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
310 g of HeP = 250 kPaT = 80 °CV = ?

What volume V does 310 g of He occupy at 250 kPa and 80 °C? Treat it as an ideal gas.

m = 310 gM = 4.003 g/molP = 250 kPaT = 80 °CR = 8.314 J/(mol·K)
V = L
2.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 350 kPa, 20 L, 30 °C to 400 kPa and 150 °C. Find the new volume V2.

P1 = 350 kPaV1 = 20 LT1 = 30 °CP2 = 400 kPaT2 = 150 °C
V2 = L
3.Energy balanceMedium
hot sidecold sideglycol mix · 130 °C90 °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 = 2.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 130 → 90 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 120 °C60 °Cwater · 20 °C35 °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 = 3 kg/scp,h = 2.44 kJ/(kg·K)Th: 120 → 60 °Ct: 20 → 35 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 900 °CTL​ = 15 °CEQH​QL​η = 49%

A heat engine receives heat from a source at 900 °C and rejects heat to the surroundings at 15 °C. Its actual thermal efficiency is 49%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.

TH = 900 °CTL = 15 °Cη = 49%
ηII = %
6.Heat engines & refrigeratorsHard
750 °CTm​ = ?25 °CABWA​WB​

Two Carnot engines work in series. Engine A receives heat from a reservoir at 750 °C and rejects heat to an intermediate reservoir at temperature Tm. Engine B takes all that heat and rejects heat to a reservoir at 25 °C. Both engines have the same thermal efficiency. Find Tm in °C.

TH = 750 °CTL = 25 °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 = 3. 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 = 3k = 1.4
η = %
8.Gas power cyclesHard
vP1234qin​qout​

A gas-turbine plant runs on a Brayton cycle with a compressor pressure ratio of 16. Air enters the compressor at 288 K and the turbine at 1300 K. The compressor has an isentropic efficiency of 82% 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 = 16T1 = 288 KT3 = 1300 Kηc = 82%ηt = 90%k = 1.4cp = 1.005 kJ/(kg·K)
wnet = kJ/kg

3Solutions

1.Ideal gas law

Answer: V = 909.5 L

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

Moles

n = 310/4.003 = 77.44 mol
T = 80 + 273.15 = 353.1 K

Volume

V = nRT/P = 77.44·8.314·353.1/250 = 909.5 L

Answer

V = 909.5 L
2.Ideal gas law

Answer: V2 = 24.43 L

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

Absolute temperatures

T1 = 303.1 K, T2 = 423.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 20·(350/400)·(423.1/303.1) = 24.43 L

Answer

V2 = 24.43 L
3.Energy balance

Answer: tout = 58.71 °C

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

Heat released by the hot stream

Q = 2.5·3.6·(130 − 90) = 360 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 360/(3·4.18) = 58.71 °C

Answer

tout = 58.71 °C
4.Energy balance

Answer: ṁc = 6.655 kg/s

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

Heat released

Qh = 3·2.44·(120 − 60) = 439.2 kW

Heat that reaches the water

Qc = (1 − 0.05)·439.2 = 417.2 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 417.2/(4.18·15) = 6.655 kg/s

Answer

ṁc = 6.655 kg/s
5.Heat engines & refrigerators

Answer: ηII = 64.95 %

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 = 900 + 273.15 = 1173 K
TL = 15 + 273.15 = 288.1 K

Carnot efficiency

ηCarnot = 1 − 288.1/1173 = 75.44 %

Second-law efficiency

ηII = η/ηCarnot = 49/75.44 = 64.95 %

Answer

ηII = 64.95 %
6.Heat engines & refrigerators

Answer: Tm = 279.2 °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 = 750 + 273.15 = 1023 K
TL = 25 + 273.15 = 298.1 K

Equal efficiencies

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

Tm = √(1023·298.1) = 552.3 K = 279.2 °C

Answer

Tm = 279.2 °C
7.Gas power cycles

Answer: η = 56.93 %

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

(31.4 − 1)/(1.4·(3 − 1)) = 1.306

Thermal efficiency

η = 1 − 0.3299·1.306 = 56.93%

Answer

η = 56.93%
8.Gas power cycles

Answer: wnet = 216.9 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 = 288·160.2857 = 636 K
T4s = 1300/160.2857 = 588.7 K

Compressor and turbine work

wc = 1.005·(636 − 288)/0.82 = 426.5 kJ/kg
wt = 0.9·1.005·(1300 − 588.7) = 643.4 kJ/kg

Net work

wnet = 643.4 − 426.5 = 216.9 kJ/kg

Answer

wnet = 216.9 kJ/kg

Back-work ratio: wc/wt = 66.29%.

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