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 340 g of CH₄ at 15 °C. Treat it as an ideal gas and find the pressure P.

m = 340 gM = 16.04 g/molV = 20 LT = 15 °CR = 8.314 J/(mol·K)

P = kPa

2.Ideal gas lawHard

A fixed amount of N₂ goes from 100 kPa, 5 L, 10 °C to 500 kPa and 100 °C. Find the new volume V₂.

P₁ = 100 kPaV₁ = 5 LT₁ = 10 °CP₂ = 500 kPaT₂ = 100 °C

V₂ = L

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 = 1.5 kg/sc_p,h = 2.1 kJ/(kg·K)T_h: 140 → 100 °Cṁ_c = 1 kg/sc_p,c = 4.18 kJ/(kg·K)t_in = 20 °C

t_out = °C

4.Energy balanceHard

Hot glycol mix 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 kg/sc_p,h = 3.6 kJ/(kg·K)T_h: 120 → 90 °Ct: 20 → 40 °Cc_p,c = 4.18 kJ/(kg·K)loss = 5 %

ṁ_c = kg/s

5.Heat engines & refrigeratorsMedium

A reversible (Carnot) heat engine takes in 3100 kJ of heat from a reservoir at 775 °C and rejects 823 kJ per cycle. Find the temperature T_L of the cold reservoir in °C.

T_H = 775 °CQ_H = 3100 kJQ_L = 823 kJ

T_L = °C

6.Heat engines & refrigeratorsHard

Two Carnot engines work in series. Engine A receives heat from a reservoir at 1200 °C and rejects heat to an intermediate reservoir at temperature T_m. Engine B takes all that heat and rejects heat to a reservoir at 40 °C. Both engines produce the same work. Find T_m in °C.

T_H = 1200 °CT_L = 40 °C

T_m = °C

7.Gas power cyclesMedium

An ideal Diesel cycle has a compression ratio r = 18 and a cutoff ratio r_c = v₃/v₂ = 2.2. At the start of compression the air is at 290 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 = 18r_c = 2.2T₁ = 290 Kk = 1.4

T₃ = K

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 295 K and the turbine at 1300 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 = 16T₁ = 295 KT₃ = 1300 Kk = 1.4c_p = 1.005 kJ/(kg·K)

w_net = kJ/kg

3Solutions

1.Ideal gas law

Answer: P = 2539 kPa

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

Moles

n = m/M = 340/16.04 = 21.2 mol

T = 15 + 273.15 = 288.1 K

Pressure

P = nRT/V = 21.2·8.314·288.1/0.02 = 2539 kPa

Answer

P = 2539 kPa

2.Ideal gas law

Answer: V₂ = 1.318 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·(100/500)·(373.1/283.1) = 1.318 L

Answer

V₂ = 1.318 L

3.Energy balance

Answer: t_out = 50.14 °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 = 1.5·2.1·(140 − 100) = 126 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 20 + 126/(1·4.18) = 50.14 °C

Answer

t_out = 50.14 °C

4.Energy balance

Answer: ṁ_c = 1.227 kg/s

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

Heat released

Q_h = 1·3.6·(120 − 90) = 108 kW

Heat that reaches the water

Q_c = (1 − 0.05)·108 = 102.6 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 102.6/(4.18·20) = 1.227 kg/s

Answer

ṁ_c = 1.227 kg/s

5.Heat engines & refrigerators

Answer: T_L = 5.117 °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 = 775 + 273.15 = 1048 K

Reversible heat ratio

T_L = T_H·Q_L/Q_H = 1048·823/3100 = 278.3 K = 5.117 °C

Answer

T_L = 5.117 °C

6.Heat engines & refrigerators

Answer: T_m = 620 °C

Absolute temperatures

T_H = 1200 + 273.15 = 1473 K

T_L = 40 + 273.15 = 313.1 K

Equal work

W = Q·(1 − T_low/T_high) and Q/T is the same at every reservoir, so W is proportional to the temperature drop.

T_m = (1473 + 313.1)/2 = 893.2 K = 620 °C

Answer

T_m = 620 °C

7.Gas power cycles

Answer: T₃ = 2027 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 = 290·18^0.4 = 921.5 K

Heat addition at constant pressure

T₃ = r_c·T₂ = 2.2·921.5 = 2027 K

Answer

T₃ = 2027 K

8.Gas power cycles

Answer: w_net = 356.6 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 = 295·16^0.2857 = 651.4 K

T₄ = 1300/16^0.2857 = 588.7 K

Compressor and turbine work

w_c = 1.005·(651.4 − 295) = 358.2 kJ/kg

w_t = 1.005·(1300 − 588.7) = 714.8 kJ/kg

Net work

w_net = 714.8 − 358.2 = 356.6 kJ/kg

Answer

w_net = 356.6 kJ/kg

Back-work ratio: w_c/w_t = 50.11%.

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