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
100 g of N₂V = 30 LT = 15 °CP = ?

A tank of 30 L contains 100 g of N₂ at 15 °C. Treat it as an ideal gas and find the pressure P.

m = 100 gM = 28.01 g/molV = 30 LT = 15 °CR = 8.314 J/(mol·K)
P = kPa
2.Ideal gas lawHard
O₂, M = 32 g/molP = 400 kPaT = 30 °Cρ = ?

Find the density ρ of O₂ at 400 kPa and 30 °C, treating it as an ideal gas.

M = 32 g/molP = 400 kPaT = 30 °CR = 8.314 J/(mol·K)
ρ = kg/m³
3.Energy balanceMedium
hot sidecold sideglycol mix · 160 °C90 °Cwater · 20 °Ct = ? °C

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

ṁh = 3 kg/scp,h = 3.6 kJ/(kg·K)Th: 160 → 90 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 180 °C150 °Cwater · 20 °C85 °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: 180 → 150 °Ct: 20 → 85 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 28 °CTL​ = -30 °CRQ̇H​ = ?Q̇L​ = 3.5 kWẆ

A reversible refrigerator keeps a cold room at -30 °C by removing 3.5 kW of heat, and rejects heat to the surroundings at 28 °C. Find the rate Q̇H at which it rejects heat.

TL = -30 °CTH = 28 °CQ̇L = 3.5 kW
Q̇H = kW
6.Heat engines & refrigeratorsHard
TH​ = 700 °CT0​ = 25 °CTL​ = -20 °CERWQ̇H​ = 160 kWQ̇L​ = ?

A Carnot heat engine receives 160 kW of heat from a source at 700 °C and rejects heat to the surroundings at 25 °C. All of its work drives a Carnot refrigerator that removes heat from a freezer at -20 °C and also rejects heat to the surroundings at 25 °C. Find the rate Q̇L at which heat is removed from the freezer.

TH = 700 °CT0 = 25 °CTL = -20 °CQ̇H = 160 kW
Q̇L = kW
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 22 and a cutoff ratio rc = v₃/v₂ = 2. At the start of compression the air is at 300 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 heat added qin per kilogram of air, in kJ/kg.

r = 22rc = 2T1 = 300 Kk = 1.4cp = 1.005 kJ/(kg·K)
qin = kJ/kg
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 1200 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 = 16T1 = 288 KT3 = 1200 Kk = 1.4cp = 1.005 kJ/(kg·K)
wnet = kJ/kg

3Solutions

1.Ideal gas law

Answer: P = 285.1 kPa

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

Moles

n = m/M = 100/28.01 = 3.57 mol
T = 15 + 273.15 = 288.1 K

Pressure

P = nRT/V = 3.57·8.314·288.1/0.03 = 285.1 kPa

Answer

P = 285.1 kPa
2.Ideal gas law

Answer: ρ = 5.079 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 30 + 273.15 = 303.1 K

Density

ρ = 400·32/(8.314·303.1) = 5.079 kg/m³

Answer

ρ = 5.079 kg/m³
3.Energy balance

Answer: tout = 80.29 °C

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

Heat released by the hot stream

Q = 3·3.6·(160 − 90) = 756 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 756/(3·4.18) = 80.29 °C

Answer

tout = 80.29 °C
4.Energy balance

Answer: ṁc = 0.5725 kg/s

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

Heat released

Qh = 2.5·2.44·(180 − 150) = 183 kW

Heat that reaches the water

Qc = (1 − 0.15)·183 = 155.5 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 155.5/(4.18·65) = 0.5725 kg/s

Answer

ṁc = 0.5725 kg/s
5.Heat engines & refrigerators

Answer: Q̇H = 4.335 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 = 28 + 273.15 = 301.1 K
TL = -30 + 273.15 = 243.1 K

Reversible heat ratio

Q̇H = Q̇L·TH/TL = 3.5·301.1/243.1 = 4.335 kW

Answer

Q̇H = 4.335 kW

Power input: Ẇ = 0.8349 kW.

6.Heat engines & refrigerators

Answer: Q̇L = 624.3 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 = 700 + 273.15 = 973.1 K
T0 = 25 + 273.15 = 298.1 K
TL = -20 + 273.15 = 253.1 K

Engine work

η = 1 − 298.1/973.1 = 0.6936
Ẇ = 0.6936·160 = 111 kW

Heat removed by the refrigerator

COPR = 253.1/(298.1 − 253.1) = 5.626
Q̇L = 5.626·111 = 624.3 kW

Answer

Q̇L = 624.3 kW
7.Gas power cycles

Answer: qin = 1038 kJ/kg

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)).

Temperatures

T2 = 300·220.4 = 1033 K
T3 = 2·1033 = 2066 K

Heat added at constant pressure

qin = 1.005·(2066 − 1033) = 1038 kJ/kg

Answer

qin = 1038 kJ/kg
8.Gas power cycles

Answer: wnet = 310.2 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 = 288·160.2857 = 636 K
T4 = 1200/160.2857 = 543.4 K

Compressor and turbine work

wc = 1.005·(636 − 288) = 349.7 kJ/kg
wt = 1.005·(1200 − 543.4) = 659.8 kJ/kg

Net work

wnet = 659.8 − 349.7 = 310.2 kJ/kg

Answer

wnet = 310.2 kJ/kg

Back-work ratio: wc/wt = 53%.

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