Lecture 9: Gas Power Cycles

High-Performance Internal Combustion: The Bugatti Veyron Benchmark

Modern internal combustion power systems achieve extraordinary energy conversion through advanced thermodynamic design. A premier baseline in high-performance automotive engineering is established by the Bugatti Veyron supercar.

Technical Specifications of the Bugatti Veyron

• General Information:

  • Vehicle Type: SuperCar

  • Base Price: $\$1,700,000.00$

  • Fuel Tank Capacity: $25\,\text{gal.}\;(94.6\,\text{l})$

  • Fuel Economy: $8\,\text{city} \;/\; 13\,\text{hwy mpg}$

• Dimensions:

  • Length: $174.2\,\text{in.}\;(442\,\text{cm})$

  • Width: $77.9\,\text{in.}\;(198\,\text{cm})$

  • Height: $47\,\text{in.}\;(119\,\text{cm})$

  • Curb Weight: $4162\,\text{lbs}\;(1888\,\text{kg})$

• Powertrain and Performance:

  • Engine Architecture: $8.0\,\text{L}\;\text{W16-Cylinder}$, Quad Turbocharger

  • Displacement: $7993\,\text{cc}$

  • Maximum Power Output: $1001\,\text{HP}$

  • Peak Engine Speed: $6000\,\text{RPM}$

  • Peak Torque: $923\,\text{lb-ft}$ between $2200\text{--}5500\,\text{RPM}$

  • Transmission: $7\text{-speed DSG sequential}$

  • Acceleration ($0\text{--}60\,\text{mph}$): $2.5\,\text{seconds}$

  • Acceleration ($0\text{--}100\,\text{mph}$): $5.1\,\text{seconds}$

  • Quarter Mile Performance: $10.2\,\text{seconds}\;\text{@}\;142.9\,\text{mph}$

  • Terminal Velocity / Maximum Speed: $253\,\text{mph}\;(407\,\text{km/h})$

Fundamental Principles and Air-Standard Assumptions

Thermodynamic cycles converting chemical energy to thermal expansion and mechanical shaft power adhere to the unified progression:

$\text{Combustion} \longrightarrow \text{Pressure} \longrightarrow \text{Work}$

Actual internal combustion systems present substantial operational complexities, including chemical transformations during combustion, turbulent non-equilibrium fluid dynamics, heat losses through cylinder boundaries, and open-system fluid exchanges during intake and exhaust strokes. To analyze these devices quantitatively, simplified idealized models are applied.

The Air-Standard Assumptions

1. Continuous Ideal Working Fluid: The working fluid is modeled as air, continuously circulating within a closed thermodynamic loop and behaving at all times as an ideal gas.

2. Internal Reversibility: All mechanical and thermal processes composing the cycle are considered internally reversible.

3. External Heat Addition: The actual chemical combustion process is replaced by a heat-addition process from an external thermal reservoir.

4. External Heat Rejection: The open exhaust and intake strokes are replaced by a heat-rejection process that restores the working fluid back to its original state.

Cold-Air-Standard Assumptions

When the ideal air-standard assumptions are further simplified by assuming constant specific heat values ($c_p$, $c_v$) evaluated at ambient room temperature ($25^\circ\text{C}$ or $298\,\text{K} \approx 300\,\text{K}$), the analysis is termed a cold-air-standard analysis.

An air-standard cycle denotes any theoretical thermodynamic power cycle operating under these idealizations.

Terminology and Geometry of Reciprocating Engines

Reciprocating engines utilize a piston translating back and forth within a rigid cylindrical housing.

Primary Geometric Parameters

• Top Dead Center (TDC): The position of the piston at its minimum distance from the cylinder head, forming the minimum contained volume within the cylinder ($V_{\text{min}}$).

• Bottom Dead Center (BDC): The position of the piston at its maximum distance from the cylinder head, forming the maximum contained volume within the cylinder ($V_{\text{max}}$).

• Bore: The inner diameter of the cylinder bore.

• Stroke: The linear distance traveled by the piston from TDC to BDC.

• Displacement Volume ($V_d$): The volume swept out by the piston face during a single stroke:

$V_d = V_{\text{max}} - V_{\text{min}} = V_{\text{BDC}} - V_{\text{TDC}}$

• Clearance Volume ($V_c$): The volume remaining above the piston head when it reaches TDC:

$V_c = V_{\text{min}} = V_{\text{TDC}}$

• Compression Ratio ($r$): The non-dimensional ratio of maximum cylinder volume to minimum cylinder volume:

$r = \frac{V_{\text{max}}}{V_{\text{min}}} = \frac{V_{\text{BDC}}}{V_{\text{TDC}}} = \frac{v_{\text{max}}}{v_{\text{min}}} = \frac{v_1}{v_2}$

Mean Effective Pressure (MEP)

The Mean Effective Pressure ($\text{MEP}$) is a fictitious, constant pressure that, if exerted against the piston throughout the power expansion stroke from TDC to BDC, would deliver the exact net work output produced over the entire actual thermodynamic cycle.

