Thermodynamics (ME-302) - Unit 3 Short Notes

UNIT 3: THERMODYNAMICS (ME-302)

Exam-Focused Short Notes | Based on RGPV Past Papers (2022-2025)

I. FUNDAMENTAL LAWS & THEIR IMPLICATIONS

First Law for Closed Systems (Non-Flow Process)

• Statement: Energy cannot be created or destroyed, only transformed. For a closed system undergoing a cycle, ∮ δQ = ∮ δW. For a process, Q - W = ΔU (Sign Convention: Q = heat added to system, W = work done by system).

$$Q = \Delta U + W$$

• Path vs. Point Functions:

  • Path Functions: Q (Heat) and W (Work) depend on the process path.

  • Point Functions (Properties): U (Internal Energy), H (Enthalpy), P, V, T, S (Entropy) depend only on the state.

• Polytropic Process (pV^n = C):

  • Work Done:

$$W = \frac{p_2 V_2 - p_1 V_1}{1 - n} = \frac{mR(T_2 - T_1)}{1 - n}$$

    (For `n=1`, isothermal: `W = mRT \ln(V_2/V_1)`).

*   **Heat Transfer:** `Q = ΔU + W = mC_v(T_2 - T_1) + W`.

*   **Finding `n`:** Use `(p_1V_1^n = p_2V_2^n)` → `n = \frac{\ln(p_2/p_1)}{\ln(V_1/V_2)}`.

> [!TIP] Common Pitfall: Forgetting sign convention. `W` is work done *by* gas. Compression work is negative (work done *on* gas).

Second Law of Thermodynamics

• Statements:

  • Kelvin-Planck: "It is impossible to construct a device that, operating in a cycle, will produce no effect other than the extraction of heat from a single reservoir and the performance of an equivalent amount of work."

  • Clausius: "It is impossible to construct a device that, operating in a cycle, will produce no effect other than the transfer of heat from a cooler to a hotter body."

  • Equivalence: Violation of one implies violation of the other.

• Clausius Inequality:

$$\oint \frac{\delta Q}{T} \leq 0$$

*   `=` for a **reversible cycle**.

*   `<` for an **irreversible cycle**.

*   **Significance:** It is the **mathematical statement of the second law**. It provides a criterion for process reversibility and defines entropy as a property.

• Reversible vs. Irreversible Processes:

  • Reversible: Quasi-static, no friction, no unrestrained expansion, no finite temperature differences. Idealized limit.

  • Irreversible: Friction, unrestrained expansion, heat transfer across finite ΔT, mixing, chemical reactions. All real processes are irreversible.

  • Causes: Friction, inelastic deformation, heat transfer across ΔT, diffusion, chemical reactions.

Entropy

• Definition: A measure of molecular disorder/randomness. A property of the system.

$$ds = \left( \frac{\delta Q}{T} \right)_{rev}$$

For any process, `ΔS ≥ ∫(δQ/T)`.

• Entropy Change Calculation:

  • For Ideal Gas (with variable C_p):

$$\Delta S = m \int_{T_1}^{T_2} \frac{C_p(T)}{T} dT - mR \ln \left( \frac{p_2}{p_1} \right)$$

    If `C_p = a + bT + cT^2`, integrate term by term.

*   **For Pure Substances (Steam):** Use steam tables. `ΔS = s₂ - s₁`.

• Principle of Increase of Entropy:

$$dS_{universe} = dS_{system} + dS_{surroundings} \geq 0$$

*   `=` for reversible process/isolated system.

*   `>` for irreversible process. **Entropy is generated (`S_gen > 0`) in irreversible processes.**

Availability (Exergy) & Irreversibility

• Availability Function (Exergy): Maximum useful work obtainable as a system comes to equilibrium with a dead state (environment at T₀, P₀).

  • For a closed system: Exergy, ψ = (U - U₀) + P₀(V - V₀) - T₀(S - S₀).

  • For an open system (flow exergy): ψ = (H - H₀) - T₀(S - S₀) + \frac{1}{2}(V^2) + gz.

