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 370 g of CH₄ occupy at 150 kPa and 100 °C? Treat it as an ideal gas.

m = 370 gM = 16.04 g/molP = 150 kPaT = 100 °CR = 8.314 J/(mol·K)

V = L

2.Ideal gas lawHard

Find the density ρ of O₂ at 150 kPa and 180 °C, treating it as an ideal gas.

M = 32 g/molP = 150 kPaT = 180 °CR = 8.314 J/(mol·K)

ρ = kg/m³

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: 120 → 80 °Cṁ_c = 3 kg/sc_p,c = 4.18 kJ/(kg·K)t_in = 25 °C

t_out = °C

4.Energy balanceHard

Hot ethanol 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 = 1 kg/sc_p,h = 2.44 kJ/(kg·K)T_h: 130 → 60 °Ct: 15 → 65 °Cc_p,c = 4.18 kJ/(kg·K)loss = 15 %

ṁ_c = kg/s

5.Heat engines & refrigeratorsMedium

A reversible (Carnot) heat engine takes in 2000 kJ of heat from a reservoir at 800 °C and rejects 537 kJ per cycle. Find the temperature T_L of the cold reservoir in °C.

T_H = 800 °CQ_H = 2000 kJQ_L = 537 kJ

T_L = °C

6.Heat engines & refrigeratorsHard

Two Carnot engines work in series. Engine A receives heat from a reservoir at 1000 °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 30 °C. Both engines produce the same work. Find T_m in °C.

T_H = 1000 °CT_L = 30 °C

T_m = °C

7.Gas power cyclesMedium

An ideal Diesel cycle has a compression ratio r = 14 and a cutoff ratio r_c = v₃/v₂ = 2.2. 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 = 14r_c = 2.2T₁ = 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 12. 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 = 12T₁ = 300 KT₃ = 1500 Kk = 1.4c_p = 1.005 kJ/(kg·K)

w_net = kJ/kg

3Solutions

1.Ideal gas law

Answer: V = 477.1 L

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

Moles

n = 370/16.04 = 23.07 mol

T = 100 + 273.15 = 373.1 K

Volume

V = nRT/P = 23.07·8.314·373.1/150 = 477.1 L

Answer

V = 477.1 L

2.Ideal gas law

Answer: ρ = 1.274 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)

T = 180 + 273.15 = 453.1 K

Density

ρ = 150·32/(8.314·453.1) = 1.274 kg/m³

Answer

ρ = 1.274 kg/m³

3.Energy balance

Answer: t_out = 30.74 °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·(120 − 80) = 72 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 25 + 72/(3·4.18) = 30.74 °C

Answer

t_out = 30.74 °C

4.Energy balance

Answer: ṁ_c = 0.6946 kg/s

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

Heat released

Q_h = 1·2.44·(130 − 60) = 170.8 kW

Heat that reaches the water

Q_c = (1 − 0.15)·170.8 = 145.2 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 145.2/(4.18·50) = 0.6946 kg/s

Answer

ṁ_c = 0.6946 kg/s

5.Heat engines & refrigerators

Answer: T_L = 14.99 °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 = 800 + 273.15 = 1073 K

Reversible heat ratio

T_L = T_H·Q_L/Q_H = 1073·537/2000 = 288.1 K = 14.99 °C

Answer

T_L = 14.99 °C

6.Heat engines & refrigerators

Answer: T_m = 515 °C

Absolute temperatures

T_H = 1000 + 273.15 = 1273 K

T_L = 30 + 273.15 = 303.1 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 = (1273 + 303.1)/2 = 788.2 K = 515 °C

Answer

T_m = 515 °C

7.Gas power cycles

Answer: q_in = 1040 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 = 300·14^0.4 = 862.1 K

T₃ = 2.2·862.1 = 1897 K

Heat added at constant pressure

q_in = 1.005·(1897 − 862.1) = 1040 kJ/kg

Answer

q_in = 1040 kJ/kg

8.Gas power cycles

Answer: w_net = 454.6 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 = 300·12^0.2857 = 610.2 K

T₄ = 1500/12^0.2857 = 737.5 K

Compressor and turbine work

w_c = 1.005·(610.2 − 300) = 311.7 kJ/kg

w_t = 1.005·(1500 − 737.5) = 766.3 kJ/kg

Net work

w_net = 766.3 − 311.7 = 454.6 kJ/kg

Answer

w_net = 454.6 kJ/kg

Back-work ratio: w_c/w_t = 40.68%.

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