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 gypsum board and 150 mm of wood, with an area of 4 m². The outer faces are at 24 °C and 10 °C. Find the heat flow Q̇ through the wall.
L1 = 50 mm, k1 = 0.17 W/(m·K)L2 = 150 mm, k2 = 0.17 W/(m·K)A = 4 m²T1 = 24 °CT3 = 10 °C
Q̇ = W
2.Heat conductionHard
A wall of 100 mm brick and 25 mm expanded polystyrene separates a room at 21 °C from outside air at -8 °C. The convection coefficients are 10 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 4 m². Find the overall heat transfer coefficient U of the wall.
L1 = 100 mm, k1 = 0.72 W/(m·K)L2 = 25 mm, k2 = 0.035 W/(m·K)A = 4 m²hi = 10 W/(m²·K)ho = 30 W/(m²·K)Troom = 21 °CTout = -8 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A shell-and-tube heat exchanger (one shell pass, two tube passes) must transfer 760 kW. The hot fluid cools from 165 to 80 °C and the cold fluid heats from 15 to 55 °C. The LMTD correction factor for this arrangement is F = 0.88, and U = 1000 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 760 kWF = 0.88U = 1000 W/(m²·K)Th: 165 → 80 °CTc: 15 → 55 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (2 kg/s, cp = 2.3 kJ/(kg·K)) cools from 170 to 85 °C in a counter-flow exchanger. It heats 3 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 15 °C. With U = 800 W/(m²·K), find the heat transfer area A needed.
ṁh = 2 kg/scp,h = 2.3 kJ/(kg·K)Th: 170 → 85 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 15 °CU = 800 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 47.6 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) = (50/1000)/(0.17·4) = 0.07353 K/W
R2 = L2/(k2A) = (150/1000)/(0.17·4) = 0.2206 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 14/0.2941 = 47.6 W
Answer
Q̇ = 47.6 W
2.Heat conduction
Answer: U = 1.014 W/(m²·K)
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/(10·4) = 0.025 K/W
R1 = L1/(k1A) = (100/1000)/(0.72·4) = 0.03472 K/W
R2 = L2/(k2A) = (25/1000)/(0.035·4) = 0.1786 K/W
Ro = 1/(hoA) = 1/(30·4) = 0.008333 K/W
ΣR = 0.2466 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(4·0.2466) = 1.014 W/(m²·K)
Answer
U = 1.014 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 10.1 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 = 165 − 55 = 110 K
ΔT₂ = Th,out − Tc,in = 80 − 15 = 65 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (110 − 65)/ln(110/65) = 85.54 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 760000/(1000·0.88·85.54) = 10.1 m²
Answer
A = 10.1 m²
4.Heat exchangers (LMTD)
Answer: A = 5.179 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) = 2·2.3·85 = 391 kW
Tc,out = 15 + 391/(3·4.18) = 46.18 °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 = 170 − 46.18 = 123.8 K
ΔT₂ = Th,out − Tc,in = 85 − 15 = 70 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (123.8 − 70)/ln(123.8/70) = 94.37 K
Area
A = Q̇/(U·ΔTlm) = 391000/(800·94.37) = 5.179 m²
Answer
A = 5.179 m²
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