Thermodynamics · Chapter 9
Thermodynamics · Chapter 9

Gas Power Cycles

Gas power cycles connect component models into repeatable state sequences. Air-standard cycles reveal how compression ratio, pressure ratio, temperature limits, and component irreversibility govern net work, efficiency, and practical design tradeoffs.

Learning objectives

  • Map Otto, Diesel, Stirling, Ericsson, Brayton, and jet cycles by process and component.
  • Apply air-standard assumptions while stating where real engines depart from them.
  • Determine state temperatures, heat transfers, work, thermal efficiency, and cycle closure.
  • Distinguish ideal isentropic states from actual component states and supplied efficiencies.
  • Audit mass, energy, entropy, exergy, choking branches, and uncertainty before design conclusions.
9–1

Basic Considerations in Power-Cycle Analysis

Choose the cycle boundary

A closed cycle returns its working fluid to the initial state, so property changes vanish and \(W_{net,out}=Q_{net,in}\). An open gas turbine uses steady component control volumes; connecting their outlet and inlet states creates the cycle model.

Physical meaning

Area enclosed on P–v or T–s diagrams represents net work or reversible net heat. Larger area is useful only if pressure, temperature, materials, emissions, and component limits remain acceptable.

9–2

The Carnot Cycle and Its Value in Engineering

Carnot efficiency \(1-T_L/T_H\) is a reversible ceiling between two reservoirs, not a component model. Gas cycles exchange heat over temperature ranges, so compare using suitable average boundary temperatures and never exceed a compatible Carnot bound.

9–3

Air-Standard Assumptions

Air-standard analysis treats air as an ideal gas circulating in a closed loop, replaces combustion and exhaust by external heat transfer, and idealizes internal processes as reversible. Cold-air-standard additionally fixes specific heats; variable-property tables are preferable across large temperature ranges.

9–4

An Overview of Reciprocating Engines

Geometry and performance
\[r=\frac{V_{BDC}}{V_{TDC}},\qquad MEP=\frac{W_{net}}{V_{BDC}-V_{TDC}}\]

Compression ratio is geometric; pressure ratio is not a substitute. Mean effective pressure compares displacement utilization, while brake efficiency includes mechanical losses that ideal cycle efficiency omits.

9–5

Otto Cycle: Ideal Cycle for Spark-Ignition Engines

The sequence is isentropic compression, constant-volume heat addition, isentropic expansion, and constant-volume heat rejection. For supplied constant k, \(\eta=1-r^{1-k}\); real compression ratio is limited by knock and thermal-mechanical constraints.

EXAMPLE 1Screening an ideal Otto cycle
An air-standard Otto cycle has compression ratio 9.00 and uses supplied constant k=1.40. Determine its ideal thermal efficiency and explain why this is not a vehicle efficiency.

Recognition strategy

The constant-volume heat-addition cycle and compression ratio identify the cold-air-standard Otto efficiency. No property interpolation is needed because k is explicitly supplied.

Given information

  • Closed fixed-mass air model completing a cycle; states close so net property changes are zero.
  • All state temperatures and pressures would be absolute; only dimensionless ratios are needed here.
  • Supplied ideal-gas constant \(k=c_p/c_v=1.40\); mass basis cancels in efficiency.
  • Heat in and work out are positive; exact arithmetic within an approximate air-standard model; cycle KE and PE changes are neglected.

Model and interpretation

Apply \(\eta_{th}=1-r^{1-k}\). This benchmark assumes internally reversible compression and expansion and does not supply combustion, friction, pumping, or heat-loss efficiencies.

Step-by-step solution
1. Classify. Identify the Otto sequence: isentropic compression, constant-volume heating, isentropic expansion, and constant-volume cooling.
2. Formulate. For constant specific heats, \(\eta_{th}=1-1/r^{k-1}\), with \(r=9.00\).
3. Evaluate. \(9.00^{0.40}=2.408\), so the rejected-to-added heat ratio is approximately \(1/2.408=0.4153\).
4. Calculate. The ideal efficiency is \(1-0.4153=0.5847\), or \(58.5\%\) to three significant figures.

Verification

Units: Compression ratio and k are dimensionless, so thermal efficiency is correctly dimensionless.

Signs and direction: Positive efficiency below one means positive net work output and positive net heat input over the closed cycle.

Plausibility: The value remains below a Carnot limit for compatible temperature extremes; energy closes as \(W_{net}=Q_{in}-Q_{out}\), and ideal processes have zero entropy generation.

