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 gypsum board and 100 mm of concrete, with an area of 15 m². The outer faces are at 16 °C and -10 °C. Find the temperature Ti at the interface between the two layers.
L1 = 100 mm, k1 = 0.17 W/(m·K)L2 = 100 mm, k2 = 1.4 W/(m·K)A = 15 m²T1 = 16 °CT3 = -10 °C
Ti = °C
2.Heat conductionHard
A wall of 50 mm brick and 150 mm mineral wool separates a room at 20 °C from outside air at 0 °C. The convection coefficients are 10 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 15 m². Find the temperature Ts of the inside surface of the wall.
L1 = 50 mm, k1 = 0.72 W/(m·K)L2 = 150 mm, k2 = 0.04 W/(m·K)A = 15 m²hi = 10 W/(m²·K)ho = 20 W/(m²·K)Troom = 20 °CTout = 0 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 150 kW. The hot fluid cools from 150 to 100 °C and the cold fluid heats from 15 to 80 °C. The overall heat transfer coefficient is 200 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 150 kWU = 200 W/(m²·K)Th: 150 → 100 °CTc: 15 → 80 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (2 kg/s, cp = 2.5 kJ/(kg·K)) cools from 125 to 40 °C in a counter-flow exchanger. It heats 2 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 20 °C. With U = 300 W/(m²·K), find the heat transfer area A needed.
ṁh = 2 kg/scp,h = 2.5 kJ/(kg·K)Th: 125 → 40 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 20 °CU = 300 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = −7.185 °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) = (100/1000)/(0.17·15) = 0.03922 K/W
R2 = L2/(k2A) = (100/1000)/(1.4·15) = 0.004762 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 26/0.04398 = 591.2 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 16 − 591.2·0.03922 = −7.185 °C
Answer
Ti = −7.185 °C
2.Heat conduction
Answer: Ts = 19.5 °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.
Resistances in series
Ri = 1/(hiA) = 1/(10·15) = 0.006667 K/W
R1 = L1/(k1A) = (50/1000)/(0.72·15) = 0.00463 K/W
R2 = L2/(k2A) = (150/1000)/(0.04·15) = 0.25 K/W
Ro = 1/(hoA) = 1/(20·15) = 0.003333 K/W
ΣR = 0.2646 K/W
Heat loss
Q̇ = (20 − (0))/0.2646 = 75.58 W
Ts = Troom − Q̇·Ri = 20 − 75.58·0.006667 = 19.5 °C
Answer
Ts = 19.5 °C
3.Heat exchangers (LMTD)
Answer: A = 9.708 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 = 150 − 80 = 70 K
ΔT₂ = Th,out − Tc,in = 100 − 15 = 85 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (70 − 85)/ln(70/85) = 77.26 K
Area
A = Q̇/(U·ΔTlm) = 150000/(200·77.26) = 9.708 m²
Answer
A = 9.708 m²
4.Heat exchangers (LMTD)
Answer: A = 41.31 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) = 2·2.5·85 = 425 kW
Tc,out = 20 + 425/(2·4.18) = 70.84 °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 = 125 − 70.84 = 54.16 K
ΔT₂ = Th,out − Tc,in = 40 − 20 = 20 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (54.16 − 20)/ln(54.16/20) = 34.29 K
Area
A = Q̇/(U·ΔTlm) = 425000/(300·34.29) = 41.31 m²
Answer
A = 41.31 m²
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