Thermodynamics 1: cram pack

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

New problems Back to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 20 g of O₂ at 0 °C. Treat it as an ideal gas and find the pressure P.

m = 20 gM = 32 g/molV = 10 LT = 0 °CR = 8.314 J/(mol·K)

P = kPa

2.Ideal gas lawHard

A fixed amount of CH₄ goes from 300 kPa, 10 L, 15 °C to 550 kPa and 90 °C. Find the new volume V₂.

P₁ = 300 kPaV₁ = 10 LT₁ = 15 °CP₂ = 550 kPaT₂ = 90 °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 = 1.5 kg/sc_p,h = 2.1 kJ/(kg·K)T_h: 150 → 110 °Cṁ_c = 2 kg/sc_p,c = 4.18 kJ/(kg·K)t_in = 15 °C

t_out = °C

4.Energy balanceHard

Hot ethanol 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 = 1 kg/sc_p,h = 2.44 kJ/(kg·K)T_h: 170 → 90 °Ct: 20 → 60 °Cc_p,c = 4.18 kJ/(kg·K)loss = 5 %

ṁ_c = kg/s

5.Heat engines & refrigeratorsMedium

A reversible (Carnot) heat engine takes in 500 kJ of heat from a reservoir at 775 °C and rejects 149 kJ per cycle. Find the temperature T_L of the cold reservoir in °C.

T_H = 775 °CQ_H = 500 kJQ_L = 149 kJ

T_L = °C

6.Heat engines & refrigeratorsHard

A heat engine takes in 3100 kJ of heat from a reservoir at 700 °C, produces 1705 kJ of work and rejects the rest to a reservoir at 35 °C, per cycle. Find the total entropy generated S_gen per cycle.

T_H = 700 °CT_L = 35 °CQ_H = 3100 kJW = 1705 kJ

S_gen = kJ/K

7.Gas power cyclesMedium

An ideal Diesel cycle has a compression ratio r = 19 and a cutoff ratio r_c = v₃/v₂ = 2. At the start of compression the air is at 300 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 temperature T₃ at the end of heat addition, in K.

r = 19r_c = 2T₁ = 300 Kk = 1.4

T₃ = K

8.Gas power cyclesHard

An ideal Brayton cycle operates with a compressor pressure ratio of 12. 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_p = 12k = 1.4

η = %

3Solutions

1.Ideal gas law

Answer: P = 141.9 kPa

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

Moles

n = m/M = 20/32 = 0.625 mol

T = 0 + 273.15 = 273.1 K

Pressure

P = nRT/V = 0.625·8.314·273.1/0.01 = 141.9 kPa

Answer

P = 141.9 kPa

2.Ideal gas law

Answer: V₂ = 6.874 L

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

Absolute temperatures

T₁ = 288.1 K, T₂ = 363.1 K

Combined gas law

V₂ = V₁·(P₁/P₂)·(T₂/T₁) = 10·(300/550)·(363.1/288.1) = 6.874 L

Answer

V₂ = 6.874 L

3.Energy balance

Answer: t_out = 30.07 °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 = 1.5·2.1·(150 − 110) = 126 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 15 + 126/(2·4.18) = 30.07 °C

Answer

t_out = 30.07 °C

4.Energy balance

Answer: ṁ_c = 1.109 kg/s

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

Heat released

Q_h = 1·2.44·(170 − 90) = 195.2 kW

Heat that reaches the water

Q_c = (1 − 0.05)·195.2 = 185.4 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 185.4/(4.18·40) = 1.109 kg/s

Answer

ṁ_c = 1.109 kg/s

5.Heat engines & refrigerators

Answer: T_L = 39.2 °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 = 775 + 273.15 = 1048 K

Reversible heat ratio

T_L = T_H·Q_L/Q_H = 1048·149/500 = 312.3 K = 39.2 °C

Answer

T_L = 39.2 °C

6.Heat engines & refrigerators

Answer: S_gen = 1.341 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 = 3100 − 1705 = 1395 kJ

Absolute temperatures

T_H = 700 + 273.15 = 973.1 K

T_L = 35 + 273.15 = 308.1 K

Entropy changes of the reservoirs

ΔS_H = −3100/973.1 = −3.186 kJ/K

ΔS_L = 1395/308.1 = 4.527 kJ/K

S_gen = 4.527 − 3.186 = 1.341 kJ/K

Answer

S_gen = 1.341 kJ/K

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

7.Gas power cycles

Answer: T₃ = 1948 K

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)).

End of compression

T₂ = 300·19^0.4 = 974.1 K

Heat addition at constant pressure

T₃ = r_c·T₂ = 2·974.1 = 1948 K

Answer

T₃ = 1948 K

8.Gas power cycles

Answer: η = 50.83 %

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.

Thermal efficiency

η = 1 − 1/12^0.2857 = 1 − 1/2.034 = 50.83%

Answer

η = 50.83%

ExamProblems · examproblems.com/courses/thermodynamics-1