• Dead State: When system is in thermodynamic equilibrium with the environment (T=T₀, P=P₀). Exergy = 0.

• Irreversibility (I) / Lost Work: I = W_{max,rev} - W_{actual}. I = T₀ S_{gen}.

  [!TIP] Steel Block Cooling Problem (Classic): Available energy = m C_p [(T - T₀) - T₀ \ln(T/T₀)].

Third Law of Thermodynamics (Nernst Heat Theorem)

• Statement: "The entropy of a perfect crystal at absolute zero is zero."

• Significance:

  1. Provides a reference point for absolute entropy values (S→0 as T→0 K for pure crystalline substances).

  2. It is impossible to reach absolute zero temperature in a finite number of steps.

  3. As T→0 K, heat capacities of substances approach zero.

II. HEAT ENGINES, REFRIGERATORS & CYCLES

Performance Parameters

• Heat Engine: Converts heat to work.

  • Thermal Efficiency: η = \frac{W_{net}}{Q_{in}} = 1 - \frac{Q_{out}}{Q_{in}}.

• Refrigerator/Heat Pump: Transfers heat from low to high temperature using work.

  • COP (Coefficient of Performance):

    • Refrigerator: COP_R = \frac{Q_L}{W_{net}}.

    • Heat Pump: COP_HP = \frac{Q_H}{W_{net}} = COP_R + 1.

• Carnot Cycle (Reversible): Consists of two reversible isothermal and two reversible adiabatic processes.

  • Efficiency: η_{Carnot} = 1 - \frac{T_L}{T_H} (T in Kelvin).

  • COP of Reversible Refrigerator: COP_{R,Carnot} = \frac{T_L}{T_H - T_L} (Maximum possible COP between T_L and T_H).

  [!TIP] Proof of Maximum COP: Assume a refrigerator with COP > Carnot COP. Couple it with a Carnot heat engine. Net effect would violate Clausius statement (transfer heat from cold to hot without external work).

Air-Standard Cycles (Ideal Cycles)

Assumptions: Working fluid is air (ideal gas, constant γ and C_p), no friction, no heat loss, processes are internally reversible.

• Otto Cycle (Constant Volume Heat Addition): Processes: 1-2 (Isentropic compression), 2-3 (Isochoric heat addition), 3-4 (Isentropic expansion), 4-1 (Isochoric heat rejection).

$$\eta_{Otto} = 1 - \frac{1}{r^{\gamma-1}}$$

where `r = V₁/V₂` (Compression ratio).

• Diesel Cycle (Constant Pressure Heat Addition): Processes: 1-2 (Isentropic compression), 2-3 (Isobaric heat addition), 3-4 (Isentropic expansion), 4-1 (Isochoric heat rejection).

$$\eta_{Diesel} = 1 - \frac{1}{r^{\gamma-1}} \left[ \frac{\rho^\gamma - 1}{\gamma(\rho - 1)} \right]$$

where `ρ = V₃/V₂` (Cut-off ratio).

• Dual Cycle (Mixed): Heat addition partly at constant volume, partly at constant pressure.

• Brayton (Gas Turbine) Cycle: Processes: 1-2 (Isentropic compression), 2-3 (Isobaric heat addition), 3-4 (Isentropic expansion), 4-1 (Isobaric heat rejection).

$$\eta_{Brayton} = 1 - \frac{1}{r_p^{(\gamma-1)/\gamma}}$$

where `r_p = P₂/P₁` (Pressure ratio).

Comparative Efficiency

For same T_max and T_min:

η_{Carnot} > η_{Otto} > η_{Diesel} > η_{Dual} > η_{Brayton} (for typical r and ρ).

[!TIP] Key Insight: Otto has highest efficiency for same compression ratio because heat addition at constant volume gives higher T_max (thus higher η). Diesel allows higher compression ratio without knocking, so can have higher η than Otto at high r.

Combined Cycles (Reversible Power Cycle + Reversible Heat Pump)

• Problem: Reversible heat engine (HE) operates between T₁ (source) and T₂. It drives a reversible refrigerator (RP) operating between T₄ (sink) and T₃. T₂ = T₃ (common reservoir).

