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 concrete and 100 mm of mineral wool, with an area of 20 m². The outer faces are at 22 °C and -3 °C. Find the heat flow Q̇ through the wall.
L1 = 200 mm, k1 = 1.4 W/(m·K)L2 = 100 mm, k2 = 0.04 W/(m·K)A = 20 m²T1 = 22 °CT3 = -3 °C
Q̇ = W
2.Heat conductionHard
A wall of 200 mm glass and 50 mm mineral wool separates a room at 21 °C from outside air at -14 °C. The convection coefficients are 7.7 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 6 m². Find the temperature Ts of the inside surface of the wall.
L1 = 200 mm, k1 = 0.96 W/(m·K)L2 = 50 mm, k2 = 0.04 W/(m·K)A = 6 m²hi = 7.7 W/(m²·K)ho = 20 W/(m²·K)Troom = 21 °CTout = -14 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
In a condenser, steam condenses at a constant 110 °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 10 to 25 °C. The overall heat transfer coefficient is 1500 W/(m²·K). Find the heat transfer area A needed.
Ts = 110 °Cṁ = 3 kg/scp = 4.18 kJ/(kg·K)Tc: 10 → 25 °CU = 1500 W/(m²·K)
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1.5 kg/s, cp = 2.1 kJ/(kg·K)) cools from 155 to 75 °C in a counter-flow exchanger. It heats 1.5 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 20 °C. With U = 300 W/(m²·K), find the heat transfer area A needed.
ṁh = 1.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 155 → 75 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 20 °CU = 300 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 189.2 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.
Layer resistances
R1 = L1/(k1A) = (200/1000)/(1.4·20) = 0.007143 K/W
R2 = L2/(k2A) = (100/1000)/(0.04·20) = 0.125 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 25/0.1321 = 189.2 W
Answer
Q̇ = 189.2 W
2.Heat conduction
Answer: Ts = 18.23 °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) = (200/1000)/(0.96·6) = 0.03472 K/W
R2 = L2/(k2A) = (50/1000)/(0.04·6) = 0.2083 K/W
Ro = 1/(hoA) = 1/(20·6) = 0.008333 K/W
ΣR = 0.273 K/W
Heat loss
Q̇ = (21 − (-14))/0.273 = 128.2 W
Ts = Troom − Q̇·Ri = 21 − 128.2·0.02165 = 18.23 °C
Answer
Ts = 18.23 °C
3.Heat exchangers (LMTD)
Answer: A = 1.359 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·(25 − 10) = 188.1 kW
End temperature differences
ΔT₁ = 110 − 10 = 100 K
ΔT₂ = 110 − 25 = 85 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (100 − 85)/ln(100/85) = 92.3 K
Area
A = Q̇/(U·ΔTlm) = 188100/(1500·92.3) = 1.359 m²
Answer
A = 1.359 m²
4.Heat exchangers (LMTD)
Answer: A = 11.49 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) = 1.5·2.1·80 = 252 kW
Tc,out = 20 + 252/(1.5·4.18) = 60.19 °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 = 155 − 60.19 = 94.81 K
ΔT₂ = Th,out − Tc,in = 75 − 20 = 55 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (94.81 − 55)/ln(94.81/55) = 73.11 K
Area
A = Q̇/(U·ΔTlm) = 252000/(300·73.11) = 11.49 m²
Answer
A = 11.49 m²
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