Heat conduction
- Fourier’s law
- Q̇ = kAΔT/L
- Conduction resistance
- R = L/(kA)
- Convection resistance
- R = 1/(hA)
- Resistances in series
- Q̇ = ΔTtotal/ΣR
- U-value
- U = 1/(ΣR·A)
Heat Transfer: cram pack
Formula sheets, the key method steps for every topic and 4 practice problems, with full solutions at the back.
1Formulas and methods
Heat exchangers (LMTD)
- Heat duty
- Q = U·A·ΔTlm
- Log mean temperature difference
- ΔTlm = (ΔT1 − ΔT2)/ln(ΔT1/ΔT2)
- Stream balance
- Q = ṁcp(Tin − Tout)
2Practice problems
1.Heat conductionMedium
A wall is made of 250 mm of brick and 75 mm of expanded polystyrene, with an area of 15 m². The outer faces are at 15 °C and 4 °C. Find the temperature Ti at the interface between the two layers.
L1 = 250 mm, k1 = 0.72 W/(m·K)L2 = 75 mm, k2 = 0.035 W/(m·K)A = 15 m²T1 = 15 °CT3 = 4 °C
Ti = °C
2.Heat conductionHard
A wall of 100 mm mineral wool and 100 mm wood separates a room at 20 °C from outside air at -12 °C. The convection coefficients are 10 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 4 m². Find the overall heat transfer coefficient U of the wall.
L1 = 100 mm, k1 = 0.04 W/(m·K)L2 = 100 mm, k2 = 0.17 W/(m·K)A = 4 m²hi = 10 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = -12 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 100 kW. The hot fluid cools from 115 to 75 °C and the cold fluid heats from 25 to 75 °C. The overall heat transfer coefficient is 800 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 100 kWU = 800 W/(m²·K)Th: 115 → 75 °CTc: 25 → 75 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a counter-flow double-pipe heat exchanger, hot oil (0.8 kg/s, cp = 2.3 kJ/(kg·K)) flows in the inner tube and cools from 160 to 95 °C. Water (0.5 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 20 °C. The inner tube has a diameter of 25 mm, and U = 150 W/(m²·K) is based on this diameter. Find the tube length L needed.
ṁh = 0.8 kg/scp,h = 2.3 kJ/(kg·K)Th: 160 → 95 °Cṁc = 0.5 kg/sTc,in = 20 °CD = 25 mmU = 150 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Ti = 13.47 °C
Thermal resistance: steady heat flow through layers in series behaves like current through resistors in series: Q̇ = ΔT/ΣR, with R = L/(kA) for a conducting layer and R = 1/(hA) for a convecting surface. The temperature drop across each layer is proportional to its resistance.
Layer resistances
R1 = L1/(k1A) = (250/1000)/(0.72·15) = 0.02315 K/W
R2 = L2/(k2A) = (75/1000)/(0.035·15) = 0.1429 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 11/0.166 = 66.26 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 15 − 66.26·0.02315 = 13.47 °C
Answer
Ti = 13.47 °C
2.Heat conduction
Answer: U = 0.3098 W/(m²·K)
Thermal resistance: steady heat flow through layers in series behaves like current through resistors in series: Q̇ = ΔT/ΣR, with R = L/(kA) for a conducting layer and R = 1/(hA) for a convecting surface. The temperature drop across each layer is proportional to its resistance.
Resistances in series
Ri = 1/(hiA) = 1/(10·4) = 0.025 K/W
R1 = L1/(k1A) = (100/1000)/(0.04·4) = 0.625 K/W
R2 = L2/(k2A) = (100/1000)/(0.17·4) = 0.1471 K/W
Ro = 1/(hoA) = 1/(25·4) = 0.01 K/W
ΣR = 0.8071 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(4·0.8071) = 0.3098 W/(m²·K)
Answer
U = 0.3098 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 2.789 m²
LMTD method: the temperature difference between the streams changes along the exchanger, so the rate equation uses its logarithmic mean: Q̇ = U·A·ΔTlm, with ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂), where ΔT₁ and ΔT₂ are the differences at the two ends. Counter-flow gives a larger ΔTlm than parallel flow for the same inlet and outlet temperatures.
End temperature differences
In counter-flow the hot inlet faces the cold outlet, and the hot outlet faces the cold inlet.
ΔT₁ = Th,in − Tc,out = 115 − 75 = 40 K
ΔT₂ = Th,out − Tc,in = 75 − 25 = 50 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (40 − 50)/ln(40/50) = 44.81 K
Area
A = Q̇/(U·ΔTlm) = 100000/(800·44.81) = 2.789 m²
Answer
A = 2.789 m²
4.Heat exchangers (LMTD)
Answer: L = 128.8 m
LMTD method: the temperature difference between the streams changes along the exchanger, so the rate equation uses its logarithmic mean: Q̇ = U·A·ΔTlm, with ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂), where ΔT₁ and ΔT₂ are the differences at the two ends. Counter-flow gives a larger ΔTlm than parallel flow for the same inlet and outlet temperatures.
Heat duty and water outlet temperature
Q̇ = 0.8·2.3·(160 − 95) = 119.6 kW
Tc,out = 20 + 119.6/(0.5·4.18) = 77.22 °C
End temperature differences
In counter-flow the hot inlet faces the cold outlet, and the hot outlet faces the cold inlet.
ΔT₁ = Th,in − Tc,out = 160 − 77.22 = 82.78 K
ΔT₂ = Th,out − Tc,in = 95 − 20 = 75 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (82.78 − 75)/ln(82.78/75) = 78.82 K
Area and length
A = 119600/(150·78.82) = 10.12 m²
L = A/(πD) = 10.12/(π·0.025) = 128.8 m
Answer
L = 128.8 m
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