1. Fundamental Concepts and Definitions
1.1 System, Surroundings, Properties, State, Process, Cycle
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System: Region of interest; surroundings: everything external.
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Properties: Macroscopic characteristics (e.g., $p$, $V$, $T$). State: Condition defined by properties.
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Process: Path of state changes. Cycle: Sequence returning to initial state.
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Path functions (heat $Q$, work $W$) vs. properties (state functions like $U$, $H$).
1.2 Temperature Scales
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Thermodynamic (Kelvin) scale: Based on Carnot cycle; absolute zero = 0 K.
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Conversions: $$\displaystyle T(\text{K}) = T(°C) + 273.15 $$.
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New scale (e.g., °N): Linear relation $$\displaystyle T_N = aT_C + b $$; solve using fixed points.
1.3 Pressure, Volume, Density
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Pressure $p$: Force per unit area. Volume $V$: Space occupied. Density $$\displaystyle \rho = m/V $$.
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Gauge vs. absolute pressure: $$\displaystyle p_{\text{abs}} = p_{\text{gauge}} + p_{\text{atm}} $$.
1.4 Energy Forms
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Internal energy $U$: Energy within system. Enthalpy $$\displaystyle H = U + pV $$.
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Kinetic energy $$\displaystyle KE = \frac{1}{2}mv^2 $$, potential energy $$\displaystyle PE = mgh $$.
1.5 Heat and Work
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Heat $Q$: Energy transfer due to temperature difference.
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Work $W$: Energy transfer by force acting through distance.
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Both path functions; depend on process, not state.
1.6 Gas Laws
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Boyle’s law: $$\displaystyle pV = \text{constant} $$ (T constant).
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Charles’ law: $$\displaystyle V/T = \text{constant} $$ (p constant).
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Ideal gas equation: $$\displaystyle pV = mRT $$ or $$\displaystyle pv = RT $$ (specific).
1.7 Constant Volume Gas Thermometer
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Principle: Gas at constant volume; pressure $\propto$ temperature.
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Preferred over constant pressure: Volume change due to expansion of bulb/capillary negligible → higher accuracy.
[!TIP]
Constant volume thermometer is primary standard for temperature measurement.
2. First Law of Thermodynamics
2.1 Statements
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Closed system: $$\displaystyle \Delta U = Q - W $$ (or $$\displaystyle Q = \Delta U + W $$, $W$ = work done by system).
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Open system (SFEE):
$$ \dot{m}\left(h_1 + \frac{V_1^2}{2} + gz_1\right) + \dot{Q} = \dot{m}\left(h_2 + \frac{V_2^2}{2} + gz_2\right) + \dot{W}_s $$
2.2 Limitations
- Cannot predict process direction or quality of energy (e.g., work vs. heat).
2.3 Energy Balance for Closed Systems
$$ \boxed{Q = \Delta U + W} $$
- $$\displaystyle W = \int p \, dV $$ for quasi-static process.
2.4 Steady Flow Energy Equation (SFEE) Applications
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Turbine: $$\displaystyle \dot{W}_s = \dot{m}(h_1 - h_2) $$ (adiabatic, negligible KE/PE).
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Compressor: $$\displaystyle \dot{W}_s = \dot{m}(h_2 - h_1) $$.
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Heat exchanger: $$\displaystyle \dot{Q} = \dot{m}(h_2 - h_1) $$ (no work).
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Throttling: $$\displaystyle h_1 = h_2 $$ (adiabatic, no work).
2.5 Specific Heat Capacities
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$$\displaystyle C_v = \left(\frac{\partial u}{\partial T}\right)_v $$, $$\displaystyle C_p = \left(\frac{\partial h}{\partial T}\right)_p $$.
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Ideal gas: $$\displaystyle C_p - C_v = R $$.
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Mean specific heat: $$\displaystyle C_{p,\text{m}} = \frac{h_2 - h_1}{T_2 - T_1} $$.
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Variable $$\displaystyle C_p $$: If $$\displaystyle C_p = a + bT $$, then $$\displaystyle \Delta h = a(T_2 - T_1) + \frac{b}{2}(T_2^2 - T_1^2) $$.
