Thermodynamics 1: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

Ideal gas law

Ideal gas law: pV = nRT
Gas constant: R = 8.314 J/(mol·K)
Combined gas law: p₁V₁/T₁ = p₂V₂/T₂
Moles: n = m/M
Temperature: T[K] = T[°C] + 273.15

Energy balance

Sensible heat: Q = ṁ·c_p·ΔT
Latent heat: Q = ṁ·h_fg
Mixing: Σṁc_pT_in = Σṁc_pT_out
Water: c_p ≈ 4.18 kJ/(kg·K)

Heat engines & refrigerators

Efficiency: η = W/Q_H = 1 − Q_C/Q_H
Carnot efficiency: η_C = 1 − T_C/T_H (kelvin)
Refrigerator COP: COP_R = Q_C/W
Heat pump COP: COP_HP = Q_H/W = COP_R + 1
Energy balance: Q_H = W + Q_C

Gas power cycles

Isentropic, volume ratio: T₂/T₁ = r^k−1
Isentropic, pressure ratio: T₂/T₁ = r_p^(k−1)/k
Otto efficiency: η = 1 − 1/r^k−1
Diesel efficiency: η = 1 − (1/r^k−1)·(r_c^k − 1)/(k(r_c − 1))
Brayton efficiency: η = 1 − 1/r_p^(k−1)/k
Heat added: Otto: c_v(T₃ − T₂) Diesel, Brayton: c_p(T₃ − T₂)
Brayton net work: w = c_p[(T₃ − T₄) − (T₂ − T₁)]
Isentropic efficiencies: w_c = w_c,s/η_c, w_t = η_t·w_t,s
Cold air: k = 1.4, c_p = 1.005, c_v = 0.718 kJ/(kg·K)

2Practice problems

1.Ideal gas lawMedium

What volume V does 30 g of O₂ occupy at 300 kPa and 100 °C? Treat it as an ideal gas.

m = 30 gM = 32 g/molP = 300 kPaT = 100 °CR = 8.314 J/(mol·K)

V = L

2.Ideal gas lawHard

Find the density ρ of CO₂ at 150 kPa and 30 °C, treating it as an ideal gas.

M = 44.01 g/molP = 150 kPaT = 30 °CR = 8.314 J/(mol·K)

ρ = kg/m³

3.Energy balanceMedium

Hot light oil cools water in a counter-flow heat exchanger with no heat loss. Find the water outlet temperature t_out.

ṁ_h = 0.5 kg/sc_p,h = 2.1 kJ/(kg·K)T_h: 150 → 90 °Cṁ_c = 2 kg/sc_p,c = 4.18 kJ/(kg·K)t_in = 30 °C

t_out = °C

4.Energy balanceHard

Hot glycol mix heats water in an exchanger that loses 5% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁ_c needed.

ṁ_h = 2.5 kg/sc_p,h = 3.6 kJ/(kg·K)T_h: 150 → 100 °Ct: 25 → 70 °Cc_p,c = 4.18 kJ/(kg·K)loss = 5 %

ṁ_c = kg/s

5.Heat engines & refrigeratorsMedium

A heat engine receives heat from a source at 850 °C and rejects heat to the surroundings at 25 °C. Its actual thermal efficiency is 48%. Find its second-law efficiency η_II, the ratio of the actual efficiency to the Carnot efficiency.

T_H = 850 °CT_L = 25 °Cη = 48%

η_II = %

6.Heat engines & refrigeratorsHard

A heat engine takes in 4200 kJ of heat from a reservoir at 300 °C, produces 1134 kJ of work and rejects the rest to a reservoir at 40 °C, per cycle. Find the total entropy generated S_gen per cycle.

T_H = 300 °CT_L = 40 °CQ_H = 4200 kJW = 1134 kJ

S_gen = kJ/K

7.Gas power cyclesMedium

An ideal Diesel cycle has a compression ratio r = 14 and a cutoff ratio r_c = v₃/v₂ = 1.5. At the start of compression the air is at 305 K. Use the cold-air-standard assumptions: air is an ideal gas with constant specific heats, k = 1.4, c_p = 1.005 kJ/(kg·K) and c_v = 0.718 kJ/(kg·K). Find the heat added q_in per kilogram of air, in kJ/kg.

r = 14r_c = 1.5T₁ = 305 Kk = 1.4c_p = 1.005 kJ/(kg·K)

q_in = kJ/kg

8.Gas power cyclesHard

A gas-turbine plant runs on a Brayton cycle with a compressor pressure ratio of 8. Air enters the compressor at 295 K and the turbine at 1100 K. Both machines are isentropic. Use the cold-air-standard assumptions: air is an ideal gas with constant specific heats, k = 1.4, c_p = 1.005 kJ/(kg·K) and c_v = 0.718 kJ/(kg·K). Find the net work output w_net per kilogram of air, in kJ/kg.

r_p = 8T₁ = 295 KT₃ = 1100 Kk = 1.4c_p = 1.005 kJ/(kg·K)

w_net = kJ/kg

3Solutions

1.Ideal gas law

Answer: V = 9.695 L

Ideal gas law: PV = nRT, with n = m/M.

