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 200 mm of mineral wool and 100 mm of gypsum board, with an area of 20 m². The outer faces are at 17 °C and -4 °C. Find the temperature Ti at the interface between the two layers.
L1 = 200 mm, k1 = 0.04 W/(m·K)L2 = 100 mm, k2 = 0.17 W/(m·K)A = 20 m²T1 = 17 °CT3 = -4 °C
Ti = °C
2.Heat conductionHard
A wall of 50 mm expanded polystyrene and 100 mm glass separates a room at 21 °C from outside air at 1 °C. The convection coefficients are 7.7 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 4 m². Find the overall heat transfer coefficient U of the wall.
L1 = 50 mm, k1 = 0.035 W/(m·K)L2 = 100 mm, k2 = 0.96 W/(m·K)A = 4 m²hi = 7.7 W/(m²·K)ho = 20 W/(m²·K)Troom = 21 °CTout = 1 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 420 kW. The hot fluid cools from 180 to 60 °C and the cold fluid heats from 25 to 160 °C. The overall heat transfer coefficient is 300 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 420 kWU = 300 W/(m²·K)Th: 180 → 60 °CTc: 25 → 160 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1.5 kg/s, cp = 2.3 kJ/(kg·K)) cools from 90 to 65 °C in a counter-flow exchanger. It heats 1.5 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 20 °C. With U = 500 W/(m²·K), find the heat transfer area A needed.
ṁh = 1.5 kg/scp,h = 2.3 kJ/(kg·K)Th: 90 → 65 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 20 °CU = 500 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = −1.789 °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) = (200/1000)/(0.04·20) = 0.25 K/W
R2 = L2/(k2A) = (100/1000)/(0.17·20) = 0.02941 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 21/0.2794 = 75.16 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 17 − 75.16·0.25 = −1.789 °C
Answer
Ti = −1.789 °C
2.Heat conduction
Answer: U = 0.5839 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/(7.7·4) = 0.03247 K/W
R1 = L1/(k1A) = (50/1000)/(0.035·4) = 0.3571 K/W
R2 = L2/(k2A) = (100/1000)/(0.96·4) = 0.02604 K/W
Ro = 1/(hoA) = 1/(20·4) = 0.0125 K/W
ΣR = 0.4282 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(4·0.4282) = 0.5839 W/(m²·K)
Answer
U = 0.5839 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 52.23 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 − 160 = 20 K
ΔT₂ = Th,out − Tc,in = 60 − 25 = 35 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (20 − 35)/ln(20/35) = 26.8 K
Area
A = Q̇/(U·ΔTlm) = 420000/(300·26.8) = 52.23 m²
Answer
A = 52.23 m²
4.Heat exchangers (LMTD)
Answer: A = 3.422 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) = 1.5·2.3·25 = 86.25 kW
Tc,out = 20 + 86.25/(1.5·4.18) = 33.76 °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 = 90 − 33.76 = 56.24 K
ΔT₂ = Th,out − Tc,in = 65 − 20 = 45 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (56.24 − 45)/ln(56.24/45) = 50.41 K
Area
A = Q̇/(U·ΔTlm) = 86250/(500·50.41) = 3.422 m²
Answer
A = 3.422 m²
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