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 250 mm of glass and 150 mm of gypsum board, with an area of 6 m². The outer faces are at 20 °C and 8 °C. Find the temperature Ti at the interface between the two layers.
L1 = 250 mm, k1 = 0.96 W/(m·K)L2 = 150 mm, k2 = 0.17 W/(m·K)A = 6 m²T1 = 20 °CT3 = 8 °C
Ti = °C
2.Heat conductionHard
A wall of 150 mm expanded polystyrene and 100 mm brick separates a room at 20 °C from outside air at -6 °C. The convection coefficients are 10 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 6 m². Find the heat loss Q̇ through the wall.
L1 = 150 mm, k1 = 0.035 W/(m·K)L2 = 100 mm, k2 = 0.72 W/(m·K)A = 6 m²hi = 10 W/(m²·K)ho = 20 W/(m²·K)Troom = 20 °CTout = -6 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 460 kW. Its heat transfer area is 10 m². The hot fluid cools from 100 to 40 °C and the cold fluid heats from 15 to 55 °C. Find the overall heat transfer coefficient U.
Q̇ = 460 kWA = 10 m²Th: 100 → 40 °CTc: 15 → 55 °C
U = W/(m²·K)
4.Heat exchangers (LMTD)Hard
Hot oil (1 kg/s, cp = 2.5 kJ/(kg·K)) cools from 100 to 40 °C in a counter-flow exchanger. It heats 1.5 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 20 °C. With U = 1500 W/(m²·K), find the heat transfer area A needed.
ṁh = 1 kg/scp,h = 2.5 kJ/(kg·K)Th: 100 → 40 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 20 °CU = 1500 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = 17.27 °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) = (250/1000)/(0.96·6) = 0.0434 K/W
R2 = L2/(k2A) = (150/1000)/(0.17·6) = 0.1471 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 12/0.1905 = 63 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 20 − 63·0.0434 = 17.27 °C
Answer
Ti = 17.27 °C
2.Heat conduction
Answer: Q̇ = 34.1 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)/(0.035·6) = 0.7143 K/W
R2 = L2/(k2A) = (100/1000)/(0.72·6) = 0.02315 K/W
Ro = 1/(hoA) = 1/(20·6) = 0.008333 K/W
ΣR = 0.7624 K/W
Heat loss
Q̇ = (20 − (-6))/0.7624 = 34.1 W
Answer
Q̇ = 34.1 W
3.Heat exchangers (LMTD)
Answer: U = 1352 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 = 100 − 55 = 45 K
ΔT₂ = Th,out − Tc,in = 40 − 15 = 25 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (45 − 25)/ln(45/25) = 34.03 K
Overall coefficient
U = Q̇/(A·ΔTlm) = 460000/(10·34.03) = 1352 W/(m²·K)
Answer
U = 1352 W/(m²·K)
4.Heat exchangers (LMTD)
Answer: A = 2.858 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) = 1·2.5·60 = 150 kW
Tc,out = 20 + 150/(1.5·4.18) = 43.92 °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 = 100 − 43.92 = 56.08 K
ΔT₂ = Th,out − Tc,in = 40 − 20 = 20 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (56.08 − 20)/ln(56.08/20) = 34.99 K
Area
A = Q̇/(U·ΔTlm) = 150000/(1500·34.99) = 2.858 m²
Answer
A = 2.858 m²
ExamProblems · examproblems.com/courses/heat-transfer