Moles

n = 30/32 = 0.9375 mol

T = 100 + 273.15 = 373.1 K

Volume

V = nRT/P = 0.9375·8.314·373.1/300 = 9.695 L

Answer

V = 9.695 L

2.Ideal gas law

Answer: ρ = 2.619 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)

T = 30 + 273.15 = 303.1 K

Density

ρ = 150·44.01/(8.314·303.1) = 2.619 kg/m³

Answer

ρ = 2.619 kg/m³

3.Energy balance

Answer: t_out = 37.54 °C

Exchanger energy balance: with no losses, ṁ_hc_p,h(T_h,in − T_h,out) = ṁ_cc_p,c(t_out − t_in).

Heat released by the hot stream

Q = 0.5·2.1·(150 − 90) = 63 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 30 + 63/(2·4.18) = 37.54 °C

Answer

t_out = 37.54 °C

4.Energy balance

Answer: ṁ_c = 2.273 kg/s

Energy balance with losses: heat released = heat absorbed + heat lost.

Heat released

Q_h = 2.5·3.6·(150 − 100) = 450 kW

Heat that reaches the water

Q_c = (1 − 0.05)·450 = 427.5 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 427.5/(4.18·45) = 2.273 kg/s

Answer

ṁ_c = 2.273 kg/s

5.Heat engines & refrigerators

Answer: η_II = 65.35 %

Second-law limits: no cycle working between reservoirs at absolute temperatures T_H and T_L can beat the reversible (Carnot) cycle: η_max = 1 − T_L/T_H, COP_R,max = T_L/(T_H − T_L) and COP_HP,max = T_H/(T_H − T_L). Energy is always conserved: Q_H = W + Q_L.

Absolute temperatures

T_H = 850 + 273.15 = 1123 K

T_L = 25 + 273.15 = 298.1 K

Carnot efficiency

η_Carnot = 1 − 298.1/1123 = 73.45 %

Second-law efficiency

η_II = η/η_Carnot = 48/73.45 = 65.35 %

Answer

η_II = 65.35 %

6.Heat engines & refrigerators

Answer: S_gen = 2.463 kJ/K

Entropy generation: over a full cycle the working fluid has no net entropy change. The hot reservoir loses Q_H/T_H and the cold reservoir gains Q_L/T_L. Their sum is the entropy generated, which is zero for a reversible cycle and positive for every real one.

Heat rejected

Q_L = Q_H − W = 4200 − 1134 = 3066 kJ

Absolute temperatures

T_H = 300 + 273.15 = 573.1 K

T_L = 40 + 273.15 = 313.1 K

Entropy changes of the reservoirs

ΔS_H = −4200/573.1 = −7.328 kJ/K

ΔS_L = 3066/313.1 = 9.791 kJ/K

S_gen = 9.791 − 7.328 = 2.463 kJ/K

Answer

S_gen = 2.463 kJ/K

S_gen > 0, so the engine is possible but irreversible.

7.Gas power cycles

Answer: q_in = 440.4 kJ/kg

Diesel cycle (compression-ignition engine): 1→2 isentropic compression, 2→3 heat addition at constant pressure, 3→4 isentropic expansion, 4→1 heat rejection at constant volume. With compression ratio r = v₁/v₂ and cutoff ratio r_c = v₃/v₂: T₂ = T₁·r^k−1, T₃ = r_c·T₂, q_in = c_p·(T₃ − T₂) and η = 1 − (1/r^k−1)·(r_c^k − 1)/(k·(r_c − 1)).

Temperatures

T₂ = 305·14^0.4 = 876.5 K

T₃ = 1.5·876.5 = 1315 K

Heat added at constant pressure

q_in = 1.005·(1315 − 876.5) = 440.4 kJ/kg

Answer

q_in = 440.4 kJ/kg

8.Gas power cycles

Answer: w_net = 254.6 kJ/kg

Brayton cycle (gas turbine): 1→2 isentropic compression, 2→3 heat addition at constant pressure, 3→4 isentropic expansion in the turbine, 4→1 heat rejection at constant pressure. With pressure ratio r_p = P₂/P₁: T₂ = T₁·r_p^(k−1)/k, T₄ = T₃/r_p^(k−1)/k and η = 1 − 1/r_p^(k−1)/k. Net work w = c_p·[(T₃ − T₄) − (T₂ − T₁)]. With isentropic efficiencies, the compressor needs w_s/η_c and the turbine gives η_t·w_s.

Isentropic temperatures

T₂ = 295·8^0.2857 = 534.4 K

T₄ = 1100/8^0.2857 = 607.2 K

Compressor and turbine work

w_c = 1.005·(534.4 − 295) = 240.6 kJ/kg

w_t = 1.005·(1100 − 607.2) = 495.2 kJ/kg

Net work

w_net = 495.2 − 240.6 = 254.6 kJ/kg

Answer

w_net = 254.6 kJ/kg

Back-work ratio: w_c/w_t = 48.58%.

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