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 50 mm of expanded polystyrene and 150 mm of gypsum board, with an area of 6 m². The outer faces are at 25 °C and 6 °C. Find the temperature Ti at the interface between the two layers.
L1 = 50 mm, k1 = 0.035 W/(m·K)L2 = 150 mm, k2 = 0.17 W/(m·K)A = 6 m²T1 = 25 °CT3 = 6 °C
Ti = °C
2.Heat conductionHard
A wall of 250 mm gypsum board and 100 mm mineral wool separates a room at 22 °C from outside air at -10 °C. The convection coefficients are 6 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 15 m². Find the temperature Ts of the inside surface of the wall.
L1 = 250 mm, k1 = 0.17 W/(m·K)L2 = 100 mm, k2 = 0.04 W/(m·K)A = 15 m²hi = 6 W/(m²·K)ho = 25 W/(m²·K)Troom = 22 °CTout = -10 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 330 kW. The hot fluid cools from 175 to 40 °C and the cold fluid heats from 15 to 65 °C. The overall heat transfer coefficient is 500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 330 kWU = 500 W/(m²·K)Th: 175 → 40 °CTc: 15 → 65 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a parallel-flow double-pipe heat exchanger, hot oil (0.5 kg/s, cp = 2.3 kJ/(kg·K)) flows in the inner tube and cools from 155 to 105 °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 50 mm, and U = 500 W/(m²·K) is based on this diameter. Find the tube length L needed.
ṁh = 0.5 kg/scp,h = 2.3 kJ/(kg·K)Th: 155 → 105 °Cṁc = 1.5 kg/sTc,in = 15 °CD = 50 mmU = 500 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Ti = 13.25 °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) = (50/1000)/(0.035·6) = 0.2381 K/W
R2 = L2/(k2A) = (150/1000)/(0.17·6) = 0.1471 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 19/0.3852 = 49.33 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 25 − 49.33·0.2381 = 13.25 °C
Answer
Ti = 13.25 °C
2.Heat conduction
Answer: Ts = 20.72 °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/(6·15) = 0.01111 K/W
R1 = L1/(k1A) = (250/1000)/(0.17·15) = 0.09804 K/W
R2 = L2/(k2A) = (100/1000)/(0.04·15) = 0.1667 K/W
Ro = 1/(hoA) = 1/(25·15) = 0.002667 K/W
ΣR = 0.2785 K/W
Heat loss
Q̇ = (22 − (-10))/0.2785 = 114.9 W
Ts = Troom − Q̇·Ri = 22 − 114.9·0.01111 = 20.72 °C
Answer
Ts = 20.72 °C
3.Heat exchangers (LMTD)
Answer: A = 11.5 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 = 175 − 65 = 110 K
ΔT₂ = Th,out − Tc,in = 40 − 15 = 25 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (110 − 25)/ln(110/25) = 57.37 K
Area
A = Q̇/(U·ΔTlm) = 330000/(500·57.37) = 11.5 m²
Answer
A = 11.5 m²
4.Heat exchangers (LMTD)
Answer: L = 6.796 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 = 0.5·2.3·(155 − 105) = 57.5 kW
Tc,out = 15 + 57.5/(1.5·4.18) = 24.17 °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 = 155 − 15 = 140 K
ΔT₂ = Th,out − Tc,out = 105 − 24.17 = 80.83 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (140 − 80.83)/ln(140/80.83) = 107.7 K
Area and length
A = 57500/(500·107.7) = 1.068 m²
L = A/(πD) = 1.068/(π·0.05) = 6.796 m
Answer
L = 6.796 m
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