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
120 g of O₂P = 100 kPaT = 25 °CV = ?

What volume V does 120 g of O₂ occupy at 100 kPa and 25 °C? Treat it as an ideal gas.

m = 120 gM = 32 g/molP = 100 kPaT = 25 °CR = 8.314 J/(mol·K)
V = L
2.Ideal gas lawHard
State 1State 2

A fixed amount of CH₄ goes from 250 kPa, 10 L, 30 °C to 450 kPa and 50 °C. Find the new volume V2.

P1 = 250 kPaV1 = 10 LT1 = 30 °CP2 = 450 kPaT2 = 50 °C
V2 = L
3.Energy balanceMedium
hot sidecold sideglycol mix · 140 °C110 °Cwater · 15 °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: 140 → 110 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 140 °C60 °Cwater · 15 °C80 °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 = 2 kg/scp,h = 2.44 kJ/(kg·K)Th: 140 → 60 °Ct: 15 → 80 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 35 °CTL​ = 0 °CRQ̇H​ = ?Q̇L​ = 0.5 kWẆ

A reversible refrigerator keeps a cold room at 0 °C by removing 0.5 kW of heat, and rejects heat to the surroundings at 35 °C. Find the rate Q̇H at which it rejects heat.

TL = 0 °CTH = 35 °CQ̇L = 0.5 kW
Q̇H = kW
6.Heat engines & refrigeratorsHard
TH​ = 22 °CTL​ = 2 °CHPQ̇H​ = 7 kWQ̇L​Ẇ = ?

A heat pump keeps a house at 22 °C while it is 2 °C outside, delivering 7 kW of heat to the house. Find the minimum power input Ẇmin.

TH = 22 °CTL = 2 °CQ̇H = 7 kW
Ẇmin = kW
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 17 and a cutoff ratio rc = v₃/v₂ = 1.8. 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 = 17rc = 1.8T1 = 295 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 15. Air enters the compressor at 288 K and the turbine at 1400 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 = 15T1 = 288 KT3 = 1400 Kk = 1.4cp = 1.005 kJ/(kg·K)
wnet = kJ/kg

3Solutions

1.Ideal gas law

Answer: V = 92.96 L

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

Moles

n = 120/32 = 3.75 mol
T = 25 + 273.15 = 298.1 K

Volume

V = nRT/P = 3.75·8.314·298.1/100 = 92.96 L

Answer

V = 92.96 L
2.Ideal gas law

Answer: V2 = 5.922 L

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

Absolute temperatures

T1 = 303.1 K, T2 = 323.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 10·(250/450)·(323.1/303.1) = 5.922 L

Answer

V2 = 5.922 L
3.Energy balance

Answer: tout = 58.06 °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·(140 − 110) = 270 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 270/(1.5·4.18) = 58.06 °C

Answer

tout = 58.06 °C
4.Energy balance

Answer: ṁc = 1.365 kg/s

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

Heat released

Qh = 2·2.44·(140 − 60) = 390.4 kW

Heat that reaches the water

Qc = (1 − 0.05)·390.4 = 370.9 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 370.9/(4.18·65) = 1.365 kg/s

Answer

ṁc = 1.365 kg/s
5.Heat engines & refrigerators

Answer: Q̇H = 0.5641 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 = 0 + 273.15 = 273.1 K

Reversible heat ratio

Q̇H = Q̇L·TH/TL = 0.5·308.1/273.1 = 0.5641 kW

Answer

Q̇H = 0.5641 kW

Power input: Ẇ = 0.06407 kW.

6.Heat engines & refrigerators

Answer: Ẇmin = 0.4743 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 = 22 + 273.15 = 295.1 K
TL = 2 + 273.15 = 275.1 K

Best possible COP

COPHP,max = TH/(TH − TL) = 295.1/(295.1 − 275.1) = 14.76

Minimum power

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

Ẇmin = Q̇H/COP = 7/14.76 = 0.4743 kW

Answer

Ẇmin = 0.4743 kW
7.Gas power cycles

Answer: T3 = 1649 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 = 295·170.4 = 916.2 K

Heat addition at constant pressure

T3 = rc·T2 = 1.8·916.2 = 1649 K

Answer

T3 = 1649 K
8.Gas power cycles

Answer: wnet = 419.9 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

T2 = 288·150.2857 = 624.3 K
T4 = 1400/150.2857 = 645.8 K

Compressor and turbine work

wc = 1.005·(624.3 − 288) = 338 kJ/kg
wt = 1.005·(1400 − 645.8) = 758 kJ/kg

Net work

wnet = 758 − 338 = 419.9 kJ/kg

Answer

wnet = 419.9 kJ/kg

Back-work ratio: wc/wt = 44.6%.

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