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

What volume V does 320 g of He occupy at 600 kPa and 80 °C? Treat it as an ideal gas.

m = 320 gM = 4.003 g/molP = 600 kPaT = 80 °CR = 8.314 J/(mol·K)

V = L

2.Ideal gas lawHard

A fixed amount of O₂ goes from 300 kPa, 20 L, 15 °C to 150 kPa and 90 °C. Find the new volume V₂.

P₁ = 300 kPaV₁ = 20 LT₁ = 15 °CP₂ = 150 kPaT₂ = 90 °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 = 2 kg/sc_p,h = 3.6 kJ/(kg·K)T_h: 140 → 60 °Cṁ_c = 1.5 kg/sc_p,c = 4.18 kJ/(kg·K)t_in = 15 °C

t_out = °C

4.Energy balanceHard

Hot light oil 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 = 2.1 kJ/(kg·K)T_h: 150 → 80 °Ct: 30 → 70 °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 2 °C by removing 4.5 kW of heat, and rejects heat to the surroundings at 30 °C. Find the rate Q̇_H at which it rejects heat.

T_L = 2 °CT_H = 30 °CQ̇_L = 4.5 kW

Q̇_H = kW

6.Heat engines & refrigeratorsHard

A heat pump keeps a house at 21 °C while it is -10 °C outside, delivering 5 kW of heat to the house. Find the minimum power input Ẇ_min.

T_H = 21 °CT_L = -10 °CQ̇_H = 5 kW

Ẇ_min = kW

7.Gas power cyclesMedium

An ideal Diesel cycle has a compression ratio r = 16 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 = 16r_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 8. Air enters the compressor at 300 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 = 8T₁ = 300 KT₃ = 1500 Kk = 1.4c_p = 1.005 kJ/(kg·K)

w_net = kJ/kg

3Solutions

1.Ideal gas law

Answer: V = 391.2 L

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

Moles

n = 320/4.003 = 79.94 mol

T = 80 + 273.15 = 353.1 K

Volume

V = nRT/P = 79.94·8.314·353.1/600 = 391.2 L

Answer

V = 391.2 L

2.Ideal gas law

Answer: V₂ = 50.41 L

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

Absolute temperatures

T₁ = 288.1 K, T₂ = 363.1 K

Combined gas law

V₂ = V₁·(P₁/P₂)·(T₂/T₁) = 20·(300/150)·(363.1/288.1) = 50.41 L

Answer

V₂ = 50.41 L

3.Energy balance

Answer: t_out = 106.9 °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 = 2·3.6·(140 − 60) = 576 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 15 + 576/(1.5·4.18) = 106.9 °C

Answer

t_out = 106.9 °C

4.Energy balance

Answer: ṁ_c = 1.253 kg/s

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

Heat released

Q_h = 1.5·2.1·(150 − 80) = 220.5 kW

Heat that reaches the water

Q_c = (1 − 0.05)·220.5 = 209.5 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 209.5/(4.18·40) = 1.253 kg/s

Answer

ṁ_c = 1.253 kg/s

5.Heat engines & refrigerators

Answer: Q̇_H = 4.958 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 = 2 + 273.15 = 275.1 K

Reversible heat ratio

Q̇_H = Q̇_L·T_H/T_L = 4.5·303.1/275.1 = 4.958 kW

Answer

Q̇_H = 4.958 kW

Power input: Ẇ = 0.4579 kW.

6.Heat engines & refrigerators

Answer: Ẇ_min = 0.5269 kW

Absolute temperatures

T_H = 21 + 273.15 = 294.1 K

T_L = -10 + 273.15 = 263.1 K

Best possible COP

COP_HP,max = T_H/(T_H − T_L) = 294.1/(294.1 − 263.1) = 9.489

Minimum power

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

Ẇ_min = Q̇_H/COP = 5/9.489 = 0.5269 kW

Answer

Ẇ_min = 0.5269 kW

7.Gas power cycles

Answer: q_in = 914 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 = 300·16^0.4 = 909.4 K

T₃ = 2·909.4 = 1819 K

Heat added at constant pressure

q_in = 1.005·(1819 − 909.4) = 914 kJ/kg

Answer

q_in = 914 kJ/kg

8.Gas power cycles

Answer: w_net = 430.6 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 = 300·8^0.2857 = 543.4 K

T₄ = 1500/8^0.2857 = 828.1 K

Compressor and turbine work

w_c = 1.005·(543.4 − 300) = 244.7 kJ/kg

w_t = 1.005·(1500 − 828.1) = 675.3 kJ/kg

Net work

w_net = 675.3 − 244.7 = 430.6 kJ/kg

Answer

w_net = 430.6 kJ/kg

Back-work ratio: w_c/w_t = 36.23%.

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