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 5 L contains 470 g of CO₂ at 95 °C. Treat it as an ideal gas and find the pressure P.

m = 470 gM = 44.01 g/molV = 5 LT = 95 °CR = 8.314 J/(mol·K)

P = kPa

2.Ideal gas lawHard

A fixed amount of N₂ goes from 200 kPa, 2 L, 20 °C to 700 kPa and 90 °C. Find the new volume V₂.

P₁ = 200 kPaV₁ = 2 LT₁ = 20 °CP₂ = 700 kPaT₂ = 90 °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 = 3 kg/sc_p,h = 2.1 kJ/(kg·K)T_h: 130 → 70 °Cṁ_c = 2 kg/sc_p,c = 4.18 kJ/(kg·K)t_in = 10 °C

t_out = °C

4.Energy balanceHard

Hot glycol mix 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 = 3.6 kJ/(kg·K)T_h: 120 → 80 °Ct: 25 → 80 °Cc_p,c = 4.18 kJ/(kg·K)loss = 15 %

ṁ_c = kg/s

5.Heat engines & refrigeratorsMedium

A heat engine works between a hot reservoir at 525 °C and a cold reservoir at 15 °C. It takes in 1000 kJ of heat. Find the most work W_max it could possibly produce.

T_H = 525 °CT_L = 15 °CQ_H = 1000 kJ

W_max = kJ

6.Heat engines & refrigeratorsHard

A refrigerator keeps its cold space at -25 °C in a kitchen at 32 °C, removing heat from the cold space at 4 kW. Find the minimum power input Ẇ_min.

T_H = 32 °CT_L = -25 °CQ̇_L = 4 kW

Ẇ_min = kW

7.Gas power cyclesMedium

An ideal Diesel cycle has a compression ratio r = 22 and a cutoff ratio r_c = v₃/v₂ = 1.5. 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 = 22r_c = 1.5T₁ = 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 12. Air enters the compressor at 290 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 = 12T₁ = 290 KT₃ = 1300 Kk = 1.4c_p = 1.005 kJ/(kg·K)

w_net = kJ/kg

3Solutions

1.Ideal gas law

Answer: P = 6537 kPa

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

Moles

n = m/M = 470/44.01 = 10.68 mol

T = 95 + 273.15 = 368.1 K

Pressure

P = nRT/V = 10.68·8.314·368.1/0.005 = 6537 kPa

Answer

P = 6537 kPa

2.Ideal gas law

Answer: V₂ = 0.7079 L

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

Absolute temperatures

T₁ = 293.1 K, T₂ = 363.1 K

Combined gas law

V₂ = V₁·(P₁/P₂)·(T₂/T₁) = 2·(200/700)·(363.1/293.1) = 0.7079 L

Answer

V₂ = 0.7079 L

3.Energy balance

Answer: t_out = 55.22 °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 = 3·2.1·(130 − 70) = 378 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 10 + 378/(2·4.18) = 55.22 °C

Answer

t_out = 55.22 °C

4.Energy balance

Answer: ṁ_c = 1.331 kg/s

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

Heat released

Q_h = 2.5·3.6·(120 − 80) = 360 kW

Heat that reaches the water

Q_c = (1 − 0.15)·360 = 306 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 306/(4.18·55) = 1.331 kg/s

Answer

ṁ_c = 1.331 kg/s

5.Heat engines & refrigerators

Answer: W_max = 639 kJ

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 = 525 + 273.15 = 798.1 K

T_L = 15 + 273.15 = 288.1 K

Carnot efficiency

η_max = 1 − T_L/T_H = 1 − 288.1/798.1 = 63.9 %

Maximum work

W_max = η_max·Q_H = 0.639·1000 = 639 kJ

Answer

W_max = 639 kJ

6.Heat engines & refrigerators

Answer: Ẇ_min = 0.9188 kW

Absolute temperatures

T_H = 32 + 273.15 = 305.1 K

T_L = -25 + 273.15 = 248.1 K

Best possible COP

COP_R,max = T_L/(T_H − T_L) = 248.1/(305.1 − 248.1) = 4.354

Minimum power

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

Ẇ_min = Q̇_L/COP = 4/4.354 = 0.9188 kW

Answer

Ẇ_min = 0.9188 kW

7.Gas power cycles

Answer: q_in = 519.1 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 = 300·22^0.4 = 1033 K

T₃ = 1.5·1033 = 1549 K

Heat added at constant pressure

q_in = 1.005·(1549 − 1033) = 519.1 kJ/kg

Answer

q_in = 519.1 kJ/kg

8.Gas power cycles

Answer: w_net = 362.8 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 = 290·12^0.2857 = 589.8 K

T₄ = 1300/12^0.2857 = 639.2 K

Compressor and turbine work

w_c = 1.005·(589.8 − 290) = 301.3 kJ/kg

w_t = 1.005·(1300 − 639.2) = 664.1 kJ/kg

Net work

w_net = 664.1 − 301.3 = 362.8 kJ/kg

Answer

w_net = 362.8 kJ/kg

Back-work ratio: w_c/w_t = 45.37%.

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