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
80 g of ArP = 350 kPaT = 90 °CV = ?

What volume V does 80 g of Ar occupy at 350 kPa and 90 °C? Treat it as an ideal gas.

m = 80 gM = 39.95 g/molP = 350 kPaT = 90 °CR = 8.314 J/(mol·K)
V = L
2.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 100 kPa, 2 L, 25 °C to 750 kPa and 130 °C. Find the new volume V2.

P1 = 100 kPaV1 = 2 LT1 = 25 °CP2 = 750 kPaT2 = 130 °C
V2 = L
3.Energy balanceMedium
hot sidecold sidelight oil · 150 °C90 °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 = 1 kg/scp,h = 2.1 kJ/(kg·K)Th: 150 → 90 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 140 °C90 °Cwater · 20 °C50 °C · ṁ = ?15% lost

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 = 3 kg/scp,h = 2.44 kJ/(kg·K)Th: 140 → 90 °Ct: 20 → 50 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Heat engines & refrigeratorsMedium
TH​ = 25 °CTL​ = 5 °CRQ̇H​ = ?Q̇L​ = 4.5 kWẆ

A reversible refrigerator keeps a cold room at 5 °C by removing 4.5 kW of heat, and rejects heat to the surroundings at 25 °C. Find the rate Q̇H at which it rejects heat.

TL = 5 °CTH = 25 °CQ̇L = 4.5 kW
Q̇H = kW
6.Heat engines & refrigeratorsHard
TH​ = 25 °CTL​ = -20 °CRQ̇H​Q̇L​ = 1.5 kWẆ = ?

A refrigerator keeps its cold space at -20 °C in a kitchen at 25 °C, removing heat from the cold space at 1.5 kW. Find the minimum power input Ẇmin.

TH = 25 °CTL = -20 °CQ̇L = 1.5 kW
Ẇmin = kW
7.Gas power cyclesMedium
vP1234qin​qout​

An ideal Diesel cycle has a compression ratio r = 20 and a cutoff ratio rc = v₃/v₂ = 2.5. At the start of compression the air is at 295 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 = 20rc = 2.5T1 = 295 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 6. 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 = 6T1 = 290 KT3 = 1200 Kk = 1.4cp = 1.005 kJ/(kg·K)
wnet = kJ/kg

3Solutions

1.Ideal gas law

Answer: V = 17.27 L

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

Moles

n = 80/39.95 = 2.003 mol
T = 90 + 273.15 = 363.1 K

Volume

V = nRT/P = 2.003·8.314·363.1/350 = 17.27 L

Answer

V = 17.27 L
2.Ideal gas law

Answer: V2 = 0.3606 L

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

Absolute temperatures

T1 = 298.1 K, T2 = 403.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 2·(100/750)·(403.1/298.1) = 0.3606 L

Answer

V2 = 0.3606 L
3.Energy balance

Answer: tout = 50.1 °C

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

Heat released by the hot stream

Q = 1·2.1·(150 − 90) = 126 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 126/(1.5·4.18) = 50.1 °C

Answer

tout = 50.1 °C
4.Energy balance

Answer: ṁc = 2.481 kg/s

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

Heat released

Qh = 3·2.44·(140 − 90) = 366 kW

Heat that reaches the water

Qc = (1 − 0.15)·366 = 311.1 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 311.1/(4.18·30) = 2.481 kg/s

Answer

ṁc = 2.481 kg/s
5.Heat engines & refrigerators

Answer: Q̇H = 4.824 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 = 5 + 273.15 = 278.1 K

Reversible heat ratio

Q̇H = Q̇L·TH/TL = 4.5·298.1/278.1 = 4.824 kW

Answer

Q̇H = 4.824 kW

Power input: Ẇ = 0.3236 kW.

6.Heat engines & refrigerators

Answer: Ẇmin = 0.2666 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 = 1.5/5.626 = 0.2666 kW

Answer

Ẇmin = 0.2666 kW
7.Gas power cycles

Answer: qin = 1474 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 = 295·200.4 = 977.8 K
T3 = 2.5·977.8 = 2444 K

Heat added at constant pressure

qin = 1.005·(2444 − 977.8) = 1474 kJ/kg

Answer

qin = 1474 kJ/kg
8.Gas power cycles

Answer: wnet = 288.4 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 = 290·60.2857 = 483.9 K
T4 = 1200/60.2857 = 719.2 K

Compressor and turbine work

wc = 1.005·(483.9 − 290) = 194.8 kJ/kg
wt = 1.005·(1200 − 719.2) = 483.2 kJ/kg

Net work

wnet = 483.2 − 194.8 = 288.4 kJ/kg

Answer

wnet = 288.4 kJ/kg

Back-work ratio: wc/wt = 40.32%.

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