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·cp·ΔT
Latent heat
Q = ṁ·hfg
Mixing
ΣṁcpTin = ΣṁcpTout
Water
cp ≈ 4.18 kJ/(kg·K)

Heat engines & refrigerators

Efficiency
η = W/QH = 1 − QC/QH
Carnot efficiency
ηC = 1 − TC/TH (kelvin)
Refrigerator COP
COPR = QC/W
Heat pump COP
COPHP = QH/W = COPR + 1
Energy balance
QH = W + QC

Gas power cycles

Isentropic, volume ratio
T2/T1 = rk−1
Isentropic, pressure ratio
T2/T1 = rp(k−1)/k
Otto efficiency
η = 1 − 1/rk−1
Diesel efficiency
η = 1 − (1/rk−1)·(rck − 1)/(k(rc − 1))
Brayton efficiency
η = 1 − 1/rp(k−1)/k
Heat added
Otto: cv(T3 − T2) Diesel, Brayton: cp(T3 − T2)
Brayton net work
w = cp[(T3 − T4) − (T2 − T1)]
Isentropic efficiencies
wc = wc,s/ηc, wt = ηt·wt,s
Cold air
k = 1.4, cp = 1.005, cv = 0.718 kJ/(kg·K)

2Practice problems

1.Ideal gas lawMedium
70 g of O₂P = 550 kPaT = 25 °CV = ?

What volume V does 70 g of O₂ occupy at 550 kPa and 25 °C? Treat it as an ideal gas.

m = 70 gM = 32 g/molP = 550 kPaT = 25 °CR = 8.314 J/(mol·K)
V = L
2.Ideal gas lawHard
Ar, M = 39.95 g/molP = 400 kPaT = 150 °Cρ = ?

Find the density ρ of Ar at 400 kPa and 150 °C, treating it as an ideal gas.

M = 39.95 g/molP = 400 kPaT = 150 °CR = 8.314 J/(mol·K)
ρ = kg/m³
3.Energy balanceMedium
hot sidecold sideethanol · 120 °C70 °Cwater · 15 °Ct = ? °C

Hot ethanol cools water in a counter-flow heat exchanger with no heat loss. Find the water outlet temperature tout.

ṁh = 0.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 120 → 70 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 130 °C70 °Cwater · 20 °C40 °C · ṁ = ?15% lost

Hot light oil heats water in an exchanger that loses 15% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁc needed.

ṁh = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 130 → 70 °Ct: 20 → 40 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 775 °CTL​ = 10 °CEQH​ = 700 kJQL​W = ?

A heat engine works between a hot reservoir at 775 °C and a cold reservoir at 10 °C. It takes in 700 kJ of heat. Find the most work Wmax it could possibly produce.

TH = 775 °CTL = 10 °CQH = 700 kJ
Wmax = kJ
6.Heat engines & refrigeratorsHard
TH​ = 35 °CTL​ = 2 °CRQ̇H​Q̇L​ = 1 kWẆ = ?

A refrigerator keeps its cold space at 2 °C in a kitchen at 35 °C, removing heat from the cold space at 1 kW. Find the minimum power input Ẇmin.

TH = 35 °CTL = 2 °CQ̇L = 1 kW
Ẇmin = kW
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 16 and a cutoff ratio rc = v₃/v₂ = 1.5. At the start of compression the air is at 295 K. Use the cold-air-standard assumptions: air is an ideal gas with constant specific heats, k = 1.4, cp = 1.005 kJ/(kg·K) and cv = 0.718 kJ/(kg·K). Find the heat added qin per kilogram of air, in kJ/kg.

r = 16rc = 1.5T1 = 295 Kk = 1.4cp = 1.005 kJ/(kg·K)
qin = kJ/kg
8.Gas power cyclesHard
vP1234qin​qout​

An ideal Brayton cycle operates with a compressor pressure ratio of 14. Use the cold-air-standard assumptions: air is an ideal gas with constant specific heats, k = 1.4, cp = 1.005 kJ/(kg·K) and cv = 0.718 kJ/(kg·K). Find its thermal efficiency η in percent.

