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
370 g of ArP = 700 kPaT = 10 °CV = ?

What volume V does 370 g of Ar occupy at 700 kPa and 10 °C? Treat it as an ideal gas.

m = 370 gM = 39.95 g/molP = 700 kPaT = 10 °CR = 8.314 J/(mol·K)
V = L
2.Ideal gas lawHard
State 1State 2

A fixed amount of O₂ goes from 400 kPa, 20 L, 20 °C to 350 kPa and 240 °C. Find the new volume V2.

P1 = 400 kPaV1 = 20 LT1 = 20 °CP2 = 350 kPaT2 = 240 °C
V2 = L
3.Energy balanceMedium
hot sidecold sidelight oil · 170 °C80 °Cwater · 10 °Ct = ? °C

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

ṁh = 1 kg/scp,h = 2.1 kJ/(kg·K)Th: 170 → 80 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 150 °C60 °Cwater · 15 °C35 °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 = 3 kg/scp,h = 2.1 kJ/(kg·K)Th: 150 → 60 °Ct: 15 → 35 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 425 °CTL​ = 15 °CEQH​QL​W

A heat engine works between a hot reservoir at 425 °C and a cold reservoir at 15 °C. Find the highest possible thermal efficiency ηmax.

TH = 425 °CTL = 15 °C
ηmax = %
6.Heat engines & refrigeratorsHard
TH​ = 900 °CT0​ = 25 °CTL​ = -5 °CERWQ̇H​ = 90 kWQ̇L​ = ?

A Carnot heat engine receives 90 kW of heat from a source at 900 °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 -5 °C and also rejects heat to the surroundings at 25 °C. Find the rate Q̇L at which heat is removed from the freezer.

TH = 900 °CT0 = 25 °CTL = -5 °CQ̇H = 90 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 = 2.5. 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 = 14rc = 2.5T1 = 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 16. Air enters the compressor at 295 K and the turbine at 1500 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 = 16T1 = 295 KT3 = 1500 Kk = 1.4cp = 1.005 kJ/(kg·K)
wnet = kJ/kg

3Solutions

1.Ideal gas law

Answer: V = 31.15 L

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

Moles

n = 370/39.95 = 9.262 mol
T = 10 + 273.15 = 283.1 K

Volume

V = nRT/P = 9.262·8.314·283.1/700 = 31.15 L

Answer

V = 31.15 L
2.Ideal gas law

Answer: V2 = 40.01 L

Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.

Absolute temperatures

T1 = 293.1 K, T2 = 513.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 20·(400/350)·(513.1/293.1) = 40.01 L

Answer

V2 = 40.01 L
3.Energy balance

Answer: tout = 100.4 °C

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

Heat released by the hot stream

Q = 1·2.1·(170 − 80) = 189 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 189/(0.5·4.18) = 100.4 °C

Answer

tout = 100.4 °C
4.Energy balance

Answer: ṁc = 5.765 kg/s

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

Heat released

Qh = 3·2.1·(150 − 60) = 567 kW

Heat that reaches the water

Qc = (1 − 0.15)·567 = 482 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 482/(4.18·20) = 5.765 kg/s

Answer

ṁc = 5.765 kg/s
5.Heat engines & refrigerators

Answer: ηmax = 58.73 %

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 = 425 + 273.15 = 698.1 K
TL = 15 + 273.15 = 288.1 K

Carnot efficiency

ηmax = 1 − TL/TH = 1 − 288.1/698.1 = 58.73 %

Answer

ηmax = 58.73 %
6.Heat engines & refrigerators

Answer: Q̇L = 600 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 = 900 + 273.15 = 1173 K
T0 = 25 + 273.15 = 298.1 K
TL = -5 + 273.15 = 268.1 K

Engine work

η = 1 − 298.1/1173 = 0.7459
Ẇ = 0.7459·90 = 67.13 kW

Heat removed by the refrigerator

COPR = 268.1/(298.1 − 268.1) = 8.938
Q̇L = 8.938·67.13 = 600 kW

Answer

Q̇L = 600 kW
7.Gas power cycles

Answer: qin = 1321 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 = 305·140.4 = 876.5 K
T3 = 2.5·876.5 = 2191 K

Heat added at constant pressure

qin = 1.005·(2191 − 876.5) = 1321 kJ/kg

Answer

qin = 1321 kJ/kg
8.Gas power cycles

Answer: wnet = 466.6 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

T2 = 295·160.2857 = 651.4 K
T4 = 1500/160.2857 = 679.3 K

Compressor and turbine work

wc = 1.005·(651.4 − 295) = 358.2 kJ/kg
wt = 1.005·(1500 − 679.3) = 824.8 kJ/kg

Net work

wnet = 824.8 − 358.2 = 466.6 kJ/kg

Answer

wnet = 466.6 kJ/kg

Back-work ratio: wc/wt = 43.43%.

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