• Derivation:

  • HE: η = 1 - T₂/T₁ = W/Q₁ → W = Q₁(1 - T₂/T₁).

  • RP: COP = T₄/(T₃ - T₄) = Q₄/W → Q₄ = W * T₄/(T₃ - T₄).

  • Energy balance at T₂/T₃: Q₂ + Q₃ = Q₁ (HE rejects Q₂, RP rejects Q₃).

  • Using Q₂/Q₁ = T₂/T₁ (Carnot) and Q₃/Q₄ = T₃/T₄ (Carnot), solve for Q₂/Q₁.

$$\boxed{\frac{Q_2}{Q_1} = \frac{T_2(T_3 - T_4)}{T_1 T_3 - T_2 T_4}}$$

> [!TIP] This is a **very high frequency** derivation. Practice the energy balance and Carnot relations step-by-step.

III. PURE SUBSTANCES & STEAM TABLES

Pure Substance & Phase Change

• Definition: A substance of uniform chemical composition throughout (e.g., pure water, nitrogen, dry air).

• Phases: Solid, Liquid, Vapor (Gas).

• Phase Change Processes: Melting (fusion), Vaporization (evaporation/boiling), Sublimation.

• Formation of Steam (at Constant P):

  1. Compressed Liquid (Subcooled): T < T_sat at given P.

  2. Saturated Liquid (f): T = T_sat, x=0. Start of vaporization.

  3. Wet Steam (Mixture): T = T_sat, 0 < x < 1. Liquid + Vapor coexist.

  4. Saturated Vapor (g): T = T_sat, x=1. End of vaporization.

  5. Superheated Vapor: T > T_sat at given P.

• Key Terms:

  • Sensible Heat: Heat added causing ΔT (compressed liquid, superheated vapor).

  • Latent Heat (Enthalpy of Evaporation, h_fg): Heat added at constant T during phase change (h_g - h_f).

• Graphical Representation: See T-s, P-v, P-h (Mollier) diagrams. Saturation curves separate single-phase and two-phase regions.

  [!DIAGRAM: CANVAS]

  T-s Diagram for Water:

  • Saturation dome (liquid-vapor).

  • Left of dome: compressed liquid (approx. constant s).

  • Right of dome: superheated vapor.

  • Inside dome: wet region (constant T, s increases with x).

  • Critical point: top of dome (T_c, P_c), v_f = v_g.

  • Triple point: line where solid, liquid, vapor coexist (T_t, P_t).

Critical Point & Triple Point

• Critical Point (T_c, P_c, v_c): State where liquid and vapor phases become indistinguishable. v_f = v_g, h_f = h_g. For water: T_c = 374°C, P_c = 22.1 bar.

• Triple Point (T_t, P_t): Unique T and P where all three phases coexist in equilibrium. For water: T_t = 0.01°C, P_t = 0.00611 bar. Only one T for triple point.

• P-V-T Surface: 3D surface showing P vs. V vs. T. The surface has a "nose" (critical point) and a "tongue" (liquid-vapor separation). The triple point is where solid, liquid, vapor surfaces meet.

Steam Properties & Tables

• States:

  • Saturated: P and T are dependent (given one, other fixed). Properties: v_f, v_g, h_f, h_g, s_f, s_g, u_f, u_g.

  • Superheated: P and T independent. Use superheated tables.

• Dryness Fraction (x): Mass fraction of vapor in wet steam.

$$x = \frac{m_g}{m_{total}}$$

`0 ≤ x ≤ 1`. `x=0` (saturated liquid), `x=1` (saturated vapor).

