MEC-230 · Lesson 12 of 12 · 26 min

Introduction to power and refrigeration cycles

Every idea in this course comes together in the thermodynamic cycles that generate most of the world's electricity and provide most of its cooling. A steam power plant is a heat engine built from four steady-flow devices: a pump, a boiler, a turbine and a condenser. A refrigerator or heat pump is the same set of devices arranged to run the other way, with a compressor and an expansion valve. In this final lesson we analyze the ideal Rankine cycle, the model for steam power plants, then see how real plants depart from it and how cogeneration plants such as Northgate's use the same hardware more completely. We finish with the vapor-compression refrigeration cycle, which drives Northgate's chillers and its sewer heat pump.

Why not a Carnot cycle?

Lesson 10 showed that the Carnot cycle is the most efficient possible between two temperatures, so it is natural to ask why power plants do not use it. With steam, a Carnot cycle could run entirely under the dome: isothermal boiling, isentropic expansion, isothermal condensation, isentropic compression. Two practical problems kill it. The isentropic expansion would end at low quality, and water droplets erode turbine blades. Worse, the compression would have to handle a wet mixture of liquid and vapor, which no practical compressor or pump can do efficiently. The Rankine cycle, devised in the 1850s, fixes both problems: it superheats the steam before the turbine and condenses it completely before the pump.

The ideal Rankine cycle

The ideal Rankine cycle has four internally reversible processes. (1-2) Isentropic compression in a pump: saturated liquid from the condenser is pumped to boiler pressure. (2-3) Constant-pressure heat addition in the boiler: the compressed liquid is heated, boiled and superheated. (3-4) Isentropic expansion in a turbine, producing work. (4-1) Constant-pressure heat rejection in the condenser, which returns the fluid to saturated liquid. Each component is a steady-flow device, analyzed with the equations of Lesson 8.

Two parts. On the left, a schematic loop: a pump at bottom left sends water up to a boiler on the left side, the boiler sends steam to a turbine at the top right, the turbine exhausts to a condenser at the bottom right, and the condenser returns liquid to the pump. Arrows show heat in at the boiler, work out at the turbine, heat out at the condenser and work in at the pump. States 1 to 4 are numbered at each connection. On the right, a temperature-entropy diagram with the vapor dome. State 1 is saturated liquid at the bottom left of the dome, state 2 just above it, the heating line rises along the 4 megapascal line through the dome to state 3 at 400 degrees in the superheated region, a vertical line drops from 3 to state 4 inside the dome at 10 kilopascals, and a horizontal line returns to state 1.
The ideal Rankine cycle: schematic and T-s diagram. The area enclosed on the T-s diagram is the net work.

On a T-s diagram the cycle encloses an area equal to the net work, and the area under the heat-addition line equals \(q_{\text{in}}\). The steady-flow energy equation for each component, with negligible kinetic and potential energy changes, gives

\[w_{\text{pump}} = h_2 - h_1 = v_1(P_2 - P_1) \qquad q_{\text{in}} = h_3 - h_2 \qquad w_{\text{turb}} = h_3 - h_4 \qquad q_{\text{out}} = h_4 - h_1\]
\[\eta_{th} = \frac{w_{\text{net}}}{q_{\text{in}}} = \frac{w_{\text{turb}} - w_{\text{pump}}}{q_{\text{in}}} = 1 - \frac{q_{\text{out}}}{q_{\text{in}}}\]

Worked problem

An ideal Rankine cycle with Northgate steam conditions

Suppose Northgate's boilers fed a condensing turbine instead of the district main. Steam enters the turbine at 4 MPa and 400 °C (\(h_3 = 3214.5\), \(s_3 = 6.7714\)) and is condensed at 10 kPa (\(h_f = 191.81\), \(h_{fg} = 2392.1\), \(s_f = 0.6492\), \(s_{fg} = 7.4996\), \(v_f = 0.001010\)). Find the thermal efficiency of the ideal cycle.

