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 290 g of He occupy at 400 kPa and 60 °C? Treat it as an ideal gas.

m = 290 gM = 4.003 g/molP = 400 kPaT = 60 °CR = 8.314 J/(mol·K)

V = L

2.Ideal gas lawHard

A fixed amount of N₂ goes from 200 kPa, 5 L, 10 °C to 750 kPa and 100 °C. Find the new volume V₂.

P₁ = 200 kPaV₁ = 5 LT₁ = 10 °CP₂ = 750 kPaT₂ = 100 °C

V₂ = L

3.Energy balanceMedium

Hot ethanol 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 = 2.44 kJ/(kg·K)T_h: 160 → 80 °Cṁ_c = 1.5 kg/sc_p,c = 4.18 kJ/(kg·K)t_in = 30 °C

t_out = °C

4.Energy balanceHard

Hot ethanol heats water in an exchanger that loses 15% 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: 150 → 100 °Ct: 15 → 85 °Cc_p,c = 4.18 kJ/(kg·K)loss = 15 %

ṁ_c = kg/s

5.Heat engines & refrigeratorsMedium

A reversible (Carnot) heat engine takes in 700 kJ of heat from a reservoir at 400 °C and rejects 326 kJ per cycle. Find the temperature T_L of the cold reservoir in °C.

T_H = 400 °CQ_H = 700 kJQ_L = 326 kJ

T_L = °C

6.Heat engines & refrigeratorsHard

A refrigerator keeps its cold space at -10 °C in a kitchen at 25 °C, removing heat from the cold space at 1.5 kW. Find the minimum power input Ẇ_min.

T_H = 25 °CT_L = -10 °CQ̇_L = 1.5 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. 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 heat added q_in per kilogram of air, in kJ/kg.

r = 19r_c = 2T₁ = 300 Kk = 1.4c_p = 1.005 kJ/(kg·K)

q_in = kJ/kg

8.Gas power cyclesHard

An ideal Brayton cycle operates with a compressor pressure ratio of 16. 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 = 16k = 1.4

η = %

3Solutions

1.Ideal gas law

Answer: V = 501.7 L

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

Moles

n = 290/4.003 = 72.45 mol

T = 60 + 273.15 = 333.1 K

Volume

V = nRT/P = 72.45·8.314·333.1/400 = 501.7 L

Answer

V = 501.7 L

2.Ideal gas law

Answer: V₂ = 1.757 L

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

Absolute temperatures

T₁ = 283.1 K, T₂ = 373.1 K

Combined gas law

V₂ = V₁·(P₁/P₂)·(T₂/T₁) = 5·(200/750)·(373.1/283.1) = 1.757 L

Answer

V₂ = 1.757 L

3.Energy balance

Answer: t_out = 123.4 °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 = 3·2.44·(160 − 80) = 585.6 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 30 + 585.6/(1.5·4.18) = 123.4 °C

Answer

t_out = 123.4 °C

4.Energy balance

Answer: ṁ_c = 0.886 kg/s

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

Heat released

Q_h = 2.5·2.44·(150 − 100) = 305 kW

Heat that reaches the water

Q_c = (1 − 0.15)·305 = 259.3 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 259.3/(4.18·70) = 0.886 kg/s

Answer

ṁ_c = 0.886 kg/s

5.Heat engines & refrigerators

Answer: T_L = 40.35 °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 = 400 + 273.15 = 673.1 K

Reversible heat ratio

T_L = T_H·Q_L/Q_H = 673.1·326/700 = 313.5 K = 40.35 °C

Answer

T_L = 40.35 °C

6.Heat engines & refrigerators

Answer: Ẇ_min = 0.1995 kW

Absolute temperatures

T_H = 25 + 273.15 = 298.1 K

T_L = -10 + 273.15 = 263.1 K

Best possible COP

COP_R,max = T_L/(T_H − T_L) = 263.1/(298.1 − 263.1) = 7.519

Minimum power

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

Ẇ_min = Q̇_L/COP = 1.5/7.519 = 0.1995 kW

Answer

Ẇ_min = 0.1995 kW

7.Gas power cycles

Answer: q_in = 979 kJ/kg

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)).

Temperatures

T₂ = 300·19^0.4 = 974.1 K

T₃ = 2·974.1 = 1948 K

Heat added at constant pressure

q_in = 1.005·(1948 − 974.1) = 979 kJ/kg

Answer

q_in = 979 kJ/kg

8.Gas power cycles

Answer: η = 54.71 %

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 − 1/16^0.2857 = 1 − 1/2.208 = 54.71%

Answer

η = 54.71%

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