• Property Calculations for Wet Steam:

$$\boxed{v = v_f + x(v_g - v_f)}$$

$$\boxed{h = h_f + x(h_{fg})}$$

$$\boxed{s = s_f + x(s_{fg})}$$

$$\boxed{u = u_f + x(u_{fg}) \quad \text{or} \quad u = h - Pv}$$

• Using Steam Tables:

  1. Identify state (saturated, superheated, compressed liquid).

  2. If saturated, find T or P in saturated table, get f and g values, use x if wet.

  3. If superheated, use P and T to interpolate in superheated table.

  4. Interpolation: Usually linear in T or P. For property Y between Y₁ at T₁ and Y₂ at T₂:

$$Y = Y_1 + \frac{(T - T_1)}{(T_2 - T_1)} (Y_2 - Y_1)$$

> [!TIP] **Common Pitfall:** For superheated steam at high `P`, sometimes `v` is given instead of `T`. Use `P` and `v` to find state (may need to check both saturated and superheated tables).

Mollier Chart (h-s Diagram)

• Construction: Plot of h vs s. Lines of constant P (isobars) and constant T (isotherms).

• Significance:

  1. Turbine Analysis: Expansion in turbine is approximately isentropic (s=const). Drop in h (Δh) gives work output w_turbine = h₁ - h₂ (neglecting KE/PE).

  2. Throttling (isenthalpic): h₁ = h₂.

• Why Isobars Diverge: For an ideal gas, (∂h/∂s)_P = T. As s increases (more disorder), T generally increases for a given P. The slope (∂h/∂s)_P = T increases with s (since T increases along an isobar in superheated region), so isobars bend and diverge.

  [!TIP] Short Note: Mollier chart is preferred for steam power plant analysis because turbine work is directly readable as vertical distance.

Wet Steam & Ideal Gas Law

• Does wet steam obey ideal gas laws? NO.

  • Reason: During phase change, P and T are dependent (P = P_sat(T)). Adding heat at constant T changes x but not P. Ideal gas law (PV=mRT) would require P to change with V at constant T, which is not the case in the wet region.

  • In Superheated Region: At low pressures, steam behaves approximately as an ideal gas.

IV. IDEAL GASES & GAS MIXTURES

Ideal Gas Equation of State

$$PV = mRT \quad \text{or} \quad Pv = RT \quad \text{or} \quad PV = nR_uT$$

• R = Specific gas constant (J/kg.K), R_u = Universal gas constant (8.314 J/mol.K).

• R = R_u / M, where M = Molecular weight (kg/kmol).

• Boyle's Law: P ∝ 1/V at constant T.

• Charles' Law: V ∝ T at constant P.

P-V-T Relationships for Mixtures (Dalton & Amagat)

• Dalton's Law of Partial Pressures (for ideal gas mixtures):

  Total pressure P is sum of partial pressures of individual components.

$$P = \sum_{i=1}^{n} P_i \quad \text{where} \quad P_i = y_i P$$

`y_i` = Mole fraction of component `i`.

• Amagat's Law of Partial Volumes (for ideal gas mixtures):

  Total volume V is sum of volumes each component would occupy at mixture T and P.

$$V = \sum_{i=1}^{n} V_i \quad \text{where} \quad V_i = y_i V$$

• Importance: Simplifies analysis of gas mixtures. Properties like U, H depend only on T for ideal gases, so mixture properties are mass/mole-weighted averages.

Thermodynamic Properties of Gas Mixtures

• Internal Energy & Enthalpy (Ideal Gas Mixture):

$$U_{mix} = \sum_{i} m_i u_i(T) \quad ; \quad H_{mix} = \sum_{i} m_i h_i(T)$$

*   **Key Point:** For ideal gases, `u` and `h` are **functions of temperature only**. Therefore, mixture `u` and `h` are independent of total pressure `P` and depend only on `T` and composition.

• Specific Heats of Mixtures:

  • Mass Basis: c_{v,mix} = \sum y_i c_{v,i} ; c_{p,mix} = \sum y_i c_{p,i}.

  • Molar Basis: C_{v,mix} = \sum y_i C_{v,i} ; C_{p,mix} = \sum y_i C_{p,i}.

  [!TIP] Remember: For ideal gases, C_p - C_v = R (per unit mass) or C_{p,m} - C_{v,m} = R_u (per mole). This holds for the mixture as well: C_{p,mix} - C_{v,mix} = R_{mix}.

Real Gases

• van der Waals Equation (Conceptual): (P + a/V²)(V - b) = RT accounts for intermolecular forces (a) and molecular volume (b).