Show solution
  1. Pump: \(w_p = 0.001010(4000 - 10) = 4.03\ \text{kJ/kg}\); \(h_2 = 191.81 + 4.03 = 195.84\ \text{kJ/kg}\).
  2. Turbine exit: \(x_4 = (6.7714 - 0.6492)/7.4996 = 0.816\); \(h_4 = 191.81 + 0.816(2392.1) = 2144.5\ \text{kJ/kg}\).
  3. \(w_t = 3214.5 - 2144.5 = 1070.0\ \text{kJ/kg}\); \(q_{\text{in}} = 3214.5 - 195.84 = 3018.7\ \text{kJ/kg}\).
  4. \[\eta_{th} = \frac{1070.0 - 4.03}{3018.7} = 0.353\]

Answer: About 35 percent. The Carnot efficiency between 400 °C and the condensing temperature of 45.8 °C is 52.6 percent; the Rankine cycle falls short mainly because most of its heat is added well below 400 °C, while the water is being warmed and boiled at 250 °C. The pump work is only 0.4 percent of the turbine work, a back work ratio that shows why liquid pumping makes the Rankine cycle so attractive.

Real cycles and how to improve them

Real pumps and turbines are not isentropic. With the isentropic efficiencies of Lesson 11, the turbine work falls and the pump work rises. Pressure drops in the boiler and piping, and heat losses, cut performance further. The program below repeats the ideal analysis and then applies efficiencies of 85 percent to both the turbine and the pump: the thermal efficiency drops from 35.3 to 30.0 percent.

The ideal and actual Rankine cycle, cogeneration utilization at Northgate, and an ideal vapor-compression cycle.
# The Rankine cycle, cogeneration at Northgate, and a compression refrigeration cycle
# Steam properties: 10 kPa saturation and 4 MPa, 400 C, rounded from IAPWS-based tables
hf, hfg, sf, sfg, vf = 191.81, 2392.1, 0.6492, 7.4996, 0.001010   # 10 kPa
h3, s3 = 3214.5, 6.7714                                          # 4 MPa, 400 C
P_boiler, P_cond = 4000.0, 10.0

def rankine(eta_t=1.0, eta_p=1.0):
    h1 = hf
    wp = vf * (P_boiler - P_cond) / eta_p
    h2 = h1 + wp
    x4s = (s3 - sf) / sfg
    h4s = hf + x4s * hfg
    wt = eta_t * (h3 - h4s)
    h4 = h3 - wt
    qin, qout = h3 - h2, h4 - h1
    return wp, wt, qin, qout, (wt - wp) / qin, x4s, h4

for label, et, ep in (("ideal", 1.0, 1.0), ("actual (0.85, 0.85)", 0.85, 0.85)):
    wp, wt, qin, qout, eta, x4s, h4 = rankine(et, ep)
    print(f"{label:20s} wp = {wp:5.2f}  wt = {wt:7.1f}  qin = {qin:7.1f}  qout = {qout:7.1f}"
          f"  eta = {eta:.3f}")
print(f"isentropic turbine exit quality x4s = {rankine()[5]:.3f}; back work ratio = {rankine()[0] / rankine()[1]:.4f}")
print(f"Carnot limit between 400 C and 45.8 C: {1 - (45.81 + 273.15) / (400 + 273.15):.3f}")

# Cogeneration at Northgate: back-pressure turbine exhausting to the 1 MPa main
m, h_fw = 12.0, 445.1                       # kg/s; feedwater enthalpy after the pump
Q_boiler = m * (3214.5 - h_fw)
fuel = Q_boiler / 0.85                      # boiler efficiency 85 %
power = 3227 * 0.96                         # shaft power x generator efficiency
heat = (m - 0.48) * (2943.1 - 356.0) - 1000 # steam to buildings, less about 1 MW of losses
print(f"\nboiler {Q_boiler:.0f} kW, fuel {fuel:.0f} kW, power {power:.0f} kW, heat delivered {heat:.0f} kW")
print(f"utilization factor = {(power + heat) / fuel:.3f}")

