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
Thermodynamics 1: cram pack
Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.
1Formulas and methods
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
What volume V does 320 g of N₂ occupy at 650 kPa and 25 °C? Treat it as an ideal gas.
m = 320 gM = 28.01 g/molP = 650 kPaT = 25 °CR = 8.314 J/(mol·K)
V = L
2.Ideal gas lawHard
Find the density ρ of N₂ at 450 kPa and 0 °C, treating it as an ideal gas.
M = 28.01 g/molP = 450 kPaT = 0 °CR = 8.314 J/(mol·K)
ρ = kg/m³
3.Energy balanceMedium
Hot light oil cools water in a counter-flow heat exchanger with no heat loss. Find the water outlet temperature tout.
ṁh = 1.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 140 → 80 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
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 → 100 °Ct: 20 → 90 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
A reversible (Carnot) heat engine takes in 3500 kJ of heat from a reservoir at 325 °C and rejects 1628 kJ per cycle. Find the temperature TL of the cold reservoir in °C.
TH = 325 °CQH = 3500 kJQL = 1628 kJ
TL = °C
6.Heat engines & refrigeratorsHard
A refrigerator keeps its cold space at -20 °C in a kitchen at 25 °C, removing heat from the cold space at 2 kW. Find the minimum power input Ẇmin.
TH = 25 °CTL = -20 °CQ̇L = 2 kW
Ẇmin = kW
7.Gas power cyclesMedium
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 305 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 = 2T1 = 305 Kk = 1.4
T3 = K
8.Gas power cyclesHard
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 1100 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 = 18T1 = 300 KT3 = 1100 Kk = 1.4cp = 1.005 kJ/(kg·K)
wnet = kJ/kg
3Solutions
1.Ideal gas law
Answer: V = 43.57 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 320/28.01 = 11.42 mol
T = 25 + 273.15 = 298.1 K
Volume
V = nRT/P = 11.42·8.314·298.1/650 = 43.57 L
Answer
V = 43.57 L
2.Ideal gas law
Answer: ρ = 5.55 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 0 + 273.15 = 273.1 K
Density
ρ = 450·28.01/(8.314·273.1) = 5.55 kg/m³
Answer
ρ = 5.55 kg/m³
3.Energy balance
Answer: tout = 75.22 °C
Exchanger energy balance: with no losses, ṁhcp,h(Th,in − Th,out) = ṁccp,c(tout − tin).
Heat released by the hot stream
Q = 1.5·2.1·(140 − 80) = 189 kW
Heat picked up by the water
tout = tin + Q/(ṁccp,c) = 30 + 189/(1·4.18) = 75.22 °C
Answer
tout = 75.22 °C
4.Energy balance
Answer: ṁc = 1.418 kg/s
Energy balance with losses: heat released = heat absorbed + heat lost.
Heat released
Qh = 2.5·2.44·(180 − 100) = 488 kW
Heat that reaches the water
Qc = (1 − 0.15)·488 = 414.8 kW
Water flow rate
ṁc = Qc/(cp,c·Δt) = 414.8/(4.18·70) = 1.418 kg/s
Answer
ṁc = 1.418 kg/s
5.Heat engines & refrigerators
Answer: TL = 5.075 °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 temperature of the source
TH = 325 + 273.15 = 598.1 K
Reversible heat ratio
TL = TH·QL/QH = 598.1·1628/3500 = 278.2 K = 5.075 °C
Answer
TL = 5.075 °C
6.Heat engines & refrigerators
Answer: Ẇmin = 0.3555 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 = 25 + 273.15 = 298.1 K
TL = -20 + 273.15 = 253.1 K
Best possible COP
COPR,max = TL/(TH − TL) = 253.1/(298.1 − 253.1) = 5.626
Minimum power
The least work is needed by a reversible machine, which has the highest COP.
Ẇmin = Q̇L/COP = 2/5.626 = 0.3555 kW
Answer
Ẇmin = 0.3555 kW
7.Gas power cycles
Answer: T3 = 2100 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 = 305·220.4 = 1050 K
Heat addition at constant pressure
T3 = rc·T2 = 2·1050 = 2100 K
Answer
T3 = 2100 K
8.Gas power cycles
Answer: wnet = 234.4 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 = 300·180.2857 = 685.1 K
T4 = 1100/180.2857 = 481.7 K
Compressor and turbine work
wc = 1.005·(685.1 − 300) = 387.1 kJ/kg
wt = 1.005·(1100 − 481.7) = 621.4 kJ/kg
Net work
wnet = 621.4 − 387.1 = 234.4 kJ/kg
Answer
wnet = 234.4 kJ/kg
Back-work ratio: wc/wt = 62.28%.
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