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
100 g of N₂P = 700 kPaT = 95 °CV = ?

What volume V does 100 g of N₂ occupy at 700 kPa and 95 °C? Treat it as an ideal gas.

m = 100 gM = 28.01 g/molP = 700 kPaT = 95 °CR = 8.314 J/(mol·K)
V = L
2.Ideal gas lawHard
State 1State 2

A fixed amount of CO₂ goes from 150 kPa, 10 L, 35 °C to 750 kPa and 190 °C. Find the new volume V2.

P1 = 150 kPaV1 = 10 LT1 = 35 °CP2 = 750 kPaT2 = 190 °C
V2 = L
3.Energy balanceMedium
hot sidecold sidelight oil · 180 °C100 °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 = 3 kg/scp,h = 2.1 kJ/(kg·K)Th: 180 → 100 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 140 °C90 °Cwater · 20 °C40 °C · ṁ = ?5% lost

Hot ethanol 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 = 2.44 kJ/(kg·K)Th: 140 → 90 °Ct: 20 → 40 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 550 °CTL​ = 15 °CEQH​QL​η = 32%

A heat engine receives heat from a source at 550 °C and rejects heat to the surroundings at 15 °C. Its actual thermal efficiency is 32%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.

TH = 550 °CTL = 15 °Cη = 32%
ηII = %
6.Heat engines & refrigeratorsHard
TH​ = 25 °CTL​ = -10 °CRQ̇H​Q̇L​ = 4.5 kWẆ = ?

A refrigerator keeps its cold space at -10 °C in a kitchen at 25 °C, removing heat from the cold space at 4.5 kW. Find the minimum power input Ẇmin.

TH = 25 °CTL = -10 °CQ̇L = 4.5 kW
Ẇmin = kW
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 20 and a cutoff ratio rc = v₃/v₂ = 1.8. 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 = 20rc = 1.8T1 = 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 20. Air enters the compressor at 290 K and the turbine at 1500 K. The compressor has an isentropic efficiency of 88% 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 = 20T1 = 290 KT3 = 1500 Kηc = 88%ηt = 92%k = 1.4cp = 1.005 kJ/(kg·K)
wnet = kJ/kg

3Solutions

1.Ideal gas law

Answer: V = 15.61 L

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

Moles

n = 100/28.01 = 3.57 mol
T = 95 + 273.15 = 368.1 K

Volume

V = nRT/P = 3.57·8.314·368.1/700 = 15.61 L

Answer

V = 15.61 L
2.Ideal gas law

Answer: V2 = 3.006 L

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

Absolute temperatures

T1 = 308.1 K, T2 = 463.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 10·(150/750)·(463.1/308.1) = 3.006 L

Answer

V2 = 3.006 L
3.Energy balance

Answer: tout = 90.38 °C

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

Heat released by the hot stream

Q = 3·2.1·(180 − 100) = 504 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 504/(1.5·4.18) = 90.38 °C

Answer

tout = 90.38 °C
4.Energy balance

Answer: ṁc = 0.6932 kg/s

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

Heat released

Qh = 0.5·2.44·(140 − 90) = 61 kW

Heat that reaches the water

Qc = (1 − 0.05)·61 = 57.95 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 57.95/(4.18·20) = 0.6932 kg/s

Answer

ṁc = 0.6932 kg/s
5.Heat engines & refrigerators

Answer: ηII = 49.24 %

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 = 550 + 273.15 = 823.1 K
TL = 15 + 273.15 = 288.1 K

Carnot efficiency

ηCarnot = 1 − 288.1/823.1 = 64.99 %

Second-law efficiency

ηII = η/ηCarnot = 32/64.99 = 49.24 %

Answer

ηII = 49.24 %
6.Heat engines & refrigerators

Answer: Ẇmin = 0.5985 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 = 25 + 273.15 = 298.1 K
TL = -10 + 273.15 = 263.1 K

Best possible COP

COPR,max = TL/(TH − TL) = 263.1/(298.1 − 263.1) = 7.519

Minimum power

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

Ẇmin = Q̇L/COP = 4.5/7.519 = 0.5985 kW

Answer

Ẇmin = 0.5985 kW
7.Gas power cycles

Answer: qin = 812.8 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 = 305·200.4 = 1011 K
T3 = 1.8·1011 = 1820 K

Heat added at constant pressure

qin = 1.005·(1820 − 1011) = 812.8 kJ/kg

Answer

qin = 812.8 kJ/kg
8.Gas power cycles

Answer: wnet = 349.3 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 = 290·200.2857 = 682.5 K
T4s = 1500/200.2857 = 637.3 K

Compressor and turbine work

wc = 1.005·(682.5 − 290)/0.88 = 448.3 kJ/kg
wt = 0.92·1.005·(1500 − 637.3) = 797.6 kJ/kg

Net work

wnet = 797.6 − 448.3 = 349.3 kJ/kg

Answer

wnet = 349.3 kJ/kg

Back-work ratio: wc/wt = 56.2%.

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