Topic 01
Basic Concepts & Definitions
System, properties, state, process, equilibrium, work and heat
~1 Mark
IMP
System Types
Closed / Open / Isolated
Closed: mass fixed, energy crosses boundary
Open: mass + energy both cross boundary
Isolated: neither mass nor energy crosses
Open: mass + energy both cross boundary
Isolated: neither mass nor energy crosses
Nozzle, turbine, compressor = open system. Piston-cylinder = closed system.
HOT
Work & Heat Sign Convention
Thermodynamic Sign Convention
Q positive: heat added TO the system
W positive: work done BY the system
W positive: work done BY the system
GATE always uses this convention unless stated otherwise. Many errors happen from sign confusion.
Q in = positive, W out = positive. Remember: "QIN WIN"
IMP
Boundary Work
Quasi-static (Reversible) Work
W_b = ∫P dV
For constant pressure: W = P(V₂ - V₁). For constant volume: W = 0. Area under P-V diagram = work done.
KNOW
Specific Heats
C_p, C_v and Their Relation
C_p - C_v = R
γ = C_p / C_v
C_v = R/(γ-1), C_p = γR/(γ-1)
γ = C_p / C_v
C_v = R/(γ-1), C_p = γR/(γ-1)
For air: C_p = 1.005 kJ/kg·K, C_v = 0.718 kJ/kg·K, γ = 1.4, R = 0.287 kJ/kg·K
Internal Energy
Enthalpy
H = U + PV
h = u + Pv (specific)
dh = C_p dT (ideal gas)
h = u + Pv (specific)
dh = C_p dT (ideal gas)
Enthalpy is useful for open systems (e.g. flow work included). du = C_v dT for ideal gas.
KNOW
Polytropic Process
General Process Law
PV^n = constant
n=0: Isobaric, n=1: Isothermal, n=γ: Isentropic, n=∞: Isochoric. Work: W = (P₁V₁ - P₂V₂)/(n-1) for n≠1
📊 Process Summary — P-V-T Relations (Ideal Gas)
| Process | Constant | n value | Work W | Heat Q |
|---|---|---|---|---|
| Isobaric | Pressure (P) | 0 | P(V₂-V₁) | mC_p(T₂-T₁) |
| Isochoric | Volume (V) | ∞ | 0 | mC_v(T₂-T₁) |
| Isothermal | Temperature (T) | 1 | mRT ln(V₂/V₁) | = W (ΔU=0) |
| Isentropic | Entropy (S) | γ | (P₁V₁-P₂V₂)/(γ-1) | 0 |
| Polytropic | PVⁿ | n | (P₁V₁-P₂V₂)/(n-1) | W × (γ-n)/(γ-1) |
Topic 02
First Law of Thermodynamics
Energy conservation, SFEE, NFEE, open system analysis
~2–3 Marks
HOT
Closed System (NFEE)
Non-Flow Energy Equation
Q - W = ΔU = U₂ - U₁ = mC_v(T₂ - T₁)
Q = heat added to system (kJ), W = work done by system (kJ), ΔU = change in internal energy. For a cycle: ∮dU = 0, so ∮Q = ∮W.
HOT
Open System (SFEE)
Steady Flow Energy Equation
Q - W_s = (h₂ - h₁) + (V₂² - V₁²)/2 + g(z₂ - z₁)
h = specific enthalpy (kJ/kg), V = velocity (m/s), z = elevation (m), W_s = shaft work per unit mass. All terms in kJ/kg.
Turbine: Q≈0, W_s positive (work out). Compressor: Q≈0, W_s negative (work in). Nozzle: W_s=0, ΔKE huge.
HOT
Nozzle
SFEE Applied to Nozzle
V₂ = √(2(h₁ - h₂) + V₁²)
If inlet velocity negligible: V₂ = √(2Δh). For isentropic nozzle, h drop converts entirely to KE gain.
HOT
Turbine / Compressor
Isentropic Work
W_turbine = h₁ - h₂ = C_p(T₁ - T₂)
W_comp = h₂ - h₁ = C_p(T₂ - T₁)
W_comp = h₂ - h₁ = C_p(T₂ - T₁)
For ideal gas. Actual work accounts for isentropic efficiency: η_t = W_actual/W_isentropic (turbine), η_c = W_isentropic/W_actual (compressor).
