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 expanded polystyrene and 75 mm of wood, with an area of 4 m². The outer faces are at 16 °C and 2 °C. Find the heat flow Q̇ through the wall.
L1 = 200 mm, k1 = 0.035 W/(m·K)L2 = 75 mm, k2 = 0.17 W/(m·K)A = 4 m²T1 = 16 °CT3 = 2 °C
Q̇ = W
2.Heat conductionHard
A wall of 250 mm concrete and 75 mm glass separates a room at 20 °C from outside air at 4 °C. The convection coefficients are 6 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 8 m². Find the heat loss Q̇ through the wall.
L1 = 250 mm, k1 = 1.4 W/(m·K)L2 = 75 mm, k2 = 0.96 W/(m·K)A = 8 m²hi = 6 W/(m²·K)ho = 30 W/(m²·K)Troom = 20 °CTout = 4 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 230 kW. The hot fluid cools from 95 to 40 °C and the cold fluid heats from 20 to 80 °C. The overall heat transfer coefficient is 800 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 230 kWU = 800 W/(m²·K)Th: 95 → 40 °CTc: 20 → 80 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (0.5 kg/s, cp = 2.3 kJ/(kg·K)) cools from 100 to 55 °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 = 1000 W/(m²·K), find the heat transfer area A needed.
ṁh = 0.5 kg/scp,h = 2.3 kJ/(kg·K)Th: 100 → 55 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 25 °CU = 1000 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 9.098 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)/(0.035·4) = 1.429 K/W
R2 = L2/(k2A) = (75/1000)/(0.17·4) = 0.1103 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 14/1.539 = 9.098 W
Answer
Q̇ = 9.098 W
2.Heat conduction
Answer: Q̇ = 280.3 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/(6·8) = 0.02083 K/W
R1 = L1/(k1A) = (250/1000)/(1.4·8) = 0.02232 K/W
R2 = L2/(k2A) = (75/1000)/(0.96·8) = 0.009766 K/W
Ro = 1/(hoA) = 1/(30·8) = 0.004167 K/W
ΣR = 0.05709 K/W
Heat loss
Q̇ = (20 − (4))/0.05709 = 280.3 W
Answer
Q̇ = 280.3 W
3.Heat exchangers (LMTD)
Answer: A = 16.54 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 = 95 − 80 = 15 K
ΔT₂ = Th,out − Tc,in = 40 − 20 = 20 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (15 − 20)/ln(15/20) = 17.38 K
Area
A = Q̇/(U·ΔTlm) = 230000/(800·17.38) = 16.54 m²
Answer
A = 16.54 m²
4.Heat exchangers (LMTD)
Answer: A = 1.107 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) = 0.5·2.3·45 = 51.75 kW
Tc,out = 25 + 51.75/(2·4.18) = 31.19 °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 = 100 − 31.19 = 68.81 K
ΔT₂ = Th,out − Tc,in = 55 − 25 = 30 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (68.81 − 30)/ln(68.81/30) = 46.75 K
Area
A = Q̇/(U·ΔTlm) = 51750/(1000·46.75) = 1.107 m²
Answer
A = 1.107 m²
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