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 concrete and 50 mm of brick, with an area of 20 m². The outer faces are at 20 °C and -9 °C. Find the temperature Ti at the interface between the two layers.
L1 = 200 mm, k1 = 1.4 W/(m·K)L2 = 50 mm, k2 = 0.72 W/(m·K)A = 20 m²T1 = 20 °CT3 = -9 °C
Ti = °C
2.Heat conductionHard
A wall of 50 mm gypsum board and 75 mm brick separates a room at 21 °C from outside air at -7 °C. The convection coefficients are 7.7 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 10 m². Find the temperature Ts of the inside surface of the wall.
L1 = 50 mm, k1 = 0.17 W/(m·K)L2 = 75 mm, k2 = 0.72 W/(m·K)A = 10 m²hi = 7.7 W/(m²·K)ho = 20 W/(m²·K)Troom = 21 °CTout = -7 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 360 kW. Its heat transfer area is 6 m². The hot fluid cools from 130 to 75 °C and the cold fluid heats from 30 to 100 °C. Find the overall heat transfer coefficient U.
Q̇ = 360 kWA = 6 m²Th: 130 → 75 °CTc: 30 → 100 °C
U = W/(m²·K)
4.Heat exchangers (LMTD)Hard
In a parallel-flow double-pipe heat exchanger, hot oil (0.8 kg/s, cp = 2.5 kJ/(kg·K)) flows in the inner tube and cools from 135 to 60 °C. Water (1 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 10 °C. The inner tube has a diameter of 25 mm, and U = 400 W/(m²·K) is based on this diameter. Find the tube length L needed.
ṁh = 0.8 kg/scp,h = 2.5 kJ/(kg·K)Th: 135 → 60 °Cṁc = 1 kg/sTc,in = 10 °CD = 25 mmU = 400 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Ti = 0.486 °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)/(1.4·20) = 0.007143 K/W
R2 = L2/(k2A) = (50/1000)/(0.72·20) = 0.003472 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 29/0.01062 = 2732 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 20 − 2732·0.007143 = 0.486 °C
Answer
Ti = 0.486 °C
2.Heat conduction
Answer: Ts = 14.71 °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/(7.7·10) = 0.01299 K/W
R1 = L1/(k1A) = (50/1000)/(0.17·10) = 0.02941 K/W
R2 = L2/(k2A) = (75/1000)/(0.72·10) = 0.01042 K/W
Ro = 1/(hoA) = 1/(20·10) = 0.005 K/W
ΣR = 0.05782 K/W
Heat loss
Q̇ = (21 − (-7))/0.05782 = 484.3 W
Ts = Troom − Q̇·Ri = 21 − 484.3·0.01299 = 14.71 °C
Answer
Ts = 14.71 °C
3.Heat exchangers (LMTD)
Answer: U = 1622 W/(m²·K)
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 = 130 − 100 = 30 K
ΔT₂ = Th,out − Tc,in = 75 − 30 = 45 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (30 − 45)/ln(30/45) = 36.99 K
Overall coefficient
U = Q̇/(A·ΔTlm) = 360000/(6·36.99) = 1622 W/(m²·K)
Answer
U = 1622 W/(m²·K)
4.Heat exchangers (LMTD)
Answer: L = 93.92 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 = 0.8·2.5·(135 − 60) = 150 kW
Tc,out = 10 + 150/(1·4.18) = 45.89 °C
End temperature differences
In parallel flow both inlets are at the same end, and both outlets at the other.
ΔT₁ = Th,in − Tc,in = 135 − 10 = 125 K
ΔT₂ = Th,out − Tc,out = 60 − 45.89 = 14.11 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (125 − 14.11)/ln(125/14.11) = 50.84 K
Area and length
A = 150000/(400·50.84) = 7.376 m²
L = A/(πD) = 7.376/(π·0.025) = 93.92 m
Answer
L = 93.92 m
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