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 430 g of Ar occupy at 650 kPa and 100 °C? Treat it as an ideal gas.

m = 430 gM = 39.95 g/molP = 650 kPaT = 100 °CR = 8.314 J/(mol·K)

V = L

2.Ideal gas lawHard

Find the density ρ of Ar at 550 kPa and 140 °C, treating it as an ideal gas.

M = 39.95 g/molP = 550 kPaT = 140 °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 = 2 kg/sc_p,h = 3.6 kJ/(kg·K)T_h: 160 → 110 °Cṁ_c = 3 kg/sc_p,c = 4.18 kJ/(kg·K)t_in = 25 °C

t_out = °C

4.Energy balanceHard

Hot glycol mix heats water in an exchanger that loses 10% 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: 160 → 70 °Ct: 15 → 75 °Cc_p,c = 4.18 kJ/(kg·K)loss = 10 %

ṁ_c = kg/s

5.Heat engines & refrigeratorsMedium

A reversible (Carnot) heat engine takes in 1600 kJ of heat from a reservoir at 725 °C and rejects 478 kJ per cycle. Find the temperature T_L of the cold reservoir in °C.

T_H = 725 °CQ_H = 1600 kJQ_L = 478 kJ

T_L = °C

6.Heat engines & refrigeratorsHard

A heat pump delivers 7.5 kW of heat to a heating system at 35 °C, taking heat from a source at -5 °C. Its actual COP is 35% of the Carnot COP between these temperatures. Find the electrical power Ẇ it uses.

T_H = 35 °CT_L = -5 °Cη_II = 35%Q̇_H = 7.5 kW

Ẇ = kW

7.Gas power cyclesMedium

An ideal Diesel cycle has a compression ratio r = 14 and a cutoff ratio r_c = v₃/v₂ = 3. 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 its thermal efficiency η in percent.

r = 14r_c = 3k = 1.4

η = %

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 310 K and the turbine at 1400 K. The compressor has an isentropic efficiency of 88% and the turbine 85%. 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 = 310 KT₃ = 1400 Kη_c = 88%η_t = 85%k = 1.4c_p = 1.005 kJ/(kg·K)

w_net = kJ/kg

3Solutions

1.Ideal gas law

Answer: V = 51.37 L

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

Moles

n = 430/39.95 = 10.76 mol

T = 100 + 273.15 = 373.1 K

Volume

V = nRT/P = 10.76·8.314·373.1/650 = 51.37 L

Answer

V = 51.37 L

2.Ideal gas law

Answer: ρ = 6.397 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)

T = 140 + 273.15 = 413.1 K

Density

ρ = 550·39.95/(8.314·413.1) = 6.397 kg/m³

Answer

ρ = 6.397 kg/m³

3.Energy balance

Answer: t_out = 53.71 °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·(160 − 110) = 360 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 25 + 360/(3·4.18) = 53.71 °C

Answer

t_out = 53.71 °C

4.Energy balance

Answer: ṁ_c = 1.163 kg/s

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

Heat released

Q_h = 1·3.6·(160 − 70) = 324 kW

Heat that reaches the water

Q_c = (1 − 0.1)·324 = 291.6 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 291.6/(4.18·60) = 1.163 kg/s

Answer

ṁ_c = 1.163 kg/s

5.Heat engines & refrigerators

Answer: T_L = 25.05 °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 = 725 + 273.15 = 998.1 K

Reversible heat ratio

T_L = T_H·Q_L/Q_H = 998.1·478/1600 = 298.2 K = 25.05 °C

Answer

T_L = 25.05 °C

6.Heat engines & refrigerators

Answer: Ẇ = 2.782 kW

Absolute temperatures

T_H = 35 + 273.15 = 308.1 K

T_L = -5 + 273.15 = 268.1 K

Carnot and actual COP

COP_max = 308.1/(308.1 − 268.1) = 7.704

COP = 0.35·7.704 = 2.696

Power input

Ẇ = Q̇_H/COP = 7.5/2.696 = 2.782 kW

Answer

Ẇ = 2.782 kW

7.Gas power cycles

Answer: η = 54.57 %

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)).

Cutoff term

(3^1.4 − 1)/(1.4·(3 − 1)) = 1.306

Thermal efficiency

η = 1 − 0.348·1.306 = 54.57%

Answer

η = 54.57%

8.Gas power cycles

Answer: w_net = 248.5 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

T_2s = 310·8^0.2857 = 561.5 K

T_4s = 1400/8^0.2857 = 772.9 K

Compressor and turbine work

w_c = 1.005·(561.5 − 310)/0.88 = 287.3 kJ/kg

w_t = 0.85·1.005·(1400 − 772.9) = 535.7 kJ/kg

Net work

w_net = 535.7 − 287.3 = 248.5 kJ/kg

Answer

w_net = 248.5 kJ/kg

Back-work ratio: w_c/w_t = 53.62%.

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