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 200 mm of glass and 75 mm of wood, with an area of 20 m². The outer faces are at 21 °C and 6 °C. Find the heat flow Q̇ through the wall.
L1 = 200 mm, k1 = 0.96 W/(m·K)L2 = 75 mm, k2 = 0.17 W/(m·K)A = 20 m²T1 = 21 °CT3 = 6 °C
Q̇ = W
2.Heat conductionHard
A wall of 150 mm concrete and 75 mm wood separates a room at 21 °C from outside air at -7 °C. The convection coefficients are 7.7 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 4 m². Find the overall heat transfer coefficient U of the wall.
L1 = 150 mm, k1 = 1.4 W/(m·K)L2 = 75 mm, k2 = 0.17 W/(m·K)A = 4 m²hi = 7.7 W/(m²·K)ho = 20 W/(m²·K)Troom = 21 °CTout = -7 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 220 kW. The hot fluid cools from 170 to 150 °C and the cold fluid heats from 30 to 150 °C. The overall heat transfer coefficient is 1500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 220 kWU = 1500 W/(m²·K)Th: 170 → 150 °CTc: 30 → 150 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a counter-flow double-pipe heat exchanger, hot oil (0.3 kg/s, cp = 2.5 kJ/(kg·K)) flows in the inner tube and cools from 110 to 80 °C. Water (0.8 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 15 °C. The inner tube has a diameter of 20 mm, and U = 200 W/(m²·K) is based on this diameter. Find the tube length L needed.
ṁh = 0.3 kg/scp,h = 2.5 kJ/(kg·K)Th: 110 → 80 °Cṁc = 0.8 kg/sTc,in = 15 °CD = 20 mmU = 200 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Q̇ = 461.9 W
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) = (200/1000)/(0.96·20) = 0.01042 K/W
R2 = L2/(k2A) = (75/1000)/(0.17·20) = 0.02206 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 15/0.03248 = 461.9 W
Answer
Q̇ = 461.9 W
2.Heat conduction
Answer: U = 1.373 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/(7.7·4) = 0.03247 K/W
R1 = L1/(k1A) = (150/1000)/(1.4·4) = 0.02679 K/W
R2 = L2/(k2A) = (75/1000)/(0.17·4) = 0.1103 K/W
Ro = 1/(hoA) = 1/(20·4) = 0.0125 K/W
ΣR = 0.182 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(4·0.182) = 1.373 W/(m²·K)
Answer
U = 1.373 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 2.628 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 = 170 − 150 = 20 K
ΔT₂ = Th,out − Tc,in = 150 − 30 = 120 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (20 − 120)/ln(20/120) = 55.81 K
Area
A = Q̇/(U·ΔTlm) = 220000/(1500·55.81) = 2.628 m²
Answer
A = 2.628 m²
4.Heat exchangers (LMTD)
Answer: L = 23.55 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 = 0.3·2.5·(110 − 80) = 22.5 kW
Tc,out = 15 + 22.5/(0.8·4.18) = 21.73 °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 = 110 − 21.73 = 88.27 K
ΔT₂ = Th,out − Tc,in = 80 − 15 = 65 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (88.27 − 65)/ln(88.27/65) = 76.04 K
Area and length
A = 22500/(200·76.04) = 1.479 m²
L = A/(πD) = 1.479/(π·0.02) = 23.55 m
Answer
L = 23.55 m
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