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 CO₂P = 100 kPaT = 80 °CV = ?

What volume V does 400 g of CO₂ occupy at 100 kPa and 80 °C? Treat it as an ideal gas.

m = 400 gM = 44.01 g/molP = 100 kPaT = 80 °CR = 8.314 J/(mol·K)
V = L
2.Ideal gas lawHard
CH₄, M = 16.04 g/molP = 450 kPaT = 170 °Cρ = ?

Find the density ρ of CH₄ at 450 kPa and 170 °C, treating it as an ideal gas.

M = 16.04 g/molP = 450 kPaT = 170 °CR = 8.314 J/(mol·K)
ρ = kg/m³
3.Energy balanceMedium
hot sidecold sidelight oil · 180 °C130 °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 = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 180 → 130 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 120 °C80 °Cwater · 30 °C55 °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 = 1.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 120 → 80 °Ct: 30 → 55 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 35 °CTL​ = 0 °CRQ̇H​ = ?Q̇L​ = 4.5 kWẆ

A reversible refrigerator keeps a cold room at 0 °C by removing 4.5 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 = 4.5 kW
Q̇H = kW
6.Heat engines & refrigeratorsHard
TH​ = 22 °CTL​ = -5 °CHPQ̇H​ = 9.5 kWQ̇L​Ẇ = ?

A heat pump delivers 9.5 kW of heat to a heating system at 22 °C, taking heat from a source at -5 °C. Its actual COP is 30% of the Carnot COP between these temperatures. Find the electrical power Ẇ it uses.

TH = 22 °CTL = -5 °CηII = 30%Q̇H = 9.5 kW
Ẇ = kW
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 16 and a cutoff ratio rc = v₃/v₂ = 3. 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 = 16rc = 3k = 1.4
η = %
8.Gas power cyclesHard
vP1234qin​qout​

A gas-turbine plant runs on a Brayton cycle with a compressor pressure ratio of 14. Air enters the compressor at 290 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 = 14T1 = 290 KT3 = 1200 Kk = 1.4cp = 1.005 kJ/(kg·K)
wnet = kJ/kg

3Solutions

1.Ideal gas law

Answer: V = 266.9 L

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

Moles

n = 400/44.01 = 9.089 mol
T = 80 + 273.15 = 353.1 K

Volume

V = nRT/P = 9.089·8.314·353.1/100 = 266.9 L

Answer

V = 266.9 L
2.Ideal gas law

Answer: ρ = 1.959 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 170 + 273.15 = 443.1 K

Density

ρ = 450·16.04/(8.314·443.1) = 1.959 kg/m³

Answer

ρ = 1.959 kg/m³
3.Energy balance

Answer: tout = 36.28 °C

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

Heat released by the hot stream

Q = 0.5·2.1·(180 − 130) = 52.5 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 52.5/(2·4.18) = 36.28 °C

Answer

tout = 36.28 °C
4.Energy balance

Answer: ṁc = 1.964 kg/s

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

Heat released

Qh = 1.5·3.6·(120 − 80) = 216 kW

Heat that reaches the water

Qc = (1 − 0.05)·216 = 205.2 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 205.2/(4.18·25) = 1.964 kg/s

Answer

ṁc = 1.964 kg/s
5.Heat engines & refrigerators

Answer: Q̇H = 5.077 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 = 4.5·308.1/273.1 = 5.077 kW

Answer

Q̇H = 5.077 kW

Power input: Ẇ = 0.5766 kW.

6.Heat engines & refrigerators

Answer: Ẇ = 2.897 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 = -5 + 273.15 = 268.1 K

Carnot and actual COP

COPmax = 295.1/(295.1 − 268.1) = 10.93
COP = 0.3·10.93 = 3.279

Power input

Ẇ = Q̇H/COP = 9.5/3.279 = 2.897 kW

Answer

Ẇ = 2.897 kW
7.Gas power cycles

Answer: η = 56.93 %

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

(31.4 − 1)/(1.4·(3 − 1)) = 1.306

Thermal efficiency

η = 1 − 0.3299·1.306 = 56.93%

Answer

η = 56.93%
8.Gas power cycles

Answer: wnet = 310.6 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 = 290·140.2857 = 616.4 K
T4 = 1200/140.2857 = 564.6 K

Compressor and turbine work

wc = 1.005·(616.4 − 290) = 328 kJ/kg
wt = 1.005·(1200 − 564.6) = 638.6 kJ/kg

Net work

wnet = 638.6 − 328 = 310.6 kJ/kg

Answer

wnet = 310.6 kJ/kg

Back-work ratio: wc/wt = 51.37%.

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