2.6 Thermodynamic Processes
| Process | Condition | $pV$ relation | $TV$ relation | Work $W$ | Heat $Q$ |
|---|---|---|---|---|---|
| Isothermal | $$\displaystyle T = \text{const} $$ | $$\displaystyle pV = \text{const} $$ | — | $$\displaystyle mRT \ln\frac{V_2}{V_1} $$ | $$\displaystyle = W $$ |
| Adiabatic reversible | $$\displaystyle Q=0 $$ | $$\displaystyle pV^\gamma = \text{const} $$ | $$\displaystyle TV^{\gamma-1}=\text{const} $$ | $$\displaystyle \frac{p_1V_1 - p_2V_2}{\gamma-1} $$ | 0 |
| Isobaric | $$\displaystyle p = \text{const} $$ | $$\displaystyle V/T = \text{const} $$ | — | $$\displaystyle p(V_2 - V_1) $$ | $$\displaystyle mC_p\Delta T $$ |
| Isochoric | $$\displaystyle V = \text{const} $$ | $$\displaystyle p/T = \text{const} $$ | — | 0 | $$\displaystyle mC_v\Delta T $$ |
| Polytropic | $$\displaystyle pV^n = \text{const} $$ | — | — | $$\displaystyle \frac{p_2V_2 - p_1V_1}{1-n} $$ | $\Delta U + W$ |
2.7 Polytropic Processes
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Work: $$\displaystyle W = \frac{p_2V_2 - p_1V_1}{1-n} $$ for $n \neq 1$; for $$\displaystyle n=1 $$, $$\displaystyle W = mRT \ln\frac{V_2}{V_1} $$.
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Heat: $$\displaystyle Q = mC_v(T_2 - T_1) + W $$.
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Index $n$ determination: From $$\displaystyle p_1V_1^n = p_2V_2^n $$.
2.8 Free Expansion (Joule Expansion)
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Characteristics: Adiabatic ($$\displaystyle Q=0 $$), no work ($$\displaystyle W=0 $$), against vacuum.
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Ideal gas: $$\displaystyle \Delta U = 0 \Rightarrow \Delta T = 0 $$.
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Real gas: Temperature may change (Joule-Thomson effect).
2.9 Cyclic Processes
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$$\displaystyle \Delta U_{\text{cycle}} = 0 \Rightarrow Q_{\text{net}} = W_{\text{net}} $$.
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Net work = area enclosed in $p$-$V$ diagram.
[!TIP]
For polytropic processes, remember special cases: $$\displaystyle n=0 $$ (isobaric), $$\displaystyle n=1 $$ (isothermal), $$\displaystyle n=\gamma $$ (adiabatic reversible), $$\displaystyle n=\infty $$ (isochoric).
3. Second Law of Thermodynamics
3.1 Need for Second Law
- First law quantifies energy transfer but cannot predict direction or quality (e.g., why heat doesn’t flow from cold to hot spontaneously).
3.2 Statements
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Kelvin-Planck: No cycle can convert all heat from a single reservoir into work.
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Clausius: No cycle can transfer heat from a colder to a hotter body without external work.
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Equivalence: Violation of one implies violation of the other.
3.3 Clausius Inequality
$$ \oint \frac{dQ}{T} \leq 0 $$
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Equality ($$\displaystyle = $$) for reversible cycles.
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Significance: Defines entropy $$\displaystyle dS \geq \frac{dQ}{T} $$.
3.4 Entropy
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Definition: $$\displaystyle dS = \left(\frac{dQ}{T}\right)_{\text{rev}} $$; state function.
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Increase principle: For isolated system, $\Delta S \geq 0$; equality for reversible processes.
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Entropy change calculations: For reversible process, $$\displaystyle \Delta S = \int \frac{dQ_{\text{rev}}}{T} $$; for ideal gas, $$\displaystyle \Delta s = C_v \ln\frac{T_2}{T_1} + R \ln\frac{V_2}{V_1} $$ or $$\displaystyle C_p \ln\frac{T_2}{T_1} - R \ln\frac{p_2}{p_1} $$.
3.5 Carnot Cycle
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Reversible processes: Two isothermal, two adiabatic.
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$T$-$s$ diagram: Vertical lines for adiabatics, horizontal for isotherms.
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Efficiency derivation:
$$ \eta = 1 - \frac{T_2}{T_1} $$
where $$\displaystyle T_1 $$, $$\displaystyle T_2 $$ are source and sink temperatures (absolute).
- Importance: Maximum possible efficiency between two reservoirs.
3.6 Carnot Theorem
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No engine operating between two reservoirs can be more efficient than a Carnot engine.
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All reversible engines between same reservoirs have same efficiency.
3.7 Heat Engines
- Thermal efficiency: $$\displaystyle \eta = \frac{W_{\text{net}}}{Q_{\text{in}}} = 1 - \frac{Q_{\text{out}}}{Q_{\text{in}}} $$.
