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 brick and 100 mm of wood, with an area of 6 m². The outer faces are at 23 °C and -1 °C. Find the heat flow Q̇ through the wall.
L1 = 100 mm, k1 = 0.72 W/(m·K)L2 = 100 mm, k2 = 0.17 W/(m·K)A = 6 m²T1 = 23 °CT3 = -1 °C
Q̇ = W
2.Heat conductionHard
A wall of 150 mm expanded polystyrene and 50 mm wood separates a room at 21 °C from outside air at -15 °C. The convection coefficients are 10 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 20 m². Find the overall heat transfer coefficient U of the wall.
L1 = 150 mm, k1 = 0.035 W/(m·K)L2 = 50 mm, k2 = 0.17 W/(m·K)A = 20 m²hi = 10 W/(m²·K)ho = 25 W/(m²·K)Troom = 21 °CTout = -15 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 250 kW. The hot fluid cools from 175 to 140 °C and the cold fluid heats from 15 to 30 °C. The overall heat transfer coefficient is 200 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 250 kWU = 200 W/(m²·K)Th: 175 → 140 °CTc: 15 → 30 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (2 kg/s, cp = 2.3 kJ/(kg·K)) cools from 105 to 75 °C in a counter-flow exchanger. It heats 0.5 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 15 °C. With U = 1000 W/(m²·K), find the heat transfer area A needed.
ṁh = 2 kg/scp,h = 2.3 kJ/(kg·K)Th: 105 → 75 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 15 °CU = 1000 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 198 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)/(0.72·6) = 0.02315 K/W
R2 = L2/(k2A) = (100/1000)/(0.17·6) = 0.09804 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 24/0.1212 = 198 W
Answer
Q̇ = 198 W
2.Heat conduction
Answer: U = 0.2119 W/(m²·K)
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/(10·20) = 0.005 K/W
R1 = L1/(k1A) = (150/1000)/(0.035·20) = 0.2143 K/W
R2 = L2/(k2A) = (50/1000)/(0.17·20) = 0.01471 K/W
Ro = 1/(hoA) = 1/(25·20) = 0.002 K/W
ΣR = 0.236 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(20·0.236) = 0.2119 W/(m²·K)
Answer
U = 0.2119 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 9.276 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 − 30 = 145 K
ΔT₂ = Th,out − Tc,in = 140 − 15 = 125 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (145 − 125)/ln(145/125) = 134.8 K
Area
A = Q̇/(U·ΔTlm) = 250000/(200·134.8) = 9.276 m²
Answer
A = 9.276 m²
4.Heat exchangers (LMTD)
Answer: A = 3.514 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 and water outlet temperature
Q̇ = ṁhcp,h(Th,in − Th,out) = 2·2.3·30 = 138 kW
Tc,out = 15 + 138/(0.5·4.18) = 81.03 °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 = 105 − 81.03 = 23.97 K
ΔT₂ = Th,out − Tc,in = 75 − 15 = 60 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (23.97 − 60)/ln(23.97/60) = 39.27 K
Area
A = Q̇/(U·ΔTlm) = 138000/(1000·39.27) = 3.514 m²
Answer
A = 3.514 m²
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