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
470 g of HeP = 250 kPaT = 70 °CV = ?

What volume V does 470 g of He occupy at 250 kPa and 70 °C? Treat it as an ideal gas.

m = 470 gM = 4.003 g/molP = 250 kPaT = 70 °CR = 8.314 J/(mol·K)
V = L
2.Ideal gas lawHard
N₂, M = 28.01 g/molP = 650 kPaT = 20 °Cρ = ?

Find the density ρ of N₂ at 650 kPa and 20 °C, treating it as an ideal gas.

M = 28.01 g/molP = 650 kPaT = 20 °CR = 8.314 J/(mol·K)
ρ = kg/m³
3.Energy balanceMedium
hot sidecold sidelight oil · 160 °C80 °Cwater · 30 °Ct = ? °C

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

ṁh = 2 kg/scp,h = 2.1 kJ/(kg·K)Th: 160 → 80 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 140 °C100 °Cwater · 25 °C45 °C · ṁ = ?5% lost

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 ṁc needed.

ṁh = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 140 → 100 °Ct: 25 → 45 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 350 °CTL​ = 35 °CEQH​QL​η = 35%

A heat engine receives heat from a source at 350 °C and rejects heat to the surroundings at 35 °C. Its actual thermal efficiency is 35%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.

TH = 350 °CTL = 35 °Cη = 35%
ηII = %
6.Heat engines & refrigeratorsHard
TH​ = 28 °CTL​ = 2 °CRQ̇H​Q̇L​ = 3.5 kWẆ = ?

A refrigerator keeps its cold space at 2 °C in a kitchen at 28 °C, removing heat from the cold space at 3.5 kW. Find the minimum power input Ẇmin.

TH = 28 °CTL = 2 °CQ̇L = 3.5 kW
Ẇmin = kW
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 19 and a cutoff ratio rc = v₃/v₂ = 2.5. At the start of compression the air is at 310 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 = 19rc = 2.5T1 = 310 Kk = 1.4cp = 1.005 kJ/(kg·K)
qin = kJ/kg
8.Gas power cyclesHard
vP1234qin​qout​

A gas-turbine plant runs on a Brayton cycle with a compressor pressure ratio of 14. Air enters the compressor at 300 K and the turbine at 1200 K. The compressor has an isentropic efficiency of 80% and the turbine 90%. 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 net work output wnet per kilogram of air, in kJ/kg.

rp = 14T1 = 300 KT3 = 1200 Kηc = 80%ηt = 90%k = 1.4cp = 1.005 kJ/(kg·K)
wnet = kJ/kg

3Solutions

1.Ideal gas law

Answer: V = 1340 L

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

Moles

n = 470/4.003 = 117.4 mol
T = 70 + 273.15 = 343.1 K

Volume

V = nRT/P = 117.4·8.314·343.1/250 = 1340 L

Answer

V = 1340 L
2.Ideal gas law

Answer: ρ = 7.47 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 20 + 273.15 = 293.1 K

Density

ρ = 650·28.01/(8.314·293.1) = 7.47 kg/m³

Answer

ρ = 7.47 kg/m³
3.Energy balance

Answer: tout = 56.79 °C

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

Heat released by the hot stream

Q = 2·2.1·(160 − 80) = 336 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 336/(3·4.18) = 56.79 °C

Answer

tout = 56.79 °C
4.Energy balance

Answer: ṁc = 0.4773 kg/s

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

Heat released

Qh = 0.5·2.1·(140 − 100) = 42 kW

Heat that reaches the water

Qc = (1 − 0.05)·42 = 39.9 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 39.9/(4.18·20) = 0.4773 kg/s

Answer

ṁc = 0.4773 kg/s
5.Heat engines & refrigerators

Answer: ηII = 69.24 %

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 = 350 + 273.15 = 623.1 K
TL = 35 + 273.15 = 308.1 K

Carnot efficiency

ηCarnot = 1 − 308.1/623.1 = 50.55 %

Second-law efficiency

ηII = η/ηCarnot = 35/50.55 = 69.24 %

Answer

ηII = 69.24 %
6.Heat engines & refrigerators

Answer: Ẇmin = 0.3307 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 = 28 + 273.15 = 301.1 K
TL = 2 + 273.15 = 275.1 K

Best possible COP

COPR,max = TL/(TH − TL) = 275.1/(301.1 − 275.1) = 10.58

Minimum power

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

Ẇmin = Q̇L/COP = 3.5/10.58 = 0.3307 kW

Answer

Ẇmin = 0.3307 kW
7.Gas power cycles

Answer: qin = 1517 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 = 310·190.4 = 1007 K
T3 = 2.5·1007 = 2517 K

Heat added at constant pressure

qin = 1.005·(2517 − 1007) = 1517 kJ/kg

Answer

qin = 1517 kJ/kg
8.Gas power cycles

Answer: wnet = 150.6 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 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.

Isentropic temperatures

T2s = 300·140.2857 = 637.7 K
T4s = 1200/140.2857 = 564.6 K

Compressor and turbine work

wc = 1.005·(637.7 − 300)/0.8 = 424.2 kJ/kg
wt = 0.9·1.005·(1200 − 564.6) = 574.7 kJ/kg

Net work

wnet = 574.7 − 424.2 = 150.6 kJ/kg

Answer

wnet = 150.6 kJ/kg

Back-work ratio: wc/wt = 73.8%.

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