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Thermodynamics · Chapter 10 of 10 · Intermediate

Vapor and Refrigeration Cycles

The Rankine cycle turns heat into electricity in nearly every power plant, and running its logic backwards gives the vapor-compression cycle in every fridge and air conditioner. Both come straight from the device models you already know.

Readiness check

This chapter assembles the device models into cycles. Tick only what you can do closed-notes.

Read enthalpy and entropy from a steam or refrigerant table.
Find a quality from entropy inside the dome.
Apply turbine, pump, and throttle balances.
Compute thermal efficiency and a coefficient of performance.
Interpolate between two table rows.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsRevisit device models in Chapter 5.

3 or more weak itemsReview the steam tables in Chapter 3.

The core idea

The Rankine cycle circulates water through a boiler, turbine, condenser, and pump, turning a heat input into net work. Reverse the flow of energy and the same four-component loop becomes a refrigerator.

η_th = w_net/q_in = 1 − q_out/q_inCOP_R = q_L/w_in

In the Rankine cycle, the boiler adds heat to make high-pressure steam, the turbine expands it to produce work, the condenser rejects heat to turn it back to liquid, and a pump raises the liquid pressure cheaply. Each component is a steady-flow device from Chapter 5, so the whole cycle is just four enthalpy differences. The vapor-compression refrigeration cycle uses the same four roles, compressor, condenser, throttle, and evaporator, to move heat from cold to hot, rated by its coefficient of performance.

The skill works when: you fix each state from the tables, then apply the right device balance around the loop.

The skill breaks down when: pump work is computed as if water were a gas, or the condenser-exit state is not taken as saturated liquid.

The concept. The Rankine loop: the boiler adds heat, the turbine makes work, the condenser rejects heat, and the pump returns the liquid to the boiler. The net work is the turbine work minus the small pump work.

The skills, taught in order

Five skills build the Rankine cycle, analyse it, note how it is improved, and then run the logic backwards for refrigeration.

10.1 The Rankine cycle components

Water flows through four steady-flow devices: a pump raises the liquid pressure (work in), a boiler adds heat to make superheated steam, a turbine expands it (work out), and a condenser rejects heat back to saturated liquid. Using water lets the cycle add and reject heat at nearly constant temperature during phase change, which is what makes it efficient.

10.2 Analysing the cycle

Fix the four states from the tables, then apply each device balance: pump w_p = v_f(P₂ − P₁); boiler q_in = h₃ − h₂; turbine w_t = h₃ − h₄; condenser q_out = h₄ − h₁. The thermal efficiency is η = w_net/q_in = (w_t − w_p)/q_in. Pump work is tiny because liquid is nearly incompressible.

Device	Process	Energy term
Pump	isentropic, liquid	w_p = v_f(P₂ − P₁)
Boiler	constant pressure	q_in = h₃ − h₂
Turbine	isentropic (ideal)	w_t = h₃ − h₄
Condenser	constant pressure	q_out = h₄ − h₁

10.3 Improving the Rankine cycle

Three moves raise efficiency: superheating the steam (hotter average heat-addition temperature), reheating between turbine stages (keeps the steam dry and adds area), and regeneration (using extracted steam to preheat the feedwater). All work by raising the average temperature at which heat is added.

10.4 The vapor-compression refrigeration cycle

Reverse the energy flow and the loop becomes a refrigerator: a compressor raises the refrigerant pressure (work in), a condenser rejects heat outdoors, a throttle drops the pressure (h constant), and an evaporator absorbs heat from the cold space. The throttle replaces the turbine because recovering the tiny work is not worth the cost.

10.5 Coefficient of performance

The refrigerator is rated by COP_R = q_L/w_in = (h₁ − h₄)/(h₂ − h₁), the cooling delivered per unit of compressor work, and a heat pump by COP_HP = q_H/w_in. Both exceed one and both fall short of the Carnot limit for the same temperatures.

Engineering connection: these two cycles, joined with the gas cycles of Chapter 9, cover almost all power generation, heating, and cooling in service today.

Worked example 1: an ideal Rankine cycle

An ideal Rankine cycle uses steam at the turbine inlet at 3 MPa and 350 °C (h₃ = 3115.3 kJ/kg, s₃ = 6.7427 kJ/kg·K) and condenses at 10 kPa (h_f = 191.81 kJ/kg, v_f = 0.001010 m³/kg, s_f = 0.6492, s_fg = 7.5010 kJ/kg·K, h_fg = 2392.82 kJ/kg). Find the pump work, the heat added, the net work, and the thermal efficiency.

