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
240 g of HeV = 5 LT = 100 °CP = ?

A tank of 5 L contains 240 g of He at 100 °C. Treat it as an ideal gas and find the pressure P.

m = 240 gM = 4.003 g/molV = 5 LT = 100 °CR = 8.314 J/(mol·K)
P = kPa
2.Ideal gas lawHard
CH₄, M = 16.04 g/molP = 700 kPaT = 20 °Cρ = ?

Find the density ρ of CH₄ at 700 kPa and 20 °C, treating it as an ideal gas.

M = 16.04 g/molP = 700 kPaT = 20 °CR = 8.314 J/(mol·K)
ρ = kg/m³
3.Energy balanceMedium
hot sidecold sideethanol · 140 °C100 °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: 140 → 100 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 160 °C70 °Cwater · 25 °C70 °C · ṁ = ?10% lost

Hot ethanol heats water in an exchanger that loses 10% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁc needed.

ṁh = 3 kg/scp,h = 2.44 kJ/(kg·K)Th: 160 → 70 °Ct: 25 → 70 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 30 °CTL​ = -5 °CRQ̇H​ = ?Q̇L​ = 3 kWẆ

A reversible refrigerator keeps a cold room at -5 °C by removing 3 kW of heat, and rejects heat to the surroundings at 30 °C. Find the rate Q̇H at which it rejects heat.

TL = -5 °CTH = 30 °CQ̇L = 3 kW
Q̇H = kW
6.Heat engines & refrigeratorsHard
TH​ = 32 °CTL​ = -25 °CRQ̇H​Q̇L​ = 3.5 kWẆ = ?

A refrigerator keeps its cold space at -25 °C in a kitchen at 32 °C, removing heat from the cold space at 3.5 kW. Find the minimum power input Ẇmin.

TH = 32 °CTL = -25 °CQ̇L = 3.5 kW
Ẇmin = kW
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 17 and a cutoff ratio rc = v₃/v₂ = 1.8. At the start of compression the air is at 320 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 = 17rc = 1.8T1 = 320 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 20. Air enters the compressor at 295 K and the turbine at 1400 K. Both machines are isentropic. 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 = 20T1 = 295 KT3 = 1400 Kk = 1.4cp = 1.005 kJ/(kg·K)
wnet = kJ/kg

3Solutions

1.Ideal gas law

Answer: P = 37201 kPa

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

Moles

n = m/M = 240/4.003 = 59.96 mol
T = 100 + 273.15 = 373.1 K

Pressure

P = nRT/V = 59.96·8.314·373.1/0.005 = 37201 kPa

Answer

P = 37201 kPa
2.Ideal gas law

Answer: ρ = 4.607 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 = 700·16.04/(8.314·293.1) = 4.607 kg/m³

Answer

ρ = 4.607 kg/m³
3.Energy balance

Answer: tout = 20.84 °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·(140 − 100) = 48.8 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 48.8/(2·4.18) = 20.84 °C

Answer

tout = 20.84 °C
4.Energy balance

Answer: ṁc = 3.152 kg/s

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

Heat released

Qh = 3·2.44·(160 − 70) = 658.8 kW

Heat that reaches the water

Qc = (1 − 0.1)·658.8 = 592.9 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 592.9/(4.18·45) = 3.152 kg/s

Answer

ṁc = 3.152 kg/s
5.Heat engines & refrigerators

Answer: Q̇H = 3.392 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 = 30 + 273.15 = 303.1 K
TL = -5 + 273.15 = 268.1 K

Reversible heat ratio

Q̇H = Q̇L·TH/TL = 3·303.1/268.1 = 3.392 kW

Answer

Q̇H = 3.392 kW

Power input: Ẇ = 0.3916 kW.

6.Heat engines & refrigerators

Answer: Ẇmin = 0.8039 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 = 32 + 273.15 = 305.1 K
TL = -25 + 273.15 = 248.1 K

Best possible COP

COPR,max = TL/(TH − TL) = 248.1/(305.1 − 248.1) = 4.354

Minimum power

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

Ẇmin = Q̇L/COP = 3.5/4.354 = 0.8039 kW

Answer

Ẇmin = 0.8039 kW
7.Gas power cycles

Answer: qin = 799.1 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 = 320·170.4 = 993.9 K
T3 = 1.8·993.9 = 1789 K

Heat added at constant pressure

qin = 1.005·(1789 − 993.9) = 799.1 kJ/kg

Answer

qin = 799.1 kJ/kg
8.Gas power cycles

Answer: wnet = 407.9 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

T2 = 295·200.2857 = 694.3 K
T4 = 1400/200.2857 = 594.8 K

Compressor and turbine work

wc = 1.005·(694.3 − 295) = 401.3 kJ/kg
wt = 1.005·(1400 − 594.8) = 809.2 kJ/kg

Net work

wnet = 809.2 − 401.3 = 407.9 kJ/kg

Answer

wnet = 407.9 kJ/kg

Back-work ratio: wc/wt = 49.59%.

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