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 290 g of CH₄ at 45 °C. Treat it as an ideal gas and find the pressure P.

m = 290 gM = 16.04 g/molV = 10 LT = 45 °CR = 8.314 J/(mol·K)

P = kPa

2.Ideal gas lawHard

A fixed amount of Ar goes from 100 kPa, 5 L, 35 °C to 100 kPa and 170 °C. Find the new volume V₂.

P₁ = 100 kPaV₁ = 5 LT₁ = 35 °CP₂ = 100 kPaT₂ = 170 °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: 160 → 80 °Cṁ_c = 0.5 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 = 2 kg/sc_p,h = 3.6 kJ/(kg·K)T_h: 180 → 110 °Ct: 15 → 80 °Cc_p,c = 4.18 kJ/(kg·K)loss = 5 %

ṁ_c = kg/s

5.Heat engines & refrigeratorsMedium

A reversible refrigerator keeps a cold room at -18 °C by removing 3 kW of heat, and rejects heat to the surroundings at 30 °C. Find the rate Q̇_H at which it rejects heat.

T_L = -18 °CT_H = 30 °CQ̇_L = 3 kW

Q̇_H = kW

6.Heat engines & refrigeratorsHard

A heat engine takes in 900 kJ of heat from a reservoir at 800 °C, produces 315 kJ of work and rejects the rest to a reservoir at 40 °C, per cycle. Find the total entropy generated S_gen per cycle.

T_H = 800 °CT_L = 40 °CQ_H = 900 kJW = 315 kJ

S_gen = kJ/K

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 295 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 = 295 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 15. Air enters the compressor at 295 K and the turbine at 1200 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 = 15T₁ = 295 KT₃ = 1200 Kk = 1.4c_p = 1.005 kJ/(kg·K)

w_net = kJ/kg

3Solutions

1.Ideal gas law

Answer: P = 4782 kPa

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

Moles

n = m/M = 290/16.04 = 18.08 mol

T = 45 + 273.15 = 318.1 K

Pressure

P = nRT/V = 18.08·8.314·318.1/0.01 = 4782 kPa

Answer

P = 4782 kPa

2.Ideal gas law

Answer: V₂ = 7.19 L

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

Absolute temperatures

T₁ = 308.1 K, T₂ = 443.1 K

Combined gas law

V₂ = V₁·(P₁/P₂)·(T₂/T₁) = 5·(100/100)·(443.1/308.1) = 7.19 L

Answer

V₂ = 7.19 L

3.Energy balance

Answer: t_out = 140.6 °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·(160 − 80) = 252 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 20 + 252/(0.5·4.18) = 140.6 °C

Answer

t_out = 140.6 °C

4.Energy balance

Answer: ṁ_c = 1.762 kg/s

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

Heat released

Q_h = 2·3.6·(180 − 110) = 504 kW

Heat that reaches the water

Q_c = (1 − 0.05)·504 = 478.8 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 478.8/(4.18·65) = 1.762 kg/s

Answer

ṁ_c = 1.762 kg/s

5.Heat engines & refrigerators

Answer: Q̇_H = 3.564 kW

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 = 30 + 273.15 = 303.1 K

T_L = -18 + 273.15 = 255.1 K

Reversible heat ratio

Q̇_H = Q̇_L·T_H/T_L = 3·303.1/255.1 = 3.564 kW

Answer

Q̇_H = 3.564 kW

Power input: Ẇ = 0.5644 kW.

6.Heat engines & refrigerators

Answer: S_gen = 1.029 kJ/K

Entropy generation: over a full cycle the working fluid has no net entropy change. The hot reservoir loses Q_H/T_H and the cold reservoir gains Q_L/T_L. Their sum is the entropy generated, which is zero for a reversible cycle and positive for every real one.

Heat rejected

Q_L = Q_H − W = 900 − 315 = 585 kJ

Absolute temperatures

T_H = 800 + 273.15 = 1073 K

T_L = 40 + 273.15 = 313.1 K

Entropy changes of the reservoirs

ΔS_H = −900/1073 = −0.8387 kJ/K

ΔS_L = 585/313.1 = 1.868 kJ/K

S_gen = 1.868 − 0.8387 = 1.029 kJ/K

Answer

S_gen = 1.029 kJ/K

S_gen > 0, so the engine is possible but irreversible.

7.Gas power cycles

Answer: q_in = 982.7 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 = 295·20^0.4 = 977.8 K

T₃ = 2·977.8 = 1956 K

Heat added at constant pressure

q_in = 1.005·(1956 − 977.8) = 982.7 kJ/kg

Answer

q_in = 982.7 kJ/kg

8.Gas power cycles

Answer: w_net = 303.5 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 = 295·15^0.2857 = 639.5 K

T₄ = 1200/15^0.2857 = 553.5 K

Compressor and turbine work

w_c = 1.005·(639.5 − 295) = 346.2 kJ/kg

w_t = 1.005·(1200 − 553.5) = 649.7 kJ/kg

Net work

w_net = 649.7 − 346.2 = 303.5 kJ/kg

Answer

w_net = 303.5 kJ/kg

Back-work ratio: w_c/w_t = 53.29%.

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