$\text{MEP} \times \text{Piston Area} \times \text{Stroke} = \text{MEP} \times (V_{\text{max}} - V_{\text{min}}) = W_{\text{net}}$

$\text{MEP} = \frac{W_{\text{net}}}{V_{\text{max}} - V_{\text{min}}} = \frac{w_{\text{net}}}{v_{\text{max}} - v_{\text{min}}}\;(\text{kPa})$

The Otto Cycle: Spark-Ignition Engines

The Otto cycle is the ideal thermodynamic cycle for four-stroke and two-stroke spark-ignition (SI) internal combustion engines, first built successfully by Nikolaus A. Otto in 1876.

Four-Stroke vs. Two-Stroke Mechanical Operation

• Four-Stroke Cycle:

  • $1\,\text{thermodynamic cycle} = 4\,\text{piston strokes} = 2\,\text{crankshaft revolutions}$

  • Intake Stroke: Piston moves downward from TDC to BDC with intake valve open, drawing in fresh air-fuel mixture at atmospheric pressure ($P_{\text{atm}}$).

  • Compression Stroke: Both valves close; piston travels upward toward TDC, compressing the trapped air-fuel mixture.

  • Power (Expansion) Stroke: The spark plug ignites the mixture near TDC. High-pressure combustion gases drive the piston downward to BDC, generating boundary work.

  • Exhaust Stroke: Piston travels upward with the exhaust valve open, discharging burned gases.

• Two-Stroke Cycle:

  • $1\,\text{thermodynamic cycle} = 2\,\text{piston strokes} = 1\,\text{crankshaft revolution}$

  • Replaces mechanical poppet valves with intake and exhaust ports uncovered by piston movement.

  • Fresh mixture is drawn into and pre-compressed within the crankcase as the piston travels upward. During the downstroke, exhaust gases vent through the exhaust port, and fresh charge rushes in via the intake transfer port.

Two-stroke engines exhibit lower thermal efficiency than four-stroke equivalents because of incomplete scavenging and partial loss of unburned fuel into the exhaust. However, they eliminate valve trains, resulting in mechanical simplicity, low production cost, and high power-to-weight and power-to-volume ratios.

Ideal Otto Cycle Process Breakdown

The ideal Otto cycle models the spark-ignition sequence as a closed-system cycle containing a stationary mass of air:

• Process $1\text{--}2$: Isentropic compression (reversible adiabatic compression of air as the piston moves from BDC to TDC).

• Process $2\text{--}3$: Constant-volume ($v = \text{const.}$) heat addition from an external source (simulating rapid combustion while the piston is momentarily at TDC).

• Process $3\text{--}4$: Isentropic expansion (reversible adiabatic power stroke from TDC to BDC).

• Process $4\text{--}1$: Constant-volume ($v = \text{const.}$) heat rejection (simulating cylinder blowdown and exhaust).

Energy Balance and Thermal Efficiency Formulation

Neglecting kinetic and potential energy changes, the first law applied to closed-system processes gives:

$q_{\text{in}} = u_3 - u_2 = c_v (T_3 - T_2)$

$q_{\text{out}} = u_4 - u_1 = c_v (T_4 - T_1)$

$w_{\text{net}} = q_{\text{in}} - q_{\text{out}}$

$\eta_{\text{th, Otto}} = \frac{w_{\text{net}}}{q_{\text{in}}} = 1 - \frac{q_{\text{out}}}{q_{\text{in}}} = 1 - \frac{c_v (T_4 - T_1)}{c_v (T_3 - T_2)} = 1 - \frac{T_1 \left( \frac{T_4}{T_1} - 1 \right)}{T_2 \left( \frac{T_3}{T_2} - 1 \right)}$

For isentropic processes with constant specific heats:

$\frac{T_2}{T_1} = \left( \frac{v_1}{v_2} \right)^{k-1} = r^{k-1}$

$\frac{T_3}{T_4} = \left( \frac{v_4}{v_3} \right)^{k-1} = \left( \frac{v_1}{v_2} \right)^{k-1} = r^{k-1}$

Equating both relations yields $\frac{T_2}{T_1} = \frac{T_3}{T_4}$, or equivalently $\frac{T_4}{T_1} = \frac{T_3}{T_2}$. Substituting into the efficiency equation cancels the bracketed terms:

$\eta_{\text{th, Otto}} = 1 - \frac{1}{r^{k-1}}$

where $k = \frac{c_p}{c_v}$ is the specific heat ratio of the working fluid.

Operational Limitations on Compression Ratio

• The thermal efficiency of the Otto cycle increases monotonically with compression ratio $r$ and with the specific heat ratio $k$.

• Typical compression ratios for commercial gasoline engines reside between $7\text{--}10$.

