plant. Energy from the fission reaction heats up a pool of water, and that pool of water in turn turns a turbine as it cools. To turn the turbine, a cold reservoir of water must be placed such that the turbine is in between them; this way, through the movement of This is better illustrated using an abridged form of the [[Power, Work Done and Energy#Potential Energy Curves|Sankey Diagrams,]] as shown below: ![[heatengine.png]] *Hot and cold reservoirs.* In this case, the work done by the system is what we want, and is how we measure the 'efficiency' of the system. So here, the efficiency of the heat engine would be: $\eta = \frac{W}{Q}$ Where: - $\eta$ is the **efficiency of the cycle,** - $W$ is the **work done by the gas,** - $Q$ is the **heat energy input into the gas.** This also applies to the cycle above. # The Carnot Cycle The Carnot cycle is a specific case of the thermodynamic cycle that is the **maximum possible efficiency for a thermodynamic cycle.** It consists of 2 adiabatic and isothermal changes each, which when illustrated appear as: ![[carnotcycle.png]] *Fig 2: The Carnot Cycle* So, if we start at A, the gas will experience the following transformations: 1. Firstly, it isothermally *compresses* along AB 2. Then, it adiabatically *compresses* (albeit not by much) along BC 3. Then, it isothermally *expands* along CD 4. Finally, it adiabatically expands along DA to reach A again. Now, we can find the efficiency of the Carnot Cycle through reconsidering the formula of $\eta = \frac{\mathrm{useful \ work}}{\mathrm{input \ energy}}$. We can write the work done during the cycle as a sum of all the work being done on the gas throughout the cycle, as: $W_{tot} = W_{AB} + W_{BC} + W_{CD} + W_{DA}$ ... (temp break. don't need to know derivation for test. the formula is:) $\eta = 1 - \frac{T_{C}}{T_{H}}$ Where $T_C$ is the colder temperature of the gas and $T_H$ the hotter temperature. **Remember, this formula only works for Carnot Engines!** # What's Next? Navigate back to the home page here: [[Home]] Navigate back to the physics home page here: [[The (Incomplete) Physics Almanac]] Navigate back to the thermodynamics home page here: [[Thermodynamics - A Contents Page]]