3.8 Refrigerators and Heat Pumps
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Refrigerator COP: $$\displaystyle \text{COP}_R = \frac{Q_L}{W} = \frac{Q_L}{Q_H - Q_L} $$.
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Heat pump COP: $$\displaystyle \text{COP}_{HP} = \frac{Q_H}{W} = \frac{Q_H}{Q_H - Q_L} $$.
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Carnot COP:
$$\displaystyle \text{COP}_{R,\text{Carnot}} = \frac{T_L}{T_H - T_L} $$,
$$\displaystyle \text{COP}_{HP,\text{Carnot}} = \frac{T_H}{T_H - T_L} $$.
3.9 Reversible Processes
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Conditions: Quasi-static, no friction, no unrestrained expansion.
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Proof of maximum COP: For given $$\displaystyle T_H $$, $$\displaystyle T_L $$, reversible refrigerator gives maximum COP (from Carnot theorem).
3.10 Combined Cycles
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Heat engine (efficiencies $\eta$) drives refrigerator (COP).
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Derivation:
Engine: $$\displaystyle \frac{Q_1}{Q_2} = \frac{T_1}{T_2} $$ (Carnot).
Refrigerator: $$\displaystyle \frac{Q_4}{Q_3} = \frac{T_4}{T_3} $$, and $$\displaystyle Q_3 = Q_2 + W $$, $$\displaystyle Q_4 = Q_3 - W $$.
Eliminate $W$ to get $$\displaystyle \frac{Q_2}{Q_1} $$ in terms of $$\displaystyle T_1, T_2, T_3, T_4 $$.
3.11 Availability (Exergy)
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Definition: Maximum useful work obtainable as system reaches equilibrium with environment.
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Availability function for closed system:
$$ \psi = (U - U_0) + p_0(V - V_0) - T_0(S - S_0) $$
where subscript 0 denotes environmental state ($$\displaystyle T_0 $$, $$\displaystyle p_0 $$).
- Significance: Measures quality of energy; destroyed by irreversibilities.
3.12 Third Law of Thermodynamics
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Statement: Entropy of a perfect crystal at absolute zero is zero.
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Implications:
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Absolute entropy values possible.
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Unattainability of absolute zero (infinite steps needed).
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[!TIP]
For combined cycles, remember: $$\displaystyle W_{\text{net}} = Q_1 - Q_2 = Q_3 - Q_4 $$. Use Carnot relations to eliminate $W$.
4. Properties of Pure Substances (Focus on Steam)
4.1 Pure Substance
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Definition: Homogeneous material with invariant chemical composition (e.g., water, air).
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Phases: Solid, liquid, vapor. Phase changes: melting, vaporization, sublimation.
4.2 Triple Point and Critical Point
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Triple point: Unique $p$, $T$ where solid, liquid, vapor coexist (e.g., water: 0.01°C, 0.611 kPa).
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Critical point: $$\displaystyle p_c $$, $$\displaystyle T_c $$ where liquid-vapor distinction vanishes (water: 374°C, 22.1 MPa).
4.3 $P$-$V$-$T$ Surface
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Features:
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Saturation curves: Liquid-vapor boundary (dome-shaped).
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Triple line: Solid-liquid-vapor coexistence.
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Critical point: Top of dome.
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Regions: Compressed liquid, wet mixture, superheated vapor, solid.
4.4 Steam Formation: Constant Pressure Heating
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Process:
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Compressed liquid (sensible heat ↑ $T$).
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Saturation liquid ($$\displaystyle x=0 $$).
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Phase change (latent heat, $T$, $p$ constant, $x$ ↑).
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Saturated vapor ($$\displaystyle x=1 $$).
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Superheated vapor (sensible heat ↑ $T$).
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Sensible heat: Energy to change temperature without phase change.
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Latent heat: Energy for phase change at constant $T$ ($$\displaystyle h_{fg} $$).
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Enthalpy:
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Wet steam: $$\displaystyle h = h_f + x h_{fg} $$.
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Superheated steam: $$\displaystyle h = h_g + C_p (T - T_{\text{sat}}) $$ (approx).
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4.5 Steam Tables
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Usage: Provide $v$, $h$, $s$, $u$ for saturated and superheated steam at given $p$ or $T$.
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Interpolation: Linear in $T$ for superheated; for saturated, use $p$-$T$ relation or quality.
