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 mineral wool and 150 mm of brick, with an area of 20 m². The outer faces are at 15 °C and 5 °C. Find the temperature Ti at the interface between the two layers.
L1 = 250 mm, k1 = 0.04 W/(m·K)L2 = 150 mm, k2 = 0.72 W/(m·K)A = 20 m²T1 = 15 °CT3 = 5 °C
Ti = °C
2.Heat conductionHard
A wall of 100 mm brick and 25 mm wood separates a room at 20 °C from outside air at -15 °C. The convection coefficients are 10 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 4 m². Find the overall heat transfer coefficient U of the wall.
L1 = 100 mm, k1 = 0.72 W/(m·K)L2 = 25 mm, k2 = 0.17 W/(m·K)A = 4 m²hi = 10 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = -15 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 80 kW. Its heat transfer area is 6 m². The hot fluid cools from 170 to 100 °C and the cold fluid heats from 20 to 65 °C. Find the overall heat transfer coefficient U.
Q̇ = 80 kWA = 6 m²Th: 170 → 100 °CTc: 20 → 65 °C
U = W/(m²·K)
4.Heat exchangers (LMTD)Hard
Hot oil (1.5 kg/s, cp = 2.1 kJ/(kg·K)) cools from 170 to 110 °C in a counter-flow exchanger. It heats 3 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 10 °C. With U = 200 W/(m²·K), find the heat transfer area A needed.
ṁh = 1.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 170 → 110 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 10 °CU = 200 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = 5.323 °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)/(0.04·20) = 0.3125 K/W
R2 = L2/(k2A) = (150/1000)/(0.72·20) = 0.01042 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 10/0.3229 = 30.97 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 15 − 30.97·0.3125 = 5.323 °C
Answer
Ti = 5.323 °C
2.Heat conduction
Answer: U = 2.348 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/(10·4) = 0.025 K/W
R1 = L1/(k1A) = (100/1000)/(0.72·4) = 0.03472 K/W
R2 = L2/(k2A) = (25/1000)/(0.17·4) = 0.03676 K/W
Ro = 1/(hoA) = 1/(25·4) = 0.01 K/W
ΣR = 0.1065 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(4·0.1065) = 2.348 W/(m²·K)
Answer
U = 2.348 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: U = 145 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 = 170 − 65 = 105 K
ΔT₂ = Th,out − Tc,in = 100 − 20 = 80 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (105 − 80)/ln(105/80) = 91.93 K
Overall coefficient
U = Q̇/(A·ΔTlm) = 80000/(6·91.93) = 145 W/(m²·K)
Answer
U = 145 W/(m²·K)
4.Heat exchangers (LMTD)
Answer: A = 7.805 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) = 1.5·2.1·60 = 189 kW
Tc,out = 10 + 189/(3·4.18) = 25.07 °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 = 170 − 25.07 = 144.9 K
ΔT₂ = Th,out − Tc,in = 110 − 10 = 100 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (144.9 − 100)/ln(144.9/100) = 121.1 K
Area
A = Q̇/(U·ΔTlm) = 189000/(200·121.1) = 7.805 m²
Answer
A = 7.805 m²
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