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 150 mm of glass and 150 mm of expanded polystyrene, with an area of 10 m². The outer faces are at 17 °C and -15 °C. Find the heat flow Q̇ through the wall.
L1 = 150 mm, k1 = 0.96 W/(m·K)L2 = 150 mm, k2 = 0.035 W/(m·K)A = 10 m²T1 = 17 °CT3 = -15 °C
Q̇ = W
2.Heat conductionHard
A wall of 250 mm mineral wool and 150 mm concrete separates a room at 21 °C from outside air at 0 °C. The convection coefficients are 10 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 4 m². Find the heat loss Q̇ through the wall.
L1 = 250 mm, k1 = 0.04 W/(m·K)L2 = 150 mm, k2 = 1.4 W/(m·K)A = 4 m²hi = 10 W/(m²·K)ho = 20 W/(m²·K)Troom = 21 °CTout = 0 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 210 kW. The hot fluid cools from 145 to 30 °C and the cold fluid heats from 10 to 65 °C. The overall heat transfer coefficient is 500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 210 kWU = 500 W/(m²·K)Th: 145 → 30 °CTc: 10 → 65 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a counter-flow double-pipe heat exchanger, hot oil (0.3 kg/s, cp = 2.5 kJ/(kg·K)) flows in the inner tube and cools from 110 to 55 °C. Water (0.2 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 15 °C. The inner tube has a diameter of 32 mm, and U = 200 W/(m²·K) is based on this diameter. Find the tube length L needed.
ṁh = 0.3 kg/scp,h = 2.5 kJ/(kg·K)Th: 110 → 55 °Cṁc = 0.2 kg/sTc,in = 15 °CD = 32 mmU = 200 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Q̇ = 72.04 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) = (150/1000)/(0.96·10) = 0.01563 K/W
R2 = L2/(k2A) = (150/1000)/(0.035·10) = 0.4286 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 32/0.4442 = 72.04 W
Answer
Q̇ = 72.04 W
2.Heat conduction
Answer: Q̇ = 12.91 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/(10·4) = 0.025 K/W
R1 = L1/(k1A) = (250/1000)/(0.04·4) = 1.563 K/W
R2 = L2/(k2A) = (150/1000)/(1.4·4) = 0.02679 K/W
Ro = 1/(hoA) = 1/(20·4) = 0.0125 K/W
ΣR = 1.627 K/W
Heat loss
Q̇ = (21 − (0))/1.627 = 12.91 W
Answer
Q̇ = 12.91 W
3.Heat exchangers (LMTD)
Answer: A = 9.704 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 = 145 − 65 = 80 K
ΔT₂ = Th,out − Tc,in = 30 − 10 = 20 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (80 − 20)/ln(80/20) = 43.28 K
Area
A = Q̇/(U·ΔTlm) = 210000/(500·43.28) = 9.704 m²
Answer
A = 9.704 m²
4.Heat exchangers (LMTD)
Answer: L = 47.97 m
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 = 0.3·2.5·(110 − 55) = 41.25 kW
Tc,out = 15 + 41.25/(0.2·4.18) = 64.34 °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 = 110 − 64.34 = 45.66 K
ΔT₂ = Th,out − Tc,in = 55 − 15 = 40 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (45.66 − 40)/ln(45.66/40) = 42.77 K
Area and length
A = 41250/(200·42.77) = 4.823 m²
L = A/(πD) = 4.823/(π·0.032) = 47.97 m
Answer
L = 47.97 m
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