Final result: \(\eta_{th,Otto}=0.585\), or approximately \(58.5\%\).

Engineering interpretation

Increasing compression ratio raises this ideal result, but real spark-ignition engines face knock, heat transfer, friction, emissions, and variable specific heats. Vendor data and test maps govern release decisions.

9–6

Diesel Cycle: Ideal Cycle for Compression-Ignition Engines

Cold-air-standard Diesel efficiency
\[\eta=1-\frac{1}{r^{k-1}}\frac{r_c^k-1}{k(r_c-1)}\]

The cutoff ratio \(r_c=V_3/V_2\) belongs to constant-pressure heat addition. Compare Otto and Diesel cycles only under clearly matched constraints; equal compression ratio and equal peak pressure are different comparisons.

9–7

Stirling and Ericsson Cycles

Ideal Stirling and Ericsson cycles use perfect regeneration and reversible isothermal heat exchange to attain Carnot efficiency. Real regenerators have finite effectiveness, pressure loss, dead volume, conduction, and cyclic transients, so effectiveness must be supplied or measured.

9–8

Brayton Cycle: Ideal Cycle for Gas-Turbine Engines

Brayton components are an isentropic compressor, constant-pressure heater, isentropic turbine, and constant-pressure cooler. The compressor consumes a large fraction of turbine work, making net output sensitive to both efficiencies and inlet temperature.

Ideal constant-k efficiency
\[\eta_{Brayton}=1-r_p^{-(k-1)/k}\]
sT
Figure 9–1  A Brayton cycle on a T–s diagram: heat added and rejected at constant pressure, bounded by two isentropic processes.
EXAMPLE 2Ideal Brayton state and work audit
An ideal Brayton cycle takes air at 300 K through pressure ratio 8.00 to a turbine inlet of 1200 K. Using supplied cp=1.005 kJ/(kg·K) and k=1.40, find net work and efficiency.

Recognition strategy

Two isentropic devices linked by constant-pressure heat transfer identify a cold-air-standard Brayton cycle. Calculate all four temperatures before energy transfers to expose cycle closure.

Given information

  • Steady-flow control volumes form a closed-loop ideal cycle on a 1 kg air basis.
  • Pressures are absolute and temperatures kelvin; \(r_p=P_2/P_1=P_3/P_4=8.00\).
  • Supplied ideal-gas constants: \(c_p=1.005\ \text{kJ/(kg\!\cdot\!K)}\), \(k=1.40\).
  • Work out and heat in are positive; internally reversible devices, constant properties, negligible KE and PE.

Model and interpretation

Use isentropic temperature ratios for compressor and turbine, then steady-flow enthalpy differences \(c_p\Delta T\). Distinguish component work from cycle net work.

Step-by-step solution
1. Temperature ratio. Compute \(a=r_p^{(k-1)/k}=8^{0.285714}=1.811\) for both isentropic devices.
2. Complete states. \(T_2=300a=543.4\ \text{K}\) and \(T_4=1200/a=662.6\ \text{K}\).
3. Work terms. \(w_c=c_p(T_2-T_1)=244.6\ \text{kJ/kg}\) and \(w_t=c_p(T_3-T_4)=540.1\ \text{kJ/kg}\).
4. Cycle result. \(w_{net}=295.5\ \text{kJ/kg}\); with \(q_{in}=c_p(1200-543.4)=659.9\ \text{kJ/kg}\), \(\eta=0.448\).

Verification

Units: Every cp times temperature difference is kJ/kg, and their ratio is dimensionless.

Signs and direction: Turbine work exceeds compressor work, so net work is positive under the selected output-positive convention.

Plausibility: Energy closes because \(q_{out}=c_p(662.6-300)=364.4\ \text{kJ/kg}\) and \(659.9-364.4=295.5\); ideal entropy generation is zero.

Final result: \(w_{net}=295.5\ \text{kJ/kg}\) and \(\eta_{th}=0.448\) (44.8%).

Engineering interpretation

The back-work ratio is substantial, so real compressor efficiency strongly affects plant output. A release model needs variable-property air data, pressure losses, combustion composition, turbine cooling, and vendor maps.

9–9

The Brayton Cycle with Regeneration

A regenerator is useful only when turbine exhaust is hotter than compressor discharge. Its supplied effectiveness determines recovered heat; pressure loss and added mass can offset fuel savings. Iterate states when temperature-dependent properties are used.

