12.3 Air-Standard Gas Power Cycles: Otto, Diesel, Dual & Brayton

11.3 Air-Standard Gas Power Cycles: Otto, Diesel, Dual & Brayton

Internal combustion engines and gas turbines provide essential prime mover power across mining excavation equipment, diesel-electric haul trucks, heavy dumpers, portable air compressors, and on-site emergency gas-turbine generator sets in Coal India complexes.

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1. Air-Standard Assumptions

To perform analytical thermodynamic modeling of complex IC engines and gas turbines, standard idealizations (Air-Standard Assumptions) are applied:

1. The working fluid is a fixed mass of air that behaves continuously as an ideal gas with constant specific heats evaluated at room temperature ($c_p = 1.005\text{ kJ/kg}\cdot\text{K}$, $c_v = 0.718\text{ kJ/kg}\cdot\text{K}$, $\gamma = 1.4$, $R = 0.287\text{ kJ/kg}\cdot\text{K}$).
2. All processes within the cycle are internally reversible.
3. The complex combustion process is replaced by an external heat addition process from a high-temperature source.
4. The exhaust blowdown and intake strokes are replaced by an external heat rejection process that restores the air to its initial state.

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2. Air-Standard Otto Cycle (Constant Volume Heat Addition)

The idealized thermodynamic cycle for spark-ignition (petrol/gasoline) four-stroke and two-stroke reciprocating engines.

          OTTO CYCLE: P-v DIAGRAM                  OTTO CYCLE: T-s DIAGRAM
    P ^                                      T ^
      |        3 (Peak P, T)                   |        3
      |       /|                               |       / \
      |  Q_in/ |                               | Q_in /   \ Isentropic
      |     /  |                               |     /     \ Expansion
      |    2   | Isentropic                    |    2       4
      |    |   | Expansion                     |    |       |
      |    |   4                               |    | Q_out | Q_out
      |    |  /                                |    1-------+
      |    1-+                                 +------------------------> s
      +----+---+------------------> v
          V_2 V_1

Four Cycle Processes

1. Process 1-2: Reversible adiabatic (isentropic) compression: $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$.
2. Process 2-3: Constant-volume (isochoric) heat addition: $Q_{in} = m c_v (T_3 - T_2)$.
3. Process 3-4: Reversible adiabatic (isentropic) expansion: $P_3 V_3^{\gamma} = P_4 V_4^{\gamma}$.
4. Process 4-1: Constant-volume (isochoric) heat rejection: $Q_{out} = m c_v (T_4 - T_1)$.

Geometric Parameters & Efficiency Derivation

• Compression Ratio ($r$): $r = \frac{V_1}{V_2} = \frac{V_d + V_c}{V_c}$ (Where $V_d$ = swept/displacement volume, $V_c$ = clearance volume).

• Thermal Efficiency ($\eta_{Otto}$): $\eta_{Otto} = 1 - \frac{Q_{out}}{Q_{in}} = 1 - \frac{m c_v (T_4 - T_1)}{m c_v (T_3 - T_2)} = 1 - \frac{T_1(T_4/T_1 - 1)}{T_2(T_3/T_2 - 1)}$ Since $T_2/T_1 = (V_1/V_2)^{\gamma-1} = r^{\gamma-1}$ and $T_3/T_4 = (V_4/V_3)^{\gamma-1} = r^{\gamma-1}$: $\eta_{Otto} = 1 - \frac{1}{r^{\gamma - 1}}$

Key Insights: $\eta_{Otto}$ increases with higher compression ratio $r$ and higher specific heat ratio $\gamma$ (monatomic gases like Argon have higher $\eta$ than diatomic air). In practice, $r$ is restricted to $6-10$ to avoid knocking (auto-ignition detonation).

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3. Air-Standard Diesel Cycle (Constant Pressure Heat Addition)

The idealized thermodynamic cycle for compression-ignition (diesel) reciprocating engines.