IMP
Throttling
Throttling Process (Isenthalpic)
h₁ = h₂ (enthalpy constant)
W=0, Q=0, ΔKE≈0
W=0, Q=0, ΔKE≈0
Pressure drops, temperature may drop or rise depending on fluid. For ideal gas: T constant (h=f(T) only). For real gas/vapour: Joule-Thomson effect applies.
KNOW
Heat Exchanger
Energy Balance
m_h(h_h1 - h_h2) = m_c(h_c2 - h_c1)
No work, adiabatic (Q to surroundings = 0). Hot fluid energy given = cold fluid energy received.
Memory Trick
SFEE device shortcuts for GATE:
Nozzle: KE up, h down, W=Q=0
Diffuser: KE down, h up, W=Q=0
Turbine: W_out, h drops, Q=0
Compressor: W_in, h rises, Q=0
Throttle: h same, P drops, T drops (vapour)
Boiler/Condenser: Q only, W=0
Nozzle: KE up, h down, W=Q=0
Diffuser: KE down, h up, W=Q=0
Turbine: W_out, h drops, Q=0
Compressor: W_in, h rises, Q=0
Throttle: h same, P drops, T drops (vapour)
Boiler/Condenser: Q only, W=0
Topic 03
Second Law, Entropy & Exergy
Kelvin-Planck, Clausius, entropy generation, availability
~2–3 Marks
HOT
Entropy
Clausius Inequality & Entropy
∮(δQ/T) ≤ 0
dS = δQ_rev / T
ΔS_univ = ΔS_sys + ΔS_surr ≥ 0
dS = δQ_rev / T
ΔS_univ = ΔS_sys + ΔS_surr ≥ 0
S = entropy (kJ/K). Reversible process: ΔS_univ = 0. Irreversible: ΔS_univ > 0. Universe entropy always increases or stays same.
HOT
Entropy Change
Ideal Gas Entropy Change
Δs = C_v ln(T₂/T₁) + R ln(v₂/v₁)
Δs = C_p ln(T₂/T₁) - R ln(P₂/P₁)
Δs = C_p ln(T₂/T₁) - R ln(P₂/P₁)
Both forms are equivalent. Use whichever two variables are given. Isentropic: Δs = 0.
HOT
Isentropic Relations
T-P-V Relations for Isentropic
T₂/T₁ = (P₂/P₁)^((γ-1)/γ)
T₂/T₁ = (V₁/V₂)^(γ-1)
P₁V₁^γ = P₂V₂^γ
T₂/T₁ = (V₁/V₂)^(γ-1)
P₁V₁^γ = P₂V₂^γ
These are the most-used relations for turbine and compressor analysis. Learn them by heart.
IMP
Exergy (Availability)
Closed System Exergy
Φ = (u - u₀) + P₀(v - v₀) - T₀(s - s₀)
Subscript 0 = dead state conditions (P₀, T₀ of surroundings). Exergy = max useful work extractable from a system.
Exergy is always destroyed (never created) in irreversible processes. Exergy destruction = T₀ × S_gen.
IMP
Second Law Efficiency
Isentropic & Second-Law Efficiency
η_II = W_actual / Φ_decrease
η_s,turbine = (h₁-h₂)/(h₁-h₂s)
η_s,compressor = (h₂s-h₁)/(h₂-h₁)
η_s,turbine = (h₁-h₂)/(h₁-h₂s)
η_s,compressor = (h₂s-h₁)/(h₂-h₁)
h₂s = isentropic exit enthalpy. Second law efficiency compares actual to maximum possible work.
KNOW
Helmholtz & Gibbs
Thermodynamic Potentials
A = U - TS (Helmholtz)
G = H - TS (Gibbs)
dG = V dP - S dT
G = H - TS (Gibbs)
dG = V dP - S dT
Gibbs function G is constant during phase change at constant T and P. Maxwell relations derived from these potentials.
Memory Trick — Second Law Statements
Kelvin-Planck: Cannot have a heat engine that converts ALL heat to work (operating in a cycle). Must reject some heat to a sink.
Clausius: Cannot transfer heat from cold to hot body WITHOUT external work. (Refrigerator needs power input.)
Both statements are equivalent. If you violate one, you violate both.
Clausius: Cannot transfer heat from cold to hot body WITHOUT external work. (Refrigerator needs power input.)
Both statements are equivalent. If you violate one, you violate both.
