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 O₂V = 10 LT = 30 °CP = ?

A tank of 10 L contains 50 g of O₂ at 30 °C. Treat it as an ideal gas and find the pressure P.

m = 50 gM = 32 g/molV = 10 LT = 30 °CR = 8.314 J/(mol·K)
P = kPa
2.Ideal gas lawHard
State 1State 2

A fixed amount of CO₂ goes from 300 kPa, 2 L, 20 °C to 350 kPa and 140 °C. Find the new volume V2.

P1 = 300 kPaV1 = 2 LT1 = 20 °CP2 = 350 kPaT2 = 140 °C
V2 = L
3.Energy balanceMedium
hot sidecold sideethanol · 120 °C70 °Cwater · 10 °Ct = ? °C

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

ṁh = 0.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 120 → 70 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 140 °C110 °Cwater · 25 °C85 °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 = 3 kg/scp,h = 3.6 kJ/(kg·K)Th: 140 → 110 °Ct: 25 → 85 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 675 °CTL​ = 15 °CEQH​QL​W

A heat engine works between a hot reservoir at 675 °C and a cold reservoir at 15 °C. Find the highest possible thermal efficiency ηmax.

TH = 675 °CTL = 15 °C
ηmax = %
6.Heat engines & refrigeratorsHard
TH​ = 30 °CTL​ = -25 °CRQ̇H​Q̇L​ = 1 kWẆ

A refrigerator keeps its cold space at -25 °C in a kitchen at 30 °C, removing heat from the cold space at 1 kW. Find the highest possible coefficient of performance COPmax.

TH = 30 °CTL = -25 °CQ̇L = 1 kW
COPmax =
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 22 and a cutoff ratio rc = v₃/v₂ = 2.5. At the start of compression the air is at 290 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 = 22rc = 2.5T1 = 290 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 6. Air enters the compressor at 288 K and the turbine at 1200 K. The compressor has an isentropic efficiency of 85% and the turbine 85%. 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 = 6T1 = 288 KT3 = 1200 Kηc = 85%ηt = 85%k = 1.4cp = 1.005 kJ/(kg·K)
wnet = kJ/kg

3Solutions

1.Ideal gas law

Answer: P = 393.8 kPa

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

Moles

n = m/M = 50/32 = 1.563 mol
T = 30 + 273.15 = 303.1 K

Pressure

P = nRT/V = 1.563·8.314·303.1/0.01 = 393.8 kPa

Answer

P = 393.8 kPa
2.Ideal gas law

Answer: V2 = 2.416 L

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

Absolute temperatures

T1 = 293.1 K, T2 = 413.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 2·(300/350)·(413.1/293.1) = 2.416 L

Answer

V2 = 2.416 L
3.Energy balance

Answer: tout = 19.73 °C

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

Heat released by the hot stream

Q = 0.5·2.44·(120 − 70) = 61 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 61/(1.5·4.18) = 19.73 °C

Answer

tout = 19.73 °C
4.Energy balance

Answer: ṁc = 1.227 kg/s

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

Heat released

Qh = 3·3.6·(140 − 110) = 324 kW

Heat that reaches the water

Qc = (1 − 0.05)·324 = 307.8 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 307.8/(4.18·60) = 1.227 kg/s

Answer

ṁc = 1.227 kg/s
5.Heat engines & refrigerators

Answer: ηmax = 69.61 %

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 = 675 + 273.15 = 948.1 K
TL = 15 + 273.15 = 288.1 K

Carnot efficiency

ηmax = 1 − TL/TH = 1 − 288.1/948.1 = 69.61 %

Answer

ηmax = 69.61 %
6.Heat engines & refrigerators

Answer: COPmax = 4.512

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 = 30 + 273.15 = 303.1 K
TL = -25 + 273.15 = 248.1 K

Best possible COP

COPR,max = TL/(TH − TL) = 248.1/(303.1 − 248.1) = 4.512

Answer

COPmax = 4.512
7.Gas power cycles

Answer: T3 = 2496 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 = 290·220.4 = 998.5 K

Heat addition at constant pressure

T3 = rc·T2 = 2.5·998.5 = 2496 K

Answer

T3 = 2496 K
8.Gas power cycles

Answer: wnet = 183.1 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 = 288·60.2857 = 480.5 K
T4s = 1200/60.2857 = 719.2 K

Compressor and turbine work

wc = 1.005·(480.5 − 288)/0.85 = 227.6 kJ/kg
wt = 0.85·1.005·(1200 − 719.2) = 410.7 kJ/kg

Net work

wnet = 410.7 − 227.6 = 183.1 kJ/kg

Answer

wnet = 183.1 kJ/kg

Back-work ratio: wc/wt = 55.42%.

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