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
490 g of N₂V = 10 LT = 30 °CP = ?

A tank of 10 L contains 490 g of N₂ at 30 °C. Treat it as an ideal gas and find the pressure P.

m = 490 gM = 28.01 g/molV = 10 LT = 30 °CR = 8.314 J/(mol·K)
P = kPa
2.Ideal gas lawHard
State 1State 2

A fixed amount of Ar goes from 150 kPa, 20 L, 20 °C to 500 kPa and 130 °C. Find the new volume V2.

P1 = 150 kPaV1 = 20 LT1 = 20 °CP2 = 500 kPaT2 = 130 °C
V2 = L
3.Energy balanceMedium
hot sidecold sidelight oil · 170 °C100 °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 = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 170 → 100 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 130 °C90 °Cwater · 30 °C60 °C · ṁ = ?5% lost

Hot ethanol 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 = 0.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 130 → 90 °Ct: 30 → 60 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 325 °CTL​ = ?EQH​ = 2600 kJQL​ = 1231 kJW

A reversible (Carnot) heat engine takes in 2600 kJ of heat from a reservoir at 325 °C and rejects 1231 kJ per cycle. Find the temperature TL of the cold reservoir in °C.

TH = 325 °CQH = 2600 kJQL = 1231 kJ
TL = °C
6.Heat engines & refrigeratorsHard
TH​ = 22 °CTL​ = -5 °CHPQ̇H​ = 6 kWQ̇L​Ẇ = ?

A heat pump keeps a house at 22 °C while it is -5 °C outside, delivering 6 kW of heat to the house. Find the minimum power input Ẇmin.

TH = 22 °CTL = -5 °CQ̇H = 6 kW
Ẇmin = kW
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 18 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 heat added qin per kilogram of air, in kJ/kg.

r = 18rc = 2.5T1 = 290 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 14. Air enters the compressor at 290 K and the turbine at 1400 K. The compressor has an isentropic efficiency of 80% and the turbine 85%. 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 = 1400 Kηc = 80%ηt = 85%k = 1.4cp = 1.005 kJ/(kg·K)
wnet = kJ/kg

3Solutions

1.Ideal gas law

Answer: P = 4409 kPa

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

Moles

n = m/M = 490/28.01 = 17.49 mol
T = 30 + 273.15 = 303.1 K

Pressure

P = nRT/V = 17.49·8.314·303.1/0.01 = 4409 kPa

Answer

P = 4409 kPa
2.Ideal gas law

Answer: V2 = 8.251 L

Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.

Absolute temperatures

T1 = 293.1 K, T2 = 403.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 20·(150/500)·(403.1/293.1) = 8.251 L

Answer

V2 = 8.251 L
3.Energy balance

Answer: tout = 28.79 °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·(170 − 100) = 73.5 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 73.5/(2·4.18) = 28.79 °C

Answer

tout = 28.79 °C
4.Energy balance

Answer: ṁc = 0.3697 kg/s

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

Heat released

Qh = 0.5·2.44·(130 − 90) = 48.8 kW

Heat that reaches the water

Qc = (1 − 0.05)·48.8 = 46.36 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 46.36/(4.18·30) = 0.3697 kg/s

Answer

ṁc = 0.3697 kg/s
5.Heat engines & refrigerators

Answer: TL = 10.05 °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 = 598.1·1231/2600 = 283.2 K = 10.05 °C

Answer

TL = 10.05 °C
6.Heat engines & refrigerators

Answer: Ẇmin = 0.5489 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

Best possible COP

COPHP,max = TH/(TH − TL) = 295.1/(295.1 − 268.1) = 10.93

Minimum power

The least work is needed by a reversible machine, which has the highest COP.

Ẇmin = Q̇H/COP = 6/10.93 = 0.5489 kW

Answer

Ẇmin = 0.5489 kW
7.Gas power cycles

Answer: qin = 1389 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 = 290·180.4 = 921.5 K
T3 = 2.5·921.5 = 2304 K

Heat added at constant pressure

qin = 1.005·(2304 − 921.5) = 1389 kJ/kg

Answer

qin = 1389 kJ/kg
8.Gas power cycles

Answer: wnet = 223.2 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

T2s = 290·140.2857 = 616.4 K
T4s = 1400/140.2857 = 658.7 K

Compressor and turbine work

wc = 1.005·(616.4 − 290)/0.8 = 410 kJ/kg
wt = 0.85·1.005·(1400 − 658.7) = 633.3 kJ/kg

Net work

wnet = 633.3 − 410 = 223.2 kJ/kg

Answer

wnet = 223.2 kJ/kg

Back-work ratio: wc/wt = 64.75%.

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