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
300 g of O₂V = 30 LT = 20 °CP = ?

A tank of 30 L contains 300 g of O₂ at 20 °C. Treat it as an ideal gas and find the pressure P.

m = 300 gM = 32 g/molV = 30 LT = 20 °CR = 8.314 J/(mol·K)
P = kPa
2.Ideal gas lawHard
Ar, M = 39.95 g/molP = 200 kPaT = 200 °Cρ = ?

Find the density ρ of Ar at 200 kPa and 200 °C, treating it as an ideal gas.

M = 39.95 g/molP = 200 kPaT = 200 °CR = 8.314 J/(mol·K)
ρ = kg/m³
3.Energy balanceMedium
hot sidecold sideethanol · 170 °C130 °Cwater · 10 °Ct = ? °C

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

ṁh = 1 kg/scp,h = 2.44 kJ/(kg·K)Th: 170 → 130 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 150 °C110 °Cwater · 20 °C85 °C · ṁ = ?15% lost

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 ṁc needed.

ṁh = 0.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 150 → 110 °Ct: 20 → 85 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 30 °CTL​ = 2 °CRQ̇H​ = ?Q̇L​ = 9 kWẆ

A reversible refrigerator keeps a cold room at 2 °C by removing 9 kW of heat, and rejects heat to the surroundings at 30 °C. Find the rate Q̇H at which it rejects heat.

TL = 2 °CTH = 30 °CQ̇L = 9 kW
Q̇H = kW
6.Heat engines & refrigeratorsHard
TH​ = 32 °CTL​ = 4 °CRQ̇H​Q̇L​ = 5 kWẆ

A refrigerator keeps its cold space at 4 °C in a kitchen at 32 °C, removing heat from the cold space at 5 kW. Find the highest possible coefficient of performance COPmax.

TH = 32 °CTL = 4 °CQ̇L = 5 kW
COPmax =
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 22 and a cutoff ratio rc = 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, 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 = 22rc = 3T1 = 305 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 295 K and the turbine at 1200 K. The compressor has an isentropic efficiency of 82% and the turbine 92%. 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 = 295 KT3 = 1200 Kηc = 82%ηt = 92%k = 1.4cp = 1.005 kJ/(kg·K)
wnet = kJ/kg

3Solutions

1.Ideal gas law

Answer: P = 761.6 kPa

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

Moles

n = m/M = 300/32 = 9.375 mol
T = 20 + 273.15 = 293.1 K

Pressure

P = nRT/V = 9.375·8.314·293.1/0.03 = 761.6 kPa

Answer

P = 761.6 kPa
2.Ideal gas law

Answer: ρ = 2.031 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 200 + 273.15 = 473.1 K

Density

ρ = 200·39.95/(8.314·473.1) = 2.031 kg/m³

Answer

ρ = 2.031 kg/m³
3.Energy balance

Answer: tout = 21.67 °C

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

Heat released by the hot stream

Q = 1·2.44·(170 − 130) = 97.6 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 97.6/(2·4.18) = 21.67 °C

Answer

tout = 21.67 °C
4.Energy balance

Answer: ṁc = 0.2252 kg/s

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

Heat released

Qh = 0.5·3.6·(150 − 110) = 72 kW

Heat that reaches the water

Qc = (1 − 0.15)·72 = 61.2 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 61.2/(4.18·65) = 0.2252 kg/s

Answer

ṁc = 0.2252 kg/s
5.Heat engines & refrigerators

Answer: Q̇H = 9.916 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 = 2 + 273.15 = 275.1 K

Reversible heat ratio

Q̇H = Q̇L·TH/TL = 9·303.1/275.1 = 9.916 kW

Answer

Q̇H = 9.916 kW

Power input: Ẇ = 0.9159 kW.

6.Heat engines & refrigerators

Answer: COPmax = 9.898

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 = 4 + 273.15 = 277.1 K

Best possible COP

COPR,max = TL/(TH − TL) = 277.1/(305.1 − 277.1) = 9.898

Answer

COPmax = 9.898
7.Gas power cycles

Answer: qin = 2111 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 = 305·220.4 = 1050 K
T3 = 3·1050 = 3151 K

Heat added at constant pressure

qin = 1.005·(3151 − 1050) = 2111 kJ/kg

Answer

qin = 2111 kJ/kg
8.Gas power cycles

Answer: wnet = 180.6 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 = 295·140.2857 = 627 K
T4s = 1200/140.2857 = 564.6 K

Compressor and turbine work

wc = 1.005·(627 − 295)/0.82 = 406.9 kJ/kg
wt = 0.92·1.005·(1200 − 564.6) = 587.5 kJ/kg

Net work

wnet = 587.5 − 406.9 = 180.6 kJ/kg

Answer

wnet = 180.6 kJ/kg

Back-work ratio: wc/wt = 69.26%.

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