rp = 14k = 1.4
η = %

3Solutions

1.Ideal gas law

Answer: V = 9.859 L

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

Moles

n = 70/32 = 2.188 mol
T = 25 + 273.15 = 298.1 K

Volume

V = nRT/P = 2.188·8.314·298.1/550 = 9.859 L

Answer

V = 9.859 L
2.Ideal gas law

Answer: ρ = 4.542 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 150 + 273.15 = 423.1 K

Density

ρ = 400·39.95/(8.314·423.1) = 4.542 kg/m³

Answer

ρ = 4.542 kg/m³
3.Energy balance

Answer: tout = 22.3 °C

Exchanger energy balance: with no losses, ṁhcp,h(Th,in − Th,out) = ṁccp,c(tout − tin).

Heat released by the hot stream

Q = 0.5·2.44·(120 − 70) = 61 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 61/(2·4.18) = 22.3 °C

Answer

tout = 22.3 °C
4.Energy balance

Answer: ṁc = 0.6406 kg/s

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

Heat released

Qh = 0.5·2.1·(130 − 70) = 63 kW

Heat that reaches the water

Qc = (1 − 0.15)·63 = 53.55 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 53.55/(4.18·20) = 0.6406 kg/s

Answer

ṁc = 0.6406 kg/s
5.Heat engines & refrigerators

Answer: Wmax = 510.9 kJ

Second-law limits: no cycle working between reservoirs at absolute temperatures TH and TL can beat the reversible (Carnot) cycle: ηmax = 1 − TL/TH, COPR,max = TL/(TH − TL) and COPHP,max = TH/(TH − TL). Energy is always conserved: QH = W + QL.

Absolute temperatures

TH = 775 + 273.15 = 1048 K
TL = 10 + 273.15 = 283.1 K

Carnot efficiency

ηmax = 1 − TL/TH = 1 − 283.1/1048 = 72.99 %

Maximum work

Wmax = ηmax·QH = 0.7299·700 = 510.9 kJ

Answer

Wmax = 510.9 kJ
6.Heat engines & refrigerators

Answer: Ẇmin = 0.1199 kW

Second-law limits: no cycle working between reservoirs at absolute temperatures TH and TL can beat the reversible (Carnot) cycle: ηmax = 1 − TL/TH, COPR,max = TL/(TH − TL) and COPHP,max = TH/(TH − TL). Energy is always conserved: QH = W + QL.

Absolute temperatures

TH = 35 + 273.15 = 308.1 K
TL = 2 + 273.15 = 275.1 K

Best possible COP

COPR,max = TL/(TH − TL) = 275.1/(308.1 − 275.1) = 8.338

Minimum power

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

Ẇmin = Q̇L/COP = 1/8.338 = 0.1199 kW

Answer

Ẇmin = 0.1199 kW
7.Gas power cycles

Answer: qin = 449.4 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 rc = v₃/v₂: T₂ = T₁·rk−1, T₃ = rc·T₂, qin = cp·(T₃ − T₂) and η = 1 − (1/rk−1)·(rck − 1)/(k·(rc − 1)).

Temperatures

T2 = 295·160.4 = 894.3 K
T3 = 1.5·894.3 = 1341 K

Heat added at constant pressure

qin = 1.005·(1341 − 894.3) = 449.4 kJ/kg

Answer

qin = 449.4 kJ/kg
8.Gas power cycles

Answer: η = 52.95 %

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 rp = P₂/P₁: T₂ = T₁·rp(k−1)/k, T₄ = T₃/rp(k−1)/k and η = 1 − 1/rp(k−1)/k. Net work w = cp·[(T₃ − T₄) − (T₂ − T₁)]. With isentropic efficiencies, the compressor needs ws/ηc and the turbine gives ηt·ws.

Thermal efficiency

η = 1 − 1/140.2857 = 1 − 1/2.126 = 52.95%

Answer

η = 52.95%

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