Topic 04
Ideal Gas & Properties of Steam
Equations of state, steam tables, dryness fraction, phase diagram
~1–2 Marks
HOT
Ideal Gas Law
Equation of State
PV = mRT = nR_uT
Pv = RT (specific)
Pv = RT (specific)
R = specific gas constant (kJ/kg·K), R_u = universal gas constant = 8.314 kJ/kmol·K, n = moles, v = specific volume (m³/kg)
HOT
Dryness Fraction
Quality of Wet Steam
x = m_vapour / (m_liquid + m_vapour)
h = h_f + x·h_fg
s = s_f + x·s_fg
h = h_f + x·h_fg
s = s_f + x·s_fg
x = dryness fraction (0=saturated liquid, 1=saturated vapour). h_fg = latent heat of vaporization. From steam tables directly.
Specific volume of wet steam: v = v_f + x·v_fg
IMP
Van der Waals
Real Gas Equation
(P + a/v²)(v - b) = RT
a = accounts for intermolecular attraction (Pa·m⁶/mol²), b = excluded volume (m³/mol). Compressibility factor: Z = Pv/RT (Z=1 for ideal gas).
KNOW
Gas Mixtures
Dalton's Law & Amagat's Law
P_total = ΣP_i (Dalton: const V, T)
V_total = ΣV_i (Amagat: const P, T)
R_mix = Σ(m_i/m_total)·R_i
V_total = ΣV_i (Amagat: const P, T)
R_mix = Σ(m_i/m_total)·R_i
Mole fraction: y_i = n_i/n_total. For ideal gas mixture: P_i = y_i · P_total (partial pressure).
📊 Steam Properties at Key Conditions (from Steam Tables — Memorize These)
| Condition | T (°C) | P (kPa) | h_f (kJ/kg) | h_fg (kJ/kg) | h_g (kJ/kg) |
|---|---|---|---|---|---|
| Triple point | 0.01 | 0.611 | 0.0 | 2501 | 2501 |
| Saturated @ 100°C | 100 | 101.3 | 419 | 2257 | 2676 |
| Saturated @ 200°C | 200 | 1554 | 852 | 1941 | 2793 |
| Critical point | 374.1 | 22,090 | 2099 | 0 | 2099 |
Memory Trick — Phase Diagram
T-s diagram regions: Left of the dome = Compressed liquid (subcooled). Inside the dome = Wet steam (two-phase). Right of the dome = Superheated steam. Top of dome = Critical point. The dome shape itself = saturation curve. GATE loves asking about which region a state point lies in given P and T.
Topic 05
Carnot Cycle & Reversed Carnot
Maximum efficiency, COP, heat engine, heat pump
~1–2 Marks
HOT
Heat Engine
Thermal Efficiency
η_th = W_net / Q_in = 1 - Q_L/Q_H
η_Carnot = 1 - T_L/T_H
η_Carnot = 1 - T_L/T_H
T_H, T_L = absolute temperatures of source and sink (Kelvin). Carnot gives MAXIMUM possible efficiency between two temperature reservoirs.
Always convert to Kelvin: T(K) = T(°C) + 273.15
HOT
Refrigerator COP
Coefficient of Performance
COP_R = Q_L / W_net = T_L / (T_H - T_L)
Refrigerator goal: remove heat from cold space. COP_R can be greater than 1 (no efficiency limit here). Maximum COP = Carnot COP.
HOT
Heat Pump COP
Heat Pump COP
COP_HP = Q_H / W_net = T_H / (T_H - T_L)
COP_HP = COP_R + 1
COP_HP = COP_R + 1
Heat pump goal: deliver heat to hot space. COP_HP is always > 1. The +1 relationship with COP_R is a very common GATE question.
IMP
Carnot Cycle Steps
Four Processes
1→2: Isothermal expansion (Q_H added)
2→3: Isentropic expansion (W out)
3→4: Isothermal compression (Q_L rejected)
4→1: Isentropic compression (W in)
2→3: Isentropic expansion (W out)
3→4: Isothermal compression (Q_L rejected)
4→1: Isentropic compression (W in)
All processes are reversible. Carnot = most efficient possible cycle between T_H and T_L.
Golden GATE Fact
COP_HP = COP_R + 1 — This single identity appears in GATE almost every year. If COP_R = 4, then COP_HP = 5. If η_Carnot = 40%, then T_L/T_H = 0.6. Always check whether the question is asking for refrigerator, heat pump, or heat engine — they use the same W_net but different goals.
