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 110 g of CO₂ occupy at 400 kPa and 95 °C? Treat it as an ideal gas.

m = 110 gM = 44.01 g/molP = 400 kPaT = 95 °CR = 8.314 J/(mol·K)

V = L

2.Ideal gas lawHard

Find the density ρ of O₂ at 900 kPa and 160 °C, treating it as an ideal gas.

M = 32 g/molP = 900 kPaT = 160 °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: 160 → 100 °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 = 3 kg/sc_p,h = 2.44 kJ/(kg·K)T_h: 120 → 60 °Ct: 30 → 80 °Cc_p,c = 4.18 kJ/(kg·K)loss = 5 %

ṁ_c = kg/s

5.Heat engines & refrigeratorsMedium

A reversible refrigerator keeps a cold room at -18 °C by removing 3.5 kW of heat, and rejects heat to the surroundings at 20 °C. Find the rate Q̇_H at which it rejects heat.

T_L = -18 °CT_H = 20 °CQ̇_L = 3.5 kW

Q̇_H = kW

6.Heat engines & refrigeratorsHard

A heat pump keeps a house at 21 °C while it is -5 °C outside, delivering 11 kW of heat to the house. Find the highest possible coefficient of performance COP_max.

T_H = 21 °CT_L = -5 °CQ̇_H = 11 kW

COP_max =

7.Gas power cyclesMedium

An ideal Diesel cycle has a compression ratio r = 14 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 = 14r_c = 2T₁ = 300 Kk = 1.4

T₃ = K

8.Gas power cyclesHard

An ideal Brayton cycle operates with a compressor pressure ratio of 15. 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 = 15k = 1.4

η = %

3Solutions

1.Ideal gas law

Answer: V = 19.13 L

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

Moles

n = 110/44.01 = 2.499 mol

T = 95 + 273.15 = 368.1 K

Volume

V = nRT/P = 2.499·8.314·368.1/400 = 19.13 L

Answer

V = 19.13 L

2.Ideal gas law

Answer: ρ = 7.997 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)

T = 160 + 273.15 = 433.1 K

Density

ρ = 900·32/(8.314·433.1) = 7.997 kg/m³

Answer

ρ = 7.997 kg/m³

3.Energy balance

Answer: t_out = 60.22 °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·(160 − 100) = 378 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 15 + 378/(2·4.18) = 60.22 °C

Answer

t_out = 60.22 °C

4.Energy balance

Answer: ṁ_c = 1.996 kg/s

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

Heat released

Q_h = 3·2.44·(120 − 60) = 439.2 kW

Heat that reaches the water

Q_c = (1 − 0.05)·439.2 = 417.2 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 417.2/(4.18·50) = 1.996 kg/s

Answer

ṁ_c = 1.996 kg/s

5.Heat engines & refrigerators

Answer: Q̇_H = 4.021 kW

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 = 20 + 273.15 = 293.1 K

T_L = -18 + 273.15 = 255.1 K

Reversible heat ratio

Q̇_H = Q̇_L·T_H/T_L = 3.5·293.1/255.1 = 4.021 kW

Answer

Q̇_H = 4.021 kW

Power input: Ẇ = 0.5213 kW.

6.Heat engines & refrigerators

Answer: COP_max = 11.31

Absolute temperatures

T_H = 21 + 273.15 = 294.1 K

T_L = -5 + 273.15 = 268.1 K

Best possible COP

COP_HP,max = T_H/(T_H − T_L) = 294.1/(294.1 − 268.1) = 11.31

Answer

COP_max = 11.31

7.Gas power cycles

Answer: T₃ = 1724 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 = 300·14^0.4 = 862.1 K

Heat addition at constant pressure

T₃ = r_c·T₂ = 2·862.1 = 1724 K

Answer

T₃ = 1724 K

8.Gas power cycles

Answer: η = 53.87 %

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 − 1/15^0.2857 = 1 − 1/2.168 = 53.87%

Answer

η = 53.87%

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