9–10

The Brayton Cycle with Intercooling, Reheating, and Regeneration

Intercooling reduces compressor work and reheating increases turbine work, but each alone can increase required external heat. Near-equal stage pressure ratios are an ideal starting point, not a final design after pressure losses and maps.

EXAMPLE 3Real gas-turbine component estimate
A simple gas turbine compresses air from 100 kPa absolute and 300 K to pressure ratio 6.00, then heats it to 1100 K. Given compressor efficiency 0.82 and turbine efficiency 0.86, estimate net specific work.

Recognition strategy

Supplied actual component efficiencies distinguish this from an ideal Brayton cycle. Find isentropic comparison states at the same exit pressures, then convert to actual temperatures.

Given information

  • Steady compressor, heater, and turbine control volumes; open-cycle state sequence, evaluated per kilogram of air.
  • All pressures are absolute and temperatures kelvin; pressure losses are neglected and expansion returns to 100 kPa absolute.
  • Supplied constants \(c_p=1.005\ \text{kJ/(kg\!\cdot\!K)}\), \(k=1.40\), \(\eta_c=0.82\), \(\eta_t=0.86\).
  • Adiabatic components, work out positive, mass basis, negligible KE and PE; ideal-gas constant-property approximation.

Model and interpretation

Use \(\eta_c=(T_{2s}-T_1)/(T_{2a}-T_1)\) and \(\eta_t=(T_3-T_{4a})/(T_3-T_{4s})\). Actual adiabatic devices generate entropy even though ideal comparison paths do not.

Step-by-step solution
1. Ideal states. With \(a=6^{0.285714}=1.6685\), \(T_{2s}=500.6\ \text{K}\) and \(T_{4s}=659.3\ \text{K}\).
2. Actual compressor. \(T_{2a}=300+(500.6-300)/0.82=544.6\ \text{K}\), giving \(w_c=245.8\ \text{kJ/kg}\).
3. Actual turbine. \(T_{4a}=1100-0.86(1100-659.3)=720.9\ \text{K}\), giving \(w_t=381.0\ \text{kJ/kg}\).
4. Net and heat. \(w_{net}=381.0-245.8=135.2\ \text{kJ/kg}\); \(q_{in}=1.005(1100-544.6)=558.2\ \text{kJ/kg}\).

Verification

Units: Specific work terms have kJ/kg units and efficiency ratios are dimensionless.

Signs and direction: Both inefficiencies raise compressor work or lower turbine work, so actual net work is below the corresponding ideal value.

Plausibility: The energy audit gives \(\eta_{th}=135.2/558.2=0.242\); positive component entropy generation is consistent with \(T_{2a}>T_{2s}\) and \(T_{4a}>T_{4s}\).

Final result: \(w_{net}\approx135\ \text{kJ/kg}\) and estimated \(\eta_{th}\approx24.2\%\).

Engineering interpretation

This estimate shows why ideal-cycle efficiency cannot be paired with real work values. Procurement or controls work must use certified component maps, real-gas properties, pressure losses, fuel flow, cooling bleeds, choking margins, emissions, and uncertainty.

9–11

Ideal Jet-Propulsion Cycles

Uninstalled one-dimensional thrust
\[F=\dot m(V_e-V_0)+(P_e-P_0)A_e\]

Retain pressure thrust unless the nozzle is perfectly expanded. Check whether the nozzle is choked before selecting an exit-pressure branch; shocks, variable area, inlet momentum, fuel mass, and installation drag can change thrust.

9–12

Second-Law Analysis of Gas Power Cycles

Combustion, finite-temperature heat transfer, pressure loss, mixing, and component friction generate entropy and destroy exergy. Audit each component, then sum destructions and verify against the cycle exergy balance.

Problem-solving strategy

  1. Recognize whether the boundary encloses a fixed mass, one steady component, or an entire cycle.
  2. Identify every state with absolute pressure and temperature, phase, and an authoritative property source.
  3. Write mass, energy, entropy, and when relevant exergy balances before inserting component models.
  4. Separate ideal benchmark states from actual states and iterate when properties depend on the unknown temperature.
  5. Audit units, signs, cycle closure, entropy generation, and sensitivity to supplied data.

Chapter recap

  • Map every state and process before applying a named-cycle efficiency formula.
  • Keep thermal efficiency distinct from COP, propulsive efficiency, and component isentropic efficiency.
  • Close energy around the cycle and require nonnegative entropy generation in every real component.

Before moving on

You should be able to calculate ideal and actual state sequences, explain applicability limits, test regeneration and choking conditions, and identify which vendor or property data control a professional result.