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

A tank of 10 L contains 130 g of CO₂ at 20 °C. Treat it as an ideal gas and find the pressure P.

m = 130 gM = 44.01 g/molV = 10 LT = 20 °CR = 8.314 J/(mol·K)

P = kPa

2.Ideal gas lawHard

Find the density ρ of CH₄ at 750 kPa and 70 °C, treating it as an ideal gas.

M = 16.04 g/molP = 750 kPaT = 70 °CR = 8.314 J/(mol·K)

ρ = kg/m³

3.Energy balanceMedium

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

ṁ_h = 2 kg/sc_p,h = 3.6 kJ/(kg·K)T_h: 180 → 140 °Cṁ_c = 2 kg/sc_p,c = 4.18 kJ/(kg·K)t_in = 10 °C

t_out = °C

4.Energy balanceHard

Hot glycol mix heats water in an exchanger that loses 15% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁ_c needed.

ṁ_h = 1.5 kg/sc_p,h = 3.6 kJ/(kg·K)T_h: 120 → 60 °Ct: 10 → 25 °Cc_p,c = 4.18 kJ/(kg·K)loss = 15 %

ṁ_c = kg/s

5.Heat engines & refrigeratorsMedium

A heat engine works between a hot reservoir at 350 °C and a cold reservoir at 20 °C. It takes in 3800 kJ of heat. Find the most work W_max it could possibly produce.

T_H = 350 °CT_L = 20 °CQ_H = 3800 kJ

W_max = kJ

6.Heat engines & refrigeratorsHard

A heat engine takes in 5000 kJ of heat from a reservoir at 650 °C, produces 2350 kJ of work and rejects the rest to a reservoir at 25 °C, per cycle. Find the total entropy generated S_gen per cycle.

T_H = 650 °CT_L = 25 °CQ_H = 5000 kJW = 2350 kJ

S_gen = kJ/K

7.Gas power cyclesMedium

An ideal Diesel cycle has a compression ratio r = 20 and a cutoff ratio r_c = v₃/v₂ = 1.8. 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 its thermal efficiency η in percent.

r = 20r_c = 1.8k = 1.4

η = %

8.Gas power cyclesHard

A gas-turbine plant runs on a Brayton cycle with a compressor pressure ratio of 6. Air enters the compressor at 310 K and the turbine at 1200 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 = 6T₁ = 310 KT₃ = 1200 Kk = 1.4c_p = 1.005 kJ/(kg·K)

w_net = kJ/kg

3Solutions

1.Ideal gas law

Answer: P = 719.9 kPa

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

Moles

n = m/M = 130/44.01 = 2.954 mol

T = 20 + 273.15 = 293.1 K

Pressure

P = nRT/V = 2.954·8.314·293.1/0.01 = 719.9 kPa

Answer

P = 719.9 kPa

2.Ideal gas law

Answer: ρ = 4.217 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)

T = 70 + 273.15 = 343.1 K

Density

ρ = 750·16.04/(8.314·343.1) = 4.217 kg/m³

Answer

ρ = 4.217 kg/m³

3.Energy balance

Answer: t_out = 44.45 °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 = 2·3.6·(180 − 140) = 288 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 10 + 288/(2·4.18) = 44.45 °C

Answer

t_out = 44.45 °C

4.Energy balance

Answer: ṁ_c = 4.392 kg/s

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

Heat released

Q_h = 1.5·3.6·(120 − 60) = 324 kW

Heat that reaches the water

Q_c = (1 − 0.15)·324 = 275.4 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 275.4/(4.18·15) = 4.392 kg/s

Answer

ṁ_c = 4.392 kg/s

5.Heat engines & refrigerators

Answer: W_max = 2012 kJ

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 = 350 + 273.15 = 623.1 K

T_L = 20 + 273.15 = 293.1 K

Carnot efficiency

η_max = 1 − T_L/T_H = 1 − 293.1/623.1 = 52.96 %

Maximum work

W_max = η_max·Q_H = 0.5296·3800 = 2012 kJ

Answer

W_max = 2012 kJ

6.Heat engines & refrigerators

Answer: S_gen = 3.472 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 = 5000 − 2350 = 2650 kJ

Absolute temperatures

T_H = 650 + 273.15 = 923.1 K

T_L = 25 + 273.15 = 298.1 K

Entropy changes of the reservoirs

ΔS_H = −5000/923.1 = −5.416 kJ/K

ΔS_L = 2650/298.1 = 8.888 kJ/K

S_gen = 8.888 − 5.416 = 3.472 kJ/K

Answer

S_gen = 3.472 kJ/K

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

7.Gas power cycles

Answer: η = 65.6 %

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)).

Cutoff term

(1.8^1.4 − 1)/(1.4·(1.8 − 1)) = 1.14

Thermal efficiency

η = 1 − 0.3017·1.14 = 65.6%

Answer

η = 65.6%

8.Gas power cycles

Answer: w_net = 274.9 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 = 310·6^0.2857 = 517.2 K

T₄ = 1200/6^0.2857 = 719.2 K

Compressor and turbine work

w_c = 1.005·(517.2 − 310) = 208.3 kJ/kg

w_t = 1.005·(1200 − 719.2) = 483.2 kJ/kg

Net work

w_net = 483.2 − 208.3 = 274.9 kJ/kg

Answer

w_net = 274.9 kJ/kg

Back-work ratio: w_c/w_t = 43.1%.

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