GATE Mechanical: Thermodynamics, previous-year questions solved

An intensive property does not depend on how much of the system you take. Cut a block of steel in half and each half has the same density, temperature and pressure, but half the mass, volume and energy.

Mass, energy and volume are extensive. Density is mass divided by volume, the ratio of two extensive properties, and so it is intensive. Specific volume and specific enthalpy are intensive for the same reason.

For an ideal gas, h = u + Pv = u(T) + RT, so enthalpy depends on temperature alone. States 2 and 3 lie on the same isotherm, so T₂ = T₃ and h₂ = h₃, whatever path took the gas there and whatever the pressures are.

(C) and (D) would only hold if P₂ = P₃, which the question does not say.

1. Insulated, so Q = 0. Rigid, so no boundary work; the only work is the 225 kJ done on the gas by the stirrer.
2. First law: ΔU = 225 kJ = m c_v ΔT = 1 × 0.750 × ΔT, so ΔT = 300 K and T₂ = 600 K.
3. Volume and mass are fixed, so P ∝ T: P₂ = 1 bar × 600/300 = 2 bar.

1. PVⁿ = constant: P₂ = 110 × (5/2.5)^1.2 = 110 × 2^1.2 = 252.71 kPa.
2. Boundary work for a polytropic process: W = (P₁V₁ − P₂V₂)/(n − 1) = (110 × 5 − 252.71 × 2.5)/0.2 = (550 − 631.78)/0.2 = −408.92 kJ.
3. The minus sign means work is done on the gas (compression). |W| = 408.92 kJ.

1. R = c_p − c_v = 14.56 − 10.4 = 4.16 kJ/kg·K.
2. Polytropic work: W = mR(T₁ − T₂)/(n − 1), so T₁ − T₂ = 400 × 0.3/4.16 = 28.85 K. The gas cools as it expands.
3. ΔU = m c_v(T₂ − T₁) = 10.4 × (−28.85) = -300.0 kJ.
4. First law: Q = ΔU + W = -300.0 + 400 = 100.0 kJ (heat added).

Shortcut: Q = W(γ − n)/(γ − 1) with γ = 14.56/10.4 = 1.4, giving 400 × 0.1/0.4 = 100 kJ.

1. While the piston floats, pressure stays at 140 kPa and the volume shrinks as the air cools. Height goes from 3 m to 2 m, so V falls to 2/3.
2. Temperature when the piston touches the stops: T = 623.15 × 2/3 = 415.43 K (142.3 °C). That is still above 25 °C, so the piston does reach the stops.
3. After that, volume is fixed and the rest of the cooling to 298.15 K does no work.
4. Work = P(v₁ − v₂) = R(T₁ − T) = 0.287 × (623.15 − 415.43) = 59.6 kJ/kg (done on the air).

1. Unsteady-flow energy balance for a tank that starts empty: m₂u₂ = m₂h_line − Q_loss, i.e. m₂c_vT₂ = m₂c_pT_line − 1000.
2. Ideal gas at the end: m₂T₂ = PV/R = 600 × 8/0.287 = 16724.7 kg·K.
3. Substitute: 0.718 × 16724.7 = m₂ × 1.005 × 306 − 1000, so m₂ = 42.30 kg.
4. T₂ = 16724.7/42.30 = 395 K.

Without the heat loss the answer would be γT_line = 1.4 × 306 ≈ 428 K. The incoming air's flow work becomes internal energy, so the tank ends hotter than the line.

1. Rigid and closed: specific volume stays constant. v = 0.001029 + 0.1 × (3.4053 − 0.001029) = 0.341456 m³/kg.
2. At 130 °C: x₂ = (0.341456 − 0.001070)/(0.66808 − 0.001070) = 0.5103. Half the water has now boiled.
3. Liquid volume before: 5 × 0.9 × 0.001029 = 0.004631 m³. After: 5 × (1 − 0.5103) × 0.001070 = 0.002620 m³.
4. Cross-section A = π(0.15)²/4 = 0.01767 m². Dip = (0.004631 − 0.002620)/0.01767 = 11.38 cm.

The u values are not needed; they would give the heat input.

For an ideal gas, u depends only on T, so in an isothermal process ΔU = 0 and the first law gives Q = W. Nothing is violated.

The second law (Kelvin-Planck) forbids a cycle that turns all the heat from one reservoir into work. A single expansion is not a cycle: the gas ends at a larger volume and lower pressure, and getting it back costs work. So the process is possible. (D) contradicts itself: a possible process can't violate the second law.

The Clausius inequality is ∮ δQ/T ≤ 0, an integral around a closed cycle. It holds for any cycle, reversible or not: equal to zero for a reversible cycle and less than zero for an irreversible one.

For a single process the matching statement is ΔS ≥ ∫δQ/T, which is the entropy form of the second law derived from it, not the inequality itself.

1. First law: Q = 0, and the stirrer does work on the air, so U rises and the temperature rises. For an ideal gas h = c_pT, so enthalpy increases.
2. Second law: no heat crosses the boundary, so ΔS = S_gen. Stirring is irreversible (viscous dissipation), so S_gen > 0 and entropy increases.

Both rise: (C). A rise in T at constant volume also gives Δs = c_v ln(T₂/T₁) > 0, which is the same answer another way.

1. Heat rate: 24000 kJ/min ÷ 60 = 400 kW.
2. The best possible engine between 1200 K and 300 K is a Carnot engine: η = 1 − 300/1200 = 0.75.
3. Maximum power = 0.75 × 400 = 300 kW.

(C) forgets to convert minutes to seconds.

