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 concrete and 50 mm of gypsum board, with an area of 6 m². The outer faces are at 19 °C and -4 °C. Find the heat flow Q̇ through the wall.
L1 = 250 mm, k1 = 1.4 W/(m·K)L2 = 50 mm, k2 = 0.17 W/(m·K)A = 6 m²T1 = 19 °CT3 = -4 °C
Q̇ = W
2.Heat conductionHard
A wall of 50 mm expanded polystyrene and 50 mm gypsum board separates a room at 20 °C from outside air at -10 °C. The convection coefficients are 6 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 = 50 mm, k1 = 0.035 W/(m·K)L2 = 50 mm, k2 = 0.17 W/(m·K)A = 20 m²hi = 6 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = -10 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 320 kW. The hot fluid cools from 110 to 85 °C and the cold fluid heats from 15 to 70 °C. The overall heat transfer coefficient is 1000 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 320 kWU = 1000 W/(m²·K)Th: 110 → 85 °CTc: 15 → 70 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (0.5 kg/s, cp = 2.1 kJ/(kg·K)) cools from 140 to 30 °C in a counter-flow exchanger. It heats 2 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 10 °C. With U = 300 W/(m²·K), find the heat transfer area A needed.
ṁh = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 140 → 30 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 10 °CU = 300 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 291.9 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) = (250/1000)/(1.4·6) = 0.02976 K/W
R2 = L2/(k2A) = (50/1000)/(0.17·6) = 0.04902 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 23/0.07878 = 291.9 W
Answer
Q̇ = 291.9 W
2.Heat conduction
Answer: Q̇ = 311 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/(6·20) = 0.008333 K/W
R1 = L1/(k1A) = (50/1000)/(0.035·20) = 0.07143 K/W
R2 = L2/(k2A) = (50/1000)/(0.17·20) = 0.01471 K/W
Ro = 1/(hoA) = 1/(25·20) = 0.002 K/W
ΣR = 0.09647 K/W
Heat loss
Q̇ = (20 − (-10))/0.09647 = 311 W
Answer
Q̇ = 311 W
3.Heat exchangers (LMTD)
Answer: A = 5.969 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 = 110 − 70 = 40 K
ΔT₂ = Th,out − Tc,in = 85 − 15 = 70 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (40 − 70)/ln(40/70) = 53.61 K
Area
A = Q̇/(U·ΔTlm) = 320000/(1000·53.61) = 5.969 m²
Answer
A = 5.969 m²
4.Heat exchangers (LMTD)
Answer: A = 7.043 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) = 0.5·2.1·110 = 115.5 kW
Tc,out = 10 + 115.5/(2·4.18) = 23.82 °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 = 140 − 23.82 = 116.2 K
ΔT₂ = Th,out − Tc,in = 30 − 10 = 20 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (116.2 − 20)/ln(116.2/20) = 54.67 K
Area
A = Q̇/(U·ΔTlm) = 115500/(300·54.67) = 7.043 m²
Answer
A = 7.043 m²
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