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 100 mm of expanded polystyrene and 150 mm of mineral wool, with an area of 4 m². The outer faces are at 21 °C and 9 °C. Find the heat flow Q̇ through the wall.
L1 = 100 mm, k1 = 0.035 W/(m·K)L2 = 150 mm, k2 = 0.04 W/(m·K)A = 4 m²T1 = 21 °CT3 = 9 °C
Q̇ = W
2.Heat conductionHard
A wall of 200 mm concrete and 150 mm brick separates a room at 20 °C from outside air at 4 °C. The convection coefficients are 7.7 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 8 m². Find the overall heat transfer coefficient U of the wall.
L1 = 200 mm, k1 = 1.4 W/(m·K)L2 = 150 mm, k2 = 0.72 W/(m·K)A = 8 m²hi = 7.7 W/(m²·K)ho = 30 W/(m²·K)Troom = 20 °CTout = 4 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 320 kW. Its heat transfer area is 20 m². The hot fluid cools from 140 to 65 °C and the cold fluid heats from 30 to 125 °C. Find the overall heat transfer coefficient U.
Q̇ = 320 kWA = 20 m²Th: 140 → 65 °CTc: 30 → 125 °C
U = W/(m²·K)
4.Heat exchangers (LMTD)Hard
In a counter-flow double-pipe heat exchanger, hot oil (1 kg/s, cp = 2.5 kJ/(kg·K)) flows in the inner tube and cools from 140 to 105 °C. Water (1 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 15 °C. The inner tube has a diameter of 25 mm, and U = 500 W/(m²·K) is based on this diameter. Find the tube length L needed.
ṁh = 1 kg/scp,h = 2.5 kJ/(kg·K)Th: 140 → 105 °Cṁc = 1 kg/sTc,in = 15 °CD = 25 mmU = 500 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Q̇ = 7.265 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) = (100/1000)/(0.035·4) = 0.7143 K/W
R2 = L2/(k2A) = (150/1000)/(0.04·4) = 0.9375 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 12/1.652 = 7.265 W
Answer
Q̇ = 7.265 W
2.Heat conduction
Answer: U = 1.944 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/(7.7·8) = 0.01623 K/W
R1 = L1/(k1A) = (200/1000)/(1.4·8) = 0.01786 K/W
R2 = L2/(k2A) = (150/1000)/(0.72·8) = 0.02604 K/W
Ro = 1/(hoA) = 1/(30·8) = 0.004167 K/W
ΣR = 0.0643 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(8·0.0643) = 1.944 W/(m²·K)
Answer
U = 1.944 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: U = 677.8 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 = 140 − 125 = 15 K
ΔT₂ = Th,out − Tc,in = 65 − 30 = 35 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (15 − 35)/ln(15/35) = 23.6 K
Overall coefficient
U = Q̇/(A·ΔTlm) = 320000/(20·23.6) = 677.8 W/(m²·K)
Answer
U = 677.8 W/(m²·K)
4.Heat exchangers (LMTD)
Answer: L = 23 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 = 1·2.5·(140 − 105) = 87.5 kW
Tc,out = 15 + 87.5/(1·4.18) = 35.93 °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 = 140 − 35.93 = 104.1 K
ΔT₂ = Th,out − Tc,in = 105 − 15 = 90 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (104.1 − 90)/ln(104.1/90) = 96.86 K
Area and length
A = 87500/(500·96.86) = 1.807 m²
L = A/(πD) = 1.807/(π·0.025) = 23 m
Answer
L = 23 m
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