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 80 g of CH₄ occupy at 200 kPa and 90 °C? Treat it as an ideal gas.

m = 80 gM = 16.04 g/molP = 200 kPaT = 90 °CR = 8.314 J/(mol·K)

V = L

2.Ideal gas lawHard

A fixed amount of O₂ goes from 400 kPa, 2 L, 40 °C to 150 kPa and 130 °C. Find the new volume V₂.

P₁ = 400 kPaV₁ = 2 LT₁ = 40 °CP₂ = 150 kPaT₂ = 130 °C

V₂ = L

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 → 80 °Cṁ_c = 0.5 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 = 2.5 kg/sc_p,h = 3.6 kJ/(kg·K)T_h: 150 → 80 °Ct: 20 → 80 °Cc_p,c = 4.18 kJ/(kg·K)loss = 15 %

ṁ_c = kg/s

5.Heat engines & refrigeratorsMedium

A reversible (Carnot) heat engine takes in 4900 kJ of heat from a reservoir at 700 °C and rejects 1627 kJ per cycle. Find the temperature T_L of the cold reservoir in °C.

T_H = 700 °CQ_H = 4900 kJQ_L = 1627 kJ

T_L = °C

6.Heat engines & refrigeratorsHard

A Carnot heat engine receives 80 kW of heat from a source at 650 °C and rejects heat to the surroundings at 20 °C. All of its work drives a Carnot refrigerator that removes heat from a freezer at -25 °C and also rejects heat to the surroundings at 20 °C. Find the rate Q̇_L at which heat is removed from the freezer.

T_H = 650 °CT₀ = 20 °CT_L = -25 °CQ̇_H = 80 kW

Q̇_L = kW

7.Gas power cyclesMedium

An ideal Diesel cycle has a compression ratio r = 19 and a cutoff ratio r_c = v₃/v₂ = 1.8. At the start of compression the air is at 320 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 = 19r_c = 1.8T₁ = 320 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 6. Air enters the compressor at 300 K and the turbine at 1100 K. The compressor has an isentropic efficiency of 80% and the turbine 90%. 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 = 300 KT₃ = 1100 Kη_c = 80%η_t = 90%k = 1.4c_p = 1.005 kJ/(kg·K)

w_net = kJ/kg

3Solutions

1.Ideal gas law

Answer: V = 75.29 L

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

Moles

n = 80/16.04 = 4.988 mol

T = 90 + 273.15 = 363.1 K

Volume

V = nRT/P = 4.988·8.314·363.1/200 = 75.29 L

Answer

V = 75.29 L

2.Ideal gas law

Answer: V₂ = 6.866 L

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

Absolute temperatures

T₁ = 313.1 K, T₂ = 403.1 K

Combined gas law

V₂ = V₁·(P₁/P₂)·(T₂/T₁) = 2·(400/150)·(403.1/313.1) = 6.866 L

Answer

V₂ = 6.866 L

3.Energy balance

Answer: t_out = 45.17 °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 − 80) = 73.5 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 10 + 73.5/(0.5·4.18) = 45.17 °C

Answer

t_out = 45.17 °C

4.Energy balance

Answer: ṁ_c = 2.135 kg/s

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

Heat released

Q_h = 2.5·3.6·(150 − 80) = 630 kW

Heat that reaches the water

Q_c = (1 − 0.15)·630 = 535.5 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 535.5/(4.18·60) = 2.135 kg/s

Answer

ṁ_c = 2.135 kg/s

5.Heat engines & refrigerators

Answer: T_L = 49.98 °C

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 temperature of the source

T_H = 700 + 273.15 = 973.1 K

Reversible heat ratio

T_L = T_H·Q_L/Q_H = 973.1·1627/4900 = 323.1 K = 49.98 °C

Answer

T_L = 49.98 °C

6.Heat engines & refrigerators

Answer: Q̇_L = 301.1 kW

Absolute temperatures

T_H = 650 + 273.15 = 923.1 K

T₀ = 20 + 273.15 = 293.1 K

T_L = -25 + 273.15 = 248.1 K

Engine work

η = 1 − 293.1/923.1 = 0.6824

Ẇ = 0.6824·80 = 54.6 kW

Heat removed by the refrigerator

COP_R = 248.1/(293.1 − 248.1) = 5.514

Q̇_L = 5.514·54.6 = 301.1 kW

Answer

Q̇_L = 301.1 kW

7.Gas power cycles

Answer: q_in = 835.4 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 = 320·19^0.4 = 1039 K

T₃ = 1.8·1039 = 1870 K

Heat added at constant pressure

q_in = 1.005·(1870 − 1039) = 835.4 kJ/kg

Answer

q_in = 835.4 kJ/kg

8.Gas power cycles

Answer: w_net = 146.7 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

T_2s = 300·6^0.2857 = 500.6 K

T_4s = 1100/6^0.2857 = 659.3 K

Compressor and turbine work

w_c = 1.005·(500.6 − 300)/0.8 = 251.9 kJ/kg

w_t = 0.9·1.005·(1100 − 659.3) = 398.6 kJ/kg

Net work

w_net = 398.6 − 251.9 = 146.7 kJ/kg

Answer

w_net = 146.7 kJ/kg

Back-work ratio: w_c/w_t = 63.2%.

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