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 100 mm of brick, with an area of 6 m². The outer faces are at 23 °C and -4 °C. Find the temperature Ti at the interface between the two layers.
L1 = 250 mm, k1 = 1.4 W/(m·K)L2 = 100 mm, k2 = 0.72 W/(m·K)A = 6 m²T1 = 23 °CT3 = -4 °C
Ti = °C
2.Heat conductionHard
A wall of 250 mm brick and 50 mm gypsum board separates a room at 20 °C from outside air at 3 °C. The convection coefficients are 8 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 15 m². Find the overall heat transfer coefficient U of the wall.
L1 = 250 mm, k1 = 0.72 W/(m·K)L2 = 50 mm, k2 = 0.17 W/(m·K)A = 15 m²hi = 8 W/(m²·K)ho = 30 W/(m²·K)Troom = 20 °CTout = 3 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A shell-and-tube heat exchanger (one shell pass, two tube passes) must transfer 200 kW. The hot fluid cools from 115 to 70 °C and the cold fluid heats from 25 to 105 °C. The LMTD correction factor for this arrangement is F = 0.8, and U = 300 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 200 kWF = 0.8U = 300 W/(m²·K)Th: 115 → 70 °CTc: 25 → 105 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (0.5 kg/s, cp = 2.5 kJ/(kg·K)) cools from 140 to 50 °C in a counter-flow exchanger. It heats 1.5 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 10 °C. With U = 500 W/(m²·K), find the heat transfer area A needed.
ṁh = 0.5 kg/scp,h = 2.5 kJ/(kg·K)Th: 140 → 50 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 10 °CU = 500 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = 7.812 °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)/(1.4·6) = 0.02976 K/W
R2 = L2/(k2A) = (100/1000)/(0.72·6) = 0.02315 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 27/0.05291 = 510.3 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 23 − 510.3·0.02976 = 7.812 °C
Answer
Ti = 7.812 °C
2.Heat conduction
Answer: U = 1.251 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/(8·15) = 0.008333 K/W
R1 = L1/(k1A) = (250/1000)/(0.72·15) = 0.02315 K/W
R2 = L2/(k2A) = (50/1000)/(0.17·15) = 0.01961 K/W
Ro = 1/(hoA) = 1/(30·15) = 0.002222 K/W
ΣR = 0.05331 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(15·0.05331) = 1.251 W/(m²·K)
Answer
U = 1.251 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 35.81 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 (counter-flow)
In counter-flow the hot inlet faces the cold outlet, and the hot outlet faces the cold inlet.
ΔT₁ = Th,in − Tc,out = 115 − 105 = 10 K
ΔT₂ = Th,out − Tc,in = 70 − 25 = 45 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (10 − 45)/ln(10/45) = 23.27 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 200000/(300·0.8·23.27) = 35.81 m²
Answer
A = 35.81 m²
4.Heat exchangers (LMTD)
Answer: A = 3.217 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) = 0.5·2.5·90 = 112.5 kW
Tc,out = 10 + 112.5/(1.5·4.18) = 27.94 °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 − 27.94 = 112.1 K
ΔT₂ = Th,out − Tc,in = 50 − 10 = 40 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (112.1 − 40)/ln(112.1/40) = 69.95 K
Area
A = Q̇/(U·ΔTlm) = 112500/(500·69.95) = 3.217 m²
Answer
A = 3.217 m²
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