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 150 mm of expanded polystyrene and 100 mm of glass, with an area of 8 m². The outer faces are at 23 °C and 2 °C. Find the heat flow Q̇ through the wall.
L1 = 150 mm, k1 = 0.035 W/(m·K)L2 = 100 mm, k2 = 0.96 W/(m·K)A = 8 m²T1 = 23 °CT3 = 2 °C
Q̇ = W
2.Heat conductionHard
A wall of 50 mm expanded polystyrene and 150 mm glass separates a room at 21 °C from outside air at -1 °C. The convection coefficients are 6 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 20 m². Find the heat loss Q̇ through the wall.
L1 = 50 mm, k1 = 0.035 W/(m·K)L2 = 150 mm, k2 = 0.96 W/(m·K)A = 20 m²hi = 6 W/(m²·K)ho = 20 W/(m²·K)Troom = 21 °CTout = -1 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A shell-and-tube heat exchanger (one shell pass, two tube passes) must transfer 430 kW. The hot fluid cools from 180 to 160 °C and the cold fluid heats from 20 to 140 °C. The LMTD correction factor for this arrangement is F = 0.9, and U = 500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 430 kWF = 0.9U = 500 W/(m²·K)Th: 180 → 160 °CTc: 20 → 140 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1.5 kg/s, cp = 2.5 kJ/(kg·K)) cools from 85 to 40 °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 = 1.5 kg/scp,h = 2.5 kJ/(kg·K)Th: 85 → 40 °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 = 38.27 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) = (150/1000)/(0.035·8) = 0.5357 K/W
R2 = L2/(k2A) = (100/1000)/(0.96·8) = 0.01302 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 21/0.5487 = 38.27 W
Answer
Q̇ = 38.27 W
2.Heat conduction
Answer: Q̇ = 244.2 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/(6·20) = 0.008333 K/W
R1 = L1/(k1A) = (50/1000)/(0.035·20) = 0.07143 K/W
R2 = L2/(k2A) = (150/1000)/(0.96·20) = 0.007813 K/W
Ro = 1/(hoA) = 1/(20·20) = 0.0025 K/W
ΣR = 0.09007 K/W
Heat loss
Q̇ = (21 − (-1))/0.09007 = 244.2 W
Answer
Q̇ = 244.2 W
3.Heat exchangers (LMTD)
Answer: A = 11.97 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 (counter-flow)
In counter-flow the hot inlet faces the cold outlet, and the hot outlet faces the cold inlet.
ΔT₁ = Th,in − Tc,out = 180 − 140 = 40 K
ΔT₂ = Th,out − Tc,in = 160 − 20 = 140 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (40 − 140)/ln(40/140) = 79.82 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 430000/(500·0.9·79.82) = 11.97 m²
Answer
A = 11.97 m²
4.Heat exchangers (LMTD)
Answer: A = 19.22 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) = 1.5·2.5·45 = 168.8 kW
Tc,out = 10 + 168.8/(3·4.18) = 23.46 °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 = 85 − 23.46 = 61.54 K
ΔT₂ = Th,out − Tc,in = 40 − 10 = 30 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (61.54 − 30)/ln(61.54/30) = 43.9 K
Area
A = Q̇/(U·ΔTlm) = 168750/(200·43.9) = 19.22 m²
Answer
A = 19.22 m²
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