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 190 g of Ar occupy at 800 kPa and 75 °C? Treat it as an ideal gas.

m = 190 gM = 39.95 g/molP = 800 kPaT = 75 °CR = 8.314 J/(mol·K)

V = L

2.Ideal gas lawHard

A fixed amount of Ar goes from 150 kPa, 20 L, 40 °C to 150 kPa and 130 °C. Find the new volume V₂.

P₁ = 150 kPaV₁ = 20 LT₁ = 40 °CP₂ = 150 kPaT₂ = 130 °C

V₂ = L

3.Energy balanceMedium

Hot glycol mix 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 = 3.6 kJ/(kg·K)T_h: 120 → 60 °Cṁ_c = 1 kg/sc_p,c = 4.18 kJ/(kg·K)t_in = 20 °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: 120 → 90 °Ct: 25 → 75 °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 675 °C and a cold reservoir at 35 °C. Find the highest possible thermal efficiency η_max.

T_H = 675 °CT_L = 35 °C

η_max = %

6.Heat engines & refrigeratorsHard

A Carnot heat engine receives 160 kW of heat from a source at 700 °C and rejects heat to the surroundings at 20 °C. All of its work drives a Carnot refrigerator that removes heat from a freezer at -25 °C and also rejects heat to the surroundings at 20 °C. Find the rate Q̇_L at which heat is removed from the freezer.

T_H = 700 °CT₀ = 20 °CT_L = -25 °CQ̇_H = 160 kW

Q̇_L = kW

7.Gas power cyclesMedium

An ideal Diesel cycle has a compression ratio r = 14 and a cutoff ratio r_c = v₃/v₂ = 2. 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 = 14r_c = 2k = 1.4

η = %

8.Gas power cyclesHard

An ideal Brayton cycle operates with a compressor pressure ratio of 8. 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 = 8k = 1.4

η = %

3Solutions

1.Ideal gas law

Answer: V = 17.21 L

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

Moles

n = 190/39.95 = 4.756 mol

T = 75 + 273.15 = 348.1 K

Volume

V = nRT/P = 4.756·8.314·348.1/800 = 17.21 L

Answer

V = 17.21 L

2.Ideal gas law

Answer: V₂ = 25.75 L

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

Absolute temperatures

T₁ = 313.1 K, T₂ = 403.1 K

Combined gas law

V₂ = V₁·(P₁/P₂)·(T₂/T₁) = 20·(150/150)·(403.1/313.1) = 25.75 L

Answer

V₂ = 25.75 L

3.Energy balance

Answer: t_out = 45.84 °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 = 0.5·3.6·(120 − 60) = 108 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 20 + 108/(1·4.18) = 45.84 °C

Answer

t_out = 45.84 °C

4.Energy balance

Answer: ṁ_c = 0.788 kg/s

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

Heat released

Q_h = 2.5·2.44·(120 − 90) = 183 kW

Heat that reaches the water

Q_c = (1 − 0.1)·183 = 164.7 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 164.7/(4.18·50) = 0.788 kg/s

Answer

ṁ_c = 0.788 kg/s

5.Heat engines & refrigerators

Answer: η_max = 67.5 %

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 = 675 + 273.15 = 948.1 K

T_L = 35 + 273.15 = 308.1 K

Carnot efficiency

η_max = 1 − T_L/T_H = 1 − 308.1/948.1 = 67.5 %

Answer

η_max = 67.5 %

6.Heat engines & refrigerators

Answer: Q̇_L = 616.5 kW

Absolute temperatures

T_H = 700 + 273.15 = 973.1 K

T₀ = 20 + 273.15 = 293.1 K

T_L = -25 + 273.15 = 248.1 K

Engine work

η = 1 − 293.1/973.1 = 0.6988

Ẇ = 0.6988·160 = 111.8 kW

Heat removed by the refrigerator

COP_R = 248.1/(293.1 − 248.1) = 5.514

Q̇_L = 5.514·111.8 = 616.5 kW

Answer

Q̇_L = 616.5 kW

7.Gas power cycles

Answer: η = 59.26 %

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

(2^1.4 − 1)/(1.4·(2 − 1)) = 1.171

Thermal efficiency

η = 1 − 0.348·1.171 = 59.26%

Answer

η = 59.26%

8.Gas power cycles

Answer: η = 44.8 %

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 − 1/8^0.2857 = 1 − 1/1.811 = 44.8%

Answer

η = 44.8%

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