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 N₂ occupy at 250 kPa and 85 °C? Treat it as an ideal gas.

m = 30 gM = 28.01 g/molP = 250 kPaT = 85 °CR = 8.314 J/(mol·K)

V = L

2.Ideal gas lawHard

A fixed amount of Ar goes from 350 kPa, 20 L, 15 °C to 450 kPa and 220 °C. Find the new volume V₂.

P₁ = 350 kPaV₁ = 20 LT₁ = 15 °CP₂ = 450 kPaT₂ = 220 °C

V₂ = L

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: 120 → 70 °Cṁ_c = 1.5 kg/sc_p,c = 4.18 kJ/(kg·K)t_in = 15 °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: 130 → 90 °Ct: 10 → 75 °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 375 °C and a cold reservoir at 20 °C. Find the highest possible thermal efficiency η_max.

T_H = 375 °CT_L = 20 °C

η_max = %

6.Heat engines & refrigeratorsHard

A heat engine takes in 3000 kJ of heat from a reservoir at 550 °C, produces 1170 kJ of work and rejects the rest to a reservoir at 15 °C, per cycle. Find the total entropy generated S_gen per cycle.

T_H = 550 °CT_L = 15 °CQ_H = 3000 kJW = 1170 kJ

S_gen = kJ/K

7.Gas power cyclesMedium

An ideal Diesel cycle has a compression ratio r = 22 and a cutoff ratio r_c = v₃/v₂ = 3. 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 = 22r_c = 3k = 1.4

η = %

8.Gas power cyclesHard

A gas-turbine plant runs on a Brayton cycle with a compressor pressure ratio of 15. Air enters the compressor at 310 K and the turbine at 1500 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 = 15T₁ = 310 KT₃ = 1500 Kk = 1.4c_p = 1.005 kJ/(kg·K)

w_net = kJ/kg

3Solutions

1.Ideal gas law

Answer: V = 12.76 L

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

Moles

n = 30/28.01 = 1.071 mol

T = 85 + 273.15 = 358.1 K

Volume

V = nRT/P = 1.071·8.314·358.1/250 = 12.76 L

Answer

V = 12.76 L

2.Ideal gas law

Answer: V₂ = 26.62 L

Combined gas law: for a fixed amount of ideal gas, P₁V₁/T₁ = P₂V₂/T₂.

Absolute temperatures

T₁ = 288.1 K, T₂ = 493.1 K

Combined gas law

V₂ = V₁·(P₁/P₂)·(T₂/T₁) = 20·(350/450)·(493.1/288.1) = 26.62 L

Answer

V₂ = 26.62 L

3.Energy balance

Answer: t_out = 72.42 °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·(120 − 70) = 360 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 15 + 360/(1.5·4.18) = 72.42 °C

Answer

t_out = 72.42 °C

4.Energy balance

Answer: ṁ_c = 0.6757 kg/s

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

Heat released

Q_h = 1.5·3.6·(130 − 90) = 216 kW

Heat that reaches the water

Q_c = (1 − 0.15)·216 = 183.6 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 183.6/(4.18·65) = 0.6757 kg/s

Answer

ṁ_c = 0.6757 kg/s

5.Heat engines & refrigerators

Answer: η_max = 54.77 %

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 = 375 + 273.15 = 648.1 K

T_L = 20 + 273.15 = 293.1 K

Carnot efficiency

η_max = 1 − T_L/T_H = 1 − 293.1/648.1 = 54.77 %

Answer

η_max = 54.77 %

6.Heat engines & refrigerators

Answer: S_gen = 2.706 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 = 3000 − 1170 = 1830 kJ

Absolute temperatures

T_H = 550 + 273.15 = 823.1 K

T_L = 15 + 273.15 = 288.1 K

Entropy changes of the reservoirs

ΔS_H = −3000/823.1 = −3.645 kJ/K

ΔS_L = 1830/288.1 = 6.351 kJ/K

S_gen = 6.351 − 3.645 = 2.706 kJ/K

Answer

S_gen = 2.706 kJ/K

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

7.Gas power cycles

Answer: η = 62.08 %

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

(3^1.4 − 1)/(1.4·(3 − 1)) = 1.306

Thermal efficiency

η = 1 − 0.2904·1.306 = 62.08%

Answer

η = 62.08%

8.Gas power cycles

Answer: w_net = 448.3 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 = 310·15^0.2857 = 672 K

T₄ = 1500/15^0.2857 = 691.9 K

Compressor and turbine work

w_c = 1.005·(672 − 310) = 363.8 kJ/kg

w_t = 1.005·(1500 − 691.9) = 812.1 kJ/kg

Net work

w_net = 812.1 − 363.8 = 448.3 kJ/kg

Answer

w_net = 448.3 kJ/kg

Back-work ratio: w_c/w_t = 44.8%.

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