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 50 mm of wood and 50 mm of brick, with an area of 4 m². The outer faces are at 18 °C and 1 °C. Find the heat flow Q̇ through the wall.
L1 = 50 mm, k1 = 0.17 W/(m·K)L2 = 50 mm, k2 = 0.72 W/(m·K)A = 4 m²T1 = 18 °CT3 = 1 °C
Q̇ = W
2.Heat conductionHard
A wall of 150 mm wood and 75 mm glass separates a room at 22 °C from outside air at -2 °C. The convection coefficients are 6 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 8 m². Find the overall heat transfer coefficient U of the wall.
L1 = 150 mm, k1 = 0.17 W/(m·K)L2 = 75 mm, k2 = 0.96 W/(m·K)A = 8 m²hi = 6 W/(m²·K)ho = 20 W/(m²·K)Troom = 22 °CTout = -2 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 90 kW. Its heat transfer area is 8 m². The hot fluid cools from 85 to 50 °C and the cold fluid heats from 10 to 75 °C. Find the overall heat transfer coefficient U.
Q̇ = 90 kWA = 8 m²Th: 85 → 50 °CTc: 10 → 75 °C
U = W/(m²·K)
4.Heat exchangers (LMTD)Hard
In a counter-flow double-pipe heat exchanger, hot oil (0.3 kg/s, cp = 2.3 kJ/(kg·K)) flows in the inner tube and cools from 135 to 100 °C. Water (1.5 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 10 °C. The inner tube has a diameter of 25 mm, and U = 400 W/(m²·K) is based on this diameter. Find the tube length L needed.
ṁh = 0.3 kg/scp,h = 2.3 kJ/(kg·K)Th: 135 → 100 °Cṁc = 1.5 kg/sTc,in = 10 °CD = 25 mmU = 400 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Q̇ = 187 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) = (50/1000)/(0.17·4) = 0.07353 K/W
R2 = L2/(k2A) = (50/1000)/(0.72·4) = 0.01736 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 17/0.09089 = 187 W
Answer
Q̇ = 187 W
2.Heat conduction
Answer: U = 0.8495 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/(6·8) = 0.02083 K/W
R1 = L1/(k1A) = (150/1000)/(0.17·8) = 0.1103 K/W
R2 = L2/(k2A) = (75/1000)/(0.96·8) = 0.009766 K/W
Ro = 1/(hoA) = 1/(20·8) = 0.00625 K/W
ΣR = 0.1471 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(8·0.1471) = 0.8495 W/(m²·K)
Answer
U = 0.8495 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: U = 519.9 W/(m²·K)
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 = 85 − 75 = 10 K
ΔT₂ = Th,out − Tc,in = 50 − 10 = 40 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (10 − 40)/ln(10/40) = 21.64 K
Overall coefficient
U = Q̇/(A·ΔTlm) = 90000/(8·21.64) = 519.9 W/(m²·K)
Answer
U = 519.9 W/(m²·K)
4.Heat exchangers (LMTD)
Answer: L = 7.335 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 = 0.3·2.3·(135 − 100) = 24.15 kW
Tc,out = 10 + 24.15/(1.5·4.18) = 13.85 °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 = 135 − 13.85 = 121.1 K
ΔT₂ = Th,out − Tc,in = 100 − 10 = 90 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (121.1 − 90)/ln(121.1/90) = 104.8 K
Area and length
A = 24150/(400·104.8) = 0.5761 m²
L = A/(πD) = 0.5761/(π·0.025) = 7.335 m
Answer
L = 7.335 m
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