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 glass and 75 mm of expanded polystyrene, with an area of 8 m². The outer faces are at 22 °C and -5 °C. Find the heat flow Q̇ through the wall.
L1 = 100 mm, k1 = 0.96 W/(m·K)L2 = 75 mm, k2 = 0.035 W/(m·K)A = 8 m²T1 = 22 °CT3 = -5 °C
Q̇ = W
2.Heat conductionHard
A wall of 250 mm gypsum board and 75 mm expanded polystyrene separates a room at 21 °C from outside air at 0 °C. The convection coefficients are 8 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 8 m². Find the overall heat transfer coefficient U of the wall.
L1 = 250 mm, k1 = 0.17 W/(m·K)L2 = 75 mm, k2 = 0.035 W/(m·K)A = 8 m²hi = 8 W/(m²·K)ho = 30 W/(m²·K)Troom = 21 °CTout = 0 °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 kg/s, cp = 4.18 kJ/(kg·K)) flows through the tubes and heats from 25 to 80 °C. The overall heat transfer coefficient is 3500 W/(m²·K). Find the heat transfer area A needed.
Ts = 140 °Cṁ = 1 kg/scp = 4.18 kJ/(kg·K)Tc: 25 → 80 °CU = 3500 W/(m²·K)
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (2 kg/s, cp = 2.3 kJ/(kg·K)) cools from 175 to 130 °C in a counter-flow exchanger. It heats 3 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 10 °C. With U = 200 W/(m²·K), find the heat transfer area A needed.
ṁh = 2 kg/scp,h = 2.3 kJ/(kg·K)Th: 175 → 130 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 10 °CU = 200 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 96.13 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)/(0.96·8) = 0.01302 K/W
R2 = L2/(k2A) = (75/1000)/(0.035·8) = 0.2679 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 27/0.2809 = 96.13 W
Answer
Q̇ = 96.13 W
2.Heat conduction
Answer: U = 0.2651 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/(8·8) = 0.01563 K/W
R1 = L1/(k1A) = (250/1000)/(0.17·8) = 0.1838 K/W
R2 = L2/(k2A) = (75/1000)/(0.035·8) = 0.2679 K/W
Ro = 1/(hoA) = 1/(30·8) = 0.004167 K/W
ΣR = 0.4715 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(8·0.4715) = 0.2651 W/(m²·K)
Answer
U = 0.2651 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 0.777 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 = 1·4.18·(80 − 25) = 229.9 kW
End temperature differences
ΔT₁ = 140 − 25 = 115 K
ΔT₂ = 140 − 80 = 60 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (115 − 60)/ln(115/60) = 84.54 K
Area
A = Q̇/(U·ΔTlm) = 229900/(3500·84.54) = 0.777 m²
Answer
A = 0.777 m²
4.Heat exchangers (LMTD)
Answer: A = 7.739 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) = 2·2.3·45 = 207 kW
Tc,out = 10 + 207/(3·4.18) = 26.51 °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 = 175 − 26.51 = 148.5 K
ΔT₂ = Th,out − Tc,in = 130 − 10 = 120 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (148.5 − 120)/ln(148.5/120) = 133.7 K
Area
A = Q̇/(U·ΔTlm) = 207000/(200·133.7) = 7.739 m²
Answer
A = 7.739 m²
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