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 wood and 150 mm of brick, with an area of 4 m². The outer faces are at 17 °C and 4 °C. Find the heat flow Q̇ through the wall.
L1 = 100 mm, k1 = 0.17 W/(m·K)L2 = 150 mm, k2 = 0.72 W/(m·K)A = 4 m²T1 = 17 °CT3 = 4 °C
Q̇ = W
2.Heat conductionHard
A wall of 100 mm expanded polystyrene and 75 mm glass separates a room at 20 °C from outside air at -8 °C. The convection coefficients are 7.7 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 20 m². Find the heat loss Q̇ through the wall.
L1 = 100 mm, k1 = 0.035 W/(m·K)L2 = 75 mm, k2 = 0.96 W/(m·K)A = 20 m²hi = 7.7 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = -8 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 200 kW. The hot fluid cools from 115 to 60 °C and the cold fluid heats from 15 to 40 °C. The overall heat transfer coefficient is 800 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 200 kWU = 800 W/(m²·K)Th: 115 → 60 °CTc: 15 → 40 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (2 kg/s, cp = 2.1 kJ/(kg·K)) cools from 165 to 120 °C in a counter-flow exchanger. It heats 3 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 30 °C. With U = 200 W/(m²·K), find the heat transfer area A needed.
ṁh = 2 kg/scp,h = 2.1 kJ/(kg·K)Th: 165 → 120 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 30 °CU = 200 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 65.28 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)/(0.17·4) = 0.1471 K/W
R2 = L2/(k2A) = (150/1000)/(0.72·4) = 0.05208 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 13/0.1991 = 65.28 W
Answer
Q̇ = 65.28 W
2.Heat conduction
Answer: Q̇ = 180.3 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/(7.7·20) = 0.006494 K/W
R1 = L1/(k1A) = (100/1000)/(0.035·20) = 0.1429 K/W
R2 = L2/(k2A) = (75/1000)/(0.96·20) = 0.003906 K/W
Ro = 1/(hoA) = 1/(25·20) = 0.002 K/W
ΣR = 0.1553 K/W
Heat loss
Q̇ = (20 − (-8))/0.1553 = 180.3 W
Answer
Q̇ = 180.3 W
3.Heat exchangers (LMTD)
Answer: A = 4.257 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 = 115 − 40 = 75 K
ΔT₂ = Th,out − Tc,in = 60 − 15 = 45 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (75 − 45)/ln(75/45) = 58.73 K
Area
A = Q̇/(U·ΔTlm) = 200000/(800·58.73) = 4.257 m²
Answer
A = 4.257 m²
4.Heat exchangers (LMTD)
Answer: A = 9.065 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) = 2·2.1·45 = 189 kW
Tc,out = 30 + 189/(3·4.18) = 45.07 °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 = 165 − 45.07 = 119.9 K
ΔT₂ = Th,out − Tc,in = 120 − 30 = 90 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (119.9 − 90)/ln(119.9/90) = 104.2 K
Area
A = Q̇/(U·ΔTlm) = 189000/(200·104.2) = 9.065 m²
Answer
A = 9.065 m²
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