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 200 mm of brick and 100 mm of glass, with an area of 12 m². The outer faces are at 19 °C and -4 °C. Find the temperature Ti at the interface between the two layers.
L1 = 200 mm, k1 = 0.72 W/(m·K)L2 = 100 mm, k2 = 0.96 W/(m·K)A = 12 m²T1 = 19 °CT3 = -4 °C
Ti = °C
2.Heat conductionHard
A wall of 150 mm gypsum board and 100 mm concrete separates a room at 21 °C from outside air at 4 °C. The convection coefficients are 8 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 4 m². Find the heat loss Q̇ through the wall.
L1 = 150 mm, k1 = 0.17 W/(m·K)L2 = 100 mm, k2 = 1.4 W/(m·K)A = 4 m²hi = 8 W/(m²·K)ho = 20 W/(m²·K)Troom = 21 °CTout = 4 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A shell-and-tube heat exchanger (one shell pass, two tube passes) must transfer 170 kW. The hot fluid cools from 100 to 75 °C and the cold fluid heats from 15 to 75 °C. The LMTD correction factor for this arrangement is F = 0.9, and U = 200 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 170 kWF = 0.9U = 200 W/(m²·K)Th: 100 → 75 °CTc: 15 → 75 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1 kg/s, cp = 2.1 kJ/(kg·K)) cools from 85 to 50 °C in a counter-flow exchanger. It heats 3 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 15 °C. With U = 1500 W/(m²·K), find the heat transfer area A needed.
ṁh = 1 kg/scp,h = 2.1 kJ/(kg·K)Th: 85 → 50 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 15 °CU = 1500 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = 2.273 °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) = (200/1000)/(0.72·12) = 0.02315 K/W
R2 = L2/(k2A) = (100/1000)/(0.96·12) = 0.008681 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 23/0.03183 = 722.6 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 19 − 722.6·0.02315 = 2.273 °C
Answer
Ti = 2.273 °C
2.Heat conduction
Answer: Q̇ = 60.24 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/(8·4) = 0.03125 K/W
R1 = L1/(k1A) = (150/1000)/(0.17·4) = 0.2206 K/W
R2 = L2/(k2A) = (100/1000)/(1.4·4) = 0.01786 K/W
Ro = 1/(hoA) = 1/(20·4) = 0.0125 K/W
ΣR = 0.2822 K/W
Heat loss
Q̇ = (21 − (4))/0.2822 = 60.24 W
Answer
Q̇ = 60.24 W
3.Heat exchangers (LMTD)
Answer: A = 23.62 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 = 100 − 75 = 25 K
ΔT₂ = Th,out − Tc,in = 75 − 15 = 60 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (25 − 60)/ln(25/60) = 39.98 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 170000/(200·0.9·39.98) = 23.62 m²
Answer
A = 23.62 m²
4.Heat exchangers (LMTD)
Answer: A = 1.019 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) = 1·2.1·35 = 73.5 kW
Tc,out = 15 + 73.5/(3·4.18) = 20.86 °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 − 20.86 = 64.14 K
ΔT₂ = Th,out − Tc,in = 50 − 15 = 35 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (64.14 − 35)/ln(64.14/35) = 48.11 K
Area
A = Q̇/(U·ΔTlm) = 73500/(1500·48.11) = 1.019 m²
Answer
A = 1.019 m²
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