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 5 L contains 210 g of N₂ at 25 °C. Treat it as an ideal gas and find the pressure P.

m = 210 gM = 28.01 g/molV = 5 LT = 25 °CR = 8.314 J/(mol·K)

P = kPa

2.Ideal gas lawHard

Find the density ρ of He at 350 kPa and 120 °C, treating it as an ideal gas.

M = 4.003 g/molP = 350 kPaT = 120 °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 = 1 kg/sc_p,h = 2.1 kJ/(kg·K)T_h: 160 → 80 °Cṁ_c = 1 kg/sc_p,c = 4.18 kJ/(kg·K)t_in = 10 °C

t_out = °C

4.Energy balanceHard

Hot light oil heats water in an exchanger that loses 10% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁ_c needed.

ṁ_h = 2 kg/sc_p,h = 2.1 kJ/(kg·K)T_h: 170 → 80 °Ct: 30 → 55 °Cc_p,c = 4.18 kJ/(kg·K)loss = 10 %

ṁ_c = kg/s

5.Heat engines & refrigeratorsMedium

A heat engine works between a hot reservoir at 750 °C and a cold reservoir at 40 °C. Find the highest possible thermal efficiency η_max.

T_H = 750 °CT_L = 40 °C

η_max = %

6.Heat engines & refrigeratorsHard

A refrigerator keeps its cold space at -18 °C in a kitchen at 30 °C, removing heat from the cold space at 5 kW. Find the minimum power input Ẇ_min.

T_H = 30 °CT_L = -18 °CQ̇_L = 5 kW

Ẇ_min = kW

7.Gas power cyclesMedium

An ideal Diesel cycle has a compression ratio r = 18 and a cutoff ratio r_c = v₃/v₂ = 2.5. At the start of compression the air is at 310 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 = 18r_c = 2.5T₁ = 310 Kk = 1.4

T₃ = K

8.Gas power cyclesHard

A gas-turbine plant runs on a Brayton cycle with a compressor pressure ratio of 10. Air enters the compressor at 300 K and the turbine at 1500 K. The compressor has an isentropic efficiency of 88% and the turbine 88%. 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 = 10T₁ = 300 KT₃ = 1500 Kη_c = 88%η_t = 88%k = 1.4c_p = 1.005 kJ/(kg·K)

w_net = kJ/kg

3Solutions

1.Ideal gas law

Answer: P = 3717 kPa

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

Moles

n = m/M = 210/28.01 = 7.497 mol

T = 25 + 273.15 = 298.1 K

Pressure

P = nRT/V = 7.497·8.314·298.1/0.005 = 3717 kPa

Answer

P = 3717 kPa

2.Ideal gas law

Answer: ρ = 0.4286 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)

T = 120 + 273.15 = 393.1 K

Density

ρ = 350·4.003/(8.314·393.1) = 0.4286 kg/m³

Answer

ρ = 0.4286 kg/m³

3.Energy balance

Answer: t_out = 50.19 °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 = 1·2.1·(160 − 80) = 168 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 10 + 168/(1·4.18) = 50.19 °C

Answer

t_out = 50.19 °C

4.Energy balance

Answer: ṁ_c = 3.256 kg/s

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

Heat released

Q_h = 2·2.1·(170 − 80) = 378 kW

Heat that reaches the water

Q_c = (1 − 0.1)·378 = 340.2 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 340.2/(4.18·25) = 3.256 kg/s

Answer

ṁ_c = 3.256 kg/s

5.Heat engines & refrigerators

Answer: η_max = 69.39 %

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 = 750 + 273.15 = 1023 K

T_L = 40 + 273.15 = 313.1 K

Carnot efficiency

η_max = 1 − T_L/T_H = 1 − 313.1/1023 = 69.39 %

Answer

η_max = 69.39 %

6.Heat engines & refrigerators

Answer: Ẇ_min = 0.9406 kW

Absolute temperatures

T_H = 30 + 273.15 = 303.1 K

T_L = -18 + 273.15 = 255.1 K

Best possible COP

COP_R,max = T_L/(T_H − T_L) = 255.1/(303.1 − 255.1) = 5.316

Minimum power

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

Ẇ_min = Q̇_L/COP = 5/5.316 = 0.9406 kW

Answer

Ẇ_min = 0.9406 kW

7.Gas power cycles

Answer: T₃ = 2463 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 = 310·18^0.4 = 985.1 K

Heat addition at constant pressure

T₃ = r_c·T₂ = 2.5·985.1 = 2463 K

Answer

T₃ = 2463 K

8.Gas power cycles

Answer: w_net = 320.6 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 = 300·10^0.2857 = 579.2 K

T_4s = 1500/10^0.2857 = 776.9 K

Compressor and turbine work

w_c = 1.005·(579.2 − 300)/0.88 = 318.9 kJ/kg

w_t = 0.88·1.005·(1500 − 776.9) = 639.5 kJ/kg

Net work

w_net = 639.5 − 318.9 = 320.6 kJ/kg

Answer

w_net = 320.6 kJ/kg

Back-work ratio: w_c/w_t = 49.86%.

ExamProblems · examproblems.com/courses/thermodynamics-1