Figure 1. The ideal Rankine cycle on a T-s diagram: heat added 2 to 3, isentropic expansion 3 to 4, heat rejected 4 to 1, and a near-vertical pump step 1 to 2. The enclosed area is the net work.

ProblemFind w_p, q_in, w_net, and η for the Rankine cycle in Figure 1.
Given / findTurbine inlet 3 MPa, 350 °C (h₃, s₃ given); condenser 10 kPa (saturation data given). Find pump work, heat added, net work, efficiency.
AssumptionsIdeal cycle: isentropic pump and turbine, constant-pressure boiler and condenser, saturated liquid leaving the condenser.
ModelState 1 is saturated liquid at 10 kPa; the pump raises it to 3 MPa; the turbine exit follows from s₄ = s₃; then apply the four device balances.
Equationsw_p = v_f(P₂ − P₁), h₂ = h₁ + w_px₄ = (s₃ − s_f)/s_fg, h₄ = h_f + x₄ h_fgη = (w_t − w_p)/q_in
Solvew_p = 0.001010 × (3000 − 10) = 3.02 kJ/kg, so h₂ = 191.81 + 3.02 = 194.83 kJ/kg. x₄ = (6.7427 − 0.6492)/7.5010 = 0.812, h₄ = 191.81 + 0.812 × 2392.82 = 2135.6 kJ/kg. q_in = h₃ − h₂ = 3115.3 − 194.83 = 2920.4 kJ/kg. w_t = h₃ − h₄ = 979.6 kJ/kg, so w_net = 979.6 − 3.02 = 976.6 kJ/kg. η = 976.6/2920.4 = 0.334.
CheckThe Carnot limit between 350 °C (623 K) and 45.81 °C (319 K) is 1 − 319/623 = 0.488, well above 0.334, as required. The pump work is under one percent of the turbine work, confirming why pumping a liquid is cheap.
ConclusionA real plant raises this efficiency with superheat, reheat, and regeneration, but the skeleton is exactly these four enthalpy differences.

Result. w_p = 3.02 kJ/kg, q_in = 2920.4 kJ/kg, w_net = 976.6 kJ/kg, η = 0.334.

Worked example 2: a vapor-compression refrigerator

An ideal R-134a refrigeration cycle evaporates at −20 °C (saturated vapor: h₁ = 386.08 kJ/kg, s₁ = 1.7395 kJ/kg·K) and condenses at 800 kPa, leaving as saturated liquid (h₃ = 243.7 kJ/kg). The compressor is isentropic; at 800 kPa the superheated table gives h = 415.72 kJ/kg at s = 1.7150 and h = 424.86 kJ/kg at s = 1.7446 kJ/kg·K. Find the compressor work, the refrigeration effect, and the coefficient of performance.

Figure 2. The vapor-compression cycle: the compressor lifts the refrigerant to condenser pressure, heat is rejected outdoors, a throttle drops the pressure, and the evaporator absorbs heat from the cold space.

ProblemFind w_in, q_L, and COP_R for the refrigerator in Figure 2.
Given / findState 1 (sat vapor, −20 °C): h₁ = 386.08, s₁ = 1.7395; state 3 (sat liquid, 800 kPa): h₃ = 243.7; throttle gives h₄ = h₃; compressor isentropic with the 800 kPa data given. Find w_in, q_L, COP_R.
AssumptionsIdeal cycle: isentropic compressor, saturated states at evaporator and condenser exits, constant-enthalpy throttle.
ModelFind h₂ by interpolating the 800 kPa superheated table at s₂ = s₁; then q_L = h₁ − h₄, w_in = h₂ − h₁, and COP_R = q_L/w_in.
Equationsh₂ = 415.72 + [(1.7395 − 1.7150)/(1.7446 − 1.7150)](424.86 − 415.72)q_L = h₁ − h₄, w_in = h₂ − h₁COP_R = q_L/w_in
SolveThe interpolation fraction is 0.0245/0.0296 = 0.828, so h₂ = 415.72 + 0.828 × 9.14 = 423.3 kJ/kg. q_L = 386.08 − 243.7 = 142.4 kJ/kg. w_in = 423.3 − 386.08 = 37.2 kJ/kg. COP_R = 142.4/37.2 = 3.83.
CheckThe heat rejected q_H = h₂ − h₃ = 179.6 kJ/kg equals q_L + w_in = 142.4 + 37.2, closing the energy balance. The Carnot COP between −20 °C and 31.3 °C is 4.93, so 3.83 sits below it, as it must.
ConclusionThe cycle delivers 3.8 units of cooling per unit of work, and the throttle, not a turbine, is what makes it simple and cheap. Smaller temperature lifts give higher COP.