• In spark-ignition engines, $r$ is constrained by autoignition (engine knock), which occurs when the temperature and pressure of the unburned air-fuel mixture during the compression stroke exceed self-ignition thresholds prior to spark plug firing, causing uncontrolled detonation and mechanical engine damage.

Detailed Analysis: Example 9-2 (The Ideal Otto Cycle)

Problem Statement

An ideal Otto cycle has a compression ratio of $r = 8$. At the beginning of the compression process, air is at $P_1 = 100\,\text{kPa}$ and $T_1 = 17^\circ\text{C} = 290\,\text{K}$, and $q_{\text{in}} = 800\,\text{kJ/kg}$ of heat is transferred to air during the constant-volume heat-addition process. Accounting for the variation of specific heats of air with temperature, determine:

1. The maximum temperature ($T_3$) and maximum pressure ($P_3$) occurring during the cycle.

2. The net work output ($w_{\text{net}}$).

3. The thermal efficiency ($\eta_{\text{th}}$).

4. The mean effective pressure ($\text{MEP}$) for the cycle.

Governing Assumptions

• Air-standard assumptions apply.

• Kinetic and potential energy changes are negligible ($\Delta \text{ke} = \Delta \text{pe} = 0$).

• Specific heat variation with temperature is evaluated using ideal gas air property tables (Table A-17).

Step-by-Step Solution

1. Evaluation of State 1 and State 2 Properties

At $T_1 = 290\,\text{K}$, from Table A-17:

$u_1 = 206.91\,\text{kJ/kg}$

$v_{r1} = 676.1$

For isentropic compression Process $1\text{--}2$:

$\frac{v_{r2}}{v_{r1}} = \frac{v_2}{v_1} = \frac{1}{r} \implies v_{r2} = \frac{v_{r1}}{r} = \frac{676.1}{8} = 84.51$

Interpolating in Table A-17 at $v_{r2} = 84.51$ yields:

$T_2 = 652.4\,\text{K}$

$u_2 = 475.11\,\text{kJ/kg}$

Using the ideal gas relation across Process $1\text{--}2$:

$P_2 = P_1 \left( \frac{T_2}{T_1} \right) \left( \frac{v_1}{v_2} \right) = (100\,\text{kPa}) \left( \frac{652.4\,\text{K}}{290\,\text{K}} \right) (8) = 1799.7\,\text{kPa} = 1.7997\,\text{MPa}$

2. Evaluation of State 3 (Maximum Temperature and Pressure)

For constant-volume heat-addition Process $2\text{--}3$:

$q_{\text{in}} = u_3 - u_2$

$800\,\text{kJ/kg} = u_3 - 475.11\,\text{kJ/kg}$

$u_3 = 1275.11\,\text{kJ/kg}$

From Table A-17 at $u_3 = 1275.11\,\text{kJ/kg}$:

$T_3 = 1575.1\,\text{K}\;(2300^\circ\text{F})$

$v_{r3} = 6.108$

Since $v_3 = v_2$, the maximum pressure is:

$P_3 = P_2 \left( \frac{T_3}{T_2} \right) \left( \frac{v_2}{v_3} \right) = (1.7997\,\text{MPa}) \left( \frac{1575.1\,\text{K}}{652.4\,\text{K}} \right) (1) = 4.345\,\text{MPa}$

3. Evaluation of State 4 and Net Work Output

For isentropic expansion Process $3\text{--}4$:

$\frac{v_{r4}}{v_{r3}} = \frac{v_4}{v_3} = r \implies v_{r4} = r \times v_{r3} = (8)(6.108) = 48.864$

From Table A-17 at $v_{r4} = 48.864$:

$T_4 = 795.6\,\text{K}$

$u_4 = 588.74\,\text{kJ/kg}$

For constant-volume heat rejection Process $4\text{--}1$:

$q_{\text{out}} = u_4 - u_1 = 588.74\,\text{kJ/kg} - 206.91\,\text{kJ/kg} = 381.83\,\text{kJ/kg}$

The net work output is:

$w_{\text{net}} = q_{\text{in}} - q_{\text{out}} = 800\,\text{kJ/kg} - 381.83\,\text{kJ/kg} = 418.17\,\text{kJ/kg}$

4. Thermal Efficiency

$\eta_{\text{th}} = \frac{w_{\text{net}}}{q_{\text{in}}} = \frac{418.17\,\text{kJ/kg}}{800\,\text{kJ/kg}} = 0.523 = 52.3\%$

Cold-Air-Standard Comparison:

$\eta_{\text{th, Otto}} = 1 - \frac{1}{r^{k-1}} = 1 - (8)^{1 - 1.4} = 0.565 = 56.5\%$

The cold-air-standard value differs significantly from $52.3\%$, demonstrating that temperature variation in specific heat values cannot be disregarded when accuracy is required.

