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 370 g of N₂ at 60 °C. Treat it as an ideal gas and find the pressure P.

m = 370 gM = 28.01 g/molV = 5 LT = 60 °CR = 8.314 J/(mol·K)

P = kPa

2.Ideal gas lawHard

A fixed amount of Ar goes from 400 kPa, 20 L, 15 °C to 200 kPa and 100 °C. Find the new volume V₂.

P₁ = 400 kPaV₁ = 20 LT₁ = 15 °CP₂ = 200 kPaT₂ = 100 °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: 120 → 90 °Cṁ_c = 0.5 kg/sc_p,c = 4.18 kJ/(kg·K)t_in = 15 °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 = 3 kg/sc_p,h = 3.6 kJ/(kg·K)T_h: 160 → 80 °Ct: 25 → 90 °Cc_p,c = 4.18 kJ/(kg·K)loss = 15 %

ṁ_c = kg/s

5.Heat engines & refrigeratorsMedium

A heat engine receives heat from a source at 400 °C and rejects heat to the surroundings at 35 °C. Its actual thermal efficiency is 41%. Find its second-law efficiency η_II, the ratio of the actual efficiency to the Carnot efficiency.

T_H = 400 °CT_L = 35 °Cη = 41%

η_II = %

6.Heat engines & refrigeratorsHard

A heat pump keeps a house at 22 °C while it is -10 °C outside, delivering 15 kW of heat to the house. Find the highest possible coefficient of performance COP_max.

T_H = 22 °CT_L = -10 °CQ̇_H = 15 kW

COP_max =

7.Gas power cyclesMedium

An ideal Diesel cycle has a compression ratio r = 18 and a cutoff ratio r_c = v₃/v₂ = 3. At the start of compression the air is at 305 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 temperature T₃ at the end of heat addition, in K.

r = 18r_c = 3T₁ = 305 Kk = 1.4

T₃ = K

8.Gas power cyclesHard

A gas-turbine plant runs on a Brayton cycle with a compressor pressure ratio of 10. Air enters the compressor at 310 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 = 10T₁ = 310 KT₃ = 1200 Kk = 1.4c_p = 1.005 kJ/(kg·K)

w_net = kJ/kg

3Solutions

1.Ideal gas law

Answer: P = 7318 kPa

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

Moles

n = m/M = 370/28.01 = 13.21 mol

T = 60 + 273.15 = 333.1 K

Pressure

P = nRT/V = 13.21·8.314·333.1/0.005 = 7318 kPa

Answer

P = 7318 kPa

2.Ideal gas law

Answer: V₂ = 51.8 L

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

Absolute temperatures

T₁ = 288.1 K, T₂ = 373.1 K

Combined gas law

V₂ = V₁·(P₁/P₂)·(T₂/T₁) = 20·(400/200)·(373.1/288.1) = 51.8 L

Answer

V₂ = 51.8 L

3.Energy balance

Answer: t_out = 118.3 °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·(120 − 90) = 216 kW

Heat picked up by the water

t_out = t_in + Q/(ṁ_cc_p,c) = 15 + 216/(0.5·4.18) = 118.3 °C

Answer

t_out = 118.3 °C

4.Energy balance

Answer: ṁ_c = 2.703 kg/s

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

Heat released

Q_h = 3·3.6·(160 − 80) = 864 kW

Heat that reaches the water

Q_c = (1 − 0.15)·864 = 734.4 kW

Water flow rate

ṁ_c = Q_c/(c_p,c·Δt) = 734.4/(4.18·65) = 2.703 kg/s

Answer

ṁ_c = 2.703 kg/s

5.Heat engines & refrigerators

Answer: η_II = 75.61 %

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 = 400 + 273.15 = 673.1 K

T_L = 35 + 273.15 = 308.1 K

Carnot efficiency

η_Carnot = 1 − 308.1/673.1 = 54.22 %

Second-law efficiency

η_II = η/η_Carnot = 41/54.22 = 75.61 %

Answer

η_II = 75.61 %

6.Heat engines & refrigerators

Answer: COP_max = 9.223

Absolute temperatures

T_H = 22 + 273.15 = 295.1 K

T_L = -10 + 273.15 = 263.1 K

Best possible COP

COP_HP,max = T_H/(T_H − T_L) = 295.1/(295.1 − 263.1) = 9.223

Answer

COP_max = 9.223

7.Gas power cycles

Answer: T₃ = 2908 K

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)).

End of compression

T₂ = 305·18^0.4 = 969.2 K

Heat addition at constant pressure

T₃ = r_c·T₂ = 3·969.2 = 2908 K

Answer

T₃ = 2908 K

8.Gas power cycles

Answer: w_net = 291.4 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 = 310·10^0.2857 = 598.5 K

T₄ = 1200/10^0.2857 = 621.5 K

Compressor and turbine work

w_c = 1.005·(598.5 − 310) = 290 kJ/kg

w_t = 1.005·(1200 − 621.5) = 581.4 kJ/kg

Net work

w_net = 581.4 − 290 = 291.4 kJ/kg

Answer

w_net = 291.4 kJ/kg

Back-work ratio: w_c/w_t = 49.88%.

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