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 50 mm of expanded polystyrene, with an area of 10 m². The outer faces are at 19 °C and -11 °C. Find the temperature Ti at the interface between the two layers.
L1 = 100 mm, k1 = 1.4 W/(m·K)L2 = 50 mm, k2 = 0.035 W/(m·K)A = 10 m²T1 = 19 °CT3 = -11 °C
Ti = °C
2.Heat conductionHard
A wall of 50 mm mineral wool and 100 mm brick separates a room at 21 °C from outside air at -2 °C. The convection coefficients are 7.7 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 = 50 mm, k1 = 0.04 W/(m·K)L2 = 100 mm, k2 = 0.72 W/(m·K)A = 20 m²hi = 7.7 W/(m²·K)ho = 25 W/(m²·K)Troom = 21 °CTout = -2 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 310 kW. The hot fluid cools from 175 to 35 °C and the cold fluid heats from 15 to 75 °C. The overall heat transfer coefficient is 500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 310 kWU = 500 W/(m²·K)Th: 175 → 35 °CTc: 15 → 75 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1.5 kg/s, cp = 2.3 kJ/(kg·K)) cools from 165 to 110 °C in a counter-flow exchanger. It heats 1 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 30 °C. With U = 800 W/(m²·K), find the heat transfer area A needed.
ṁh = 1.5 kg/scp,h = 2.3 kJ/(kg·K)Th: 165 → 110 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 30 °CU = 800 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = 17.57 °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)/(1.4·10) = 0.007143 K/W
R2 = L2/(k2A) = (50/1000)/(0.035·10) = 0.1429 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 30/0.15 = 200 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 19 − 200·0.007143 = 17.57 °C
Answer
Ti = 17.57 °C
2.Heat conduction
Answer: U = 0.6415 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/(7.7·20) = 0.006494 K/W
R1 = L1/(k1A) = (50/1000)/(0.04·20) = 0.0625 K/W
R2 = L2/(k2A) = (100/1000)/(0.72·20) = 0.006944 K/W
Ro = 1/(hoA) = 1/(25·20) = 0.002 K/W
ΣR = 0.07794 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(20·0.07794) = 0.6415 W/(m²·K)
Answer
U = 0.6415 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 12.47 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 − 75 = 100 K
ΔT₂ = Th,out − Tc,in = 35 − 15 = 20 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (100 − 20)/ln(100/20) = 49.71 K
Area
A = Q̇/(U·ΔTlm) = 310000/(500·49.71) = 12.47 m²
Answer
A = 12.47 m²
4.Heat exchangers (LMTD)
Answer: A = 2.8 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) = 1.5·2.3·55 = 189.7 kW
Tc,out = 30 + 189.7/(1·4.18) = 75.39 °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 = 165 − 75.39 = 89.61 K
ΔT₂ = Th,out − Tc,in = 110 − 30 = 80 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (89.61 − 80)/ln(89.61/80) = 84.71 K
Area
A = Q̇/(U·ΔTlm) = 189750/(800·84.71) = 2.8 m²
Answer
A = 2.8 m²
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