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 concrete and 25 mm of brick, with an area of 15 m². The outer faces are at 16 °C and 1 °C. Find the temperature Ti at the interface between the two layers.
L1 = 100 mm, k1 = 1.4 W/(m·K)L2 = 25 mm, k2 = 0.72 W/(m·K)A = 15 m²T1 = 16 °CT3 = 1 °C
Ti = °C
2.Heat conductionHard
A wall of 250 mm gypsum board and 75 mm mineral wool separates a room at 21 °C from outside air at 1 °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 = 250 mm, k1 = 0.17 W/(m·K)L2 = 75 mm, k2 = 0.04 W/(m·K)A = 4 m²hi = 8 W/(m²·K)ho = 30 W/(m²·K)Troom = 21 °CTout = 1 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 20 kW. The hot fluid cools from 105 to 40 °C and the cold fluid heats from 15 to 60 °C. The overall heat transfer coefficient is 200 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 20 kWU = 200 W/(m²·K)Th: 105 → 40 °CTc: 15 → 60 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a parallel-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 140 to 65 °C. Water (1.5 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 20 °C. The inner tube has a diameter of 25 mm, and U = 500 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: 140 → 65 °Cṁc = 1.5 kg/sTc,in = 20 °CD = 25 mmU = 500 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Ti = 5.907 °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)/(1.4·15) = 0.004762 K/W
R2 = L2/(k2A) = (25/1000)/(0.72·15) = 0.002315 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 15/0.007077 = 2120 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 16 − 2120·0.004762 = 5.907 °C
Answer
Ti = 5.907 °C
2.Heat conduction
Answer: Q̇ = 22.83 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) = (250/1000)/(0.17·4) = 0.3676 K/W
R2 = L2/(k2A) = (75/1000)/(0.04·4) = 0.4688 K/W
Ro = 1/(hoA) = 1/(30·4) = 0.008333 K/W
ΣR = 0.876 K/W
Heat loss
Q̇ = (21 − (1))/0.876 = 22.83 W
Answer
Q̇ = 22.83 W
3.Heat exchangers (LMTD)
Answer: A = 2.939 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 = 105 − 60 = 45 K
ΔT₂ = Th,out − Tc,in = 40 − 15 = 25 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (45 − 25)/ln(45/25) = 34.03 K
Area
A = Q̇/(U·ΔTlm) = 20000/(200·34.03) = 2.939 m²
Answer
A = 2.939 m²
4.Heat exchangers (LMTD)
Answer: L = 16.97 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·(140 − 65) = 47.25 kW
Tc,out = 20 + 47.25/(1.5·4.18) = 27.54 °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 = 140 − 20 = 120 K
ΔT₂ = Th,out − Tc,out = 65 − 27.54 = 37.46 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (120 − 37.46)/ln(120/37.46) = 70.9 K
Area and length
A = 47250/(500·70.9) = 1.333 m²
L = A/(πD) = 1.333/(π·0.025) = 16.97 m
Answer
L = 16.97 m
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