5. Mean Effective Pressure

$v_1 = \frac{R T_1}{P_1} = \frac{(0.287\,\text{kPa}\cdot\text{m}^3/\text{kg}\cdot\text{K})(290\,\text{K})}{100\,\text{kPa}} = 0.832\,\text{m}^3/\text{kg}$

$v_2 = \frac{v_1}{r} = \frac{0.832\,\text{m}^3/\text{kg}}{8}$

$\text{MEP} = \frac{w_{\text{net}}}{v_1 \left(1 - \frac{1}{r}\right)} = \frac{418.17\,\text{kJ/kg}}{(0.832\,\text{m}^3/\text{kg}) \left(1 - \frac{1}{8}\right)} \left( \frac{1\,\text{kPa}\cdot\text{m}^3}{1\,\text{kJ}} \right) = 574\,\text{kPa}$

A constant pressure of $574\,\text{kPa}$ maintained during the expansion stroke produces an identical net work output to the full cycle.

The Diesel Cycle: Compression-Ignition Engines

The Diesel cycle is the idealized thermodynamic cycle for compression-ignition (CI) reciprocating engines, conceived by Rudolf Diesel in the 1890s.

Distinctions from Spark-Ignition Systems

• Air alone is inducted and compressed during the compression stroke. Fuel is injected directly into the cylinder near the completion of the stroke.

• Because the compressed air reaches temperatures well above the fuel's autoignition temperature, self-ignition occurs spontaneously upon injection, eliminating the spark plug.

• Pre-ignition is inherently prevented, permitting significantly higher compression ratios, typically $r = 12\text{--}24$.

Process Sequence of the Ideal Diesel Cycle

• Process $1\text{--}2$: Isentropic compression.

• Process $2\text{--}3$: Constant-pressure ($P = \text{const.}$) heat addition.

• Process $3\text{--}4$: Isentropic expansion.

• Process $4\text{--}1$: Constant-volume ($v = \text{const.}$) heat rejection.

The Cutoff Ratio ($r_c$)

The cutoff ratio ($r_c$) is defined as the ratio of cylinder volumes after and before the constant-pressure combustion process:

$r_c = \frac{V_3}{V_2} = \frac{v_3}{v_2}$

Energy Analysis and Thermal Efficiency Formulation

During constant-pressure Process $2\text{--}3$, boundary work is produced concurrently with heat addition:

$q_{\text{in}} - w_{b,\text{out}} = u_3 - u_2$

$q_{\text{in}} = P_2 (v_3 - v_2) + (u_3 - u_2) = h_3 - h_2 = c_p (T_3 - T_2)$

Heat rejection during constant-volume Process $4\text{--}1$ is governed by:

$q_{\text{out}} = u_4 - u_1 = c_v (T_4 - T_1)$

The thermal efficiency is therefore:

$\eta_{\text{th, Diesel}} = \frac{w_{\text{net}}}{q_{\text{in}}} = 1 - \frac{q_{\text{out}}}{q_{\text{in}}} = 1 - \frac{c_v (T_4 - T_1)}{c_p (T_3 - T_2)} = 1 - \frac{T_4 - T_1}{k (T_3 - T_2)} = 1 - \frac{T_1 \left( \frac{T_4}{T_1} - 1 \right)}{k T_2 \left( \frac{T_3}{T_2} - 1 \right)}$

Expressing temperature ratios in terms of compression ratio $r$ and cutoff ratio $r_c$ under cold-air-standard conditions:

$\eta_{\text{th, Diesel}} = 1 - \frac{1}{r^{k-1}} \left[ \frac{r_c^k - 1}{k (r_c - 1)} \right]$

Otto vs. Diesel Efficiency Comparison

• The bracketed term $\left[ \frac{r_c^k - 1}{k (r_c - 1)} \right]$ is strictly greater than $1$ for any $r_c > 1$, approaching unity as $r_c \rightarrow 1$.

• Consequently, at the same compression ratio, the Otto cycle possesses a higher thermal efficiency than the Diesel cycle:

$\eta_{\text{th, Otto}} > \eta_{\text{th, Diesel}}$

• However, because diesel engines are not knock-limited, they operate at much higher compression ratios ($r = 12\text{--}24$) compared to spark-ignition engines ($r = 7\text{--}10$). In practical applications, the efficiency of diesel engines exceeds that of gasoline engines.

The Stirling and Ericsson Cycles: Regeneration and Carnot Equivalence

The Carnot cycle achieves maximum theoretical thermal efficiency between two operating temperatures ($T_H$ and $T_L$), but involves isothermal compression and expansion processes that produce very small net work outputs per cycle and require impractical peak pressures.

The Stirling and Ericsson cycles overcome these power-density limitations by introducing regeneration, while retaining Carnot thermal efficiency.

$\eta_{\text{th, Stirling}} = \eta_{\text{th, Ericsson}} = \eta_{\text{th, Carnot}} = 1 - \frac{T_L}{T_H}$

The Concept of Regeneration

A regenerator is an internal energy-storage component that absorbs thermal energy from the working fluid during one phase of the cycle and transfers that identical thermal energy back to the fluid during another phase, requiring zero net energy exchange across the system boundary.