• P-V-T Surface: Unlike ideal gas, real gas surface has a liquid-vapor coexistence region (similar to pure substance) and a critical point. At low T, isotherms show a "hump" (van der Waals loop) indicating phase change.

V. COMBUSTION & THERMOCHEMISTRY

Fuel Analysis & Stoichiometry

• Ultimate (Proximate) Analysis: Gives mass percentages of C, H₂, O₂, N₂, S, Ash, Moisture.

• Balancing Combustion Equation (e.g., Octane C₈H₁₈):

  C₈H₁₈ + a(O₂ + 3.76N₂) → bCO₂ + cH₂O + dN₂

  • C balance: b = 8

  • H balance: 2c = 18 → c = 9

  • O balance: 2a = 2b + c → 2a = 16 + 9 = 25 → a = 12.5

  • Stoichiometric Equation: C₈H₁₈ + 12.5(O₂ + 3.76N₂) → 8CO₂ + 9H₂O + 47N₂.

• Stoichiometric Air-Fuel Ratio (AFR_st):

  • Mass Basis: AFR_st = \frac{(32 + 3.76×28) × a}{\text{Fuel Mass}} (per kg fuel, a from balanced eqn).

  • For C₈H₁₈: AFR_st = 12.5 × (32 + 3.76×28) / 114 ≈ 14.7 kg air/kg fuel.

• Excess Air (%): Actual air supplied > Stoichiometric air.

  • m_{air,actual} = (1 + \frac{excess}{100}) × m_{air,stoich}.

  • Flue Gas Analysis (Volumetric): Use actual a in balanced equation. Find mole fractions of CO₂, H₂O, O₂ (from excess air), N₂.

Enthalpy of Combustion

• Enthalpy of Formation (ΔH_f): Enthalpy change when 1 mole of compound is formed from its elements in their standard states (at 1 bar, 298 K). Elements in standard state have ΔH_f = 0.

• Enthalpy of Reaction (ΔH_R): Enthalpy change when reaction occurs as written.

$$\Delta H_R = \sum_{products} n_p (\Delta H_f)_p - \sum_{reactants} n_r (\Delta H_f)_r$$

*   `n` = stoichiometric coefficients.

*   **First Law for Reacting Systems (SFEE):** For a combustion chamber at steady state:

$$\dot{Q} - \dot{W}_{shaft} = \dot{m}_{fuel} \left[ \sum_{out} y_i h_i - \sum_{in} y_i h_i \right]$$

    Often `Q=0`, `W_shaft=0`, so `H_{products} = H_{reactants}` (for adiabatic, no work).

Adiabatic Flame Temperature (AFT)

• Definition: Temperature of products when combustion is complete, adiabatic, and no dissociation.

• Calculation:

  1. Write balanced combustion equation with actual air (including excess N₂ and O₂).

  2. Apply First Law (SFEE) for adiabatic process: H_{reactants} = H_{products}.

  3. H_{reactants} = n_fuel * h_fuel(T_ref) + n_air * [y_O2 h_O2(T_ref) + y_N2 h_N2(T_ref)].

  4. H_{products} = n_CO2 h_CO2(T_AFT) + n_H2O h_H2O(T_AFT) + n_O2 h_O2(T_AFT) + n_N2 h_N2(T_AFT).

  5. Solve for T_AFT iteratively using h = h°(T) + [h - h°] (from tables or C_p integrals). Usually requires trial and error.

  [!TIP] Key: Assume all products at same T_AFT. Use C_p averages or tables for enthalpy calculation.

Actual vs. Theoretical Combustion

• Theoretical (Ideal): Complete combustion, no dissociation, adiabatic, no heat loss, no pressure drop.

• Actual:

  • Incomplete Combustion: CO, C (soot), unburnt HC.

  • Dissociation: At high T, CO₂, H₂O, O₂ dissociate into CO, H₂, O, OH, etc. Reduces T_max and NOx formation.

  • Heat Loss: To walls, uninsulated surfaces.

  • Pressure Drop: Friction in ducts.

