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
290 g of N₂V = 50 LT = 65 °CP = ?

A tank of 50 L contains 290 g of N₂ at 65 °C. Treat it as an ideal gas and find the pressure P.

m = 290 gM = 28.01 g/molV = 50 LT = 65 °CR = 8.314 J/(mol·K)
P = kPa
2.Ideal gas lawHard
State 1State 2

A fixed amount of N₂ goes from 300 kPa, 10 L, 40 °C to 600 kPa and 70 °C. Find the new volume V2.

P1 = 300 kPaV1 = 10 LT1 = 40 °CP2 = 600 kPaT2 = 70 °C
V2 = L
3.Energy balanceMedium
hot sidecold sideethanol · 170 °C70 °Cwater · 15 °Ct = ? °C

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

ṁh = 2 kg/scp,h = 2.44 kJ/(kg·K)Th: 170 → 70 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 160 °C60 °Cwater · 10 °C65 °C · ṁ = ?5% lost

Hot glycol mix 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 = 3.6 kJ/(kg·K)Th: 160 → 60 °Ct: 10 → 65 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 450 °CTL​ = 15 °CEQH​QL​η = 42%

A heat engine receives heat from a source at 450 °C and rejects heat to the surroundings at 15 °C. Its actual thermal efficiency is 42%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.

TH = 450 °CTL = 15 °Cη = 42%
ηII = %
6.Heat engines & refrigeratorsHard
1150 °CTm​ = ?40 °CABWA​WB​

Two Carnot engines work in series. Engine A receives heat from a reservoir at 1150 °C and rejects heat to an intermediate reservoir at temperature Tm. Engine B takes all that heat and rejects heat to a reservoir at 40 °C. Both engines have the same thermal efficiency. Find Tm in °C.

TH = 1150 °CTL = 40 °C
Tm = °C
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 22 and a cutoff ratio rc = v₃/v₂ = 2. 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 = 22rc = 2k = 1.4
η = %
8.Gas power cyclesHard
vP1234qin​qout​

A gas-turbine plant runs on a Brayton cycle with a compressor pressure ratio of 10. Air enters the compressor at 300 K and the turbine at 1200 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 = 10T1 = 300 KT3 = 1200 Kk = 1.4cp = 1.005 kJ/(kg·K)
wnet = kJ/kg

3Solutions

1.Ideal gas law

Answer: P = 582.1 kPa

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

Moles

n = m/M = 290/28.01 = 10.35 mol
T = 65 + 273.15 = 338.1 K

Pressure

P = nRT/V = 10.35·8.314·338.1/0.05 = 582.1 kPa

Answer

P = 582.1 kPa
2.Ideal gas law

Answer: V2 = 5.479 L

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

Absolute temperatures

T1 = 313.1 K, T2 = 343.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 10·(300/600)·(343.1/313.1) = 5.479 L

Answer

V2 = 5.479 L
3.Energy balance

Answer: tout = 131.7 °C

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

Heat released by the hot stream

Q = 2·2.44·(170 − 70) = 488 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 488/(1·4.18) = 131.7 °C

Answer

tout = 131.7 °C
4.Energy balance

Answer: ṁc = 0.7438 kg/s

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

Heat released

Qh = 0.5·3.6·(160 − 60) = 180 kW

Heat that reaches the water

Qc = (1 − 0.05)·180 = 171 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 171/(4.18·55) = 0.7438 kg/s

Answer

ṁc = 0.7438 kg/s
5.Heat engines & refrigerators

Answer: ηII = 69.82 %

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 = 450 + 273.15 = 723.1 K
TL = 15 + 273.15 = 288.1 K

Carnot efficiency

ηCarnot = 1 − 288.1/723.1 = 60.15 %

Second-law efficiency

ηII = η/ηCarnot = 42/60.15 = 69.82 %

Answer

ηII = 69.82 %
6.Heat engines & refrigerators

Answer: Tm = 394.4 °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 temperatures

TH = 1150 + 273.15 = 1423 K
TL = 40 + 273.15 = 313.1 K

Equal efficiencies

1 − Tm/TH = 1 − TL/Tm gives Tm² = TH·TL.

Tm = √(1423·313.1) = 667.6 K = 394.4 °C

Answer

Tm = 394.4 °C
7.Gas power cycles

Answer: η = 66 %

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

(21.4 − 1)/(1.4·(2 − 1)) = 1.171

Thermal efficiency

η = 1 − 0.2904·1.171 = 66%

Answer

η = 66%
8.Gas power cycles

Answer: wnet = 300.7 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 = 300·100.2857 = 579.2 K
T4 = 1200/100.2857 = 621.5 K

Compressor and turbine work

wc = 1.005·(579.2 − 300) = 280.6 kJ/kg
wt = 1.005·(1200 − 621.5) = 581.4 kJ/kg

Net work

wnet = 581.4 − 280.6 = 300.7 kJ/kg

Answer

wnet = 300.7 kJ/kg

Back-work ratio: wc/wt = 48.27%.

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