6 Thermodynamic Cycles

Learn how thermodynamic cycles return a working fluid to its initial state, how diagrams and energy balances describe them, and how power and refrigeration cycles are evaluated.

What makes a cycle

A is a sequence of processes that returns a working substance to its starting state. Since internal energy is a property of state, its net change over the complete cycle is zero:

ΔUcycle=0.\Delta U_{\text{cycle}}=0.

Cycles can produce useful work, as in a power plant, or use work to move heat, as in a refrigerator. To analyze a cycle, track the working fluid’s states, energy transfers, and direction of operation.

Reading cycle diagrams

Cycle diagrams show the working fluid’s path through a sequence of states. Each point represents a state, and each curve represents a process.

On a , the signed area enclosed by the cycle gives net boundary work:

Wnet=∮P dV.W_{\text{net}}=\oint P\,dV.

A clockwise loop generally indicates positive net work done by the working fluid, while a counterclockwise loop indicates net work input.

A is also useful. For an internally reversible process, the area under a path represents heat transfer:

Q=∫T dS.Q=\int T\,dS.

Diagrams help visualize a cycle, but energy balances and property data are needed to calculate its performance.

Energy balances

Using the convention that heat entering the system and work done by the system are positive, the first law for a closed system is

ΔU=Q−W.\Delta U=Q-W.

Over a complete cycle, the system returns to its starting state and ΔU=0\Delta U=0. Therefore, net heat transfer equals net work:

Qnet=Wnet.Q_{\text{net}}=W_{\text{net}}.

Here, QnetQ_{\text{net}} and WnetW_{\text{net}} refer to the net heat and work over all processes in the cycle. For a , net work output equals heat supplied minus heat rejected. For a refrigeration cycle, net work input equals heat rejected to the warm surroundings minus heat absorbed from the cooled space.

For steady-flow devices such as turbines, compressors, pumps, and heat exchangers, energy balances are usually applied to each component. In an introductory analysis, changes in kinetic and potential energy are often neglected. A turbine typically produces work as the fluid’s enthalpy decreases; a compressor or pump requires work to raise the fluid’s pressure.

Power cycles and efficiency

A receives heat, turns part of that energy into net work, and rejects the rest to a cooler environment. Its simplified energy balance is

Wnet,out=Qin−Qout.W_{\text{net,out}}=Q_{\text{in}}-Q_{\text{out}}.

compares net work output with heat input:

η=Wnet,outQin=1−QoutQin.\eta=\frac{W_{\text{net,out}}}{Q_{\text{in}}}=1-\frac{Q_{\text{out}}}{Q_{\text{in}}}.

For example, a cycle that receives 1000 kJ1000\ \text{kJ} of heat and rejects 600 kJ600\ \text{kJ} produces 400 kJ400\ \text{kJ} of net work. Its is 0.400.40, or 40%40\%. By the second law, a cycle operating between hot and cold surroundings cannot convert all supplied heat into work; some heat must be rejected.

The is a common model for steam power plants. Its pump raises liquid pressure, its boiler adds heat, its turbine produces work as steam expands, and its condenser rejects heat and returns the fluid to liquid. Real devices are not perfectly reversible, so their actual work output and efficiency are below ideal predictions.

Refrigeration and heat pumps

A refrigerator or heat pump uses work input to transfer heat from a colder region to a warmer one. The energy balance is

Qout=Qin+Win,Q_{\text{out}}=Q_{\text{in}}+W_{\text{in}},

where QinQ_{\text{in}} is heat absorbed from the cold space and QoutQ_{\text{out}} is heat delivered to the warm surroundings.

A refrigerator is judged by how much heat it removes, while a heat pump is judged by how much heat it supplies. Their coefficients of performance are

COPrefrigerator=QinWin,COPheat pump=QoutWin.\mathrm{COP}_{\text{refrigerator}}=\frac{Q_{\text{in}}}{W_{\text{in}}}, \qquad \mathrm{COP}_{\text{heat pump}}=\frac{Q_{\text{out}}}{W_{\text{in}}}.

For example, if a refrigerator removes 300 kJ300\ \text{kJ} of heat while using 100 kJ100\ \text{kJ} of work, its is 33. A can exceed 11 because the device moves heat rather than converting work entirely into heat.

The common circulates refrigerant through a compressor, condenser, expansion device, and evaporator. The refrigerant absorbs heat at the evaporator and releases it at the condenser.