Heat engines, power cycles and thermal efficiency

Close the cycle before counting net work

A thermodynamic cycle returns the working substance to its initial state. Its changes in internal energy and entropy over the full cycle are zero. Heat and work are transfers along the route and can have nonzero cycle totals. For a two-reservoir heat engine, use positive magnitudes QH absorbed, QC rejected and Wnet delivered: Wnet=QH−QC.

Thermal efficiency ηth=Wnet/QH. It compares net work with heat supplied to the chosen engine cycle, not gross turbine output with fuel flow or a whole facility’s electricity consumption. A compressor or pump within the cycle uses part of the generated work. The net output must subtract such requirements on a consistent boundary.

An ideal cycle is a model of a route

The ideal Rankine cycle represents pumping liquid, heating and vaporising it, expanding vapor through a turbine and condensing it again. Its pump and turbine idealisations simplify component calculations; practical pressure drops and irreversible expansion reduce performance. A real plant also has auxiliary demands and operating constraints that lie beyond a basic cycle sketch.

The Carnot cycle instead provides a reversible two-reservoir benchmark: reversible heat exchange at two temperatures connected by reversible adiabatic stages. For fixed reservoir temperatures TH and TC, ηmax=1−TC/TH, using kelvin. This is a limit for the specified heat-engine setting, not a promise that a chosen practical cycle can reach it or a universal efficiency formula for fuel cells and heat pumps.

Follow a numbered liquid–vapor return route

Label the loop 1→2→3→4→1. At 1, low-pressure liquid leaves the condenser. The pump takes work in to raise its pressure to state 2. Heating at high pressure then takes it through vaporisation to the turbine inlet, state 3. Expansion produces turbine work and reaches the lower-pressure outlet, state 4. Heat rejection in the condenser returns that stream to liquid state 1. The condenser restores the state needed by the pump; omitting it leaves an open route rather than a repeatable cycle.

For a nearly incompressible liquid, raising pressure involves much less volume displacement per kilogram than compressing a low-density vapor through a comparable pressure rise. This is why the liquid pumping stage matters to the Rankine route. It still consumes work. The route is not the exact Carnot sequence: high-pressure heating need not all occur at one temperature, and real losses/auxiliaries are additional questions.

The table is a supplied enthalpy bookkeeping model, with h in kJ/kg on one reference: h1=100, h2=103, h3=203, h4=163. The liquid/vapor labels are illustrative state assumptions; no real fluid, steam-table pressure/temperature pair or plant capability is established. Assume steady components, negligible kinetic/potential changes, adiabatic pump/turbine and no shaft work in the heater or condenser. The energy arrows below state positive input/output magnitudes.

Use both laws to assess a proposed result

Returning the working fluid to its starting state does not return both reservoirs to theirs. Heat rejected to the cold reservoir carries away entropy. A cyclic engine whose only net effect is turning heat from one reservoir completely into work would satisfy a simple energy total but violate the second law.

Finite temperature differences, friction and pressure losses generate entropy. Increasing a reservoir temperature can raise the reversible ceiling, but materials, heat transfer and emissions can introduce other constraints. A meaningful comparison states source and sink conditions, gross versus net output, and which auxiliary inputs are included. The diagram illustrates a bounded energy account rather than a component design.

Test an engine against energy and the reversible ceiling

For a constructed cycle, QH=100 kJ at 600 K and Wnet=35 kJ. The rejected amount must be QC=100−35=65 kJ, giving ηth=35/100=35%. The reversible ceiling is 1−300/600=50%, so the result is below that ceiling.

The two reservoirs change entropy by −100/600+65/300=0.05 kJ/K=50 J/K. The working fluid returns to its state, so this is the combined generation. A proposed70 kJ work with only30 kJ rejection would conserve energy but exceed the50% ceiling and give negative combined entropy change.

The separate numbered-route ledger gives net work 40−3=37 kJ/kg. Heat input minus heat rejection gives 100−63=37 kJ/kg as an independent closure. Its thermal efficiency is 37/100=37%; 40/100=40% counts turbine output before paying for the pump. This illustrative component account is separate from the earlier 600 K/300 K reservoir case. The enthalpy values alone supply no reservoir temperatures or entropy changes with which to certify its second-law feasibility.

Try the reasoning

1. Why can gross turbine output exceed net cycle output?

   Show the answer and reasoning

   The pump or compressor and any included auxiliaries require part of that work. Net output subtracts applicable inputs.

2. Does a cyclic fluid have zero heat transfer because ΔU=0?

   Show the answer and reasoning

   No. Net heat transfer equals net work over the full cycle; individual heat inputs and outputs can be substantial.

3. Can Celsius temperatures be used in1−TC/TH?

   Show the answer and reasoning

   No. The Carnot temperature ratio requires absolute kelvin temperatures.

4. A different supplied route receives 100 kJ/kg heat, produces 40 kJ/kg turbine work and needs 5 kJ/kg pump work. With no other transfers/storage, find net work, heat rejection and thermal efficiency.

   Show the answer and reasoning

   Net work is 40−5=35 kJ/kg. Rejection must be 100−35=65 kJ/kg, and thermal efficiency is 35%. These values close the first-law account; compatible properties and entropy evidence are still needed to assess a real route.

5. Why cannot the turbine outlet simply be declared the next pump inlet without examining its state?

   Show the answer and reasoning

   The liquid pump model needs a liquid inlet. The numbered route rejects heat and restores liquid state 1 before pumping; declaring a return arrow does not provide that state change or its heat rejection. A different compressor-based route would need its own component and work model.

What to carry forward

Count the complete route and net work. Apply the Carnot limit only to the stated reservoirs and heat-engine boundary.