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 440 g of CH₄ occupy at 750 kPa and 85 °C? Treat it as an ideal gas.

m = 440 gM = 16.04 g/molP = 750 kPaT = 85 °CR = 8.314 J/(mol·K)

V = L

2.Ideal gas lawHard

Find the density ρ of CH₄ at 950 kPa and 60 °C, treating it as an ideal gas.

M = 16.04 g/molP = 950 kPaT = 60 °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 = 0.5 kg/sc_p,h = 2.1 kJ/(kg·K)T_h: 140 → 90 °Cṁ_c = 2 kg/sc_p,c = 4.18 kJ/(kg·K)t_in = 10 °C

t_out = °C

4.Energy balanceHard

Hot ethanol 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 = 2.5 kg/sc_p,h = 2.44 kJ/(kg·K)T_h: 130 → 90 °Ct: 20 → 70 °Cc_p,c = 4.18 kJ/(kg·K)loss = 10 %

ṁ_c = kg/s

5.Heat engines & refrigeratorsMedium

A reversible (Carnot) heat engine takes in 2500 kJ of heat from a reservoir at 425 °C and rejects 996 kJ per cycle. Find the temperature T_L of the cold reservoir in °C.

T_H = 425 °CQ_H = 2500 kJQ_L = 996 kJ

T_L = °C

6.Heat engines & refrigeratorsHard

A heat pump keeps a house at 20 °C while it is -5 °C outside, delivering 9 kW of heat to the house. Find the minimum power input Ẇ_min.

T_H = 20 °CT_L = -5 °CQ̇_H = 9 kW

Ẇ_min = kW

7.Gas power cyclesMedium

An ideal Diesel cycle has a compression ratio r = 19 and a cutoff ratio r_c = v₃/v₂ = 2.5. At the start of compression the air is at 305 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 = 2.5T₁ = 305 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 290 K and the turbine at 1300 K. The compressor has an isentropic efficiency of 80% and the turbine 92%. 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 = 290 KT₃ = 1300 Kη_c = 80%η_t = 92%k = 1.4c_p = 1.005 kJ/(kg·K)

w_net = kJ/kg

3Solutions

1.Ideal gas law

Answer: V = 108.9 L

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

Moles

n = 440/16.04 = 27.43 mol

T = 85 + 273.15 = 358.1 K

Volume

V = nRT/P = 27.43·8.314·358.1/750 = 108.9 L

Answer

V = 108.9 L

2.Ideal gas law

Answer: ρ = 5.501 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)

T = 60 + 273.15 = 333.1 K

Density

ρ = 950·16.04/(8.314·333.1) = 5.501 kg/m³

Answer

ρ = 5.501 kg/m³

3.Energy balance

Answer: t_out = 16.28 °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·(140 − 90) = 52.5 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 10 + 52.5/(2·4.18) = 16.28 °C

Answer

t_out = 16.28 °C

4.Energy balance

Answer: ṁ_c = 1.051 kg/s

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

Heat released

Q_h = 2.5·2.44·(130 − 90) = 244 kW

Heat that reaches the water

Q_c = (1 − 0.1)·244 = 219.6 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 219.6/(4.18·50) = 1.051 kg/s

Answer

ṁ_c = 1.051 kg/s

5.Heat engines & refrigerators

Answer: T_L = 4.993 °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 = 425 + 273.15 = 698.1 K

Reversible heat ratio

T_L = T_H·Q_L/Q_H = 698.1·996/2500 = 278.1 K = 4.993 °C

Answer

T_L = 4.993 °C

6.Heat engines & refrigerators

Answer: Ẇ_min = 0.7675 kW

Absolute temperatures

T_H = 20 + 273.15 = 293.1 K

T_L = -5 + 273.15 = 268.1 K

Best possible COP

COP_HP,max = T_H/(T_H − T_L) = 293.1/(293.1 − 268.1) = 11.73

Minimum power

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

Ẇ_min = Q̇_H/COP = 9/11.73 = 0.7675 kW

Answer

Ẇ_min = 0.7675 kW

7.Gas power cycles

Answer: T₃ = 2476 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 = 305·19^0.4 = 990.4 K

Heat addition at constant pressure

T₃ = r_c·T₂ = 2.5·990.4 = 2476 K

Answer

T₃ = 2476 K

8.Gas power cycles

Answer: w_net = 240.4 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

T_2s = 290·10^0.2857 = 559.9 K

T_4s = 1300/10^0.2857 = 673.3 K

Compressor and turbine work

w_c = 1.005·(559.9 − 290)/0.8 = 339.1 kJ/kg

w_t = 0.92·1.005·(1300 − 673.3) = 579.4 kJ/kg

Net work

w_net = 579.4 − 339.1 = 240.4 kJ/kg

Answer

w_net = 240.4 kJ/kg

Back-work ratio: w_c/w_t = 58.52%.

ExamProblems · examproblems.com/courses/thermodynamics-1