Result. w_in = 37.2 kJ/kg, q_L = 142.4 kJ/kg, COP_R = 3.83.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Pump work as a gas	Huge, wrong pump work	"Is the fluid a liquid here?"	Use w_p = v_f(P₂ − P₁) for the nearly incompressible liquid.
Wrong condenser-exit state	h₁ taken as a mixture	"Does the condenser deliver saturated liquid?"	In the ideal cycle, state 1 is saturated liquid at the condenser pressure.
Turbine instead of throttle	Refrigerator with a turbine expander	"Is the expansion device a throttle?"	Refrigeration uses a throttle, where h is constant, not a turbine.
COP above Carnot	COP_R exceeds T_L/(T_H − T_L)	"Is my COP under the Carnot value?"	No real cycle can beat the Carnot COP for its temperatures.

Practice ladder

Level 1 · Direct skill

A Rankine turbine takes steam at h₃ = 3115.3 kJ/kg and exhausts it at h₄ = 2135.6 kJ/kg. Ignoring pump work, find the turbine work and the heat added if h₂ = 191.8 kJ/kg.

Show answer

w_t = h₃ − h₄ = 979.7 kJ/kg. q_in = h₃ − h₂ = 3115.3 − 191.8 = 2923.5 kJ/kg. The rough efficiency 979.7/2923.5 = 0.335 matches the full result, since pump work is negligible.

Level 2 · Mixed concept

The refrigerator of Worked Example 2 must provide 3 kW of cooling. What compressor power does it need?

Show answer

Ẇ = Q̇_L/COP_R = 3/3.83 = 0.78 kW. The mass flow would be ṁ = Q̇_L/q_L = 3/142.4 = 0.021 kg/s.

Level 3 · Independent problem

The same R-134a cycle is used as a heat pump for the warm space. Using q_H = 179.6 kJ/kg and w_in = 37.2 kJ/kg, find its COP.

Show answer

COP_HP = q_H/w_in = 179.6/37.2 = 4.83, which is COP_R + 1 = 3.83 + 1, exactly as expected for the same cycle.

Level 4 · Transfer to real engineering

Explain why power plants superheat the steam before the turbine, using both efficiency and the wetness of the steam leaving the turbine.

What good work looks like

Superheating raises the average temperature of heat addition, lifting efficiency, and it keeps the turbine-exit quality high so less liquid forms, protecting the blades from erosion.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I used v_f for pump work and saturated liquid at the condenser exit."

"Give me four cycle states; I will name the device that connects each pair."

"What is the cycle efficiency?" Assembling the four enthalpy differences yourself is the skill.

"Compute the COP." Forming q_L/w_in from the states is the point.

Portfolio task

Analyse a real power plant or refrigerator at a chosen pair of pressures. Fix the four states from the tables, compute the efficiency or COP, and compare it with the Carnot limit.

Must include: four located states, the four device balances, and an efficiency or COP below the Carnot value.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Name the four Rankine components and their roles.

Pump (raise liquid pressure), boiler (add heat), turbine (produce work), condenser (reject heat).

2. Why is pump work small?

Liquid is nearly incompressible, so w_p = v_f(P₂ − P₁) is tiny next to the turbine work.

3. How does the refrigeration cycle differ from Rankine?

Energy flows the other way: a compressor and throttle replace the turbine and pump, moving heat from cold to hot.

4. Write the refrigerator COP.

COP_R = q_L/w_in = (h₁ − h₄)/(h₂ − h₁).

5. Name three ways to improve a Rankine cycle.

Superheat, reheat, and regeneration, all raising the average heat-addition temperature.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the Rankine efficiency and the COP from a blank page.

+3 daysAnalyse a cycle at new pressures.

+7 daysRevisit the whole study pattern from the course home.

+30 daysConnect these cycles to Heat Transfer and Energy Systems.

Textbook mapping

Item	Mapping
Primary source	Borgnakke and Sonntag, Fundamentals of Thermodynamics, Chapter 11 (Power and Refrigeration Systems, Vapor Working Fluids) and Appendix B
Cross-reference	Çengel and Boles, Ch. 10 and 11 · Moran, Ch. 8 and 10
Core topics	10.1 Rankine components · 10.2 Cycle analysis · 10.3 Improvements · 10.4 Vapor-compression refrigeration · 10.5 Coefficient of performance
Engineering connection	The Rankine and refrigeration cycles run nearly all power generation, heating, and cooling.
Read next	Return to the course home, then continue to Heat Transfer.