         DIESEL CYCLE: P-v DIAGRAM                DIESEL CYCLE: T-s DIAGRAM
    P ^                                      T ^
      |      2  Q_in  3                        |          3
      |      *--------* (P = C)                |         / \
      |     /          \                       |   Q_in /   \ Isentropic
      |    /            \ Isentropic           |       /     \ Expansion
      |   /              \ Expansion           |      2       4
      |  1                4                    |      |       |
      |  +----------------+                    |      1-------+
      +--+---+------------+-------> v          +------------------------> s
        V_2 V_3          V_1

Four Cycle Processes

1. Process 1-2: Isentropic compression ($s_1 = s_2$): $T_2 = T_1 r^{\gamma-1}$.
2. Process 2-3: Constant-pressure (isobaric) heat addition: $Q_{in} = m c_p (T_3 - T_2)$.
3. Process 3-4: Isentropic expansion ($s_3 = s_4$): $T_4 = T_3 (V_3/V_4)^{\gamma-1}$.
4. Process 4-1: Constant-volume (isochoric) heat rejection: $Q_{out} = m c_v (T_4 - T_1)$.

Governing Definitions & Thermal Efficiency

• Cut-off Ratio ($r_c$): Ratio of cylinder volume after heat addition to before heat addition: $r_c = \frac{V_3}{V_2} = \frac{T_3}{T_2}$

• Expansion Ratio ($r_e$): $r_e = \frac{V_4}{V_3} = \frac{V_1}{V_3} = \frac{r}{r_c}$

• Diesel Thermal Efficiency ($\eta_{Diesel}$): $\eta_{Diesel} = 1 - \frac{Q_{out}}{Q_{in}} = 1 - \frac{m c_v (T_4 - T_1)}{m c_p (T_3 - T_2)} = 1 - \frac{1}{r^{\gamma - 1}} \left[ \frac{r_c^{\gamma} - 1}{\gamma(r_c - 1)} \right]$

Critical Note: Since $\frac{r_c^{\gamma} - 1}{\gamma(r_c - 1)} > 1$ for all $r_c > 1$, the Diesel cycle efficiency is lower than the Otto cycle at the same compression ratio. However, because diesel engines compress pure air (no fuel during compression), they operate at much higher compression ratios ($r = 14-22$) without knocking, resulting in higher actual operational efficiency.

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4. Dual Combustion Cycle (Sabathe / Limited Pressure Cycle)

In modern high-speed diesel and heavy haul truck engines, fuel injection begins before TDC, causing heat addition to occur partly at constant volume and partly at constant pressure.

• Pressure Ratio (Explosion Ratio, $r_p$ or $\alpha$): $r_p = \frac{P_3}{P_2}$
• Cut-off Ratio ($r_c$): $r_c = \frac{V_4}{V_3}$

$\eta_{Dual} = 1 - \frac{1}{r^{\gamma - 1}} \left[ \frac{r_p r_c^{\gamma} - 1}{(r_p - 1) + \gamma r_p (r_c - 1)} \right]$

• When $r_c = 1 \implies$ Otto cycle equation.
• When $r_p = 1 \implies$ Diesel cycle equation.

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5. Comparative Performance Analysis of Gas Power Cycles

+-------------------------------------------------------------------------+
|                CYCLE EFFICIENCY COMPARISONS SUMMARY                     |
+------------------------------------+------------------------------------+
| Comparison Condition               | Efficiency Ranking                 |
+------------------------------------+------------------------------------+
| Same Compression Ratio ($r$) and   | $\eta_{Otto} > \eta_{Dual} >       |
| Same Heat Input ($Q_{in}$)         | \eta_{Diesel}$                     |
+------------------------------------+------------------------------------+
| Same Maximum Peak Pressure         | $\eta_{Diesel} > \eta_{Dual} >     |
| ($P_{max}$) & Same Heat Input      | \eta_{Otto}$                       |
+------------------------------------+------------------------------------+
| Same Maximum Peak Pressure         | $\eta_{Diesel} > \eta_{Dual} >     |
| ($P_{max}$) & Maximum Temp ($T_{max}$) | \eta_{Otto}$                   |
+------------------------------------+------------------------------------+
| Same Maximum Peak Pressure         | $\eta_{Diesel} > \eta_{Dual} >     |
| ($P_{max}$) & Same Work Output ($W$) | \eta_{Otto}$                     |
+------------------------------------+------------------------------------+

Mean Effective Pressure (MEP)

The Mean Effective Pressure is a fictitious constant pressure that, if exerted on the piston during the entire power stroke, would produce the identical net work output:

$\text{MEP} = \frac{W_{net}}{V_d} = \frac{W_{net}}{V_1 - V_2} = \frac{W_{net}}{V_1(1 - 1/r)}$

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6. The Joule / Brayton Gas Turbine Cycle

The fundamental air-standard cycle for open and closed gas turbine power plants.