Topic 06
Rankine Cycle (Steam Power Plant)
Ideal and actual Rankine, reheat, regeneration, efficiency
~2–3 Marks
🔄 Ideal Rankine Cycle — Four Processes
Process 1→2
Pump
Isentropic Compression
→
Process 2→3
Boiler
Constant Pressure Heat Add
→
Process 3→4
Turbine
Isentropic Expansion
→
Process 4→1
Condenser
Constant Pressure Heat Reject
HOT
Rankine Efficiency
Thermal Efficiency
η = (W_T - W_P) / Q_boiler
= (h₃ - h₄) - (h₂ - h₁) / (h₃ - h₂)
= (h₃ - h₄) - (h₂ - h₁) / (h₃ - h₂)
Pump work W_P = v₁(P₂ - P₁) ≈ small. Turbine work W_T = h₃ - h₄ dominates. States 1,2,3,4 correspond to cycle diagram above.
HOT
Pump Work
Isentropic Pump Work
W_P = v₁(P₂ - P₁) = h₂ - h₁
h₂ = h₁ + v₁(P₂ - P₁)
h₂ = h₁ + v₁(P₂ - P₁)
v₁ = specific volume at condenser exit (saturated liquid). Pump work is usually 1–2% of turbine work for steam cycles.
IMP
Reheat Rankine
With Reheat (Improved Cycle)
η_reheat = (W_T1 + W_T2 - W_P) / (Q_boiler + Q_reheat)
Reheat increases work output and improves dryness fraction at LP turbine exit. Reduces blade erosion.
Reheat raises average heat addition temperature → increases efficiency.
IMP
Regenerative Rankine
Open Feed Water Heater
m·h₆ + (1-m)·h₂ = 1·h₃
→ m = (h₃ - h₂)/(h₆ - h₂)
→ m = (h₃ - h₂)/(h₆ - h₂)
m = mass fraction extracted from turbine. Regeneration raises average heat addition temperature, improving efficiency.
KNOW
Back Work Ratio
BWR Definition
BWR = W_pump / W_turbine
For Rankine cycle: BWR ≈ 0.5–2% (very low — pump work is tiny). For Brayton: BWR ≈ 40–80% (compressor takes large fraction of turbine output).
IMP
Rankine vs Carnot
Why Rankine Efficiency is Less
η_Rankine < η_Carnot always
Heat addition in boiler is not isothermal (except for phase change part). Irreversibilities in pump/turbine further reduce efficiency. Methods to improve: superheat, reheat, regeneration, higher boiler pressure.
Topic 07
Brayton Cycle (Gas Turbine)
Ideal Brayton, pressure ratio, intercooling, reheating, regeneration
~2–3 Marks
🔄 Ideal Brayton Cycle — Four Processes
1 → 2
Compressor
Isentropic Compression
→
2 → 3
Combustor
Constant Pressure Heat Add
→
3 → 4
Turbine
Isentropic Expansion
→
4 → 1
Exhaust/Cooler
Constant Pressure Heat Reject
HOT
Brayton Efficiency
Thermal Efficiency (Ideal)
η_Brayton = 1 - T₁/T₂ = 1 - (1/r_p)^((γ-1)/γ)
r_p = P₂/P₁ = pressure ratio. Efficiency increases with pressure ratio. T₂/T₁ = (r_p)^((γ-1)/γ) from isentropic relation.
HOT
Temperature Ratios
Isentropic Temperature Relations
T₂/T₁ = (P₂/P₁)^((γ-1)/γ) = r_p^((γ-1)/γ)
T₃/T₄ = r_p^((γ-1)/γ) [same ratio]
T₃/T₄ = r_p^((γ-1)/γ) [same ratio]
For air: (γ-1)/γ = 0.4/1.4 = 2/7 ≈ 0.286. Compressor and turbine share the same pressure ratio (isentropic).
HOT
Optimal Pressure Ratio
For Maximum Net Work
r_p,opt = (T₃/T₁)^(γ/(2(γ-1)))
Maximum net work output occurs at this pressure ratio. For maximum efficiency, higher r_p is always better. These are two different optima.
GATE often tests: "efficiency increases with r_p, but net work is maximum at r_p,opt".
IMP
Regeneration
Effectiveness of Regenerator
ε = (T₅ - T₂)/(T₄ - T₂)
T₅ = air temperature after regenerator, T₄ = turbine exhaust temp, T₂ = compressor exit temp. Regeneration is effective only when T₄ > T₂ (i.e., low pressure ratio).
