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 concrete and 75 mm of brick, with an area of 8 m². The outer faces are at 22 °C and 7 °C. Find the temperature Ti at the interface between the two layers.
L1 = 150 mm, k1 = 1.4 W/(m·K)L2 = 75 mm, k2 = 0.72 W/(m·K)A = 8 m²T1 = 22 °CT3 = 7 °C
Ti = °C
2.Heat conductionHard
A wall of 200 mm wood and 50 mm gypsum board separates a room at 20 °C from outside air at -3 °C. The convection coefficients are 10 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 12 m². Find the heat loss Q̇ through the wall.
L1 = 200 mm, k1 = 0.17 W/(m·K)L2 = 50 mm, k2 = 0.17 W/(m·K)A = 12 m²hi = 10 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = -3 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 60 kW. The hot fluid cools from 180 to 140 °C and the cold fluid heats from 15 to 105 °C. The overall heat transfer coefficient is 800 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 60 kWU = 800 W/(m²·K)Th: 180 → 140 °CTc: 15 → 105 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1.5 kg/s, cp = 2.3 kJ/(kg·K)) cools from 160 to 85 °C in a counter-flow exchanger. It heats 2 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 20 °C. With U = 800 W/(m²·K), find the heat transfer area A needed.
ṁh = 1.5 kg/scp,h = 2.3 kJ/(kg·K)Th: 160 → 85 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 20 °CU = 800 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = 14.39 °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) = (150/1000)/(1.4·8) = 0.01339 K/W
R2 = L2/(k2A) = (75/1000)/(0.72·8) = 0.01302 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 15/0.02641 = 567.9 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 22 − 567.9·0.01339 = 14.39 °C
Answer
Ti = 14.39 °C
2.Heat conduction
Answer: Q̇ = 171.4 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/(10·12) = 0.008333 K/W
R1 = L1/(k1A) = (200/1000)/(0.17·12) = 0.09804 K/W
R2 = L2/(k2A) = (50/1000)/(0.17·12) = 0.02451 K/W
Ro = 1/(hoA) = 1/(25·12) = 0.003333 K/W
ΣR = 0.1342 K/W
Heat loss
Q̇ = (20 − (-3))/0.1342 = 171.4 W
Answer
Q̇ = 171.4 W
3.Heat exchangers (LMTD)
Answer: A = 0.7662 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 = 180 − 105 = 75 K
ΔT₂ = Th,out − Tc,in = 140 − 15 = 125 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (75 − 125)/ln(75/125) = 97.88 K
Area
A = Q̇/(U·ΔTlm) = 60000/(800·97.88) = 0.7662 m²
Answer
A = 0.7662 m²
4.Heat exchangers (LMTD)
Answer: A = 3.799 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) = 1.5·2.3·75 = 258.8 kW
Tc,out = 20 + 258.8/(2·4.18) = 50.95 °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 = 160 − 50.95 = 109 K
ΔT₂ = Th,out − Tc,in = 85 − 20 = 65 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (109 − 65)/ln(109/65) = 85.13 K
Area
A = Q̇/(U·ΔTlm) = 258750/(800·85.13) = 3.799 m²
Answer
A = 3.799 m²
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