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 100 mm of concrete and 25 mm of brick, with an area of 4 m². The outer faces are at 25 °C and -12 °C. Find the temperature Ti at the interface between the two layers.
L1 = 100 mm, k1 = 1.4 W/(m·K)L2 = 25 mm, k2 = 0.72 W/(m·K)A = 4 m²T1 = 25 °CT3 = -12 °C
Ti = °C
2.Heat conductionHard
A wall of 100 mm brick and 25 mm gypsum board separates a room at 20 °C from outside air at -1 °C. The convection coefficients are 10 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 10 m². Find the heat loss Q̇ through the wall.
L1 = 100 mm, k1 = 0.72 W/(m·K)L2 = 25 mm, k2 = 0.17 W/(m·K)A = 10 m²hi = 10 W/(m²·K)ho = 30 W/(m²·K)Troom = 20 °CTout = -1 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
In a condenser, steam condenses at a constant 130 °C on the outside of the tubes. Cooling water (3 kg/s, cp = 4.18 kJ/(kg·K)) flows through the tubes and heats from 15 to 50 °C. The overall heat transfer coefficient is 3000 W/(m²·K). Find the heat transfer area A needed.
Ts = 130 °Cṁ = 3 kg/scp = 4.18 kJ/(kg·K)Tc: 15 → 50 °CU = 3000 W/(m²·K)
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1 kg/s, cp = 2.5 kJ/(kg·K)) cools from 105 to 70 °C in a counter-flow exchanger. It heats 2 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 25 °C. With U = 300 W/(m²·K), find the heat transfer area A needed.
ṁh = 1 kg/scp,h = 2.5 kJ/(kg·K)Th: 105 → 70 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 25 °CU = 300 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = 0.1028 °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) = (100/1000)/(1.4·4) = 0.01786 K/W
R2 = L2/(k2A) = (25/1000)/(0.72·4) = 0.008681 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 37/0.02654 = 1394 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 25 − 1394·0.01786 = 0.1028 °C
Answer
Ti = 0.1028 °C
2.Heat conduction
Answer: Q̇ = 500.9 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/(10·10) = 0.01 K/W
R1 = L1/(k1A) = (100/1000)/(0.72·10) = 0.01389 K/W
R2 = L2/(k2A) = (25/1000)/(0.17·10) = 0.01471 K/W
Ro = 1/(hoA) = 1/(30·10) = 0.003333 K/W
ΣR = 0.04193 K/W
Heat loss
Q̇ = (20 − (-1))/0.04193 = 500.9 W
Answer
Q̇ = 500.9 W
3.Heat exchangers (LMTD)
Answer: A = 1.517 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 from the water side
Q̇ = 3·4.18·(50 − 15) = 438.9 kW
End temperature differences
ΔT₁ = 130 − 15 = 115 K
ΔT₂ = 130 − 50 = 80 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (115 − 80)/ln(115/80) = 96.44 K
Area
A = Q̇/(U·ΔTlm) = 438900/(3000·96.44) = 1.517 m²
Answer
A = 1.517 m²
4.Heat exchangers (LMTD)
Answer: A = 5.173 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) = 1·2.5·35 = 87.5 kW
Tc,out = 25 + 87.5/(2·4.18) = 35.47 °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 = 105 − 35.47 = 69.53 K
ΔT₂ = Th,out − Tc,in = 70 − 25 = 45 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (69.53 − 45)/ln(69.53/45) = 56.38 K
Area
A = Q̇/(U·ΔTlm) = 87500/(300·56.38) = 5.173 m²
Answer
A = 5.173 m²
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