# Ideal compression refrigeration cycle, R-134a, evaporating at -10 C and condensing at 40 C (rounded values)
h1, h2, h3r = 244.5, 278.0, 108.3
qL, w = h1 - h3r, h2 - h1
print(f"\nrefrigeration: qL = {qL:.1f} kJ/kg, w = {w:.1f} kJ/kg, COP_R = {qL / w:.2f}, COP_HP = {(qL + w) / w:.2f}")
print(f"Carnot COP_R between -10 C and 40 C = {263.15 / 50:.2f}")
Output
ideal                wp =  4.03  wt =  1069.9  qin =  3018.7  qout =  1952.8  eta = 0.353
actual (0.85, 0.85)  wp =  4.74  wt =   909.4  qin =  3017.9  qout =  2113.2  eta = 0.300
isentropic turbine exit quality x4s = 0.816; back work ratio = 0.0038
Carnot limit between 400 C and 45.8 C: 0.526

boiler 33233 kW, fuel 39097 kW, power 3098 kW, heat delivered 28803 kW
utilization factor = 0.816

refrigeration: qL = 136.2 kJ/kg, w = 33.5 kJ/kg, COP_R = 4.07, COP_HP = 5.07
Carnot COP_R between -10 C and 40 C = 5.26

Three changes raise the efficiency of a Rankine plant, all by raising the average temperature at which heat is added or lowering the temperature at which it is rejected. Lowering the condenser pressure lowers the rejection temperature; large plants condense at 5 to 10 kPa, limited by the temperature of the cooling water. Superheating the steam to higher temperatures raises the average heat-addition temperature and also dries the turbine exhaust; modern plants reach 600 °C or more, limited by the creep strength of steels. Raising the boiler pressure raises the boiling temperature; supercritical plants operate above 22 MPa. Two further refinements, reheat (expanding in stages and reheating the steam between them) and regeneration (heating the feedwater with steam extracted from the turbine), are the subject of a later course on power cycles. Northgate's deaerator, heated with steam from the main, is a simple form of regenerative feedwater heating.

Cogeneration: using the rejected heat

Even the best power stations reject more than half their fuel energy to cooling water at temperatures too low to be useful. Cogeneration, or combined heat and power, places the heat rejection at a temperature useful to someone: the Northgate turbine exhausts at 1 MPa and about 250 °C, hot enough to heat forty buildings. The price is less electricity per kilogram of steam, since the turbine expands only to 1 MPa instead of 10 kPa. The reward is that almost all the fuel energy is used. The performance of a cogeneration plant is measured by its utilization factor, the useful output (electricity plus delivered heat) divided by the fuel energy input:

\[\epsilon_u = \frac{\dot{W}_{\text{net}} + \dot{Q}_{\text{useful}}}{\dot{Q}_{\text{fuel}}}\]

Worked problem

Utilization factor of the Northgate plant

At full load the boilers deliver 12 kg/s of steam at 4 MPa and 400 °C from feedwater at \(h = 445.1\ \text{kJ/kg}\), with a boiler efficiency of 85 percent. The turbine generator produces 3.10 MW of electricity. The buildings receive steam at 1 MPa and 250 °C, apart from the 0.48 kg/s used in the deaerator, and return condensate at 85 °C; about 1 MW is lost from the distribution piping. Find the utilization factor.

Show solution
  1. Boiler output: \(12(3214.5 - 445.1) = 33{,}233\ \text{kW}\). Fuel input: \(33{,}233/0.85 = 39{,}100\ \text{kW}\).
  2. Heat delivered: \((12 - 0.48)(2943.1 - 356.0) - 1000 = 28{,}800\ \text{kW}\).
  3. \[\epsilon_u = \frac{3100 + 28{,}800}{39{,}100} = 0.816\]

Answer: About 82 percent of the fuel energy is used, either as electricity or as heat in the buildings, against perhaps 50 to 60 percent for a modern gas-fired power station alone. The remaining 18 percent is lost mostly up the boiler stacks and from the pipes. That number is the starting point for every argument about Northgate's future, from heat-pump conversion to lower-temperature distribution.

