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 wood and 75 mm of brick, with an area of 4 m². The outer faces are at 17 °C and -6 °C. Find the heat flow Q̇ through the wall.
L1 = 100 mm, k1 = 0.17 W/(m·K)L2 = 75 mm, k2 = 0.72 W/(m·K)A = 4 m²T1 = 17 °CT3 = -6 °C
Q̇ = W
2.Heat conductionHard
A wall of 150 mm wood and 75 mm expanded polystyrene separates a room at 22 °C from outside air at -9 °C. The convection coefficients are 8 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 15 m². Find the heat loss Q̇ through the wall.
L1 = 150 mm, k1 = 0.17 W/(m·K)L2 = 75 mm, k2 = 0.035 W/(m·K)A = 15 m²hi = 8 W/(m²·K)ho = 20 W/(m²·K)Troom = 22 °CTout = -9 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
In a condenser, steam condenses at a constant 105 °C on the outside of the tubes. Cooling water (5 kg/s, cp = 4.18 kJ/(kg·K)) flows through the tubes and heats from 25 to 65 °C. The overall heat transfer coefficient is 3500 W/(m²·K). Find the heat transfer area A needed.
Ts = 105 °Cṁ = 5 kg/scp = 4.18 kJ/(kg·K)Tc: 25 → 65 °CU = 3500 W/(m²·K)
A = m²
4.Heat exchangers (LMTD)Hard
In a counter-flow double-pipe heat exchanger, hot oil (0.5 kg/s, cp = 2.1 kJ/(kg·K)) flows in the inner tube and cools from 120 to 60 °C. Water (0.2 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 10 °C. The inner tube has a diameter of 40 mm, and U = 150 W/(m²·K) is based on this diameter. Find the tube length L needed.
ṁh = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 120 → 60 °Cṁc = 0.2 kg/sTc,in = 10 °CD = 40 mmU = 150 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Q̇ = 132.9 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.
Layer resistances
R1 = L1/(k1A) = (100/1000)/(0.17·4) = 0.1471 K/W
R2 = L2/(k2A) = (75/1000)/(0.72·4) = 0.02604 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 23/0.1731 = 132.9 W
Answer
Q̇ = 132.9 W
2.Heat conduction
Answer: Q̇ = 145.3 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/(8·15) = 0.008333 K/W
R1 = L1/(k1A) = (150/1000)/(0.17·15) = 0.05882 K/W
R2 = L2/(k2A) = (75/1000)/(0.035·15) = 0.1429 K/W
Ro = 1/(hoA) = 1/(20·15) = 0.003333 K/W
ΣR = 0.2133 K/W
Heat loss
Q̇ = (22 − (-9))/0.2133 = 145.3 W
Answer
Q̇ = 145.3 W
3.Heat exchangers (LMTD)
Answer: A = 4.139 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 = 5·4.18·(65 − 25) = 836 kW
End temperature differences
ΔT₁ = 105 − 25 = 80 K
ΔT₂ = 105 − 65 = 40 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (80 − 40)/ln(80/40) = 57.71 K
Area
A = Q̇/(U·ΔTlm) = 836000/(3500·57.71) = 4.139 m²
Answer
A = 4.139 m²
4.Heat exchangers (LMTD)
Answer: L = 79.86 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 = 0.5·2.1·(120 − 60) = 63 kW
Tc,out = 10 + 63/(0.2·4.18) = 85.36 °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 = 120 − 85.36 = 34.64 K
ΔT₂ = Th,out − Tc,in = 60 − 10 = 50 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (34.64 − 50)/ln(34.64/50) = 41.85 K
Area and length
A = 63000/(150·41.85) = 10.04 m²
L = A/(πD) = 10.04/(π·0.04) = 79.86 m
Answer
L = 79.86 m
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