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 brick, with an area of 6 m². The outer faces are at 24 °C and 6 °C. Find the heat flow Q̇ through the wall.
L1 = 50 mm, k1 = 0.17 W/(m·K)L2 = 50 mm, k2 = 0.72 W/(m·K)A = 6 m²T1 = 24 °CT3 = 6 °C
Q̇ = W
2.Heat conductionHard
A wall of 100 mm concrete and 75 mm mineral wool separates a room at 20 °C from outside air at 5 °C. The convection coefficients are 8 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 10 m². Find the heat loss Q̇ through the wall.
L1 = 100 mm, k1 = 1.4 W/(m·K)L2 = 75 mm, k2 = 0.04 W/(m·K)A = 10 m²hi = 8 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = 5 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 150 kW. The hot fluid cools from 90 to 65 °C and the cold fluid heats from 30 to 70 °C. The overall heat transfer coefficient is 800 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 150 kWU = 800 W/(m²·K)Th: 90 → 65 °CTc: 30 → 70 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a counter-flow double-pipe heat exchanger, hot oil (0.5 kg/s, cp = 2.3 kJ/(kg·K)) flows in the inner tube and cools from 140 to 75 °C. Water (0.8 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 15 °C. The inner tube has a diameter of 40 mm, and U = 250 W/(m²·K) is based on this diameter. Find the tube length L needed.
ṁh = 0.5 kg/scp,h = 2.3 kJ/(kg·K)Th: 140 → 75 °Cṁc = 0.8 kg/sTc,in = 15 °CD = 40 mmU = 250 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Q̇ = 297.1 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.
Layer resistances
R1 = L1/(k1A) = (50/1000)/(0.17·6) = 0.04902 K/W
R2 = L2/(k2A) = (50/1000)/(0.72·6) = 0.01157 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 18/0.06059 = 297.1 W
Answer
Q̇ = 297.1 W
2.Heat conduction
Answer: Q̇ = 71.04 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/(8·10) = 0.0125 K/W
R1 = L1/(k1A) = (100/1000)/(1.4·10) = 0.007143 K/W
R2 = L2/(k2A) = (75/1000)/(0.04·10) = 0.1875 K/W
Ro = 1/(hoA) = 1/(25·10) = 0.004 K/W
ΣR = 0.2111 K/W
Heat loss
Q̇ = (20 − (5))/0.2111 = 71.04 W
Answer
Q̇ = 71.04 W
3.Heat exchangers (LMTD)
Answer: A = 6.995 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 = 90 − 70 = 20 K
ΔT₂ = Th,out − Tc,in = 65 − 30 = 35 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (20 − 35)/ln(20/35) = 26.8 K
Area
A = Q̇/(U·ΔTlm) = 150000/(800·26.8) = 6.995 m²
Answer
A = 6.995 m²
4.Heat exchangers (LMTD)
Answer: L = 29.96 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 = 0.5·2.3·(140 − 75) = 74.75 kW
Tc,out = 15 + 74.75/(0.8·4.18) = 37.35 °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 = 140 − 37.35 = 102.6 K
ΔT₂ = Th,out − Tc,in = 75 − 15 = 60 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (102.6 − 60)/ln(102.6/60) = 79.42 K
Area and length
A = 74750/(250·79.42) = 3.765 m²
L = A/(πD) = 3.765/(π·0.04) = 29.96 m
Answer
L = 29.96 m
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