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Key columns: $$\displaystyle v_f $$, $$\displaystyle v_g $$, $$\displaystyle h_f $$, $$\displaystyle h_g $$, $$\displaystyle h_{fg} $$, $$\displaystyle s_f $$, $$\displaystyle s_g $$, $$\displaystyle s_{fg} $$, $$\displaystyle u_f $$, $$\displaystyle u_g $$, $$\displaystyle u_{fg} $$.
4.6 Mollier Chart ($h$-$s$ Diagram)
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Usage: Graphical determination of steam properties; constant pressure lines diverge.
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Divergence of isobars: From $$\displaystyle dh = Tds + vdp $$, at constant $p$, $$\displaystyle dh = Tds $$. As $s$ increases, $T$ increases → slope increases → isobars diverge.
DiagramSEARCH: Mollier chart h-s diagram steam
4.7 Dryness Fraction (Quality) $x$
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Definition: $$\displaystyle x = \frac{\text{mass of vapor}}{\text{total mass}} $$; $0 \leq x \leq 1$.
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Measurement:
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Separating calorimeter: Separates liquid; $$\displaystyle x = \frac{m_v}{m_v + m_l} $$.
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Throttling calorimeter: Throttle to low pressure; measure $T$, $p$ after throttle; $$\displaystyle h_1 = h_2 $$ → find $x$.
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Significance: Indicates steam quality; high $x$ preferred for turbines to avoid erosion.
4.8 $T$-$s$ Diagram for Water
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Wet region: Between saturated liquid and vapor lines.
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Dry saturated steam: On vapor line ($$\displaystyle x=1 $$).
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Superheated steam: Above vapor line.
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Compressed liquid: Left of saturated liquid line.
DiagramSEARCH: T-s diagram water steam phases
4.9 Properties of Wet Steam
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$$\displaystyle v = v_f + x v_{fg} $$
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$$\displaystyle h = h_f + x h_{fg} $$
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$$\displaystyle s = s_f + x s_{fg} $$
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$$\displaystyle u = u_f + x u_{fg} $$
4.10 Properties of Superheated Steam
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From steam tables at given $p$, $T$ (interpolate if needed).
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Alternatively, use $$\displaystyle h = h_g + C_p (T - T_{\text{sat}}) $$, $$\displaystyle s = s_g + C_p \ln\frac{T}{T_{\text{sat}}} $$ (approx, $$\displaystyle C_p $$ constant).
4.11 Wet Steam and Ideal Gas Law
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Does not obey ideal gas law because it is a two-phase mixture; specific volume depends strongly on $x$.
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Ideal gas law applies only to superheated steam (low $p$, high $T$).
4.12 Rigid Vessel Problem Example
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Given: Rigid vessel ($$\displaystyle V = \text{const} $$), mass $m$, initial $p$, find properties.
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Steps:
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Compute $$\displaystyle v = V/m $$.
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From $p$ and $v$, locate state in steam tables: if $$\displaystyle v_f < v < v_g $$ at given $p$, then wet.
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Find $$\displaystyle x = \frac{v - v_f}{v_{fg}} $$.
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Then $$\displaystyle h = h_f + x h_{fg} $$, $$\displaystyle s = s_f + x s_{fg} $$, $$\displaystyle u = u_f + x u_{fg} $$, $$\displaystyle T = T_{\text{sat}}(p) $$.
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Heating to dry saturated: $$\displaystyle x=1 $$, $v$ constant → find $$\displaystyle p_2 $$ such that $$\displaystyle v = v_g(p_2) $$. Then $$\displaystyle Q = m(u_2 - u_1) $$ (since $$\displaystyle W=0 $$).
[!TIP]
In rigid vessel problems, volume constant → $v$ constant. Use $v$ and $x$ to find final pressure when $$\displaystyle x=1 $$.
5. Ideal Gases and Thermodynamic Processes
5.1 Ideal Gas Equation
$$ \boxed{pV = mRT \quad \text{or} \quad pv = RT} $$
- $$\displaystyle R = \frac{R_{\text{univ}}}{M} $$; for air, $$\displaystyle R = 0.287\ \text{kJ/kg·K} $$.
5.2 Specific Heats and $\gamma$
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$$\displaystyle C_p - C_v = R $$
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$$\displaystyle \gamma = \frac{C_p}{C_v} $$
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For monatomic gas: $$\displaystyle \gamma = 1.67 $$; diatomic: $\gamma \approx 1.4$.
5.3 Mean and Variable Specific Heat
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Mean: $$\displaystyle C_{p,m} = \frac{h_2 - h_1}{T_2 - T_1} $$.
