Thermodynamics 1: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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
420 g of CH₄P = 450 kPaT = 25 °CV = ?

What volume V does 420 g of CH₄ occupy at 450 kPa and 25 °C? Treat it as an ideal gas.

m = 420 gM = 16.04 g/molP = 450 kPaT = 25 °CR = 8.314 J/(mol·K)
V = L
2.Ideal gas lawHard
CH₄, M = 16.04 g/molP = 300 kPaT = 70 °Cρ = ?

Find the density ρ of CH₄ at 300 kPa and 70 °C, treating it as an ideal gas.

M = 16.04 g/molP = 300 kPaT = 70 °CR = 8.314 J/(mol·K)
ρ = kg/m³
3.Energy balanceMedium
hot sidecold sidelight oil · 140 °C110 °Cwater · 20 °Ct = ? °C

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 → 110 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 120 °C60 °Cwater · 10 °C35 °C · ṁ = ?10% lost

Hot glycol mix heats water in an exchanger that loses 10% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁc needed.

ṁh = 1 kg/scp,h = 3.6 kJ/(kg·K)Th: 120 → 60 °Ct: 10 → 35 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 625 °CTL​ = 30 °CEQH​QL​W

A heat engine works between a hot reservoir at 625 °C and a cold reservoir at 30 °C. Find the highest possible thermal efficiency ηmax.

TH = 625 °CTL = 30 °C
ηmax = %
6.Heat engines & refrigeratorsHard
TH​ = 800 °CT0​ = 25 °CTL​ = -20 °CERWQ̇H​ = 50 kWQ̇L​ = ?

A Carnot heat engine receives 50 kW of heat from a source at 800 °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 = 800 °CT0 = 25 °CTL = -20 °CQ̇H = 50 kW
Q̇L = kW
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 16 and a cutoff ratio rc = v₃/v₂ = 2.5. At the start of compression the air is at 290 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 = 16rc = 2.5T1 = 290 Kk = 1.4
T3 = K
8.Gas power cyclesHard
vP1234qin​qout​

An ideal Brayton cycle operates with a compressor pressure ratio of 15. 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.

rp = 15k = 1.4
η = %

3Solutions

1.Ideal gas law

Answer: V = 144.2 L

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

Moles

n = 420/16.04 = 26.18 mol
T = 25 + 273.15 = 298.1 K

Volume

V = nRT/P = 26.18·8.314·298.1/450 = 144.2 L

Answer

V = 144.2 L
2.Ideal gas law

Answer: ρ = 1.687 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 70 + 273.15 = 343.1 K

Density

ρ = 300·16.04/(8.314·343.1) = 1.687 kg/m³

Answer

ρ = 1.687 kg/m³
3.Energy balance

Answer: tout = 65.22 °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 − 110) = 94.5 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 94.5/(0.5·4.18) = 65.22 °C

Answer

tout = 65.22 °C
4.Energy balance

Answer: ṁc = 1.86 kg/s

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

Heat released

Qh = 1·3.6·(120 − 60) = 216 kW

Heat that reaches the water

Qc = (1 − 0.1)·216 = 194.4 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 194.4/(4.18·25) = 1.86 kg/s

Answer

ṁc = 1.86 kg/s
5.Heat engines & refrigerators

Answer: ηmax = 66.25 %

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 = 625 + 273.15 = 898.1 K
TL = 30 + 273.15 = 303.1 K

Carnot efficiency

ηmax = 1 − TL/TH = 1 − 303.1/898.1 = 66.25 %

Answer

ηmax = 66.25 %
6.Heat engines & refrigerators

Answer: Q̇L = 203.1 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 = 800 + 273.15 = 1073 K
T0 = 25 + 273.15 = 298.1 K
TL = -20 + 273.15 = 253.1 K

Engine work

η = 1 − 298.1/1073 = 0.7222
Ẇ = 0.7222·50 = 36.11 kW

Heat removed by the refrigerator

COPR = 253.1/(298.1 − 253.1) = 5.626
Q̇L = 5.626·36.11 = 203.1 kW

Answer

Q̇L = 203.1 kW
7.Gas power cycles

Answer: T3 = 2198 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 = 290·160.4 = 879.1 K

Heat addition at constant pressure

T3 = rc·T2 = 2.5·879.1 = 2198 K

Answer

T3 = 2198 K
8.Gas power cycles

Answer: η = 53.87 %

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.

Thermal efficiency

η = 1 − 1/150.2857 = 1 − 1/2.168 = 53.87%

Answer

η = 53.87%

ExamProblems · examproblems.com/courses/thermodynamics-1