IMP
Compressor Efficiency
Isentropic Efficiency
η_c = W_c,ideal / W_c,actual = (h₂s-h₁)/(h₂-h₁)
η_T = W_T,actual / W_T,ideal = (h₃-h₄)/(h₃-h₄s)
η_T = W_T,actual / W_T,ideal = (h₃-h₄)/(h₃-h₄s)
η_c typically 80–90%. η_T typically 85–92%. Both reduce cycle efficiency and net work.
KNOW
Back Work Ratio
Brayton BWR
BWR = W_compressor / W_turbine ≈ 40–80%
Much higher than Rankine (0.5–2%). This is why isentropic efficiency of compressor and turbine critically affects gas turbine performance.
Brayton vs Rankine — Key Differences for GATE
Brayton (gas turbine): Working fluid = gas (air), remains gaseous throughout, high BWR (40–80%), used in aircraft and power plants, regeneration is beneficial at low pressure ratios only.
Rankine (steam turbine): Working fluid = steam (changes phase), very low BWR (0.5–2%), pump work negligible, regeneration always beneficial, used in conventional power plants.
Rankine (steam turbine): Working fluid = steam (changes phase), very low BWR (0.5–2%), pump work negligible, regeneration always beneficial, used in conventional power plants.
Topic 08
IC Engine Cycles
Otto, Diesel, Dual (mixed) cycle, engine performance parameters
~2–3 Marks
HOT
Otto Cycle
Petrol Engine — Efficiency
η_Otto = 1 - 1/r^(γ-1)
r = compression ratio = V₁/V₂ = V_BDC/V_TDC. Heat addition at constant volume (isochoric). Typical r = 8–12 for petrol engines.
Higher compression ratio = higher efficiency. Knocking limits r in petrol engines.
HOT
Diesel Cycle
Diesel Engine — Efficiency
η_Diesel = 1 - (1/r^(γ-1)) · (r_c^γ - 1)/(γ(r_c - 1))
r_c = cut-off ratio = V₃/V₂ (volume at end of combustion / clearance volume). Heat addition at constant pressure (isobaric).
At same r: η_Otto > η_Diesel. At same peak pressure: η_Diesel > η_Otto. GATE loves this comparison!
HOT
Dual Cycle
Mixed Cycle — Efficiency
η_Dual = 1 - (1/r^(γ-1)) · (α·r_p^γ - 1) / ((α-1) + α·γ(r_c-1))
α = pressure ratio during constant volume part, r_p = pressure ratio P₃/P₂. Dual cycle lies between Otto and Diesel.
IMP
Engine Performance
Mean Effective Pressure (MEP)
MEP = W_net / (V_max - V_min)
BP = MEP × L × A × N/k
BP = MEP × L × A × N/k
MEP in kPa, L = stroke, A = bore area, N = rpm, k = 2 for 4-stroke, 1 for 2-stroke. Higher MEP = more powerful engine for same size.
IMP
Efficiencies
Indicated, Brake, Mechanical
η_mech = BP / IP
η_bth = BP / (m_f · CV)
η_ith = IP / (m_f · CV)
η_bth = BP / (m_f · CV)
η_ith = IP / (m_f · CV)
BP = brake power, IP = indicated power, CV = calorific value of fuel, m_f = fuel mass flow rate. η_bth = η_ith × η_mech.
KNOW
Volumetric Efficiency
Charge Breathing
η_v = V_actual / V_displacement
Ratio of actual air inducted to swept volume. Supercharging and turbocharging increase η_v above 100% (by compressing intake air). Higher η_v = more fuel can burn = more power.
📊 Otto vs Diesel vs Dual — Quick Comparison
| Parameter | Otto Cycle | Diesel Cycle | Dual Cycle |
|---|---|---|---|
| Heat addition | Isochoric (const V) | Isobaric (const P) | Isochoric + Isobaric |
| Engine type | Petrol (SI) | Diesel (CI) | Modern high-speed diesel |
| Compression ratio | 8–12 | 16–22 | 14–20 |
| Efficiency @ same r | Highest | Lowest | Between |
| Efficiency @ same peak P & T | Lowest | Highest | Between |
Topic 09
Refrigeration & Heat Pump Cycles
VCR cycle, air refrigeration, refrigerants, COP calculations
~2 Marks
🔄 Vapour Compression Refrigeration (VCR) Cycle
1 → 2
Compressor
Isentropic Compression (dry vapour)
→
2 → 3
Condenser
Const. Pressure Condensation (heat rejected)
→
3 → 4
Expansion Valve
Throttling (isenthalpic, h₃=h₄)
→
4 → 1
Evaporator
Const. Pressure Evaporation (heat absorbed)
HOT
VCR COP
COP of VCR Cycle
COP = Q_L / W_comp = (h₁ - h₄) / (h₂ - h₁)
h₁ = evaporator exit (dry sat vapour), h₂ = compressor exit, h₃ = condenser exit (sat liquid), h₄ = h₃ (throttling). Values from refrigerant tables.
