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
260 g of HeP = 550 kPaT = 20 °CV = ?

What volume V does 260 g of He occupy at 550 kPa and 20 °C? Treat it as an ideal gas.

m = 260 gM = 4.003 g/molP = 550 kPaT = 20 °CR = 8.314 J/(mol·K)
V = L
2.Ideal gas lawHard
State 1State 2

A fixed amount of CO₂ goes from 300 kPa, 20 L, 30 °C to 400 kPa and 220 °C. Find the new volume V2.

P1 = 300 kPaV1 = 20 LT1 = 30 °CP2 = 400 kPaT2 = 220 °C
V2 = L
3.Energy balanceMedium
hot sidecold sideglycol mix · 130 °C70 °Cwater · 10 °Ct = ? °C

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

ṁh = 2 kg/scp,h = 3.6 kJ/(kg·K)Th: 130 → 70 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 130 °C60 °Cwater · 20 °C60 °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 = 1 kg/scp,h = 2.44 kJ/(kg·K)Th: 130 → 60 °Ct: 20 → 60 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 775 °CTL​ = 35 °CEQH​QL​W

A heat engine works between a hot reservoir at 775 °C and a cold reservoir at 35 °C. Find the highest possible thermal efficiency ηmax.

TH = 775 °CTL = 35 °C
ηmax = %
6.Heat engines & refrigeratorsHard
TH​ = 22 °CTL​ = -10 °CHPQ̇H​ = 13 kWQ̇L​Ẇ = ?

A heat pump keeps a house at 22 °C while it is -10 °C outside, delivering 13 kW of heat to the house. Find the minimum power input Ẇmin.

TH = 22 °CTL = -10 °CQ̇H = 13 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 = 2. At the start of compression the air is at 310 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 temperature T3 at the end of heat addition, in K.

r = 20rc = 2T1 = 310 Kk = 1.4
T3 = K
8.Gas power cyclesHard
vP1234qin​qout​

A gas-turbine plant runs on a Brayton cycle with a compressor pressure ratio of 18. Air enters the compressor at 290 K and the turbine at 1500 K. The compressor has an isentropic efficiency of 88% and the turbine 90%. 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 = 18T1 = 290 KT3 = 1500 Kηc = 88%ηt = 90%k = 1.4cp = 1.005 kJ/(kg·K)
wnet = kJ/kg

3Solutions

1.Ideal gas law

Answer: V = 287.8 L

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

Moles

n = 260/4.003 = 64.95 mol
T = 20 + 273.15 = 293.1 K

Volume

V = nRT/P = 64.95·8.314·293.1/550 = 287.8 L

Answer

V = 287.8 L
2.Ideal gas law

Answer: V2 = 24.4 L

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

Absolute temperatures

T1 = 303.1 K, T2 = 493.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 20·(300/400)·(493.1/303.1) = 24.4 L

Answer

V2 = 24.4 L
3.Energy balance

Answer: tout = 44.45 °C

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

Heat released by the hot stream

Q = 2·3.6·(130 − 70) = 432 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 432/(3·4.18) = 44.45 °C

Answer

tout = 44.45 °C
4.Energy balance

Answer: ṁc = 0.9705 kg/s

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

Heat released

Qh = 1·2.44·(130 − 60) = 170.8 kW

Heat that reaches the water

Qc = (1 − 0.05)·170.8 = 162.3 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 162.3/(4.18·40) = 0.9705 kg/s

Answer

ṁc = 0.9705 kg/s
5.Heat engines & refrigerators

Answer: ηmax = 70.6 %

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 = 775 + 273.15 = 1048 K
TL = 35 + 273.15 = 308.1 K

Carnot efficiency

ηmax = 1 − TL/TH = 1 − 308.1/1048 = 70.6 %

Answer

ηmax = 70.6 %
6.Heat engines & refrigerators

Answer: Ẇmin = 1.409 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 = 22 + 273.15 = 295.1 K
TL = -10 + 273.15 = 263.1 K

Best possible COP

COPHP,max = TH/(TH − TL) = 295.1/(295.1 − 263.1) = 9.223

Minimum power

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

Ẇmin = Q̇H/COP = 13/9.223 = 1.409 kW

Answer

Ẇmin = 1.409 kW
7.Gas power cycles

Answer: T3 = 2055 K

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)).

End of compression

T2 = 310·200.4 = 1027 K

Heat addition at constant pressure

T3 = rc·T2 = 2·1027 = 2055 K

Answer

T3 = 2055 K
8.Gas power cycles

Answer: wnet = 337.5 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 = 290·180.2857 = 662.3 K
T4s = 1500/180.2857 = 656.8 K

Compressor and turbine work

wc = 1.005·(662.3 − 290)/0.88 = 425.2 kJ/kg
wt = 0.9·1.005·(1500 − 656.8) = 762.7 kJ/kg

Net work

wnet = 762.7 − 425.2 = 337.5 kJ/kg

Answer

wnet = 337.5 kJ/kg

Back-work ratio: wc/wt = 55.75%.

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