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 250 mm of glass and 100 mm of wood, with an area of 6 m². The outer faces are at 19 °C and 0 °C. Find the temperature Ti at the interface between the two layers.
L1 = 250 mm, k1 = 0.96 W/(m·K)L2 = 100 mm, k2 = 0.17 W/(m·K)A = 6 m²T1 = 19 °CT3 = 0 °C
Ti = °C
2.Heat conductionHard
A wall of 100 mm mineral wool and 75 mm gypsum board separates a room at 21 °C from outside air at -8 °C. The convection coefficients are 6 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 20 m². Find the heat loss Q̇ through the wall.
L1 = 100 mm, k1 = 0.04 W/(m·K)L2 = 75 mm, k2 = 0.17 W/(m·K)A = 20 m²hi = 6 W/(m²·K)ho = 30 W/(m²·K)Troom = 21 °CTout = -8 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
In a condenser, steam condenses at a constant 140 °C on the outside of the tubes. Cooling water (1.5 kg/s, cp = 4.18 kJ/(kg·K)) flows through the tubes and heats from 10 to 55 °C. The overall heat transfer coefficient is 3500 W/(m²·K). Find the heat transfer area A needed.
Ts = 140 °Cṁ = 1.5 kg/scp = 4.18 kJ/(kg·K)Tc: 10 → 55 °CU = 3500 W/(m²·K)
A = m²
4.Heat exchangers (LMTD)Hard
In a parallel-flow double-pipe heat exchanger, hot oil (0.3 kg/s, cp = 2.3 kJ/(kg·K)) flows in the inner tube and cools from 90 to 60 °C. Water (1.5 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 15 °C. The inner tube has a diameter of 32 mm, and U = 150 W/(m²·K) is based on this diameter. Find the tube length L needed.
ṁh = 0.3 kg/scp,h = 2.3 kJ/(kg·K)Th: 90 → 60 °Cṁc = 1.5 kg/sTc,in = 15 °CD = 32 mmU = 150 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Ti = 13.17 °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) = (250/1000)/(0.96·6) = 0.0434 K/W
R2 = L2/(k2A) = (100/1000)/(0.17·6) = 0.09804 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 19/0.1414 = 134.3 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 19 − 134.3·0.0434 = 13.17 °C
Answer
Ti = 13.17 °C
2.Heat conduction
Answer: Q̇ = 184.6 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/(6·20) = 0.008333 K/W
R1 = L1/(k1A) = (100/1000)/(0.04·20) = 0.125 K/W
R2 = L2/(k2A) = (75/1000)/(0.17·20) = 0.02206 K/W
Ro = 1/(hoA) = 1/(30·20) = 0.001667 K/W
ΣR = 0.1571 K/W
Heat loss
Q̇ = (21 − (-8))/0.1571 = 184.6 W
Answer
Q̇ = 184.6 W
3.Heat exchangers (LMTD)
Answer: A = 0.7611 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 from the water side
Q̇ = 1.5·4.18·(55 − 10) = 282.1 kW
End temperature differences
ΔT₁ = 140 − 10 = 130 K
ΔT₂ = 140 − 55 = 85 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (130 − 85)/ln(130/85) = 105.9 K
Area
A = Q̇/(U·ΔTlm) = 282150/(3500·105.9) = 0.7611 m²
Answer
A = 0.7611 m²
4.Heat exchangers (LMTD)
Answer: L = 24.2 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 = 0.3·2.3·(90 − 60) = 20.7 kW
Tc,out = 15 + 20.7/(1.5·4.18) = 18.3 °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 = 90 − 15 = 75 K
ΔT₂ = Th,out − Tc,out = 60 − 18.3 = 41.7 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (75 − 41.7)/ln(75/41.7) = 56.73 K
Area and length
A = 20700/(150·56.73) = 2.433 m²
L = A/(πD) = 2.433/(π·0.032) = 24.2 m
Answer
L = 24.2 m
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