The Stirling Cycle

Operating in a closed system, the Stirling cycle consists of:

• Process $1\text{--}2$: Isothermal expansion at $T_H$ (heat addition $q_{\text{in}}$ from high-temperature source).

• Process $2\text{--}3$: Constant-volume ($v = \text{const.}$) regeneration (internal thermal transfer from fluid to regenerator; temperature drops from $T_H$ to $T_L$).

• Process $3\text{--}4$: Isothermal compression at $T_L$ (heat rejection $q_{\text{out}}$ to low-temperature sink).

• Process $4\text{--}1$: Constant-volume ($v = \text{const.}$) regeneration (internal thermal transfer from regenerator to fluid; temperature increases from $T_L$ to $T_H$).

The Ericsson Cycle

The Ericsson cycle operates on steady-flow equipment (compressor, turbine, heat exchangers) and matches the Stirling cycle, except the constant-volume processes are replaced with constant-pressure processes:

• Process $1\text{--}2$: Isothermal expansion at $T_H$ (in an isothermal turbine with heat addition).

• Process $2\text{--}3$: Constant-pressure ($P = \text{const.}$) regeneration (heat transfer from gas to counter-flowing stream).

• Process $3\text{--}4$: Isothermal compression at $T_L$ (in an isothermal compressor with heat rejection).

• Process $4\text{--}1$: Constant-pressure ($P = \text{const.}$) regeneration (gas preheated prior to turbine entry).

The Brayton Cycle: Gas-Turbine Power Plants

The Brayton cycle (first proposed by George Brayton in 1870) represents the standard thermodynamic cycle for gas turbines.

Open vs. Closed Cycle Modes

• Open-Cycle Gas Turbine: Fresh ambient air enters the compressor, mixes with injected fuel inside the combustion chamber, burns at constant pressure, expands through the turbine producing work, and vents to the atmosphere.

• Closed-Cycle Air-Standard Gas Turbine: Combustion is replaced by constant-pressure heat addition via an external heat exchanger, and the open exhaust is replaced by a constant-pressure heat exchanger rejecting heat to the ambient environment.

Ideal Brayton Cycle Processes

• Process $1\text{--}2$: Isentropic compression (in an axial or centrifugal compressor).

• Process $2\text{--}3$: Constant-pressure ($P = \text{const.}$) heat addition.

• Process $3\text{--}4$: Isentropic expansion (in a turbine).

• Process $4\text{--}1$: Constant-pressure ($P = \text{const.}$) heat rejection.

Governing Energy Relations

Assuming steady flow with negligible changes in kinetic and potential energy:

$w_c = \frac{\dot{W}_c}{\dot{m}} = h_2 - h_1 = c_p (T_2 - T_1)$

$w_t = \frac{\dot{W}_t}{\dot{m}} = h_3 - h_4 = c_p (T_3 - T_4)$

$q_{\text{in}} = \frac{\dot{Q}_{\text{in}}}{\dot{m}} = h_3 - h_2 = c_p (T_3 - T_2)$

$q_{\text{out}} = \frac{\dot{Q}_{\text{out}}}{\dot{m}} = h_4 - h_1 = c_p (T_4 - T_1)$

$w_{\text{net}} = w_t - w_c = q_{\text{in}} - q_{\text{out}}$

Back Work Ratio (BWR)

The back work ratio ($\text{bwr}$) is defined as the fraction of turbine work consumed to power the compressor:

$\text{bwr} = \frac{w_c}{w_t} = \frac{h_2 - h_1}{h_3 - h_4}$

• In gas turbines, compression occurs entirely in the vapor/gas phase, leading to high specific volumes and substantial pump work requirements. Consequently, typical back work ratios range from $40\%\text{--}80\%$.

• By comparison, vapor power plants (Rankine cycle) pump liquid water, resulting in back work ratios of only $1\%\text{--}2\%$.

Thermal Efficiency Formulation

Defining the pressure ratio as $r_p = \frac{P_2}{P_1} = \frac{P_3}{P_4}$:

$\eta_{\text{th, Brayton}} = \frac{w_{\text{net}}}{q_{\text{in}}} = 1 - \frac{q_{\text{out}}}{q_{\text{in}}} = 1 - \frac{c_p (T_4 - T_1)}{c_p (T_3 - T_2)} = 1 - \frac{T_1 \left( \frac{T_4}{T_1} - 1 \right)}{T_2 \left( \frac{T_3}{T_2} - 1 \right)}$

Using the isentropic pressure-temperature relationships:

$\frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{(k-1)/k} = r_p^{(k-1)/k}$

$\frac{T_3}{T_4} = \left( \frac{P_3}{P_4} \right)^{(k-1)/k} = r_p^{(k-1)/k}$

Since $\frac{T_2}{T_1} = \frac{T_3}{T_4}$, it follows that $\frac{T_4}{T_1} = \frac{T_3}{T_2}$, which reduces the efficiency expression to:

$\eta_{\text{th, Brayton}} = 1 - \frac{1}{r_p^{(k-1)/k}}$

Parametric Optimization of Pressure Ratio

For specified minimum ($T_{\text{min}} = T_1$) and maximum ($T_{\text{max}} = T_3$) cycle temperatures:

• At very low pressure ratios, net work output is limited because the pressure difference is small.

