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 expanded polystyrene and 50 mm of concrete, with an area of 12 m². The outer faces are at 17 °C and 10 °C. Find the heat flow Q̇ through the wall.
L1 = 50 mm, k1 = 0.035 W/(m·K)L2 = 50 mm, k2 = 1.4 W/(m·K)A = 12 m²T1 = 17 °CT3 = 10 °C
Q̇ = W
2.Heat conductionHard
A wall of 150 mm gypsum board and 75 mm concrete separates a room at 21 °C from outside air at -8 °C. The convection coefficients are 10 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 6 m². Find the heat loss Q̇ through the wall.
L1 = 150 mm, k1 = 0.17 W/(m·K)L2 = 75 mm, k2 = 1.4 W/(m·K)A = 6 m²hi = 10 W/(m²·K)ho = 25 W/(m²·K)Troom = 21 °CTout = -8 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 50 kW. Its heat transfer area is 6 m². The hot fluid cools from 125 to 65 °C and the cold fluid heats from 15 to 95 °C. Find the overall heat transfer coefficient U.
Q̇ = 50 kWA = 6 m²Th: 125 → 65 °CTc: 15 → 95 °C
U = W/(m²·K)
4.Heat exchangers (LMTD)Hard
Hot oil (0.5 kg/s, cp = 2.5 kJ/(kg·K)) cools from 80 to 45 °C in a counter-flow exchanger. It heats 3 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 25 °C. With U = 1000 W/(m²·K), find the heat transfer area A needed.
ṁh = 0.5 kg/scp,h = 2.5 kJ/(kg·K)Th: 80 → 45 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 25 °CU = 1000 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 57.37 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) = (50/1000)/(0.035·12) = 0.119 K/W
R2 = L2/(k2A) = (50/1000)/(1.4·12) = 0.002976 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 7/0.122 = 57.37 W
Answer
Q̇ = 57.37 W
2.Heat conduction
Answer: Q̇ = 161.7 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/(10·6) = 0.01667 K/W
R1 = L1/(k1A) = (150/1000)/(0.17·6) = 0.1471 K/W
R2 = L2/(k2A) = (75/1000)/(1.4·6) = 0.008929 K/W
Ro = 1/(hoA) = 1/(25·6) = 0.006667 K/W
ΣR = 0.1793 K/W
Heat loss
Q̇ = (21 − (-8))/0.1793 = 161.7 W
Answer
Q̇ = 161.7 W
3.Heat exchangers (LMTD)
Answer: U = 212.8 W/(m²·K)
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 = 125 − 95 = 30 K
ΔT₂ = Th,out − Tc,in = 65 − 15 = 50 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (30 − 50)/ln(30/50) = 39.15 K
Overall coefficient
U = Q̇/(A·ΔTlm) = 50000/(6·39.15) = 212.8 W/(m²·K)
Answer
U = 212.8 W/(m²·K)
4.Heat exchangers (LMTD)
Answer: A = 1.314 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) = 0.5·2.5·35 = 43.75 kW
Tc,out = 25 + 43.75/(3·4.18) = 28.49 °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 = 80 − 28.49 = 51.51 K
ΔT₂ = Th,out − Tc,in = 45 − 25 = 20 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (51.51 − 20)/ln(51.51/20) = 33.31 K
Area
A = Q̇/(U·ΔTlm) = 43750/(1000·33.31) = 1.314 m²
Answer
A = 1.314 m²
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