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
380 g of CO₂P = 550 kPaT = 20 °CV = ?

What volume V does 380 g of CO₂ occupy at 550 kPa and 20 °C? Treat it as an ideal gas.

m = 380 gM = 44.01 g/molP = 550 kPaT = 20 °CR = 8.314 J/(mol·K)
V = L
2.Ideal gas lawHard
CO₂, M = 44.01 g/molP = 600 kPaT = 80 °Cρ = ?

Find the density ρ of CO₂ at 600 kPa and 80 °C, treating it as an ideal gas.

M = 44.01 g/molP = 600 kPaT = 80 °CR = 8.314 J/(mol·K)
ρ = kg/m³
3.Energy balanceMedium
hot sidecold sideglycol mix · 150 °C60 °Cwater · 30 °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: 150 → 60 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 120 °C60 °Cwater · 30 °C75 °C · ṁ = ?10% lost

Hot glycol mix heats water in an exchanger that loses 10% 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: 120 → 60 °Ct: 30 → 75 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 800 °CTL​ = 20 °CEQH​QL​η = 51%

A heat engine receives heat from a source at 800 °C and rejects heat to the surroundings at 20 °C. Its actual thermal efficiency is 51%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.

TH = 800 °CTL = 20 °Cη = 51%
ηII = %
6.Heat engines & refrigeratorsHard
1000 °CTm​ = ?20 °CABWA​WB​

Two Carnot engines work in series. Engine A receives heat from a reservoir at 1000 °C and rejects heat to an intermediate reservoir at temperature Tm. Engine B takes all that heat and rejects heat to a reservoir at 20 °C. Both engines have the same thermal efficiency. Find Tm in °C.

TH = 1000 °CTL = 20 °C
Tm = °C
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 16 and a cutoff ratio rc = v₃/v₂ = 2.2. 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 = 16rc = 2.2k = 1.4
η = %
8.Gas power cyclesHard
vP1234qin​qout​

An ideal Brayton cycle operates with a compressor pressure ratio of 14. 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.

rp = 14k = 1.4
η = %

3Solutions

1.Ideal gas law

Answer: V = 38.26 L

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

Moles

n = 380/44.01 = 8.634 mol
T = 20 + 273.15 = 293.1 K

Volume

V = nRT/P = 8.634·8.314·293.1/550 = 38.26 L

Answer

V = 38.26 L
2.Ideal gas law

Answer: ρ = 8.994 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 80 + 273.15 = 353.1 K

Density

ρ = 600·44.01/(8.314·353.1) = 8.994 kg/m³

Answer

ρ = 8.994 kg/m³
3.Energy balance

Answer: tout = 133.3 °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·(150 − 60) = 648 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 648/(1.5·4.18) = 133.3 °C

Answer

tout = 133.3 °C
4.Energy balance

Answer: ṁc = 0.5167 kg/s

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

Heat released

Qh = 0.5·3.6·(120 − 60) = 108 kW

Heat that reaches the water

Qc = (1 − 0.1)·108 = 97.2 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 97.2/(4.18·45) = 0.5167 kg/s

Answer

ṁc = 0.5167 kg/s
5.Heat engines & refrigerators

Answer: ηII = 70.17 %

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 = 800 + 273.15 = 1073 K
TL = 20 + 273.15 = 293.1 K

Carnot efficiency

ηCarnot = 1 − 293.1/1073 = 72.68 %

Second-law efficiency

ηII = η/ηCarnot = 51/72.68 = 70.17 %

Answer

ηII = 70.17 %
6.Heat engines & refrigerators

Answer: Tm = 337.8 °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 = 1000 + 273.15 = 1273 K
TL = 20 + 273.15 = 293.1 K

Equal efficiencies

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

Tm = √(1273·293.1) = 610.9 K = 337.8 °C

Answer

Tm = 337.8 °C
7.Gas power cycles

Answer: η = 60.42 %

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

(2.21.4 − 1)/(1.4·(2.2 − 1)) = 1.2

Thermal efficiency

η = 1 − 0.3299·1.2 = 60.42%

Answer

η = 60.42%
8.Gas power cycles

Answer: η = 52.95 %

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.

Thermal efficiency

η = 1 − 1/140.2857 = 1 − 1/2.126 = 52.95%

Answer

η = 52.95%

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