Why is the thermal efficiency of the ideal Rankine cycle always less than the Carnot efficiency operating between the same maximum and minimum temperatures?
AThe Rankine cycle rejects more heat because the condenser is less efficient than an ideal isothermal heat sink
BHeat is added over a range of temperatures in the boiler, making the mean addition temperature less than T_H, unlike Carnot's isothermal heat addition at T_H
CThe pump consumes work, reducing net output in a way the Carnot cycle does not
DThe Rankine cycle uses a two-phase working fluid, which reduces the work output of the turbine
Carnot efficiency η = 1 − T_L/T_H assumes all heat is added at T_H. In the Rankine cycle, heat is added across the entire boiler process: first heating subcooled liquid to saturation temperature, then boiling, then possibly superheating — so heat enters at temperatures ranging from below T_H up to T_H. The *mean* temperature of heat addition is lower than T_H, reducing efficiency below Carnot. The T-s diagram shows this graphically: the Rankine heat addition rectangle is 'smaller' than the Carnot rectangle with the same T_H and T_L limits.
Question 2 Multiple Choice
In the ideal Rankine cycle, what is the expression for pump work per unit mass, and why is it small compared to turbine work?
Aw_pump = h₃ − h₄; it is small because steam expands at high enthalpy
Bw_pump = ν(P₂ − P₁); it is small because liquids have very low specific volume
Cw_pump = R·T·ln(P₂/P₁); it is small because the temperature ratio is modest
Dw_pump = c_p(T₂ − T₁); it is small because liquid heat capacity is low
Pump work for an incompressible fluid is w_pump = ν(P₂ − P₁), where ν is the specific volume of the liquid. Liquid water has a very small specific volume (~0.001 m³/kg), so even a large pressure rise (e.g., from 10 kPa to 10 MPa) produces relatively small pump work. Compare this to turbine work, which involves expanding compressible steam from high enthalpy at high pressure — the turbine handles far more energy per unit mass, which is why net work output is dominated by the turbine term.
Question 3 True / False
The Rankine cycle can theoretically achieve Carnot efficiency if the boiler pressure is increased to a sufficiently high value.
TTrue
FFalse
Answer: False
The Carnot efficiency is an absolute upper limit — no heat engine operating between two temperature reservoirs can exceed it. The Rankine cycle is always less efficient than Carnot because heat is added over a range of temperatures rather than isothermally at T_H. Increasing boiler pressure does raise thermal efficiency by elevating the mean temperature of heat addition, but the Rankine efficiency asymptotically approaches the Carnot limit as a bound it can never reach. Furthermore, very high pressures create practical problems (turbine blade erosion from wet steam) before the efficiency gain is fully realized.
Question 4 True / False
Superheating steam beyond the saturation point at constant boiler pressure simultaneously raises thermal efficiency and reduces moisture content at the turbine exit.
TTrue
FFalse
Answer: True
Superheating achieves two benefits at once. First, it raises the mean temperature of heat addition (since heat enters at higher temperatures during superheating), directly improving thermal efficiency. Second, the turbine exit state (State 4 on the T-s diagram) moves to the right — toward higher quality (lower moisture fraction) — because the expansion starts from a higher-enthalpy, higher-entropy state. Wet steam (high moisture) erodes turbine blades, so reducing moisture is an important practical constraint, and superheating solves both problems simultaneously.
Question 5 Short Answer
Use the T-s diagram to explain why the Rankine cycle's thermal efficiency is always less than Carnot efficiency between the same temperature limits.
Think about your answer, then reveal below.
Model answer: On the T-s diagram, Carnot efficiency corresponds to a rectangle bounded by T_H at the top, T_L at the bottom, and two vertical (isentropic) sides — all heat is added at T_H. The Rankine cycle's heat addition process traces a path that begins at low temperature (subcooled liquid), rises through the saturation dome, and only reaches T_H at the end of superheating (if present). The area under this non-rectangular heat addition path represents lower average temperature than T_H. Since efficiency depends on the mean temperature of heat addition, not just the peak, the Rankine cycle's mean T_add < T_H, giving η_Rankine < η_Carnot.
The T-s diagram makes this geometric: Carnot's heat addition is a horizontal line at T_H; Rankine's is a rising curve. The area under Rankine's curve (heat added) divided by the total area of the process gives a lower ratio than Carnot's rectangle. Any cycle that cannot add all its heat at T_H will fall short of the Carnot limit.