HOT
Refrigerating Effect
Ton of Refrigeration
1 TR = 3.517 kW = 3517 W
= 210 kJ/min = 12000 BTU/hr
= 210 kJ/min = 12000 BTU/hr
Historically: heat to melt 1 ton (2000 lb) of ice in 24 hours. Q_L(kW) / 3.517 = capacity in TR.
Remember 3.517 kW = 1 TR. Very common GATE numerical.
IMP
Air Refrigeration
Bell-Coleman Cycle COP
COP = T_L / (T_H - T_L) × 1/(r_p^((γ-1)/γ) - 1) × T_L
Simpler form: COP = T₁ / (T₂ - T₁) where T₁ = refrigerator temp, T₂ = after compression. Used in aircraft and cold storage.
KNOW
Refrigerants
Properties of Ideal Refrigerant
High latent heat, low boiling point,
high COP, non-toxic, non-flammable,
low GWP and ODP
high COP, non-toxic, non-flammable,
low GWP and ODP
R-134a (HFC) replaced R-12 (CFC — banned). R-410A used in AC. R-717 (Ammonia) = highest COP but toxic. R-744 (CO₂) = natural refrigerant, rising use.
Topic 10
Psychrometrics
Moist air properties, humidity, wet bulb, dew point, psychrometric chart
~1 Mark
HOT
Humidity
Specific Humidity (Humidity Ratio)
ω = m_v / m_a = 0.622 · P_v / (P - P_v)
ω = kg water vapour per kg dry air, P_v = partial pressure of water vapour, P = total pressure. Range: 0 (dry air) to ~0.030 (very humid).
HOT
Relative Humidity
RH Definition
φ = P_v / P_sat(T) = m_v / m_v,sat
φ = 0 for completely dry air, φ = 1 (100%) for saturated air. At saturation: dew point = dry bulb temperature.
IMP
Enthalpy of Moist Air
Specific Enthalpy
h = C_pa·T + ω(h_g)
h_g ≈ 2501 + 1.86T (kJ/kg vapour)
h_g ≈ 2501 + 1.86T (kJ/kg vapour)
C_pa = 1.005 kJ/kg·K for dry air, T in °C. The 2501 kJ/kg is latent heat at 0°C. h_g = enthalpy of saturated vapour.
KNOW
Wet Bulb Temp
Psychrometric Relation
ω = ω_wb - [C_pa(T_db - T_wb)] / h_fg,wb
T_db = dry bulb temp, T_wb = wet bulb temp. Wet bulb = lowest temp achievable by evaporative cooling at that RH. T_dp ≤ T_wb ≤ T_db always.
All three temperatures equal at 100% RH. T_db - T_wb = wet bulb depression.
📊 Psychrometric Processes on h-ω Chart
| Process | Change in T_db | Change in ω | Example Device |
|---|---|---|---|
| Sensible Heating | Increases | No change | Electric heater |
| Sensible Cooling | Decreases | No change | Cooling coil (above dew point) |
| Cooling + Dehumidification | Decreases | Decreases | AC system coil |
| Humidification | No change or slight drop | Increases | Humidifier, steam injection |
| Evaporative Cooling | Decreases | Increases | Desert cooler (constant h) |
| Mixing of air streams | Between T₁ & T₂ | Between ω₁ & ω₂ | AHU mixing box |
Topic 11
GATE Previous Year Questions
Solved PYQs with step-by-step solutions — tap an option to attempt, then reveal solution
Practice!
GATE 2023 ME
Carnot Efficiency
1 Mark
A heat engine receives heat at 1000 K and rejects heat at 400 K. If the actual thermal efficiency of the engine is 50%, the ratio of actual efficiency to Carnot efficiency is:
Answer: (C) 0.833
Step 1: Carnot efficiency = 1 - T_L/T_H = 1 - 400/1000 = 1 - 0.4 =
Step 2: Actual efficiency = 50% = 0.50
Step 3: Ratio = η_actual / η_Carnot = 0.50 / 0.60 = 0.833
Key concept: Actual efficiency is always less than or equal to Carnot efficiency. The ratio (called second law efficiency) tells how close the engine is to ideal performance.