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Variable: If $$\displaystyle C_p = a + bT $$, then
$$ h_2 - h_1 = a(T_2 - T_1) + \frac{b}{2}(T_2^2 - T_1^2) $$
5.4 Thermodynamic Processes for Ideal Gases
| Process | $pV$ relation | $TV$ relation | $Tp$ relation | Work $W$ | Heat $Q$ |
|---|---|---|---|---|---|
| Isothermal | $$\displaystyle pV = \text{const} $$ | — | — | $$\displaystyle mRT \ln\frac{V_2}{V_1} $$ | $$\displaystyle = W $$ |
| Adiabatic reversible | $$\displaystyle pV^\gamma = \text{const} $$ | $$\displaystyle TV^{\gamma-1}=\text{const} $$ | $$\displaystyle T^\gamma p^{1-\gamma}=\text{const} $$ | $$\displaystyle \frac{mR(T_1 - T_2)}{\gamma-1} $$ | 0 |
| Polytropic | $$\displaystyle pV^n = \text{const} $$ | $$\displaystyle TV^{n-1}=\text{const} $$ | $$\displaystyle T^n p^{(1-n)/n}=\text{const} $$ | $$\displaystyle \frac{mR(T_1 - T_2)}{n-1} $$ | $$\displaystyle mC_v(T_2-T_1) + W $$ |
| Isobaric | $$\displaystyle V/T = \text{const} $$ | — | $$\displaystyle p = \text{const} $$ | $$\displaystyle p(V_2 - V_1) = mR(T_2 - T_1) $$ | $$\displaystyle mC_p(T_2 - T_1) $$ |
| Isochoric | $$\displaystyle p/T = \text{const} $$ | $$\displaystyle V = \text{const} $$ | — | 0 | $$\displaystyle mC_v(T_2 - T_1) $$ |
5.5 Work and Heat Calculations
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Use first law: $$\displaystyle Q = \Delta U + W $$, with $$\displaystyle \Delta U = mC_v \Delta T $$ for ideal gas.
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For polytropic, $$\displaystyle W = \frac{p_2V_2 - p_1V_1}{1-n} $$ ($n \neq 1$).
5.6 Applications
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Isothermal expansion: $$\displaystyle W = mRT \ln\frac{p_1}{p_2} $$.
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Adiabatic compression: $$\displaystyle T_2 = T_1 \left(\frac{p_2}{p_1}\right)^{(\gamma-1)/\gamma} $$.
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Polytropic: Find $n$ from $$\displaystyle p_1V_1^n = p_2V_2^n $$.
5.7 First Law Applications
- Always: $$\displaystyle Q = \Delta U + W $$; determine $$\displaystyle \Delta U = mC_v \Delta T $$, $W$ from process.
[!TIP]
For polytropic processes, if $n$ is not given, compute from initial and final states: $$\displaystyle n = \frac{\ln(p_2/p_1)}{\ln(V_1/V_2)} $$.
6. Gas Mixtures
6.1 Dalton’s and Amagat’s Laws
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Dalton’s law (partial pressures): $$\displaystyle p = \sum p_i $$, where $$\displaystyle p_i = y_i p $$, $$\displaystyle y_i $$ = mole fraction.
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Amagat’s law (partial volumes): $$\displaystyle V = \sum V_i $$, where $$\displaystyle V_i = y_i V $$ at total $p$, $T$.
6.2 $P$-$V$-$T$ Relationships
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For ideal gas mixture: $$\displaystyle pV = nRT = \left(\sum n_i\right)RT $$.
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Mixture gas constant: $$\displaystyle R_{\text{mix}} = \sum y_i R_i $$.
6.3 Specific Heat, Enthalpy, Internal Energy
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Mass-weighted averages:
$$\displaystyle c_{p,\text{mix}} = \sum y_i c_{p,i} $$,
$$\displaystyle h_{\text{mix}} = \sum y_i h_i $$,
$$\displaystyle u_{\text{mix}} = \sum y_i u_i $$.
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For constant pressure process, $$\displaystyle h_{\text{mix}} $$ change = $$\displaystyle \sum m_i h_i $$ change.
6.4 Mollier Charts for Mixtures
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Used for air-vapor mixtures (psychrometrics).
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Enthalpy per kg dry air: $$\displaystyle h = 1.005 T + \omega (2501 + 1.88 T) $$ (kJ/kg da), where $\omega$ = humidity ratio.
6.5 Adiabatic Saturation Process
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Process: Air humidified adiabatically at constant pressure (e.g., adiabatic saturator).
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Enthalpy constant: $$\displaystyle h_{\text{air+vapor}} = \text{const} $$ because $$\displaystyle Q=0 $$, negligible KE/PE.