The vapor-compression refrigeration cycle

Reverse the Rankine cycle and replace the pump and turbine with a compressor and an expansion valve, and you have the vapor-compression refrigeration cycle, used in nearly every refrigerator, air conditioner, chiller and heat pump. The ideal cycle has four processes. (1-2) Saturated vapor from the evaporator is compressed isentropically to the condenser pressure, becoming superheated. (2-3) The vapor is cooled and condensed at constant pressure, rejecting heat \(q_H\) to the warm surroundings. (3-4) The saturated liquid is throttled through an expansion valve to the evaporator pressure, partly flashing to vapor and cooling sharply; this step is isenthalpic and irreversible, and is the one deliberate departure from a reversed Carnot cycle, accepted because an expansion turbine handling a wet mixture would recover little work at great cost. (4-1) The cold mixture evaporates at constant pressure, absorbing heat \(q_L\) from the cold space.

A schematic loop of the vapor-compression cycle. A compressor at the top right receives low-pressure vapor at state 1 and delivers high-pressure vapor at state 2 to a condenser at the top left, which rejects heat Q H to the warm space. Liquid at state 3 leaves the condenser and passes down through an expansion valve to state 4, a cold low-pressure mixture, which enters an evaporator at the bottom that absorbs heat Q L from the cold space and returns vapor to the compressor. Work enters the compressor.
The vapor-compression cycle. The same loop is a refrigerator when Q_L is the purpose and a heat pump when Q_H is the purpose.
\[q_L = h_1 - h_4 \qquad w_{\text{in}} = h_2 - h_1 \qquad q_H = h_2 - h_3 \qquad h_4 = h_3\]

Worked problem

A small chiller using R-134a

An ideal vapor-compression cycle using refrigerant R-134a evaporates at −10 °C and condenses at 40 °C. From R-134a property tables, rounded: saturated vapor at −10 °C, \(h_1 \approx 244.5\ \text{kJ/kg}\); after isentropic compression to the condenser pressure, \(h_2 \approx 278.0\); saturated liquid at 40 °C, \(h_3 \approx 108.3\ \text{kJ/kg}\). Find the refrigeration effect, the work, and the COP as a refrigerator and as a heat pump.

Show solution
  1. Throttling: \(h_4 = h_3 = 108.3\ \text{kJ/kg}\).
  2. \(q_L = 244.5 - 108.3 = 136.2\ \text{kJ/kg}\); \(w_{\text{in}} = 278.0 - 244.5 = 33.5\ \text{kJ/kg}\).
  3. \(\text{COP}_R = 136.2/33.5 = 4.07\); \(\text{COP}_{HP} = (136.2 + 33.5)/33.5 = 5.07\).
  4. Carnot limit as a refrigerator: \(263.15/(313.15 - 263.15) = 5.26\).

Answer: The ideal vapor-compression cycle achieves about 77 percent of the Carnot COP, the shortfall coming from the throttling valve and the superheated discharge. Real machines, with compressor inefficiency and finite temperature differences in the heat exchangers, typically reach 45 to 60 percent of Carnot, as the sewer heat pump in Lesson 10 did.

Refrigerant choice is now shaped by environmental rules as much as by thermodynamics. R-134a has no effect on stratospheric ozone but is a potent greenhouse gas, and the Kigali Amendment to the Montreal Protocol commits Canada, the United States and most other countries to phasing down such hydrofluorocarbons. New chillers and heat pumps increasingly use low-global-warming refrigerants such as hydrofluoroolefins, ammonia, carbon dioxide or hydrocarbons. Each has different properties, pressures and safety requirements, but the cycle analysis is identical: only the property tables change.