1. Mass flow: 100 L/min × 1 kg/L ÷ 60 = 1.6667 kg/s.
2. Flow exergy of an incompressible liquid (pressure term ignored): ψ = c[(T − T₀) − T₀ ln(T/T₀)], in kelvin.
3. T = 523.15 K, T₀ = 298.15 K: (T − T₀) = 225 K and T₀ ln(T/T₀) = 167.64 K, so ψ = 4.18 × 57.36 = 239.76 kJ/kg.
4. Exergy rate = 1.6667 × 239.76 = 399.60 kW.

The stream carries 1568 kW of heat relative to ambient, but only about a quarter of it could ever become work.

1. Entropy balance, steady flow, reversible (no generation): s₂ = s₁ − q_out/T, so q_out = 500 × (6.5 − 6.3) = 100 kJ/kg.
2. Energy balance: w = h₁ − h₂ − q_out = 3500 − 2500 − 100 = 900 kJ/kg.
3. Power = 500 kg/s × 900 kJ/kg = 450 MW.

• (A) True: superheating raises the mean temperature of heat addition.
• (B) True: the turbine exhausts into the condenser, where the pressure is the saturation pressure at the condensing temperature.
• (C) True: lower condenser pressure means a lower temperature of heat rejection.
• (D) False: raising boiler pressure raises the mean heat-addition temperature and the efficiency. Lowering it does the opposite.

1. Net work = 400 + 850 − 15 = 1235 kJ/kg.
2. Heat supplied (boiler + reheater) = 1235/0.32 = 3859.375 kJ/kg.
3. Heat rejected = 3859.375 − 1235 = 2624.38 kJ/kg.

About two-thirds of the fuel heat leaves through the condenser. That's normal for a steam plant.

1. Isentropic exit: s = 6.5, so x = (6.5 − 1.1)/(7.6 − 1.1) = 0.8308.
2. h_2s = 341 + 0.8308 × (2645 − 341) = 2255.09 kJ/kg.
3. Actual work = 0.9 × (3344 − 2255.09) = 980.02 kJ/kg.
4. Power = 102 × 980.02 = 99962 kW = 100.0 MW.

In the ideal regenerative cycle the feedwater (2 → 3) flows through coils around the turbine, counter to the steam. At every temperature, the heat the expanding steam gives up equals the heat the water picks up, reversibly. So the expansion line from 4 runs parallel to the heating line 2 → 3 on the T-s diagram, and the boiler only supplies 3 → 4, all at the top temperature.

• (B) is the simple Rankine cycle: vertical, isentropic expansion with no regeneration.
• (C) has the pump raising the water to boiler temperature, which a pump can't do.
• (D) expands along the saturated-vapour line, which is a different (and also impractical) idea.

So (A). Heat is added only at the top temperature and rejected only at the bottom, so the ideal regenerative cycle reaches Carnot efficiency.

1. Regenerator balance (same c_p, same mass flow): T₃ − T₂ = T₅ − T₆ = 800 − 600 = 200 K, so T₃ = 750 K.
2. Turbine work = c_p(1250 − 800) = 450c_p. Compressor work = c_p(550 − 300) = 250c_p. Net = 200c_p.
3. Heat added in the combustion chamber = c_p(1250 − 750) = 500c_p.
4. η = 200/500 = 40%. c_p cancels, so the γ given is not needed.

Without the regenerator the heat added would be c_p(1250 − 550) = 700c_p and η = 200/700 ≈ 29%.

1. η_Otto = 1 − 1/r^γ−1, so 0.5 = 1 − 1/r^0.4 and r^0.4 = 2.
2. r = 2^1/0.4 = 2^2.5 = 5.657 ≈ 5.7.

1. Compression ratio r = (250 + 35.7)/35.7 = 8.0028.
2. Isentropic compression: T₂ = 300 × r^0.4 = 689.3 K; P₂ = 100 × r^1.4 = 1838.8 kPa.
3. Constant-volume heat addition: T₃ = T₂ + 800/0.718 = 689.3 + 1114.2 = 1803.5 K.
4. P₃ = P₂ × T₃/T₂ = 4811 kPa (about 48 bar).

1. Compressor work = 300 − 250 = 50 kJ/kg. COP = refrigerating effect/work, so refrigerating effect = 3 × 50 = 150 kJ/kg.
2. h₄ (evaporator inlet) = 250 − 150 = 100 kJ/kg.
3. At evaporator pressure h_f = 50 and h_g = 250 (the compressor inlet is saturated vapour), so x = (100 − 50)/(250 − 50) = 0.25.

1. Steady state: the heat pump must deliver the 15 kW the auditorium loses.
2. COP_{HP, Carnot} = T_H/(T_H − T_L) = 300/20 = 15.
3. Power = 15/15 = 1.00 kW.

1. Ideal (Carnot) engine between 1000 K and the 300 K room: W = 150 × (1 − 300/1000) = 105 kJ, and it rejects 45 kJ to the room.
2. Ideal heat pump from 280 K to 300 K: COP = 300/20 = 15, so it delivers 15 × 105 = 1575 kJ.
3. Total to the room = 1575 + 45 = 1620 kJ.

Burning the fuel directly in the room would give just 150 kJ. The ideal combination gives more than ten times that.

The standard psychrometric chart plots dry-bulb temperature on the horizontal axis and specific humidity (humidity ratio, kg of water vapour per kg of dry air) on the vertical axis.

Relative humidity appears as curves, wet-bulb temperature and enthalpy as sloping lines, and specific volume as steeper sloping lines. None of them is an axis.