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Psychrometric application: Find humidity ratio from inlet and outlet states.
[!TIP]
In adiabatic saturation, enthalpy of air-vapor mixture remains constant because no heat transfer and negligible kinetic/potential energy changes.
7. Thermodynamic Cycles
7.1 Air Standard Assumptions
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Ideal gas with constant $$\displaystyle C_p $$, $$\displaystyle C_v $$.
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Closed system (no mass flow).
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No friction, pressure losses.
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Heat addition/rejection instantaneous.
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No chemical reactions (except combustion cycles).
7.2 Carnot Cycle
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Processes: Isothermal expansion ($$\displaystyle T_H $$), adiabatic expansion, isothermal compression ($$\displaystyle T_L $$), adiabatic compression.
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Efficiency: $$\displaystyle \eta = 1 - \frac{T_L}{T_H} $$.
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Significance: Upper limit for any cycle between $$\displaystyle T_H $$, $$\displaystyle T_L $$.
7.3 Otto Cycle (Constant Volume Heat Addition)
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Processes:
1–2: Adiabatic compression ($$\displaystyle V_1/V_2 = r $$).
2–3: Constant volume heat addition.
3–4: Adiabatic expansion.
4–1: Constant volume heat rejection.
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Efficiency derivation:
$$ \eta = 1 - \frac{1}{r^{\gamma-1}} $$
where $$\displaystyle r = \frac{V_1}{V_2} $$ (compression ratio).
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Work output: $$\displaystyle W_{\text{net}} = Q_{\text{in}} - Q_{\text{out}} $$.
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Mean effective pressure (MEP): $$\displaystyle \text{MEP} = \frac{W_{\text{net}}}{V_1 - V_2} $$.
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Comparison: Otto efficiency depends only on $r$ and $\gamma$; higher $r$ → higher $\eta$.
7.4 Diesel Cycle (Constant Pressure Heat Addition)
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Processes:
1–2: Adiabatic compression.
2–3: Constant pressure heat addition.
3–4: Adiabatic expansion.
4–1: Constant volume heat rejection.
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Cut-off ratio: $$\displaystyle \rho = \frac{V_3}{V_2} $$.
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Efficiency derivation:
$$ \eta = 1 - \frac{1}{r^{\gamma-1}} \cdot \frac{\rho^\gamma - 1}{\gamma(\rho - 1)} $$
- Comparison: For same $r$, Diesel $\eta$ < Otto $\eta$ if $$\displaystyle \rho > 1 $$, but Diesel can operate at higher $r$ without knocking.
7.5 Dual Cycle (Mixed Cycle)
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Heat addition: Part constant volume, part constant pressure.
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Efficiency between Otto and Diesel.
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Efficiency expression:
$$ \eta = 1 - \frac{1}{r^{\gamma-1}} \cdot \frac{\rho^\gamma (\alpha^\gamma - 1)}{\gamma(\rho - 1)(\alpha - 1) + \gamma(\alpha^\gamma - 1)} $$
where $$\displaystyle \alpha = \frac{V_4}{V_3} $$ (pressure ratio at end of constant volume heat addition).
7.6 Brayton Cycle (Gas Turbine)
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Processes:
1–2: Adiabatic compression ($$\displaystyle r_p = p_2/p_1 $$).
2–3: Constant pressure heat addition.
3–4: Adiabatic expansion.
4–1: Constant pressure heat rejection.
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Efficiency:
$$ \eta = 1 - \frac{1}{r_p^{(\gamma-1)/\gamma}} $$
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Work output: $$\displaystyle W_{\text{net}} = C_p[(T_3 - T_2) - (T_2 - T_1)] $$.
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Regeneration: Use exhaust heat to preheat compressed air; improves efficiency (regenerative Brayton).
7.7 Comparison of Efficiencies
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Carnot (highest): $$\displaystyle \eta_{\text{Carnot}} = 1 - T_L/T_H $$.
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Otto: $$\displaystyle \eta_{\text{Otto}} = 1 - 1/r^{\gamma-1} $$.
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Diesel: $$\displaystyle \eta_{\text{Diesel}} = 1 - \frac{1}{r^{\gamma-1}} \cdot \frac{\rho^\gamma - 1}{\gamma(\rho - 1)} $$.
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Brayton: $$\displaystyle \eta_{\text{Brayton}} = 1 - 1/r_p^{(\gamma-1)/\gamma} $$.
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Order (typical): Carnot > Otto > Diesel > Brayton (but depends on parameters).