The course in one plant

Look back over the Northgate plant with the whole course in hand. The gauges on the morning round become absolute pressures (Lesson 1). Each energy flow is heat or work across a chosen boundary (Lesson 2). The steam's state is fixed by its phase and two properties (Lessons 3 and 4); combustion air, flue gas and instrument air are ideal gases (Lesson 5). Vessels and cylinders obey the closed-system first law (Lesson 6); the turbine, pumps, valves, deaerator and heat exchangers obey the steady-flow energy equation (Lessons 7 and 8); the air receiver charges as a transient (Lesson 9). The second law limits the turbine, the heat pump and any engine anyone offers to sell the plant (Lesson 10), and entropy shows where work potential is destroyed (Lesson 11). The cycles of this lesson tie them together. Further courses build on this base: power and refrigeration cycles in detail, gas mixtures and psychrometrics for air conditioning, combustion, heat transfer and fluid mechanics. Tomás finished his co-op term with a report recommending a larger heat pump and a lower loop temperature, every number in it traceable to the ideas in this course.

Common mistakes

  • MistakeIncluding the pump work with the wrong sign in the net work.

    Insteadw_net = w_turb − w_pump.

  • MistakeUsing the turbine inlet entropy with the wrong pressure to find h₄s.

    Insteadh₄s is at the condenser pressure with s₄ = s₃.

  • MistakeAssuming the expansion valve is isentropic.

    InsteadThrottling is isenthalpic: h₄ = h₃, and entropy rises.

  • MistakeCalling a cogeneration utilization factor a thermal efficiency.

    InsteadUtilization counts useful heat as output; thermal efficiency counts only work.

  • MistakeExpecting a real cycle to reach the Carnot efficiency.

    InsteadHeat added at varying temperatures and real irreversibilities keep real cycles well below Carnot.

Summary and key terms

  • The ideal Rankine cycle: isentropic pumping, constant-pressure heating, isentropic expansion, constant-pressure condensation.
  • η_th = (w_turb − w_pump)/q_in; pump work is a tiny fraction of turbine work.
  • Efficiency rises with higher boiler pressure and superheat temperature and lower condenser pressure.
  • Real turbines and pumps reduce efficiency according to their isentropic efficiencies.
  • Cogeneration rejects heat at a useful temperature; its utilization factor counts electricity plus useful heat.
  • The vapor-compression cycle: isentropic compression, condensation, throttling (h₄ = h₃), evaporation.
  • COP_R = (h₁ − h₄)/(h₂ − h₁); COP_HP = COP_R + 1.
  • Refrigerant choice is constrained by environmental rules; the analysis method is the same for all.
Rankine cycle
The ideal cycle for steam power plants: pump, boiler, turbine and condenser.
Back work ratio
The ratio of compressor or pump work to turbine work in a power cycle.
Superheat
Heating steam above its saturation temperature before the turbine.
Reheat
Reheating steam between turbine stages to raise efficiency and dry the exhaust.
Regeneration
Heating feedwater with steam extracted from the turbine.
Cogeneration
Combined production of electricity and useful heat from one fuel input.
Utilization factor
The ratio of electricity plus useful heat to fuel energy in a cogeneration plant.
Vapor-compression cycle
The refrigeration cycle of compressor, condenser, expansion valve and evaporator.
Expansion valve
A throttling device that lowers refrigerant pressure at constant enthalpy.
Refrigeration effect
The heat absorbed per unit mass of refrigerant in the evaporator, q_L.

Practice problems

Work each problem on paper first, then reveal the answer to check.

  1. In the ideal Rankine example, what is the heat rejected per kilogram?

    Show answer

    1952.7 kJ/kg (2144.5 − 191.8).

  2. For the ideal cycle, what steam flow gives 10 MW of net power?

    Show answer

    About 9.38 kg/s.

  3. If the turbine efficiency is 0.85, what is the turbine exit enthalpy?

    Show answer

    About 2305 kJ/kg.

  4. A refrigeration cycle has h₁ = 250, h₂ = 285 and h₃ = 105 kJ/kg. Find COP_R.

    Show answer

    4.14.

  5. A cogeneration plant burns 50 MW of fuel and delivers 12 MW of electricity and 26 MW of heat. Find its utilization factor.

    Show answer

    0.76.

  6. In the program, lower the condenser pressure's saturation data to 5 kPa (h_f = 137.75, h_fg = 2423.0, s_f = 0.4762, s_fg = 7.9176, v_f = 0.001005). What ideal efficiency results?

    Show answer

    About 0.373.

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