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 50 mm of wood and 50 mm of gypsum board, with an area of 6 m². The outer faces are at 15 °C and -11 °C. Find the temperature Ti at the interface between the two layers.
L1 = 50 mm, k1 = 0.17 W/(m·K)L2 = 50 mm, k2 = 0.17 W/(m·K)A = 6 m²T1 = 15 °CT3 = -11 °C
Ti = °C
2.Heat conductionHard
A wall of 50 mm mineral wool and 100 mm expanded polystyrene separates a room at 20 °C from outside air at -15 °C. The convection coefficients are 7.7 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 6 m². Find the temperature Ts of the inside surface of the wall.
L1 = 50 mm, k1 = 0.04 W/(m·K)L2 = 100 mm, k2 = 0.035 W/(m·K)A = 6 m²hi = 7.7 W/(m²·K)ho = 30 W/(m²·K)Troom = 20 °CTout = -15 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
In a condenser, steam condenses at a constant 105 °C on the outside of the tubes. Cooling water (4 kg/s, cp = 4.18 kJ/(kg·K)) flows through the tubes and heats from 25 to 65 °C. The overall heat transfer coefficient is 3000 W/(m²·K). Find the heat transfer area A needed.
Ts = 105 °Cṁ = 4 kg/scp = 4.18 kJ/(kg·K)Tc: 25 → 65 °CU = 3000 W/(m²·K)
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1.5 kg/s, cp = 2.5 kJ/(kg·K)) cools from 145 to 100 °C in a counter-flow exchanger. It heats 0.5 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 25 °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: 145 → 100 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 25 °CU = 1000 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = 2 °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) = (50/1000)/(0.17·6) = 0.04902 K/W
R2 = L2/(k2A) = (50/1000)/(0.17·6) = 0.04902 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 26/0.09804 = 265.2 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 15 − 265.2·0.04902 = 2 °C
Answer
Ti = 2 °C
2.Heat conduction
Answer: Ts = 18.94 °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/(7.7·6) = 0.02165 K/W
R1 = L1/(k1A) = (50/1000)/(0.04·6) = 0.2083 K/W
R2 = L2/(k2A) = (100/1000)/(0.035·6) = 0.4762 K/W
Ro = 1/(hoA) = 1/(30·6) = 0.005556 K/W
ΣR = 0.7117 K/W
Heat loss
Q̇ = (20 − (-15))/0.7117 = 49.18 W
Ts = Troom − Q̇·Ri = 20 − 49.18·0.02165 = 18.94 °C
Answer
Ts = 18.94 °C
3.Heat exchangers (LMTD)
Answer: A = 3.863 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 = 4·4.18·(65 − 25) = 668.8 kW
End temperature differences
ΔT₁ = 105 − 25 = 80 K
ΔT₂ = 105 − 65 = 40 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (80 − 40)/ln(80/40) = 57.71 K
Area
A = Q̇/(U·ΔTlm) = 668800/(3000·57.71) = 3.863 m²
Answer
A = 3.863 m²
4.Heat exchangers (LMTD)
Answer: A = 3.056 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 = 25 + 168.8/(0.5·4.18) = 105.7 °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 = 145 − 105.7 = 39.26 K
ΔT₂ = Th,out − Tc,in = 100 − 25 = 75 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (39.26 − 75)/ln(39.26/75) = 55.21 K
Area
A = Q̇/(U·ΔTlm) = 168750/(1000·55.21) = 3.056 m²
Answer
A = 3.056 m²
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