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
330 g of CO₂V = 20 LT = 35 °CP = ?

A tank of 20 L contains 330 g of CO₂ at 35 °C. Treat it as an ideal gas and find the pressure P.

m = 330 gM = 44.01 g/molV = 20 LT = 35 °CR = 8.314 J/(mol·K)
P = kPa
2.Ideal gas lawHard
CH₄, M = 16.04 g/molP = 550 kPaT = 60 °Cρ = ?

Find the density ρ of CH₄ at 550 kPa and 60 °C, treating it as an ideal gas.

M = 16.04 g/molP = 550 kPaT = 60 °CR = 8.314 J/(mol·K)
ρ = kg/m³
3.Energy balanceMedium
hot sidecold sidelight oil · 130 °C70 °Cwater · 20 °Ct = ? °C

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

ṁh = 2.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 130 → 70 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 160 °C120 °Cwater · 30 °C50 °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 = 1 kg/scp,h = 2.44 kJ/(kg·K)Th: 160 → 120 °Ct: 30 → 50 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 725 °CTL​ = ?EQH​ = 3200 kJQL​ = 892 kJW

A reversible (Carnot) heat engine takes in 3200 kJ of heat from a reservoir at 725 °C and rejects 892 kJ per cycle. Find the temperature TL of the cold reservoir in °C.

TH = 725 °CQH = 3200 kJQL = 892 kJ
TL = °C
6.Heat engines & refrigeratorsHard
TH​ = 850 °CT0​ = 25 °CTL​ = -10 °CERWQ̇H​ = 60 kWQ̇L​ = ?

A Carnot heat engine receives 60 kW of heat from a source at 850 °C and rejects heat to the surroundings at 25 °C. All of its work drives a Carnot refrigerator that removes heat from a freezer at -10 °C and also rejects heat to the surroundings at 25 °C. Find the rate Q̇L at which heat is removed from the freezer.

TH = 850 °CT0 = 25 °CTL = -10 °CQ̇H = 60 kW
Q̇L = kW
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 14 and a cutoff ratio rc = v₃/v₂ = 1.8. 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.

r = 14rc = 1.8k = 1.4
η = %
8.Gas power cyclesHard
vP1234qin​qout​

An ideal Brayton cycle operates with a compressor pressure ratio of 15. 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 = 15k = 1.4
η = %

3Solutions

1.Ideal gas law

Answer: P = 960.5 kPa

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

Moles

n = m/M = 330/44.01 = 7.498 mol
T = 35 + 273.15 = 308.1 K

Pressure

P = nRT/V = 7.498·8.314·308.1/0.02 = 960.5 kPa

Answer

P = 960.5 kPa
2.Ideal gas law

Answer: ρ = 3.185 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 60 + 273.15 = 333.1 K

Density

ρ = 550·16.04/(8.314·333.1) = 3.185 kg/m³

Answer

ρ = 3.185 kg/m³
3.Energy balance

Answer: tout = 45.12 °C

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

Heat released by the hot stream

Q = 2.5·2.1·(130 − 70) = 315 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 315/(3·4.18) = 45.12 °C

Answer

tout = 45.12 °C
4.Energy balance

Answer: ṁc = 1.051 kg/s

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

Heat released

Qh = 1·2.44·(160 − 120) = 97.6 kW

Heat that reaches the water

Qc = (1 − 0.1)·97.6 = 87.84 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 87.84/(4.18·20) = 1.051 kg/s

Answer

ṁc = 1.051 kg/s
5.Heat engines & refrigerators

Answer: TL = 5.084 °C

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 temperature of the source

TH = 725 + 273.15 = 998.1 K

Reversible heat ratio

TL = TH·QL/QH = 998.1·892/3200 = 278.2 K = 5.084 °C

Answer

TL = 5.084 °C
6.Heat engines & refrigerators

Answer: Q̇L = 331.4 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 = 850 + 273.15 = 1123 K
T0 = 25 + 273.15 = 298.1 K
TL = -10 + 273.15 = 263.1 K

Engine work

η = 1 − 298.1/1123 = 0.7345
Ẇ = 0.7345·60 = 44.07 kW

Heat removed by the refrigerator

COPR = 263.1/(298.1 − 263.1) = 7.519
Q̇L = 7.519·44.07 = 331.4 kW

Answer

Q̇L = 331.4 kW
7.Gas power cycles

Answer: η = 60.32 %

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)).

Cutoff term

(1.81.4 − 1)/(1.4·(1.8 − 1)) = 1.14

Thermal efficiency

η = 1 − 0.348·1.14 = 60.32%

Answer

η = 60.32%
8.Gas power cycles

Answer: η = 53.87 %

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 − 1/150.2857 = 1 − 1/2.168 = 53.87%

Answer

η = 53.87%

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