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
450 g of N₂V = 20 LT = 40 °CP = ?

A tank of 20 L contains 450 g of N₂ at 40 °C. Treat it as an ideal gas and find the pressure P.

m = 450 gM = 28.01 g/molV = 20 LT = 40 °CR = 8.314 J/(mol·K)
P = kPa
2.Ideal gas lawHard
CH₄, M = 16.04 g/molP = 800 kPaT = 90 °Cρ = ?

Find the density ρ of CH₄ at 800 kPa and 90 °C, treating it as an ideal gas.

M = 16.04 g/molP = 800 kPaT = 90 °CR = 8.314 J/(mol·K)
ρ = kg/m³
3.Energy balanceMedium
hot sidecold sidelight oil · 130 °C100 °Cwater · 15 °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: 130 → 100 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 130 °C100 °Cwater · 25 °C70 °C · ṁ = ?10% lost

Hot light oil 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 = 2.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 130 → 100 °Ct: 25 → 70 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 35 °CTL​ = -5 °CRQ̇H​ = ?Q̇L​ = 9 kWẆ

A reversible refrigerator keeps a cold room at -5 °C by removing 9 kW of heat, and rejects heat to the surroundings at 35 °C. Find the rate Q̇H at which it rejects heat.

TL = -5 °CTH = 35 °CQ̇L = 9 kW
Q̇H = kW
6.Heat engines & refrigeratorsHard
TH​ = 500 °CT0​ = 20 °CTL​ = -20 °CERWQ̇H​ = 70 kWQ̇L​ = ?

A Carnot heat engine receives 70 kW of heat from a source at 500 °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 -20 °C and also rejects heat to the surroundings at 20 °C. Find the rate Q̇L at which heat is removed from the freezer.

TH = 500 °CT0 = 20 °CTL = -20 °CQ̇H = 70 kW
Q̇L = kW
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 19 and a cutoff ratio rc = v₃/v₂ = 2.2. 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 = 19rc = 2.2k = 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 288 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 = 288 KT3 = 1100 Kk = 1.4cp = 1.005 kJ/(kg·K)
wnet = kJ/kg

3Solutions

1.Ideal gas law

Answer: P = 2091 kPa

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

Moles

n = m/M = 450/28.01 = 16.07 mol
T = 40 + 273.15 = 313.1 K

Pressure

P = nRT/V = 16.07·8.314·313.1/0.02 = 2091 kPa

Answer

P = 2091 kPa
2.Ideal gas law

Answer: ρ = 4.25 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 90 + 273.15 = 363.1 K

Density

ρ = 800·16.04/(8.314·363.1) = 4.25 kg/m³

Answer

ρ = 4.25 kg/m³
3.Energy balance

Answer: tout = 22.54 °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·(130 − 100) = 31.5 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 31.5/(1·4.18) = 22.54 °C

Answer

tout = 22.54 °C
4.Energy balance

Answer: ṁc = 0.7536 kg/s

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

Heat released

Qh = 2.5·2.1·(130 − 100) = 157.5 kW

Heat that reaches the water

Qc = (1 − 0.1)·157.5 = 141.8 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 141.8/(4.18·45) = 0.7536 kg/s

Answer

ṁc = 0.7536 kg/s
5.Heat engines & refrigerators

Answer: Q̇H = 10.34 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 = -5 + 273.15 = 268.1 K

Reversible heat ratio

Q̇H = Q̇L·TH/TL = 9·308.1/268.1 = 10.34 kW

Answer

Q̇H = 10.34 kW

Power input: Ẇ = 1.343 kW.

6.Heat engines & refrigerators

Answer: Q̇L = 275 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 = 500 + 273.15 = 773.1 K
T0 = 20 + 273.15 = 293.1 K
TL = -20 + 273.15 = 253.1 K

Engine work

η = 1 − 293.1/773.1 = 0.6208
Ẇ = 0.6208·70 = 43.46 kW

Heat removed by the refrigerator

COPR = 253.1/(293.1 − 253.1) = 6.329
Q̇L = 6.329·43.46 = 275 kW

Answer

Q̇L = 275 kW
7.Gas power cycles

Answer: η = 63.05 %

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

(2.21.4 − 1)/(1.4·(2.2 − 1)) = 1.2

Thermal efficiency

η = 1 − 0.308·1.2 = 63.05%

Answer

η = 63.05%
8.Gas power cycles

Answer: wnet = 249.9 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 = 288·180.2857 = 657.7 K
T4 = 1100/180.2857 = 481.7 K

Compressor and turbine work

wc = 1.005·(657.7 − 288) = 371.6 kJ/kg
wt = 1.005·(1100 − 481.7) = 621.4 kJ/kg

Net work

wnet = 621.4 − 371.6 = 249.9 kJ/kg

Answer

wnet = 249.9 kJ/kg

Back-work ratio: wc/wt = 59.79%.

ExamProblems · examproblems.com/courses/thermodynamics-1