7.8 Actual vs. Ideal Cycles
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Losses: Friction, pressure drops, heat transfer to walls, non-instantaneous combustion, valve timing, exhaust losses.
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Effect: Lower efficiency, less work output.
7.9 Air Standard Efficiency
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Efficiency based on air standard assumptions; used for comparison.
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Assumes air as ideal gas, constant specific heats, no friction, etc.
[!TIP]
For Otto, Diesel, Brayton, remember key parameters: compression ratio $r$ for Otto/Diesel, pressure ratio $$\displaystyle r_p $$ for Brayton, cut-off ratio $\rho$ for Diesel.
8. Refrigeration and Heat Pump Cycles
8.1 Reversed Carnot Cycle
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Processes: Isothermal expansion (evaporator, $$\displaystyle T_L $$), adiabatic compression, isothermal compression (condenser, $$\displaystyle T_H $$), adiabatic expansion.
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COP refrigerator:
$$ \text{COP}_R = \frac{T_L}{T_H - T_L} $$
- COP heat pump:
$$ \text{COP}_{HP} = \frac{T_H}{T_H - T_L} $$
8.2 Vapor Compression Refrigeration Cycle
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Components: Compressor, condenser, expansion valve, evaporator.
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Processes:
1–2: Adiabatic compression.
2–3: Constant pressure condensation.
3–4: Throttling ($$\displaystyle h_3 = h_4 $$).
4–1: Constant pressure evaporation.
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COP: $$\displaystyle \text{COP} = \frac{h_1 - h_4}{h_2 - h_1} $$.
8.3 Heat Pump
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Same cycle as refrigeration but for heating.
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COP: $$\displaystyle \text{COP}_{HP} = \frac{h_1 - h_4}{h_2 - h_1} + 1 $$ (since $$\displaystyle Q_H = Q_L + W $$).
8.4 Combined Systems
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Heat engine drives refrigerator/heat pump.
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Derive heat transfer ratios as in Section 3.10.
8.5 Minimum Power Requirements
- For given cooling load $$\displaystyle Q_L $$, minimum power $$\displaystyle W_{\text{min}} $$ when COP is maximum (Carnot COP):
$$ W_{\text{min}} = \frac{Q_L}{\text{COP}_{\text{Carnot}}} = Q_L \frac{T_H - T_L}{T_L} $$
[!TIP]
In vapor compression cycle, throttling is isenthalpic ($$\displaystyle h_3 = h_4 $$); use this to find state after expansion.
9. Combustion Thermodynamics
9.1 Stoichiometry
- General hydrocarbon $$\displaystyle \mathrm{C}_x\mathrm{H}_y $$:
$$ \mathrm{C}_x\mathrm{H}_y + \left(x + \frac{y}{4}\right) \mathrm{O}_2 \rightarrow x\mathrm{CO}_2 + \frac{y}{2} \mathrm{H}_2\mathrm{O} $$
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Air composition: $21\%$ $$\displaystyle \mathrm{O}_2 $$, $79\%$ $$\displaystyle \mathrm{N}_2 $$ (by volume, mole basis).
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Theoretical air: Stoichiometric $$\displaystyle \mathrm{O}_2 $$ × $$\displaystyle \frac{100}{21} $$ (moles air per mole fuel).
9.2 Air-Fuel Ratio
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Theoretical (stoichiometric) AF: Mass of air for complete combustion.
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Excess air: $$\displaystyle \text{Excess \%} = \frac{\text{actual air} - \text{theoretical air}}{\text{theoretical air}} \times 100 $$.
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Actual air: $(1 + \text{excess fraction}) \times \text{theoretical air}$.
9.3 Combustion Products Analysis
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Dry basis: Excludes water vapor; mole fractions sum to 1.
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Wet basis: Includes water vapor.
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Volumetric composition = mole fraction × 100% (for ideal gases, volume fraction = mole fraction).
9.4 Enthalpy of Formation $$\displaystyle \Delta H_f^\circ $$
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Definition: Enthalpy change when 1 mole of compound is formed from elements in their standard states at specified $T$ (usually 298 K) and $p$ (1 atm).
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Standard values: Tabulated (e.g., $$\displaystyle \Delta H_f^\circ(\mathrm{CO}_2) = -393.5\ \text{kJ/mol} $$).
-
Hess’s Law: Total enthalpy change independent of path.
