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 50 mm of gypsum board and 100 mm of brick, with an area of 20 m². The outer faces are at 20 °C and -13 °C. Find the temperature Ti at the interface between the two layers.
L1 = 50 mm, k1 = 0.17 W/(m·K)L2 = 100 mm, k2 = 0.72 W/(m·K)A = 20 m²T1 = 20 °CT3 = -13 °C
Ti = °C
2.Heat conductionHard
A wall of 150 mm concrete and 150 mm mineral wool separates a room at 20 °C from outside air at 0 °C. The convection coefficients are 8 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 4 m². Find the heat loss Q̇ through the wall.
L1 = 150 mm, k1 = 1.4 W/(m·K)L2 = 150 mm, k2 = 0.04 W/(m·K)A = 4 m²hi = 8 W/(m²·K)ho = 30 W/(m²·K)Troom = 20 °CTout = 0 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 480 kW. The hot fluid cools from 175 to 155 °C and the cold fluid heats from 15 to 125 °C. The overall heat transfer coefficient is 500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 480 kWU = 500 W/(m²·K)Th: 175 → 155 °CTc: 15 → 125 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1 kg/s, cp = 2.3 kJ/(kg·K)) cools from 90 to 70 °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 = 1 kg/scp,h = 2.3 kJ/(kg·K)Th: 90 → 70 °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: Ti = −2.415 °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) = (50/1000)/(0.17·20) = 0.01471 K/W
R2 = L2/(k2A) = (100/1000)/(0.72·20) = 0.006944 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 33/0.02165 = 1524 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 20 − 1524·0.01471 = −2.415 °C
Answer
Ti = −2.415 °C
2.Heat conduction
Answer: Q̇ = 19.92 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/(8·4) = 0.03125 K/W
R1 = L1/(k1A) = (150/1000)/(1.4·4) = 0.02679 K/W
R2 = L2/(k2A) = (150/1000)/(0.04·4) = 0.9375 K/W
Ro = 1/(hoA) = 1/(30·4) = 0.008333 K/W
ΣR = 1.004 K/W
Heat loss
Q̇ = (20 − (0))/1.004 = 19.92 W
Answer
Q̇ = 19.92 W
3.Heat exchangers (LMTD)
Answer: A = 10.98 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 = 175 − 125 = 50 K
ΔT₂ = Th,out − Tc,in = 155 − 15 = 140 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (50 − 140)/ln(50/140) = 87.41 K
Area
A = Q̇/(U·ΔTlm) = 480000/(500·87.41) = 10.98 m²
Answer
A = 10.98 m²
4.Heat exchangers (LMTD)
Answer: A = 4.822 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) = 1·2.3·20 = 46 kW
Tc,out = 30 + 46/(3·4.18) = 33.67 °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 − 33.67 = 56.33 K
ΔT₂ = Th,out − Tc,in = 70 − 30 = 40 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (56.33 − 40)/ln(56.33/40) = 47.7 K
Area
A = Q̇/(U·ΔTlm) = 46000/(200·47.7) = 4.822 m²
Answer
A = 4.822 m²
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