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 150 mm of wood and 100 mm of concrete, with an area of 15 m². The outer faces are at 20 °C and 4 °C. Find the heat flow Q̇ through the wall.
L1 = 150 mm, k1 = 0.17 W/(m·K)L2 = 100 mm, k2 = 1.4 W/(m·K)A = 15 m²T1 = 20 °CT3 = 4 °C
Q̇ = W
2.Heat conductionHard
A wall of 200 mm wood and 100 mm expanded polystyrene separates a room at 21 °C from outside air at -2 °C. The convection coefficients are 7.7 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 4 m². Find the heat loss Q̇ through the wall.
L1 = 200 mm, k1 = 0.17 W/(m·K)L2 = 100 mm, k2 = 0.035 W/(m·K)A = 4 m²hi = 7.7 W/(m²·K)ho = 30 W/(m²·K)Troom = 21 °CTout = -2 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 220 kW. The hot fluid cools from 85 to 60 °C and the cold fluid heats from 20 to 60 °C. The overall heat transfer coefficient is 1000 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 220 kWU = 1000 W/(m²·K)Th: 85 → 60 °CTc: 20 → 60 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1.5 kg/s, cp = 2.1 kJ/(kg·K)) cools from 125 to 75 °C in a counter-flow exchanger. It heats 1 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 20 °C. With U = 1000 W/(m²·K), find the heat transfer area A needed.
ṁh = 1.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 125 → 75 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 20 °CU = 1000 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 251.6 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) = (150/1000)/(0.17·15) = 0.05882 K/W
R2 = L2/(k2A) = (100/1000)/(1.4·15) = 0.004762 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 16/0.06359 = 251.6 W
Answer
Q̇ = 251.6 W
2.Heat conduction
Answer: Q̇ = 21.92 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.
Resistances in series
Ri = 1/(hiA) = 1/(7.7·4) = 0.03247 K/W
R1 = L1/(k1A) = (200/1000)/(0.17·4) = 0.2941 K/W
R2 = L2/(k2A) = (100/1000)/(0.035·4) = 0.7143 K/W
Ro = 1/(hoA) = 1/(30·4) = 0.008333 K/W
ΣR = 1.049 K/W
Heat loss
Q̇ = (21 − (-2))/1.049 = 21.92 W
Answer
Q̇ = 21.92 W
3.Heat exchangers (LMTD)
Answer: A = 6.893 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 = 85 − 60 = 25 K
ΔT₂ = Th,out − Tc,in = 60 − 20 = 40 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (25 − 40)/ln(25/40) = 31.91 K
Area
A = Q̇/(U·ΔTlm) = 220000/(1000·31.91) = 6.893 m²
Answer
A = 6.893 m²
4.Heat exchangers (LMTD)
Answer: A = 2.584 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.
Heat duty and water outlet temperature
Q̇ = ṁhcp,h(Th,in − Th,out) = 1.5·2.1·50 = 157.5 kW
Tc,out = 20 + 157.5/(1·4.18) = 57.68 °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 = 125 − 57.68 = 67.32 K
ΔT₂ = Th,out − Tc,in = 75 − 20 = 55 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (67.32 − 55)/ln(67.32/55) = 60.95 K
Area
A = Q̇/(U·ΔTlm) = 157500/(1000·60.95) = 2.584 m²
Answer
A = 2.584 m²
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