        BRAYTON CYCLE: P-v DIAGRAM               BRAYTON CYCLE: T-s DIAGRAM
    P ^                                      T ^
      |      2  Q_in  3                        |          3 (T_max)
      |      *--------* (P_2 = P_3)            |         / \
      |     /          \                       |   Q_in /   \ Isentropic
      |    /            \ Isentropic           |       /     \ Expansion
      |   /              \ Expansion           |      2       4
      |  1*--------------*4 (P_1 = P_4)        |       \     /
      |        Q_out                           |  Q_out \   /
      +--------------------------> v           +---------1-+------------> s

Brayton Cycle Thermal Efficiency Derivation

• Pressure Ratio ($r_p$): $r_p = \frac{P_2}{P_1} = \frac{P_3}{P_4}$

• Thermal Efficiency ($\eta_{Brayton}$): $\eta_{Brayton} = 1 - \frac{Q_{out}}{Q_{in}} = 1 - \frac{c_p(T_4 - T_1)}{c_p(T_3 - T_2)} = 1 - \frac{T_1(T_4/T_1 - 1)}{T_2(T_3/T_2 - 1)}$ Since $\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} = r_p^{\frac{\gamma-1}{\gamma}}$ and $\frac{T_3}{T_4} = r_p^{\frac{\gamma-1}{\gamma}}$: $\eta_{Brayton} = 1 - \frac{1}{r_p^{\frac{\gamma - 1}{\gamma}}}$

Back Work Ratio ($r_{bw}$)

$r_{bw} = \frac{W_{compressor}}{W_{turbine}} = \frac{c_p(T_2 - T_1)}{c_p(T_3 - T_4)}$ (Gas turbines have high back work ratios of $40%-60%$, compared to $<1%-2%$ in Rankine steam plants).

Optimum Pressure Ratio for Maximum Specific Net Work Output

Specific net work is: $w_{net} = w_T - w_C = c_p (T_3 - T_4) - c_p (T_2 - T_1) = c_p \left[ T_3\left(1 - \frac{1}{r_p^x}\right) - T_1(r_p^x - 1) \right]$ (Where $x = \frac{\gamma - 1}{\gamma}$). Differentiating $w_{net}$ with respect to $r_p^x$ and setting to zero:

$r_{p, opt} = \left( \frac{T_{max}}{T_{min}} \right)^{\frac{\gamma}{2(\gamma - 1)}} = \left( \frac{T_3}{T_1} \right)^{\frac{\gamma}{2(\gamma - 1)}}$

At this optimum pressure ratio: $T_2 = T_4 = \sqrt{T_1 T_3}$ $w_{net, max} = c_p \left(\sqrt{T_3} - \sqrt{T_1}\right)^2$

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7. Gas Turbine Modifications: Regeneration, Intercooling & Reheat

                      GAS TURBINE CYCLE ENHANCEMENTS
+-----------------------+-------------------------------------------------+
| Modification          | Primary Thermodynamic Effect & Purpose          |
+-----------------------+-------------------------------------------------+
| **Regeneration**      | Uses hot turbine exhaust ($T_4 > T_2$) to       |
| (Recuperation)        | preheat combustor air. Increases thermal        |
|                       | efficiency at moderate $r_p$; work output same. |
+-----------------------+-------------------------------------------------+
| **Intercooling**      | Multi-stage compression with intermediate       |
|                       | cooling. Minimizes compressor work $\int v dP$. |
|                       | Increases net work output.                      |
+-----------------------+-------------------------------------------------+
| **Reheating**         | Multi-stage expansion with intermediate heating.|
|                       | Maximizes turbine work output; increases $w_n$. |
+-----------------------+-------------------------------------------------+
| **Combined Cycle**    | Regeneration + Intercooling + Reheat approaches |
|                       | Ericsson cycle limit ($\eta \to \eta_{Carnot}$).|
+-----------------------+-------------------------------------------------+