Step 1: Carnot efficiency = 1 - T_L/T_H = 1 - 400/1000 = 1 - 0.4 =
0.60 (60%)Step 2: Actual efficiency = 50% = 0.50
Step 3: Ratio = η_actual / η_Carnot = 0.50 / 0.60 = 0.833
Key concept: Actual efficiency is always less than or equal to Carnot efficiency. The ratio (called second law efficiency) tells how close the engine is to ideal performance.
GATE 2022 ME
Otto Cycle Efficiency
1 Mark
An air-standard Otto cycle operates with a compression ratio of 8. Assuming γ = 1.4, the thermal efficiency of the cycle is approximately:
Answer: (B) 56.5%
Formula: η_Otto = 1 - 1/r^(γ-1)
= 1 - 1/(8)^(1.4-1)
= 1 - 1/(8)^0.4
Step: 8^0.4 = e^(0.4 × ln8) = e^(0.4 × 2.0794) = e^0.8318 =
η = 1 - 1/2.2974 = 1 - 0.4353 = 0.5647 ≈ 56.5%
Quick check: 8^0.4 ≈ 2.3 is a good approximation to remember for r=8, γ=1.4.
Formula: η_Otto = 1 - 1/r^(γ-1)
= 1 - 1/(8)^(1.4-1)
= 1 - 1/(8)^0.4
Step: 8^0.4 = e^(0.4 × ln8) = e^(0.4 × 2.0794) = e^0.8318 =
2.2974η = 1 - 1/2.2974 = 1 - 0.4353 = 0.5647 ≈ 56.5%
Quick check: 8^0.4 ≈ 2.3 is a good approximation to remember for r=8, γ=1.4.
GATE 2021 ME
Rankine Cycle — Pump Work
2 Marks
In an ideal Rankine cycle, steam enters the turbine at 3 MPa, 350°C (h₁ = 3115 kJ/kg, s₁ = 6.74 kJ/kg·K). It exits the condenser as saturated liquid at 10 kPa (h_f = 192 kJ/kg, v_f = 0.00101 m³/kg). The pump work input (in kJ/kg) is most nearly:
Answer: (C) ≈ 3.0 kJ/kg
Isentropic pump work: W_P = v_f × (P_high - P_low)
= 0.00101 m³/kg × (3000 - 10) kPa
= 0.00101 × 2990
= 3.02 kJ/kg
Note: (1 kPa × 1 m³/kg = 1 kJ/kg — unit conversion is direct). This confirms h₂ = h_f + W_P = 192 + 3.02 = 195.02 kJ/kg.
Pump work is very small compared to turbine work (~2900 kJ/kg at these conditions) — this is why Rankine cycle back work ratio is so low (<0.1%).
Isentropic pump work: W_P = v_f × (P_high - P_low)
= 0.00101 m³/kg × (3000 - 10) kPa
= 0.00101 × 2990
= 3.02 kJ/kg
Note: (1 kPa × 1 m³/kg = 1 kJ/kg — unit conversion is direct). This confirms h₂ = h_f + W_P = 192 + 3.02 = 195.02 kJ/kg.
Pump work is very small compared to turbine work (~2900 kJ/kg at these conditions) — this is why Rankine cycle back work ratio is so low (<0.1%).
GATE 2020 ME
Entropy Generation
2 Marks
100 kJ of heat is transferred from a reservoir at 500 K to another reservoir at 300 K. The entropy generation (in kJ/K) for this irreversible process is:
Answer: S_gen = 0.133 kJ/K
Entropy change of hot reservoir (loses heat):
ΔS_hot = -Q/T_H = -100/500 =
Entropy change of cold reservoir (gains heat):
ΔS_cold = +Q/T_L = +100/300 =
Total entropy generation = ΔS_cold + ΔS_hot
= 0.333 - 0.200 = +0.133 kJ/K
This is positive (irreversible process). If heat transfer were reversible (T_H ≈ T_L), S_gen → 0. The larger the temperature difference, the greater the irreversibility.
Entropy change of hot reservoir (loses heat):
ΔS_hot = -Q/T_H = -100/500 =
-0.20 kJ/KEntropy change of cold reservoir (gains heat):
ΔS_cold = +Q/T_L = +100/300 =
+0.333 kJ/KTotal entropy generation = ΔS_cold + ΔS_hot
= 0.333 - 0.200 = +0.133 kJ/K
This is positive (irreversible process). If heat transfer were reversible (T_H ≈ T_L), S_gen → 0. The larger the temperature difference, the greater the irreversibility.