• At very high pressure ratios, compression raises $T_2$ near $T_{\text{max}}$, drastically reducing heat input and total expansion work.

• Net work output reaches a distinct maximum at an optimum pressure ratio:

$r_p = \left( \frac{T_{\text{max}}}{T_{\text{min}}} \right)^{\frac{k}{2(k-1)}}$

Gas turbine applications encompass two primary domains: electric power generation and aircraft propulsion. In these engines, air acts both as the chemical oxidant and as an internal coolant, maintaining turbine blade temperatures within metallurgical boundaries. Air-fuel ratios exceeding $50$ by mass are common.

Brayton Cycle Component Efficiencies and Parametric Analysis (Examples 2 & 3)

Example 2: Ideal Air-Standard Brayton Cycle

Given Conditions:

• Ambient air enters compressor: $P_1 = 100\,\text{kPa}$, $T_1 = 300\,\text{K}$

• Volumetric inlet flow rate: $\dot{V}_1 = 5\,\text{m}^3/\text{s}$

• Compressor pressure ratio: $r_p = 10$

• Turbine inlet temperature: $T_3 = 1400\,\text{K}$

• Cold-air-standard properties: $k = 1.4$, $c_p = 1.005\,\text{kJ/kg}\cdot\text{K}$, $R = 0.287\,\text{kJ/kg}\cdot\text{K}$

Analysis:

1. State 2 Temperature (Compressor Exit):

$T_2 = T_1 r_p^{(k-1)/k} = (300\,\text{K}) (10)^{(1.4 - 1)/1.4} = (300\,\text{K}) (10)^{0.2857} = (300\,\text{K}) (1.9307) = 579.2\,\text{K}$

2. State 4 Temperature (Turbine Exit):

$T_4 = \frac{T_3}{r_p^{(k-1)/k}} = \frac{1400\,\text{K}}{1.9307} = 725.1\,\text{K}$

3. Cycle Thermal Efficiency:

$\eta_{\text{th}} = 1 - \frac{1}{r_p^{(k-1)/k}} = 1 - \frac{1}{1.9307} = 0.482 = 48.2\%$

4. Specific Work and Heat Transfers:

$w_c = c_p (T_2 - T_1) = (1.005\,\text{kJ/kg}\cdot\text{K}) (579.2\,\text{K} - 300\,\text{K}) = 280.6\,\text{kJ/kg}$

$w_t = c_p (T_3 - T_4) = (1.005\,\text{kJ/kg}\cdot\text{K}) (1400\,\text{K} - 725.1\,\text{K}) = 678.3\,\text{kJ/kg}$

$w_{\text{net}} = w_t - w_c = 678.3\,\text{kJ/kg} - 280.6\,\text{kJ/kg} = 397.7\,\text{kJ/kg}$

5. Back Work Ratio:

$\text{bwr} = \frac{w_c}{w_t} = \frac{280.6\,\text{kJ/kg}}{678.3\,\text{kJ/kg}} = 0.414 = 41.4\%$

6. Net Power Output in kW:

$v_1 = \frac{R T_1}{P_1} = \frac{(0.287\,\text{kPa}\cdot\text{m}^3/\text{kg}\cdot\text{K})(300\,\text{K})}{100\,\text{kPa}} = 0.861\,\text{m}^3/\text{kg}$

$\dot{m} = \frac{\dot{V}_1}{v_1} = \frac{5\,\text{m}^3/\text{s}}{0.861\,\text{m}^3/\text{kg}} = 5.807\,\text{kg/s}$

$\dot{W}_{\text{net}} = \dot{m} w_{\text{net}} = (5.807\,\text{kg/s}) (397.7\,\text{kJ/kg}) = 2309.5\,\text{kW} \approx 2.31\,\text{MW}$

Impact of Component Irreversibilities

Turbine and compressor irreversibilities (fluid friction, shock losses, blade stall) degrade overall cycle performance.

