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
50 g of HeP = 150 kPaT = 100 °CV = ?

What volume V does 50 g of He occupy at 150 kPa and 100 °C? Treat it as an ideal gas.

m = 50 gM = 4.003 g/molP = 150 kPaT = 100 °CR = 8.314 J/(mol·K)
V = L
2.Ideal gas lawHard
State 1State 2

A fixed amount of N₂ goes from 150 kPa, 20 L, 35 °C to 700 kPa and 200 °C. Find the new volume V2.

P1 = 150 kPaV1 = 20 LT1 = 35 °CP2 = 700 kPaT2 = 200 °C
V2 = L
3.Energy balanceMedium
hot sidecold sidelight oil · 170 °C60 °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 = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 170 → 60 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 140 °C80 °Cwater · 30 °C45 °C · ṁ = ?10% lost

Hot light oil 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 = 2.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 140 → 80 °Ct: 30 → 45 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 575 °CTL​ = 25 °CEQH​ = 3900 kJQL​W = ?

A heat engine works between a hot reservoir at 575 °C and a cold reservoir at 25 °C. It takes in 3900 kJ of heat. Find the most work Wmax it could possibly produce.

TH = 575 °CTL = 25 °CQH = 3900 kJ
Wmax = kJ
6.Heat engines & refrigeratorsHard
TH​ = 35 °CTL​ = 5 °CRQ̇H​Q̇L​ = 2 kWẆ = ?

A refrigerator keeps its cold space at 5 °C in a kitchen at 35 °C, removing heat from the cold space at 2 kW. Find the minimum power input Ẇmin.

TH = 35 °CTL = 5 °CQ̇L = 2 kW
Ẇmin = kW
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 18 and a cutoff ratio rc = v₃/v₂ = 1.5. At the start of compression the air is at 295 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 = 18rc = 1.5T1 = 295 Kk = 1.4
T3 = K
8.Gas power cyclesHard
vP1234qin​qout​

An ideal Brayton cycle operates with a compressor pressure ratio of 16. 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 = 16k = 1.4
η = %

3Solutions

1.Ideal gas law

Answer: V = 258.3 L

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

Moles

n = 50/4.003 = 12.49 mol
T = 100 + 273.15 = 373.1 K

Volume

V = nRT/P = 12.49·8.314·373.1/150 = 258.3 L

Answer

V = 258.3 L
2.Ideal gas law

Answer: V2 = 6.581 L

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

Absolute temperatures

T1 = 308.1 K, T2 = 473.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 20·(150/700)·(473.1/308.1) = 6.581 L

Answer

V2 = 6.581 L
3.Energy balance

Answer: tout = 37.63 °C

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

Heat released by the hot stream

Q = 0.5·2.1·(170 − 60) = 115.5 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 115.5/(1·4.18) = 37.63 °C

Answer

tout = 37.63 °C
4.Energy balance

Answer: ṁc = 4.522 kg/s

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

Heat released

Qh = 2.5·2.1·(140 − 80) = 315 kW

Heat that reaches the water

Qc = (1 − 0.1)·315 = 283.5 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 283.5/(4.18·15) = 4.522 kg/s

Answer

ṁc = 4.522 kg/s
5.Heat engines & refrigerators

Answer: Wmax = 2529 kJ

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 = 575 + 273.15 = 848.1 K
TL = 25 + 273.15 = 298.1 K

Carnot efficiency

ηmax = 1 − TL/TH = 1 − 298.1/848.1 = 64.85 %

Maximum work

Wmax = ηmax·QH = 0.6485·3900 = 2529 kJ

Answer

Wmax = 2529 kJ
6.Heat engines & refrigerators

Answer: Ẇmin = 0.2157 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 = 35 + 273.15 = 308.1 K
TL = 5 + 273.15 = 278.1 K

Best possible COP

COPR,max = TL/(TH − TL) = 278.1/(308.1 − 278.1) = 9.272

Minimum power

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

Ẇmin = Q̇L/COP = 2/9.272 = 0.2157 kW

Answer

Ẇmin = 0.2157 kW
7.Gas power cycles

Answer: T3 = 1406 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 = 295·180.4 = 937.4 K

Heat addition at constant pressure

T3 = rc·T2 = 1.5·937.4 = 1406 K

Answer

T3 = 1406 K
8.Gas power cycles

Answer: η = 54.71 %

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 − 1/160.2857 = 1 − 1/2.208 = 54.71%

Answer

η = 54.71%

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