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 brick, with an area of 6 m². The outer faces are at 22 °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 = 0.72 W/(m·K)A = 6 m²T1 = 22 °CT3 = 2 °C
Q̇ = W
2.Heat conductionHard
A wall of 100 mm expanded polystyrene and 100 mm wood separates a room at 22 °C from outside air at 3 °C. The convection coefficients are 8 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 20 m². Find the heat loss Q̇ through the wall.
L1 = 100 mm, k1 = 0.035 W/(m·K)L2 = 100 mm, k2 = 0.17 W/(m·K)A = 20 m²hi = 8 W/(m²·K)ho = 30 W/(m²·K)Troom = 22 °CTout = 3 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A shell-and-tube heat exchanger (one shell pass, two tube passes) must transfer 430 kW. The hot fluid cools from 125 to 95 °C and the cold fluid heats from 10 to 95 °C. The LMTD correction factor for this arrangement is F = 0.8, and U = 500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 430 kWF = 0.8U = 500 W/(m²·K)Th: 125 → 95 °CTc: 10 → 95 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (0.5 kg/s, cp = 2.1 kJ/(kg·K)) cools from 120 to 90 °C in a counter-flow exchanger. It heats 2 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 15 °C. With U = 200 W/(m²·K), find the heat transfer area A needed.
ṁh = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 120 → 90 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 15 °CU = 200 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 20.62 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)/(0.035·6) = 0.9524 K/W
R2 = L2/(k2A) = (75/1000)/(0.72·6) = 0.01736 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 20/0.9697 = 20.62 W
Answer
Q̇ = 20.62 W
2.Heat conduction
Answer: Q̇ = 105.4 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/(8·20) = 0.00625 K/W
R1 = L1/(k1A) = (100/1000)/(0.035·20) = 0.1429 K/W
R2 = L2/(k2A) = (100/1000)/(0.17·20) = 0.02941 K/W
Ro = 1/(hoA) = 1/(30·20) = 0.001667 K/W
ΣR = 0.1802 K/W
Heat loss
Q̇ = (22 − (3))/0.1802 = 105.4 W
Answer
Q̇ = 105.4 W
3.Heat exchangers (LMTD)
Answer: A = 20.36 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 (counter-flow)
In counter-flow the hot inlet faces the cold outlet, and the hot outlet faces the cold inlet.
ΔT₁ = Th,in − Tc,out = 125 − 95 = 30 K
ΔT₂ = Th,out − Tc,in = 95 − 10 = 85 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (30 − 85)/ln(30/85) = 52.81 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 430000/(500·0.8·52.81) = 20.36 m²
Answer
A = 20.36 m²
4.Heat exchangers (LMTD)
Answer: A = 1.801 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) = 0.5·2.1·30 = 31.5 kW
Tc,out = 15 + 31.5/(2·4.18) = 18.77 °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 = 120 − 18.77 = 101.2 K
ΔT₂ = Th,out − Tc,in = 90 − 15 = 75 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (101.2 − 75)/ln(101.2/75) = 87.46 K
Area
A = Q̇/(U·ΔTlm) = 31500/(200·87.46) = 1.801 m²
Answer
A = 1.801 m²
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