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 expanded polystyrene and 150 mm of gypsum board, with an area of 12 m². The outer faces are at 15 °C and 9 °C. Find the temperature Ti at the interface between the two layers.
L1 = 250 mm, k1 = 0.035 W/(m·K)L2 = 150 mm, k2 = 0.17 W/(m·K)A = 12 m²T1 = 15 °CT3 = 9 °C
Ti = °C
2.Heat conductionHard
A wall of 250 mm concrete and 150 mm wood separates a room at 22 °C from outside air at -7 °C. The convection coefficients are 8 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 8 m². Find the overall heat transfer coefficient U of the wall.
L1 = 250 mm, k1 = 1.4 W/(m·K)L2 = 150 mm, k2 = 0.17 W/(m·K)A = 8 m²hi = 8 W/(m²·K)ho = 25 W/(m²·K)Troom = 22 °CTout = -7 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A shell-and-tube heat exchanger (one shell pass, two tube passes) must transfer 420 kW. The hot fluid cools from 140 to 65 °C and the cold fluid heats from 30 to 130 °C. The LMTD correction factor for this arrangement is F = 0.95, and U = 200 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 420 kWF = 0.95U = 200 W/(m²·K)Th: 140 → 65 °CTc: 30 → 130 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (0.5 kg/s, cp = 2.1 kJ/(kg·K)) cools from 125 to 55 °C in a counter-flow exchanger. It heats 1.5 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 10 °C. With U = 1500 W/(m²·K), find the heat transfer area A needed.
ṁh = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 125 → 55 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 10 °CU = 1500 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = 9.66 °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)/(0.035·12) = 0.5952 K/W
R2 = L2/(k2A) = (150/1000)/(0.17·12) = 0.07353 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 6/0.6688 = 8.972 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 15 − 8.972·0.5952 = 9.66 °C
Answer
Ti = 9.66 °C
2.Heat conduction
Answer: U = 0.8157 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/(8·8) = 0.01563 K/W
R1 = L1/(k1A) = (250/1000)/(1.4·8) = 0.02232 K/W
R2 = L2/(k2A) = (150/1000)/(0.17·8) = 0.1103 K/W
Ro = 1/(hoA) = 1/(25·8) = 0.005 K/W
ΣR = 0.1532 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(8·0.1532) = 0.8157 W/(m²·K)
Answer
U = 0.8157 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 110.8 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 = 140 − 130 = 10 K
ΔT₂ = Th,out − Tc,in = 65 − 30 = 35 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (10 − 35)/ln(10/35) = 19.96 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 420000/(200·0.95·19.96) = 110.8 m²
Answer
A = 110.8 m²
4.Heat exchangers (LMTD)
Answer: A = 0.6985 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) = 0.5·2.1·70 = 73.5 kW
Tc,out = 10 + 73.5/(1.5·4.18) = 21.72 °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 = 125 − 21.72 = 103.3 K
ΔT₂ = Th,out − Tc,in = 55 − 10 = 45 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (103.3 − 45)/ln(103.3/45) = 70.15 K
Area
A = Q̇/(U·ΔTlm) = 73500/(1500·70.15) = 0.6985 m²
Answer
A = 0.6985 m²
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