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

A tank of 10 L contains 400 g of O₂ at 40 °C. Treat it as an ideal gas and find the pressure P.

m = 400 gM = 32 g/molV = 10 LT = 40 °CR = 8.314 J/(mol·K)

P = kPa

2.Ideal gas lawHard

Find the density ρ of Ar at 400 kPa and 180 °C, treating it as an ideal gas.

M = 39.95 g/molP = 400 kPaT = 180 °CR = 8.314 J/(mol·K)

ρ = kg/m³

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 = 1.5 kg/sc_p,h = 3.6 kJ/(kg·K)T_h: 160 → 80 °Cṁ_c = 1 kg/sc_p,c = 4.18 kJ/(kg·K)t_in = 25 °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 = 1.5 kg/sc_p,h = 2.44 kJ/(kg·K)T_h: 180 → 150 °Ct: 15 → 35 °Cc_p,c = 4.18 kJ/(kg·K)loss = 5 %

ṁ_c = kg/s

5.Heat engines & refrigeratorsMedium

A heat engine receives heat from a source at 750 °C and rejects heat to the surroundings at 10 °C. Its actual thermal efficiency is 54%. Find its second-law efficiency η_II, the ratio of the actual efficiency to the Carnot efficiency.

T_H = 750 °CT_L = 10 °Cη = 54%

η_II = %

6.Heat engines & refrigeratorsHard

A Carnot heat engine receives 200 kW of heat from a source at 650 °C and rejects heat to the surroundings at 27 °C. All of its work drives a Carnot refrigerator that removes heat from a freezer at -20 °C and also rejects heat to the surroundings at 27 °C. Find the rate Q̇_L at which heat is removed from the freezer.

T_H = 650 °CT₀ = 27 °CT_L = -20 °CQ̇_H = 200 kW

Q̇_L = kW

7.Gas power cyclesMedium

An ideal Diesel cycle has a compression ratio r = 15 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 = 15r_c = 2T₁ = 300 Kk = 1.4c_p = 1.005 kJ/(kg·K)

q_in = kJ/kg

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 1500 K. Both machines are isentropic. 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 = 1500 Kk = 1.4c_p = 1.005 kJ/(kg·K)

w_net = kJ/kg

3Solutions

1.Ideal gas law

Answer: P = 3254 kPa

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

Moles

n = m/M = 400/32 = 12.5 mol

T = 40 + 273.15 = 313.1 K

Pressure

P = nRT/V = 12.5·8.314·313.1/0.01 = 3254 kPa

Answer

P = 3254 kPa

2.Ideal gas law

Answer: ρ = 4.242 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)

T = 180 + 273.15 = 453.1 K

Density

ρ = 400·39.95/(8.314·453.1) = 4.242 kg/m³

Answer

ρ = 4.242 kg/m³

3.Energy balance

Answer: t_out = 128.3 °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 = 1.5·3.6·(160 − 80) = 432 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 25 + 432/(1·4.18) = 128.3 °C

Answer

t_out = 128.3 °C

4.Energy balance

Answer: ṁ_c = 1.248 kg/s

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

Heat released

Q_h = 1.5·2.44·(180 − 150) = 109.8 kW

Heat that reaches the water

Q_c = (1 − 0.05)·109.8 = 104.3 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 104.3/(4.18·20) = 1.248 kg/s

Answer

ṁ_c = 1.248 kg/s

5.Heat engines & refrigerators

Answer: η_II = 74.66 %

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 = 10 + 273.15 = 283.1 K

Carnot efficiency

η_Carnot = 1 − 283.1/1023 = 72.33 %

Second-law efficiency

η_II = η/η_Carnot = 54/72.33 = 74.66 %

Answer

η_II = 74.66 %

6.Heat engines & refrigerators

Answer: Q̇_L = 727 kW

Absolute temperatures

T_H = 650 + 273.15 = 923.1 K

T₀ = 27 + 273.15 = 300.1 K

T_L = -20 + 273.15 = 253.1 K

Engine work

η = 1 − 300.1/923.1 = 0.6749

Ẇ = 0.6749·200 = 135 kW

Heat removed by the refrigerator

COP_R = 253.1/(300.1 − 253.1) = 5.386

Q̇_L = 5.386·135 = 727 kW

Answer

Q̇_L = 727 kW

7.Gas power cycles

Answer: q_in = 890.7 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 = 300·15^0.4 = 886.3 K

T₃ = 2·886.3 = 1773 K

Heat added at constant pressure

q_in = 1.005·(1773 − 886.3) = 890.7 kJ/kg

Answer

q_in = 890.7 kJ/kg

8.Gas power cycles

Answer: w_net = 455.4 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

T₂ = 290·10^0.2857 = 559.9 K

T₄ = 1500/10^0.2857 = 776.9 K

Compressor and turbine work

w_c = 1.005·(559.9 − 290) = 271.3 kJ/kg

w_t = 1.005·(1500 − 776.9) = 726.7 kJ/kg

Net work

w_net = 726.7 − 271.3 = 455.4 kJ/kg

Answer

w_net = 455.4 kJ/kg

Back-work ratio: w_c/w_t = 37.33%.

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