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 expanded polystyrene and 25 mm of concrete, with an area of 20 m². The outer faces are at 16 °C and -2 °C. Find the temperature Ti at the interface between the two layers.
L1 = 250 mm, k1 = 0.035 W/(m·K)L2 = 25 mm, k2 = 1.4 W/(m·K)A = 20 m²T1 = 16 °CT3 = -2 °C
Ti = °C
2.Heat conductionHard
A wall of 100 mm gypsum board and 25 mm brick separates a room at 20 °C from outside air at -13 °C. The convection coefficients are 6 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 15 m². Find the heat loss Q̇ through the wall.
L1 = 100 mm, k1 = 0.17 W/(m·K)L2 = 25 mm, k2 = 0.72 W/(m·K)A = 15 m²hi = 6 W/(m²·K)ho = 30 W/(m²·K)Troom = 20 °CTout = -13 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A parallel-flow heat exchanger transfers 460 kW. The hot fluid cools from 85 to 55 °C and the cold fluid heats from 30 to 50 °C. The overall heat transfer coefficient is 200 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 460 kWU = 200 W/(m²·K)Th: 85 → 55 °CTc: 30 → 50 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1 kg/s, cp = 2.1 kJ/(kg·K)) cools from 120 to 85 °C in a counter-flow exchanger. It heats 3 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 kg/scp,h = 2.1 kJ/(kg·K)Th: 120 → 85 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 20 °CU = 1000 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = −1.955 °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)/(0.035·20) = 0.3571 K/W
R2 = L2/(k2A) = (25/1000)/(1.4·20) = 0.0008929 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 18/0.358 = 50.27 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 16 − 50.27·0.3571 = −1.955 °C
Answer
Ti = −1.955 °C
2.Heat conduction
Answer: Q̇ = 601.5 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/(6·15) = 0.01111 K/W
R1 = L1/(k1A) = (100/1000)/(0.17·15) = 0.03922 K/W
R2 = L2/(k2A) = (25/1000)/(0.72·15) = 0.002315 K/W
Ro = 1/(hoA) = 1/(30·15) = 0.002222 K/W
ΣR = 0.05486 K/W
Heat loss
Q̇ = (20 − (-13))/0.05486 = 601.5 W
Answer
Q̇ = 601.5 W
3.Heat exchangers (LMTD)
Answer: A = 110.3 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 parallel flow both inlets are at the same end, and both outlets at the other.
ΔT₁ = Th,in − Tc,in = 85 − 30 = 55 K
ΔT₂ = Th,out − Tc,out = 55 − 50 = 5 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (55 − 5)/ln(55/5) = 20.85 K
Area
A = Q̇/(U·ΔTlm) = 460000/(200·20.85) = 110.3 m²
Answer
A = 110.3 m²
4.Heat exchangers (LMTD)
Answer: A = 0.9343 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) = 1·2.1·35 = 73.5 kW
Tc,out = 20 + 73.5/(3·4.18) = 25.86 °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 = 120 − 25.86 = 94.14 K
ΔT₂ = Th,out − Tc,in = 85 − 20 = 65 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (94.14 − 65)/ln(94.14/65) = 78.67 K
Area
A = Q̇/(U·ΔTlm) = 73500/(1000·78.67) = 0.9343 m²
Answer
A = 0.9343 m²
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