$\eta_c = \frac{w_s}{w_a} = \frac{h_{2s} - h_1}{h_2 - h_1}$

$\eta_t = \frac{w_a}{w_s} = \frac{h_3 - h_4}{h_3 - h_{4s}}$

Example 3: Non-Ideal Brayton Cycle Analysis

Reconsidering Example 2 with an isentropic compressor efficiency of $\eta_c = 80\%$ and an isentropic turbine efficiency of $\eta_t = 80\%$:

1. Actual Work Inputs and Outputs:

$w_{c,a} = \frac{w_{c,s}}{\eta_c} = \frac{280.6\,\text{kJ/kg}}{0.80} = 350.75\,\text{kJ/kg}$

$w_{t,a} = \eta_t w_{t,s} = (0.80)(678.3\,\text{kJ/kg}) = 542.64\,\text{kJ/kg}$

2. Actual Net Specific Work:

$w_{\text{net}} = w_{t,a} - w_{c,a} = 542.64\,\text{kJ/kg} - 350.75\,\text{kJ/kg} = 191.89\,\text{kJ/kg}$

3. Modified Back Work Ratio:

$\text{bwr} = \frac{w_{c,a}}{w_{t,a}} = \frac{350.75\,\text{kJ/kg}}{542.64\,\text{kJ/kg}} = 0.646 = 64.6\%$

4. Actual Combustion Heat Input ($q_{\text{in}}$):

$T_{2,a} = T_1 + \frac{w_{c,a}}{c_p} = 300\,\text{K} + \frac{350.75\,\text{kJ/kg}}{1.005\,\text{kJ/kg}\cdot\text{K}} = 649.0\,\text{K}$

$q_{\text{in}} = c_p (T_3 - T_{2,a}) = (1.005\,\text{kJ/kg}\cdot\text{K}) (1400\,\text{K} - 649.0\,\text{K}) = 754.75\,\text{kJ/kg}$

5. Realized Thermal Efficiency:

$\eta_{\text{th}} = \frac{w_{\text{net}}}{q_{\text{in}}} = \frac{191.89\,\text{kJ/kg}}{754.75\,\text{kJ/kg}} = 0.254 = 25.4\%$

6. Total Net Power Output:

$\dot{W}_{\text{net}} = \dot{m} w_{\text{net}} = (5.807\,\text{kg/s}) (191.89\,\text{kJ/kg}) = 1114.3\,\text{kW}$

Component irreversibilities increase the back work ratio from $41.4\%$ to $64.6\%$, while decreasing thermal efficiency from $48.2\%$ to $25.4\%$ and halving net power output.

Advanced Gas Turbine Modifications: Regeneration, Intercooling, and Reheating

The Regenerative Brayton Cycle

In standard gas turbine cycles, the turbine exhaust temperature ($T_4$) is typically higher than the compressor discharge temperature ($T_2$). A regenerator (counter-flow heat exchanger) utilizes this temperature difference to preheat the air exiting the compressor before it enters the combustor.

• Preheating the working fluid from State 2 to State 5 using exhaust gas energy ($q_{\text{regen}}$) reduces the required fuel energy addition in the combustor ($q_{\text{in}}$).

• Because net shaft work output remains essentially unchanged while fuel input decreases, thermal efficiency increases.

Multistage Compression with Intercooling, and Multistage Expansion with Reheating

To increase net work and approach Carnot efficiency, gas turbine cycles incorporate multistage compression with intercooling and multistage expansion with reheating.

• Intercooling: Passing partially compressed air through an intercooler between compressor stages lowers fluid temperature and specific volume, reducing total compression work ($w_c$). Compressor work is minimized when intermediate pressure ratios match:

$\frac{P_2}{P_1} = \frac{P_4}{P_3}$

• Reheating: Re-expanding gas through multiple turbine stages with intermediate combustion reheating maximizes work output ($w_t$). Turbine work output is maximized when intermediate expansion pressure ratios match:

$\frac{P_6}{P_7} = \frac{P_8}{P_9}$

• Convergence to Ericsson Limits: As the number of compression and expansion stages increases, the compression and expansion curves approach isothermal behavior, while regeneration approaches isobaric heat exchange. The combined cycle then approaches the thermodynamic performance of an Ericsson cycle.

Jet Propulsion Cycles and Aircraft Engines

Aircraft gas-turbine engines operate on an open propulsion cycle where the primary useful output is linear thrust rather than net shaft work.

Principles of Jet Propulsion

• In an ideal jet propulsion cycle, net shaft work output is zero ($w_{\text{net}} = 0$). The turbine produces only enough mechanical energy to drive the compressor and internal accessories.

• Hot, high-pressure exhaust gas exiting the turbine is expanded to ambient pressure through a converging-diverging nozzle, converting thermal energy into kinetic energy ($V_{\text{exit}}$).

• Thrust ($F$): The net force propelling the aircraft forward, generated by accelerating fluid rearward:

$F = \dot{m} (V_{\text{exit}} - V_{\text{inlet}})$

• Propulsive Power ($\dot{W}_P$): The mechanical power developed by thrust moving the aircraft through air at velocity $V$:

$\dot{W}_P = F \cdot V$

• Propulsive Efficiency ($\eta_P$): The ratio of propulsive power delivered to the total rate of kinetic energy imparted to the fluid stream:

$\eta_P = \frac{\dot{W}_P}{\dot{Q}_{\text{in}}}$

Advanced Aircraft Applications: The F-35 Program

The Lockheed Martin F-35 Joint Strike Fighter program illustrates modern high-thrust military propulsion technology, utilizing variable-geometry thrust-vectoring nozzles to achieve vertical and short takeoff and landing (STOVL) capabilities.