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8. Worked Numerical Examples

Example 1: Otto Cycle Efficiency & Mean Effective Pressure (MEP)

Problem: An air-standard Otto cycle operates with a compression ratio $r = 8$. At the start of compression, $P_1 = 100\text{ kPa}, T_1 = 300\text{ K}$. The maximum cycle temperature is $T_3 = 1800\text{ K}$. Given $c_v = 0.718\text{ kJ/kg}\cdot\text{K}, R = 0.287\text{ kJ/kg}\cdot\text{K}, \gamma = 1.4$. Calculate:

1. Thermal efficiency ($\eta_{Otto}$).
2. Net work output per kg ($w_{net}$).
3. Mean Effective Pressure ($\text{MEP}$).

Solution:

1. Thermal efficiency: $\eta_{Otto} = 1 - \frac{1}{r^{\gamma - 1}} = 1 - \frac{1}{8^{1.4 - 1}} = 1 - \frac{1}{8^{0.4}} = 1 - \frac{1}{2.2974} = 1 - 0.4353 = 0.5647 \text{ (56.47\%)}$

2. Temperature at state 2: $T_2 = T_1 r^{\gamma-1} = 300 \times 2.2974 = 689.22\text{ K}$ Heat supplied ($q_{in}$): $q_{in} = c_v (T_3 - T_2) = 0.718 \times (1800 - 689.22) = 0.718 \times 1110.78 = 797.54\text{ kJ/kg}$ Net work output ($w_{net}$): $w_{net} = \eta_{Otto} \times q_{in} = 0.5647 \times 797.54 = 450.37\text{ kJ/kg}$

3. Mean Effective Pressure: $v_1 = \frac{R T_1}{P_1} = \frac{0.287 \times 300}{100} = 0.861\text{ m}^3/\text{kg}$ $v_2 = \frac{v_1}{r} = \frac{0.861}{8} = 0.1076\text{ m}^3/\text{kg}$ $v_d = v_1 - v_2 = 0.861 - 0.1076 = 0.7534\text{ m}^3/\text{kg}$ $\text{MEP} = \frac{w_{net}}{v_d} = \frac{450.37\text{ kJ/kg}}{0.7534\text{ m}^3/\text{kg}} = 597.8\text{ kPa} = 5.98\text{ bar}$

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Example 2: Brayton Cycle Optimum Pressure Ratio

Problem: In a stationary gas turbine power plant, air enters the compressor at $T_1 = 300\text{ K}$ and turbine inlet temperature is limited by blade metallurgy to $T_3 = 1200\text{ K}$. Assuming an ideal Brayton cycle with $\gamma = 1.4$, determine:

1. Optimum pressure ratio ($r_{p,opt}$) for maximum specific work output.
2. Thermal efficiency at this optimum pressure ratio.

Solution:

1. Optimum pressure ratio: $r_{p,opt} = \left(\frac{T_3}{T_1}\right)^{\frac{\gamma}{2(\gamma - 1)}} = \left(\frac{1200}{300}\right)^{\frac{1.4}{2(0.4)}} = (4)^{\frac{1.4}{0.8}} = 4^{1.75} = (2^2)^{1.75} = 2^{3.5} \approx 11.31$

2. Thermal efficiency at $r_{p,opt}$: $\eta = 1 - \frac{1}{r_{p,opt}^{\frac{\gamma-1}{\gamma}}} = 1 - \frac{1}{11.31^{\frac{0.4}{1.4}}} = 1 - \frac{1}{11.31^{0.2857}} = 1 - \frac{1}{2.0} = 0.50\text{ (50.0\%)}$ (Note: At optimum pressure ratio, $\eta = 1 - \sqrt{T_1/T_3} = 1 - \sqrt{300/1200} = 1 - 0.5 = 0.50$).