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 concrete and 25 mm of brick, with an area of 12 m². The outer faces are at 16 °C and -11 °C. Find the temperature Ti at the interface between the two layers.
L1 = 250 mm, k1 = 1.4 W/(m·K)L2 = 25 mm, k2 = 0.72 W/(m·K)A = 12 m²T1 = 16 °CT3 = -11 °C
Ti = °C
2.Heat conductionHard
A wall of 150 mm expanded polystyrene and 75 mm mineral wool separates a room at 20 °C from outside air at -9 °C. The convection coefficients are 6 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 15 m². Find the temperature Ts of the inside surface of the wall.
L1 = 150 mm, k1 = 0.035 W/(m·K)L2 = 75 mm, k2 = 0.04 W/(m·K)A = 15 m²hi = 6 W/(m²·K)ho = 30 W/(m²·K)Troom = 20 °CTout = -9 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 410 kW. The hot fluid cools from 105 to 85 °C and the cold fluid heats from 30 to 70 °C. The overall heat transfer coefficient is 1000 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 410 kWU = 1000 W/(m²·K)Th: 105 → 85 °CTc: 30 → 70 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1.5 kg/s, cp = 2.5 kJ/(kg·K)) cools from 140 to 110 °C in a counter-flow exchanger. It heats 1 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 20 °C. With U = 1000 W/(m²·K), find the heat transfer area A needed.
ṁh = 1.5 kg/scp,h = 2.5 kJ/(kg·K)Th: 140 → 110 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 20 °CU = 1000 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = −6.605 °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)/(1.4·12) = 0.01488 K/W
R2 = L2/(k2A) = (25/1000)/(0.72·12) = 0.002894 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 27/0.01777 = 1519 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 16 − 1519·0.01488 = −6.605 °C
Answer
Ti = −6.605 °C
2.Heat conduction
Answer: Ts = 19.24 °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.
Resistances in series
Ri = 1/(hiA) = 1/(6·15) = 0.01111 K/W
R1 = L1/(k1A) = (150/1000)/(0.035·15) = 0.2857 K/W
R2 = L2/(k2A) = (75/1000)/(0.04·15) = 0.125 K/W
Ro = 1/(hoA) = 1/(30·15) = 0.002222 K/W
ΣR = 0.424 K/W
Heat loss
Q̇ = (20 − (-9))/0.424 = 68.39 W
Ts = Troom − Q̇·Ri = 20 − 68.39·0.01111 = 19.24 °C
Answer
Ts = 19.24 °C
3.Heat exchangers (LMTD)
Answer: A = 9.266 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 = 105 − 70 = 35 K
ΔT₂ = Th,out − Tc,in = 85 − 30 = 55 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (35 − 55)/ln(35/55) = 44.25 K
Area
A = Q̇/(U·ΔTlm) = 410000/(1000·44.25) = 9.266 m²
Answer
A = 9.266 m²
4.Heat exchangers (LMTD)
Answer: A = 1.229 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) = 1.5·2.5·30 = 112.5 kW
Tc,out = 20 + 112.5/(1·4.18) = 46.91 °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 − 46.91 = 93.09 K
ΔT₂ = Th,out − Tc,in = 110 − 20 = 90 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (93.09 − 90)/ln(93.09/90) = 91.53 K
Area
A = Q̇/(U·ΔTlm) = 112500/(1000·91.53) = 1.229 m²
Answer
A = 1.229 m²
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