Primary Jet Engine Architectures

1. Turbojet: Air passes sequentially through a diffuser, compressor, combustor, turbine, and propelling nozzle.

2. Turbofan: Employs an oversized inlet fan driven by a low-pressure turbine, directing a large fraction of inlet air around the core engine through an outer bypass duct.

The Pratt & Whitney PW4084 Turbofan Engine:

• Thrust: $374\,\text{kN}$

• Length: $4.87\,\text{m}$

• Fan Diameter: $2.84\,\text{m}$

• Engine Mass: $6800\,\text{kg}$

• Application: High-capacity commercial passenger flight (Boeing 777)

1. Turboprop: The turbine powers an external propeller through a reduction gearbox. It moves a large mass of air with a small velocity increase, offering high fuel efficiency at speeds below Mach 0.6.

4. Ramjet: Eliminates all rotating mechanical assemblies (no compressor or turbine). Compression occurs entirely through aerodynamic deceleration (ram effect) as high-speed inlet air enters at supersonic velocities ($\text{Mach} > 3$).

Second-Law Analysis and Exergy Accounting of Gas Power Cycles

A Second-Law analysis identifies the locations, causes, and magnitudes of thermodynamic irreversibilities across cyclic components.

Exergy Destruction Formulations

• General Closed System:

$X_{\text{dest}} = T_0 S_{\text{gen}} = T_0 (\Delta S_{\text{sys}} - S_{\text{in}} + S_{\text{out}}) = T_0 \left[ (S_2 - S_1)_{\text{sys}} - \frac{Q_{\text{in}}}{T_{b,\text{in}}} + \frac{Q_{\text{out}}}{T_{b,\text{out}}} \right]\;(\text{kJ})$

• Steady-Flow Open System:

$\dot{X}_{\text{dest}} = T_0 \dot{S}_{\text{gen}} = T_0 (\dot{S}_{\text{out}} - \dot{S}_{\text{in}}) = T_0 \left( \sum_{\text{out}} \dot{m} s - \sum_{\text{in}} \dot{m} s - \frac{\dot{Q}_{\text{in}}}{T_{b,\text{in}}} + \frac{\dot{Q}_{\text{out}}}{T_{b,\text{out}}} \right)\;(\text{kW})$

• Steady-Flow Single-Inlet, Single-Exit Component:

$X_{\text{dest}} = T_0 s_{\text{gen}} = T_0 \left( s_e - s_i - \frac{q_{\text{in}}}{T_{b,\text{in}}} + \frac{q_{\text{out}}}{T_{b,\text{out}}} \right)\;(\text{kJ/kg})$

• Exergy Destruction of a Cycle:

$x_{\text{dest}} = T_0 \left( \sum \frac{q_{\text{out}}}{T_{b,\text{out}}} - \sum \frac{q_{\text{in}}}{T_{b,\text{in}}} \right)\;(\text{kJ/kg})$

• Cycle Transferring Heat Exclusively with Reservoirs at $T_H$ and $T_L$:

$x_{\text{dest}} = T_0 \left( \frac{q_{\text{out}}}{T_L} - \frac{q_{\text{in}}}{T_H} \right)\;(\text{kJ/kg})$

Nonflow and Flow Exergy Expressions

• Closed System (Nonflow) Specific Exergy ($\phi$):

$\phi = (u - u_0) - T_0 (s - s_0) + P_0 (v - v_0) + \frac{V^2}{2} + g z$

• Stream (Flow) Specific Exergy ($\psi$):

$\psi = (h - h_0) - T_0 (s - s_0) + \frac{V^2}{2} + g z$

Evaluating component exergy destruction shows that combustion chambers account for the majority of cycle irreversibility due to uncontrolled chemical reaction kinetics and large internal temperature gradients.

Combined Gas-Vapor Power Cycles

The combined gas-vapor power cycle couples a high-temperature gas-turbine (Brayton) topping cycle with a lower-temperature steam-turbine (Rankine) bottoming cycle.

Operating Principles

• The Brayton cycle operates at high peak temperatures ($T_{\text{max}} > 1400\,\text{K}$), but rejects exhaust gas at relatively high temperatures ($T_8 \approx 500\text{--}600^\circ\text{C}$).

• Instead of venting this thermal energy to the atmosphere, the gas turbine exhaust is routed through a Heat Recovery Steam Generator (HRSG), which generates superheated steam to drive a Rankine cycle.

• The combined cycle achieves higher thermal efficiencies than either cycle operating individually, with modern installations routinely exceeding $50\%$ overall thermal efficiency.