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 expanded polystyrene and 150 mm of wood, with an area of 20 m². The outer faces are at 22 °C and -9 °C. Find the temperature Ti at the interface between the two layers.
L1 = 200 mm, k1 = 0.035 W/(m·K)L2 = 150 mm, k2 = 0.17 W/(m·K)A = 20 m²T1 = 22 °CT3 = -9 °C
Ti = °C
2.Heat conductionHard
A wall of 100 mm gypsum board and 150 mm brick separates a room at 20 °C from outside air at 0 °C. The convection coefficients are 6 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 6 m². Find the overall heat transfer coefficient U of the wall.
L1 = 100 mm, k1 = 0.17 W/(m·K)L2 = 150 mm, k2 = 0.72 W/(m·K)A = 6 m²hi = 6 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = 0 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 50 kW. The hot fluid cools from 135 to 60 °C and the cold fluid heats from 15 to 55 °C. The overall heat transfer coefficient is 1500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 50 kWU = 1500 W/(m²·K)Th: 135 → 60 °CTc: 15 → 55 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a counter-flow double-pipe heat exchanger, hot oil (0.3 kg/s, cp = 2.1 kJ/(kg·K)) flows in the inner tube and cools from 100 to 65 °C. Water (1 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 15 °C. The inner tube has a diameter of 50 mm, and U = 200 W/(m²·K) is based on this diameter. Find the tube length L needed.
ṁh = 0.3 kg/scp,h = 2.1 kJ/(kg·K)Th: 100 → 65 °Cṁc = 1 kg/sTc,in = 15 °CD = 50 mmU = 200 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Ti = −4.854 °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)/(0.035·20) = 0.2857 K/W
R2 = L2/(k2A) = (150/1000)/(0.17·20) = 0.04412 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 31/0.3298 = 93.99 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 22 − 93.99·0.2857 = −4.854 °C
Answer
Ti = −4.854 °C
2.Heat conduction
Answer: U = 0.9968 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/(6·6) = 0.02778 K/W
R1 = L1/(k1A) = (100/1000)/(0.17·6) = 0.09804 K/W
R2 = L2/(k2A) = (150/1000)/(0.72·6) = 0.03472 K/W
Ro = 1/(hoA) = 1/(25·6) = 0.006667 K/W
ΣR = 0.1672 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(6·0.1672) = 0.9968 W/(m²·K)
Answer
U = 0.9968 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 0.548 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 = 135 − 55 = 80 K
ΔT₂ = Th,out − Tc,in = 60 − 15 = 45 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (80 − 45)/ln(80/45) = 60.83 K
Area
A = Q̇/(U·ΔTlm) = 50000/(1500·60.83) = 0.548 m²
Answer
A = 0.548 m²
4.Heat exchangers (LMTD)
Answer: L = 11.02 m
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 = 0.3·2.1·(100 − 65) = 22.05 kW
Tc,out = 15 + 22.05/(1·4.18) = 20.28 °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 = 100 − 20.28 = 79.72 K
ΔT₂ = Th,out − Tc,in = 65 − 15 = 50 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (79.72 − 50)/ln(79.72/50) = 63.71 K
Area and length
A = 22050/(200·63.71) = 1.73 m²
L = A/(πD) = 1.73/(π·0.05) = 11.02 m
Answer
L = 11.02 m
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