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 gypsum board and 50 mm of mineral wool, with an area of 6 m². The outer faces are at 21 °C and -10 °C. Find the temperature Ti at the interface between the two layers.
L1 = 50 mm, k1 = 0.17 W/(m·K)L2 = 50 mm, k2 = 0.04 W/(m·K)A = 6 m²T1 = 21 °CT3 = -10 °C
Ti = °C
2.Heat conductionHard
A wall of 250 mm gypsum board and 75 mm brick separates a room at 20 °C from outside air at -1 °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 = 250 mm, k1 = 0.17 W/(m·K)L2 = 75 mm, k2 = 0.72 W/(m·K)A = 20 m²hi = 10 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = -1 °C
U = W/(m²·K)
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 20 to 50 °C. The overall heat transfer coefficient is 1000 W/(m²·K). Find the heat transfer area A needed.
Ts = 140 °Cṁ = 1.5 kg/scp = 4.18 kJ/(kg·K)Tc: 20 → 50 °CU = 1000 W/(m²·K)
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (0.5 kg/s, cp = 2.1 kJ/(kg·K)) cools from 90 to 55 °C in a counter-flow exchanger. It heats 3 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 15 °C. With U = 500 W/(m²·K), find the heat transfer area A needed.
ṁh = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 90 → 55 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 15 °CU = 500 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = 15.1 °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)/(0.17·6) = 0.04902 K/W
R2 = L2/(k2A) = (50/1000)/(0.04·6) = 0.2083 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 31/0.2574 = 120.5 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 21 − 120.5·0.04902 = 15.1 °C
Answer
Ti = 15.1 °C
2.Heat conduction
Answer: U = 0.5832 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) = (250/1000)/(0.17·20) = 0.07353 K/W
R2 = L2/(k2A) = (75/1000)/(0.72·20) = 0.005208 K/W
Ro = 1/(hoA) = 1/(25·20) = 0.002 K/W
ΣR = 0.08574 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(20·0.08574) = 0.5832 W/(m²·K)
Answer
U = 0.5832 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 1.804 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·(50 − 20) = 188.1 kW
End temperature differences
ΔT₁ = 140 − 20 = 120 K
ΔT₂ = 140 − 50 = 90 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (120 − 90)/ln(120/90) = 104.3 K
Area
A = Q̇/(U·ΔTlm) = 188100/(1000·104.3) = 1.804 m²
Answer
A = 1.804 m²
4.Heat exchangers (LMTD)
Answer: A = 1.349 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) = 0.5·2.1·35 = 36.75 kW
Tc,out = 15 + 36.75/(3·4.18) = 17.93 °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 = 90 − 17.93 = 72.07 K
ΔT₂ = Th,out − Tc,in = 55 − 15 = 40 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (72.07 − 40)/ln(72.07/40) = 54.47 K
Area
A = Q̇/(U·ΔTlm) = 36750/(500·54.47) = 1.349 m²
Answer
A = 1.349 m²
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