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 brick and 75 mm of expanded polystyrene, with an area of 4 m². The outer faces are at 19 °C and 1 °C. Find the heat flow Q̇ through the wall.
L1 = 50 mm, k1 = 0.72 W/(m·K)L2 = 75 mm, k2 = 0.035 W/(m·K)A = 4 m²T1 = 19 °CT3 = 1 °C
Q̇ = W
2.Heat conductionHard
A wall of 50 mm gypsum board and 150 mm glass separates a room at 20 °C from outside air at -7 °C. The convection coefficients are 7.7 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 6 m². Find the overall heat transfer coefficient U of the wall.
L1 = 50 mm, k1 = 0.17 W/(m·K)L2 = 150 mm, k2 = 0.96 W/(m·K)A = 6 m²hi = 7.7 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = -7 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 160 kW. The hot fluid cools from 140 to 35 °C and the cold fluid heats from 10 to 70 °C. The overall heat transfer coefficient is 1000 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 160 kWU = 1000 W/(m²·K)Th: 140 → 35 °CTc: 10 → 70 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a parallel-flow double-pipe heat exchanger, hot oil (0.2 kg/s, cp = 2.3 kJ/(kg·K)) flows in the inner tube and cools from 100 to 50 °C. Water (1.5 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 25 °C. The inner tube has a diameter of 40 mm, and U = 300 W/(m²·K) is based on this diameter. Find the tube length L needed.
ṁh = 0.2 kg/scp,h = 2.3 kJ/(kg·K)Th: 100 → 50 °Cṁc = 1.5 kg/sTc,in = 25 °CD = 40 mmU = 300 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Q̇ = 32.55 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)/(0.72·4) = 0.01736 K/W
R2 = L2/(k2A) = (75/1000)/(0.035·4) = 0.5357 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 18/0.5531 = 32.55 W
Answer
Q̇ = 32.55 W
2.Heat conduction
Answer: U = 1.612 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/(7.7·6) = 0.02165 K/W
R1 = L1/(k1A) = (50/1000)/(0.17·6) = 0.04902 K/W
R2 = L2/(k2A) = (150/1000)/(0.96·6) = 0.02604 K/W
Ro = 1/(hoA) = 1/(25·6) = 0.006667 K/W
ΣR = 0.1034 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(6·0.1034) = 1.612 W/(m²·K)
Answer
U = 1.612 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 3.661 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 = 140 − 70 = 70 K
ΔT₂ = Th,out − Tc,in = 35 − 10 = 25 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (70 − 25)/ln(70/25) = 43.71 K
Area
A = Q̇/(U·ΔTlm) = 160000/(1000·43.71) = 3.661 m²
Answer
A = 3.661 m²
4.Heat exchangers (LMTD)
Answer: L = 14.29 m
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 = 0.2·2.3·(100 − 50) = 23 kW
Tc,out = 25 + 23/(1.5·4.18) = 28.67 °C
End temperature differences
In parallel flow both inlets are at the same end, and both outlets at the other.
ΔT₁ = Th,in − Tc,in = 100 − 25 = 75 K
ΔT₂ = Th,out − Tc,out = 50 − 28.67 = 21.33 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (75 − 21.33)/ln(75/21.33) = 42.69 K
Area and length
A = 23000/(300·42.69) = 1.796 m²
L = A/(πD) = 1.796/(π·0.04) = 14.29 m
Answer
L = 14.29 m
ExamProblems · examproblems.com/courses/heat-transfer