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 20 L contains 230 g of CH₄ at 100 °C. Treat it as an ideal gas and find the pressure P.

m = 230 gM = 16.04 g/molV = 20 LT = 100 °CR = 8.314 J/(mol·K)

P = kPa

2.Ideal gas lawHard

A fixed amount of He goes from 150 kPa, 10 L, 20 °C to 600 kPa and 200 °C. Find the new volume V₂.

P₁ = 150 kPaV₁ = 10 LT₁ = 20 °CP₂ = 600 kPaT₂ = 200 °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: 170 → 80 °Cṁ_c = 1 kg/sc_p,c = 4.18 kJ/(kg·K)t_in = 30 °C

t_out = °C

4.Energy balanceHard

Hot light oil 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 = 3 kg/sc_p,h = 2.1 kJ/(kg·K)T_h: 160 → 120 °Ct: 20 → 90 °Cc_p,c = 4.18 kJ/(kg·K)loss = 15 %

ṁ_c = kg/s

5.Heat engines & refrigeratorsMedium

A heat engine receives heat from a source at 600 °C and rejects heat to the surroundings at 25 °C. Its actual thermal efficiency is 49%. Find its second-law efficiency η_II, the ratio of the actual efficiency to the Carnot efficiency.

T_H = 600 °CT_L = 25 °Cη = 49%

η_II = %

6.Heat engines & refrigeratorsHard

A heat pump keeps a house at 21 °C while it is 2 °C outside, delivering 13 kW of heat to the house. Find the minimum power input Ẇ_min.

T_H = 21 °CT_L = 2 °CQ̇_H = 13 kW

Ẇ_min = kW

7.Gas power cyclesMedium

An ideal Diesel cycle has a compression ratio r = 20 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 = 20r_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 16. Air enters the compressor at 300 K and the turbine at 1300 K. The compressor has an isentropic efficiency of 82% and the turbine 85%. 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 = 16T₁ = 300 KT₃ = 1300 Kη_c = 82%η_t = 85%k = 1.4c_p = 1.005 kJ/(kg·K)

w_net = kJ/kg

3Solutions

1.Ideal gas law

Answer: P = 2224 kPa

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

Moles

n = m/M = 230/16.04 = 14.34 mol

T = 100 + 273.15 = 373.1 K

Pressure

P = nRT/V = 14.34·8.314·373.1/0.02 = 2224 kPa

Answer

P = 2224 kPa

2.Ideal gas law

Answer: V₂ = 4.035 L

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

Absolute temperatures

T₁ = 293.1 K, T₂ = 473.1 K

Combined gas law

V₂ = V₁·(P₁/P₂)·(T₂/T₁) = 10·(150/600)·(473.1/293.1) = 4.035 L

Answer

V₂ = 4.035 L

3.Energy balance

Answer: t_out = 68.76 °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·(170 − 80) = 162 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 30 + 162/(1·4.18) = 68.76 °C

Answer

t_out = 68.76 °C

4.Energy balance

Answer: ṁ_c = 0.7321 kg/s

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

Heat released

Q_h = 3·2.1·(160 − 120) = 252 kW

Heat that reaches the water

Q_c = (1 − 0.15)·252 = 214.2 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 214.2/(4.18·70) = 0.7321 kg/s

Answer

ṁ_c = 0.7321 kg/s

5.Heat engines & refrigerators

Answer: η_II = 74.41 %

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 = 600 + 273.15 = 873.1 K

T_L = 25 + 273.15 = 298.1 K

Carnot efficiency

η_Carnot = 1 − 298.1/873.1 = 65.85 %

Second-law efficiency

η_II = η/η_Carnot = 49/65.85 = 74.41 %

Answer

η_II = 74.41 %

6.Heat engines & refrigerators

Answer: Ẇ_min = 0.8397 kW

Absolute temperatures

T_H = 21 + 273.15 = 294.1 K

T_L = 2 + 273.15 = 275.1 K

Best possible COP

COP_HP,max = T_H/(T_H − T_L) = 294.1/(294.1 − 275.1) = 15.48

Minimum power

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

Ẇ_min = Q̇_H/COP = 13/15.48 = 0.8397 kW

Answer

Ẇ_min = 0.8397 kW

7.Gas power cycles

Answer: q_in = 999.3 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 = 300·20^0.4 = 994.3 K

T₃ = 2·994.3 = 1989 K

Heat added at constant pressure

q_in = 1.005·(1989 − 994.3) = 999.3 kJ/kg

Answer

q_in = 999.3 kJ/kg

8.Gas power cycles

Answer: w_net = 163.4 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 = 300·16^0.2857 = 662.5 K

T_4s = 1300/16^0.2857 = 588.7 K

Compressor and turbine work

w_c = 1.005·(662.5 − 300)/0.82 = 444.2 kJ/kg

w_t = 0.85·1.005·(1300 − 588.7) = 607.6 kJ/kg

Net work

w_net = 607.6 − 444.2 = 163.4 kJ/kg

Answer

w_net = 163.4 kJ/kg

Back-work ratio: w_c/w_t = 73.11%.

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