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
400 g of N₂P = 800 kPaT = 45 °CV = ?

What volume V does 400 g of N₂ occupy at 800 kPa and 45 °C? Treat it as an ideal gas.

m = 400 gM = 28.01 g/molP = 800 kPaT = 45 °CR = 8.314 J/(mol·K)
V = L
2.Ideal gas lawHard
State 1State 2

A fixed amount of CO₂ goes from 350 kPa, 5 L, 35 °C to 500 kPa and 180 °C. Find the new volume V2.

P1 = 350 kPaV1 = 5 LT1 = 35 °CP2 = 500 kPaT2 = 180 °C
V2 = L
3.Energy balanceMedium
hot sidecold sidelight oil · 150 °C90 °Cwater · 30 °Ct = ? °C

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

ṁh = 2 kg/scp,h = 2.1 kJ/(kg·K)Th: 150 → 90 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 130 °C60 °Cwater · 25 °C65 °C · ṁ = ?15% lost

Hot ethanol heats water in an exchanger that loses 15% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁc needed.

ṁh = 2.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 130 → 60 °Ct: 25 → 65 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 30 °CTL​ = -30 °CRQ̇H​ = ?Q̇L​ = 5.5 kWẆ

A reversible refrigerator keeps a cold room at -30 °C by removing 5.5 kW of heat, and rejects heat to the surroundings at 30 °C. Find the rate Q̇H at which it rejects heat.

TL = -30 °CTH = 30 °CQ̇L = 5.5 kW
Q̇H = kW
6.Heat engines & refrigeratorsHard
TH​ = 600 °CT0​ = 20 °CTL​ = -10 °CERWQ̇H​ = 140 kWQ̇L​ = ?

A Carnot heat engine receives 140 kW of heat from a source at 600 °C and rejects heat to the surroundings at 20 °C. All of its work drives a Carnot refrigerator that removes heat from a freezer at -10 °C and also rejects heat to the surroundings at 20 °C. Find the rate Q̇L at which heat is removed from the freezer.

TH = 600 °CT0 = 20 °CTL = -10 °CQ̇H = 140 kW
Q̇L = kW
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 14 and a cutoff ratio rc = v₃/v₂ = 1.5. 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 = 14rc = 1.5k = 1.4
η = %
8.Gas power cyclesHard
vP1234qin​qout​

A gas-turbine plant runs on a Brayton cycle with a compressor pressure ratio of 18. Air enters the compressor at 300 K and the turbine at 1400 K. The compressor has an isentropic efficiency of 88% and the turbine 92%. 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 = 18T1 = 300 KT3 = 1400 Kηc = 88%ηt = 92%k = 1.4cp = 1.005 kJ/(kg·K)
wnet = kJ/kg

3Solutions

1.Ideal gas law

Answer: V = 47.22 L

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

Moles

n = 400/28.01 = 14.28 mol
T = 45 + 273.15 = 318.1 K

Volume

V = nRT/P = 14.28·8.314·318.1/800 = 47.22 L

Answer

V = 47.22 L
2.Ideal gas law

Answer: V2 = 5.147 L

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

Absolute temperatures

T1 = 308.1 K, T2 = 453.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 5·(350/500)·(453.1/308.1) = 5.147 L

Answer

V2 = 5.147 L
3.Energy balance

Answer: tout = 50.1 °C

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

Heat released by the hot stream

Q = 2·2.1·(150 − 90) = 252 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 252/(3·4.18) = 50.1 °C

Answer

tout = 50.1 °C
4.Energy balance

Answer: ṁc = 2.171 kg/s

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

Heat released

Qh = 2.5·2.44·(130 − 60) = 427 kW

Heat that reaches the water

Qc = (1 − 0.15)·427 = 362.9 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 362.9/(4.18·40) = 2.171 kg/s

Answer

ṁc = 2.171 kg/s
5.Heat engines & refrigerators

Answer: Q̇H = 6.857 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 = 30 + 273.15 = 303.1 K
TL = -30 + 273.15 = 243.1 K

Reversible heat ratio

Q̇H = Q̇L·TH/TL = 5.5·303.1/243.1 = 6.857 kW

Answer

Q̇H = 6.857 kW

Power input: Ẇ = 1.357 kW.

6.Heat engines & refrigerators

Answer: Q̇L = 815.7 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 = 600 + 273.15 = 873.1 K
T0 = 20 + 273.15 = 293.1 K
TL = -10 + 273.15 = 263.1 K

Engine work

η = 1 − 293.1/873.1 = 0.6643
Ẇ = 0.6643·140 = 93 kW

Heat removed by the refrigerator

COPR = 263.1/(293.1 − 263.1) = 8.772
Q̇L = 8.772·93 = 815.7 kW

Answer

Q̇L = 815.7 kW
7.Gas power cycles

Answer: η = 62.02 %

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

(1.51.4 − 1)/(1.4·(1.5 − 1)) = 1.092

Thermal efficiency

η = 1 − 0.348·1.092 = 62.02%

Answer

η = 62.02%
8.Gas power cycles

Answer: wnet = 287.8 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 = 300·180.2857 = 685.1 K
T4s = 1400/180.2857 = 613 K

Compressor and turbine work

wc = 1.005·(685.1 − 300)/0.88 = 439.8 kJ/kg
wt = 0.92·1.005·(1400 − 613) = 727.6 kJ/kg

Net work

wnet = 727.6 − 439.8 = 287.8 kJ/kg

Answer

wnet = 287.8 kJ/kg

Back-work ratio: wc/wt = 60.45%.

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