• Impact on Performance: Actual AFT < Theoretical AFT. Lower AFT means lower cycle efficiency (Brayton, Otto, Diesel). Incomplete combustion reduces available energy, increases pollutants.

VI. APPLICATIONS & PROBLEM-SOLVING METHODOLOGIES

Process-Specific Calculations (Ideal Gas)

Steady Flow Energy Equation (SFEE)

$$\dot{Q} - \dot{W}_{shaft} = \dot{m} \left[ (h_2 - h_1) + \frac{(V_2^2 - V_1^2)}{2} + g(z_2 - z_1) \right]$$

• Turbine: W_shaft > 0 (output), Q≈0. w_turbine = h₁ - h₂ (ideal, isentropic).

• Compressor: W_shaft > 0 (input), Q≈0. w_comp = h₂ - h₁.

• Nozzle: W_shaft=0, Q≈0. V₂²/2 = h₁ - h₂.

• Throttle (Valve): W_shaft=0, Q≈0, ΔKE≈0, ΔPE≈0. h₁ = h₂ (isenthalpic).

• Heat Exchanger: W_shaft=0. \dot{m}_h (h_{h,in} - h_{h,out}) = \dot{m}_c (h_{c,out} - h_{c,in}) (if no loss).

• Refrigerator/Heat Pump (Cycle): COP_R = Q_L / W_{net}, COP_HP = Q_H / W_{net}. For Carnot, COP = T_L/(T_H-T_L) or T_H/(T_H-T_L).

First Law Analysis of Closed Systems (Cycles)

• For a cycle: ∮ δQ = ∮ δW → Q_net,cycle = W_net,cycle.

• Thermal Efficiency: η = W_net / Q_in.

• Example - Given Cycle: Constant volume heat addition (1-2), isothermal expansion (2-3), constant pressure cooling (3-1).

  • W_12 = 0, Q_12 = m C_v (T₂ - T₁).

  • W_23 = mRT₂ \ln(V₃/V₂), Q_23 = W_23 (ΔU=0).

  • W_31 = P₃(V₁ - V₃) = mR(T₁ - T₃), Q_31 = m C_p (T₁ - T₃).

  • Net work W_net = W_12 + W_23 + W_31. Q_in = Q_12 (if only heat added in 1-2). η = W_net / Q_12.

Systematic Procedure for Steam Problems

1. Draw Diagram: Sketch process (piston, rigid tank, throttling, etc.). Label knowns (P, T, V, m, x).

2. Identify State: Determine if state is saturated (wet/dry) or superheated. Use given P and T or P and x or V.

3. Locate on Tables:

   • If saturated (T given or P given and x known), use saturated water table.

   • If superheated (P and T given, or P and v), use superheated table.

   • If compressed liquid (T < T_sat at given P), approximate v≈v_f, h≈h_f, s≈s_f.

4. Interpolate: If property not directly listed, interpolate linearly between closest entries.

5. Apply Relations: For wet steam, use v = v_f + x v_fg, etc. For ideal gas, use PV=mRT.

6. Apply First Law: Q = ΔU + W (closed) or SFEE (open). For throttling: h₁ = h₂.

7. Check Units: Ensure consistency (kJ, kg, kPa, m³).

Final Exam Strategy:

1. Definitions First: Always state definitions clearly (Clausius, Entropy, Carnot, Dryness fraction, etc.).

2. Derivations: For cycle efficiencies, start with P-V and T-s diagrams, list processes, apply Q and W for each, then sum.

3. Steam Problems: 90% of mistakes happen in identifying the correct state and interpolation. Double-check if T and P correspond to saturation.

4. Combined Cycles & SFEE: Write separate energy balances for engine and refrigerator/pump, then link via common reservoir.

5. Combustion: Balance equation first! Then calculate AFR_st, excess air, flue gas composition. For AFT, set H_reactants = H_products.

6. Time Management: 7-mark questions require ~15-20 lines with formulas and a small diagram if relevant. 4-mark questions are concise definitions + one key formula.

All the best! Focus on understanding concepts behind formulas, not just rote memorization.