9.5 Enthalpy of Reaction $$\displaystyle \Delta H_{\text{rxn}} $$
- Relation to formation enthalpies:
$$ \Delta H_{\text{rxn}} = \sum \nu_i \Delta H_{f,i}^\circ (\text{products}) - \sum \nu_i \Delta H_{f,i}^\circ (\text{reactants}) $$
where $$\displaystyle \nu_i $$ are stoichiometric coefficients (positive for products, negative for reactants).
9.6 Adiabatic Flame Temperature
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Concept: Temperature when all heat from combustion heats products, no heat loss.
-
Calculation:
$$ \sum n_i h_i(T_{\text{ad}}) = \sum n_i h_i(T_{\text{initial}}) + \Delta H_{\text{rxn}} $$
Iterative: guess $$\displaystyle T_{\text{ad}} $$, compute $$\displaystyle h_i $$ (from tables or $$\displaystyle C_p $$ integrals), adjust until equality.
9.7 Actual vs. Theoretical Combustion
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Theoretical: Exactly stoichiometric air; complete combustion assumed.
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Actual: Often with excess air; may have incomplete combustion (CO, soot).
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Impact: Excess air lowers flame temperature, increases losses; incomplete combustion reduces efficiency, increases pollutants.
9.8 Heating Values
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Higher Heating Value (HHV): Includes latent heat of vaporization of water in products.
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Lower Heating Value (LHV): Excludes latent heat; $$\displaystyle \text{HHV} = \text{LHV} + n_{\mathrm{H}_2\mathrm{O}} \cdot h_{fg} $$.
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Significance: LHV used for engine efficiency; HHV for boilers (condensation possible).
9.9 Ultimate Analysis of Fuels
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Mass fractions: C, H, O, N, S, moisture, ash.
-
Used to compute theoretical air, products composition.
9.10 Problems
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Oxygen required: From stoichiometry.
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Air supplied: Theoretical air × $(1 + \text{excess fraction})$.
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Products composition: Include excess $$\displaystyle \mathrm{O}_2 $$, $$\displaystyle \mathrm{N}_2 $$, $$\displaystyle \mathrm{CO}_2 $$, $$\displaystyle \mathrm{H}_2\mathrm{O} $$; compute mole fractions on dry/wet basis.
[!TIP]
For combustion calculations, always balance atoms first. Use mole basis; convert mass to moles using molecular weights.
10. Advanced Topics
10.1 Real Gases
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Deviations: Significant at high pressure, low temperature.
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van der Waals equation:
$$ \left(p + \frac{a}{V^2}\right)(V - b) = RT $$
$a$ accounts for attractive forces, $b$ for molecular volume.
- Compressibility factor: $$\displaystyle Z = \frac{pV}{RT} $$; $$\displaystyle Z < 1 $$: attractive forces dominate; $$\displaystyle Z > 1 $$: repulsive forces dominate.
10.2 $P$-$V$-$T$ Surface for Real Gases
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Similar to ideal but:
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No sharp liquid-vapor boundary above critical point.
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Vapor pressure curve terminates at critical point.
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May show loops (unstable region) for some gases.
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DiagramCANVAS: 3D surface showing critical point and vapor pressure curve
10.3 Availability (Exergy) Analysis
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Closed system: $$\displaystyle \psi = (U - U_0) + p_0(V - V_0) - T_0(S - S_0) $$.
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Open system (flow exergy):
$$ \psi_{\text{flow}} = (h - h_0) - T_0(s - s_0) + \frac{V^2}{2} + gz $$
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Environmental reference state: $$\displaystyle T_0 $$, $$\displaystyle p_0 $$ (usually 298 K, 1 atm).
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Exergy destruction: $$\displaystyle T_0 \Delta S_{\text{gen}} $$ due to irreversibilities.
10.4 Third Law of Thermodynamics
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Statement: Entropy of a perfect crystal at absolute zero is zero.
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Implications:
-
Absolute entropy values obtainable by integrating $$\displaystyle C_p/T $$ from 0 K.
-
Unattainability of absolute zero (infinite steps).
-
Residual entropy for imperfect crystals (e.g., glasses).
-
[!TIP]
For real gases, use compressibility charts: $$\displaystyle Z = f(p_r, T_r) $$ where $$\displaystyle p_r = p/p_c $$, $$\displaystyle T_r = T/T_c $$.
Note: All formulas assume consistent units (SI: K, Pa, m³, J). Use $$\displaystyle R = 0.287\ \text{kJ/kg·K} $$ for air, $$\displaystyle C_p \approx 1.005\ \text{kJ/kg·K} $$, $$\displaystyle C_v \approx 0.718\ \text{kJ/kg·K} $$, $\gamma \approx 1.4$.