GATE 2019 ME
Brayton Cycle — Pressure Ratio
2 Marks
In an ideal Brayton cycle, the compressor inlet is at 300 K. The turbine inlet is at 1200 K. For maximum net work output, the optimal compressor pressure ratio (γ = 1.4) is:
Answer: (C) ≈ 11.3
For maximum net work: r_p,opt = (T₃/T₁)^(γ/2(γ-1))
= (1200/300)^(1.4 / (2 × 0.4))
= (4)^(1.4/0.8)
= (4)^1.75
= 4^1 × 4^0.75
4^0.75 = (4^3)^(1/4) = 64^0.25 = (64)^0.25 ≈ 2.828
r_p,opt = 4 × 2.828 = 11.31
Important: This pressure ratio gives maximum NET WORK. But the efficiency at this point is NOT maximum — higher r_p gives better efficiency but less net work per unit mass.
For maximum net work: r_p,opt = (T₃/T₁)^(γ/2(γ-1))
= (1200/300)^(1.4 / (2 × 0.4))
= (4)^(1.4/0.8)
= (4)^1.75
= 4^1 × 4^0.75
4^0.75 = (4^3)^(1/4) = 64^0.25 = (64)^0.25 ≈ 2.828
r_p,opt = 4 × 2.828 = 11.31
Important: This pressure ratio gives maximum NET WORK. But the efficiency at this point is NOT maximum — higher r_p gives better efficiency but less net work per unit mass.
Bonus
GATE Rank vs PSU Cutoff Table
Approximate GATE ME scores and ranks needed for top PSU recruitment — plan your target score
Plan Smart
📊 Top PSU Cutoffs — Mechanical Engineering (Historical Indicative Data)
| PSU / Organisation | Sector | Approx. GATE Score | Approx. Rank (ME) | Approx. CTC |
|---|---|---|---|---|
| IOCL (Indian Oil) | Oil & Gas | 750+ | Top 300 | 18–22 LPA |
| HPCL | Oil & Gas | 740+ | Top 350 | 16–20 LPA |
| BPCL | Oil & Gas | 730+ | Top 400 | 15–18 LPA |
| GAIL | Gas Transmission | 720+ | Top 500 | 14–18 LPA |
| NTPC | Power Generation | 720+ | Top 500 | 13–16 LPA |
| BHEL | Heavy Engineering | 700+ | Top 700 | 10–14 LPA |
| ONGC | Oil & Gas | 700+ | Top 600 | 12–16 LPA |
| PGCIL (Power Grid) | Power Transmission | 690+ | Top 800 | 11–14 LPA |
| SAIL | Steel | 660+ | Top 1200 | 9–12 LPA |
| Coal India (CIL) | Mining | 620+ | Top 2000 | 9–12 LPA |
| AAI | Aviation | 650+ | Top 1500 | 8–12 LPA |
| DRDO / HAL | Defence | 700+ | Top 700 | 8–14 LPA + benefits |
Disclaimer: Cutoffs vary every year with paper difficulty, vacancies, and category. Always check official PSU notifications and GATE score cards. Target 700+ GATE score for most options; 750+ opens almost all doors.
Strategy: GATE score is valid for 3 years. You can apply to multiple PSUs simultaneously in the same year. Some PSUs conduct additional GD/interview rounds; shortlisting is based on GATE score. M.Tech admissions at IITs/NITs also use GATE score with stipend of ~14,000/month.
Why I Built This
For Every Engineer Who Deserves a Fair Shot at GATE
I am Vaibhav, a mechanical engineer from Maharashtra. Every year, lakhs of students prepare for GATE without access to good, free, organized study material. Coaching institutes charge thousands for printed notes that are often outdated or badly formatted on a phone screen.
This Thermodynamics formula sheet took many hours to write and organize — every formula verified, every memory trick tested. It is completely free, works on your phone, and loads fast even on slow data. No login. No popup. No paywall.
If this helps you crack GATE or land a PSU job, share it with one friend who needs it too. That is all I ask.
This Thermodynamics formula sheet took many hours to write and organize — every formula verified, every memory trick tested. It is completely free, works on your phone, and loads fast even on slow data. No login. No popup. No paywall.
If this helps you crack GATE or land a PSU job, share it with one friend who needs it too. That is all I ask.