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

What volume V does 130 g of CO₂ occupy at 650 kPa and 0 °C? Treat it as an ideal gas.

m = 130 gM = 44.01 g/molP = 650 kPaT = 0 °CR = 8.314 J/(mol·K)

V = L

2.Ideal gas lawHard

Find the density ρ of O₂ at 150 kPa and 50 °C, treating it as an ideal gas.

M = 32 g/molP = 150 kPaT = 50 °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 = 3 kg/sc_p,h = 2.1 kJ/(kg·K)T_h: 150 → 110 °Cṁ_c = 1.5 kg/sc_p,c = 4.18 kJ/(kg·K)t_in = 25 °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 = 2.5 kg/sc_p,h = 2.44 kJ/(kg·K)T_h: 180 → 100 °Ct: 10 → 65 °Cc_p,c = 4.18 kJ/(kg·K)loss = 5 %

ṁ_c = kg/s

5.Heat engines & refrigeratorsMedium

A heat engine works between a hot reservoir at 775 °C and a cold reservoir at 5 °C. It takes in 4600 kJ of heat. Find the most work W_max it could possibly produce.

T_H = 775 °CT_L = 5 °CQ_H = 4600 kJ

W_max = kJ

6.Heat engines & refrigeratorsHard

A refrigerator keeps its cold space at -18 °C in a kitchen at 25 °C, removing heat from the cold space at 2.5 kW. Find the minimum power input Ẇ_min.

T_H = 25 °CT_L = -18 °CQ̇_L = 2.5 kW

Ẇ_min = kW

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 300 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 = 15T₁ = 300 KT₃ = 1100 Kk = 1.4c_p = 1.005 kJ/(kg·K)

w_net = kJ/kg

3Solutions

1.Ideal gas law

Answer: V = 10.32 L

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

Moles

n = 130/44.01 = 2.954 mol

T = 0 + 273.15 = 273.1 K

Volume

V = nRT/P = 2.954·8.314·273.1/650 = 10.32 L

Answer

V = 10.32 L

2.Ideal gas law

Answer: ρ = 1.787 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)

T = 50 + 273.15 = 323.1 K

Density

ρ = 150·32/(8.314·323.1) = 1.787 kg/m³

Answer

ρ = 1.787 kg/m³

3.Energy balance

Answer: t_out = 65.19 °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 = 3·2.1·(150 − 110) = 252 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 25 + 252/(1.5·4.18) = 65.19 °C

Answer

t_out = 65.19 °C

4.Energy balance

Answer: ṁ_c = 2.017 kg/s

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

Heat released

Q_h = 2.5·2.44·(180 − 100) = 488 kW

Heat that reaches the water

Q_c = (1 − 0.05)·488 = 463.6 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 463.6/(4.18·55) = 2.017 kg/s

Answer

ṁ_c = 2.017 kg/s

5.Heat engines & refrigerators

Answer: W_max = 3379 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 = 775 + 273.15 = 1048 K

T_L = 5 + 273.15 = 278.1 K

Carnot efficiency

η_max = 1 − T_L/T_H = 1 − 278.1/1048 = 73.46 %

Maximum work

W_max = η_max·Q_H = 0.7346·4600 = 3379 kJ

Answer

W_max = 3379 kJ

6.Heat engines & refrigerators

Answer: Ẇ_min = 0.4213 kW

Absolute temperatures

T_H = 25 + 273.15 = 298.1 K

T_L = -18 + 273.15 = 255.1 K

Best possible COP

COP_R,max = T_L/(T_H − T_L) = 255.1/(298.1 − 255.1) = 5.934

Minimum power

The least work is needed by a reversible machine, which has the highest COP.

Ẇ_min = Q̇_L/COP = 2.5/5.934 = 0.4213 kW

Answer

Ẇ_min = 0.4213 kW

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 = 243.4 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 = 300·15^0.2857 = 650.4 K

T₄ = 1100/15^0.2857 = 507.4 K

Compressor and turbine work

w_c = 1.005·(650.4 − 300) = 352.1 kJ/kg

w_t = 1.005·(1100 − 507.4) = 595.5 kJ/kg

Net work

w_net = 595.5 − 352.1 = 243.4 kJ/kg

Answer

w_net = 243.4 kJ/kg

Back-work ratio: w_c/w_t = 59.12%.

ExamProblems · examproblems.com/courses/thermodynamics-1