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 250 mm of expanded polystyrene and 50 mm of glass, with an area of 20 m². The outer faces are at 20 °C and 9 °C. Find the heat flow Q̇ through the wall.
L1 = 250 mm, k1 = 0.035 W/(m·K)L2 = 50 mm, k2 = 0.96 W/(m·K)A = 20 m²T1 = 20 °CT3 = 9 °C
Q̇ = W
2.Heat conductionHard
A wall of 250 mm concrete and 25 mm mineral wool separates a room at 22 °C from outside air at 2 °C. The convection coefficients are 8 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 10 m². Find the temperature Ts of the inside surface of the wall.
L1 = 250 mm, k1 = 1.4 W/(m·K)L2 = 25 mm, k2 = 0.04 W/(m·K)A = 10 m²hi = 8 W/(m²·K)ho = 20 W/(m²·K)Troom = 22 °CTout = 2 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A parallel-flow heat exchanger transfers 300 kW. The hot fluid cools from 110 to 85 °C and the cold fluid heats from 20 to 60 °C. The overall heat transfer coefficient is 500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 300 kWU = 500 W/(m²·K)Th: 110 → 85 °CTc: 20 → 60 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a parallel-flow double-pipe heat exchanger, hot oil (1 kg/s, cp = 2.5 kJ/(kg·K)) flows in the inner tube and cools from 100 to 70 °C. Water (1.5 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 = 150 W/(m²·K) is based on this diameter. Find the tube length L needed.
ṁh = 1 kg/scp,h = 2.5 kJ/(kg·K)Th: 100 → 70 °Cṁc = 1.5 kg/sTc,in = 15 °CD = 32 mmU = 150 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Q̇ = 30.58 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) = (250/1000)/(0.035·20) = 0.3571 K/W
R2 = L2/(k2A) = (50/1000)/(0.96·20) = 0.002604 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 11/0.3597 = 30.58 W
Answer
Q̇ = 30.58 W
2.Heat conduction
Answer: Ts = 19.45 °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/(8·10) = 0.0125 K/W
R1 = L1/(k1A) = (250/1000)/(1.4·10) = 0.01786 K/W
R2 = L2/(k2A) = (25/1000)/(0.04·10) = 0.0625 K/W
Ro = 1/(hoA) = 1/(20·10) = 0.005 K/W
ΣR = 0.09786 K/W
Heat loss
Q̇ = (22 − (2))/0.09786 = 204.4 W
Ts = Troom − Q̇·Ri = 22 − 204.4·0.0125 = 19.45 °C
Answer
Ts = 19.45 °C
3.Heat exchangers (LMTD)
Answer: A = 11.82 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 parallel flow both inlets are at the same end, and both outlets at the other.
ΔT₁ = Th,in − Tc,in = 110 − 20 = 90 K
ΔT₂ = Th,out − Tc,out = 85 − 60 = 25 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (90 − 25)/ln(90/25) = 50.74 K
Area
A = Q̇/(U·ΔTlm) = 300000/(500·50.74) = 11.82 m²
Answer
A = 11.82 m²
4.Heat exchangers (LMTD)
Answer: L = 80.66 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 = 1·2.5·(100 − 70) = 75 kW
Tc,out = 15 + 75/(1.5·4.18) = 26.96 °C
End temperature differences
In parallel flow both inlets are at the same end, and both outlets at the other.
ΔT₁ = Th,in − Tc,in = 100 − 15 = 85 K
ΔT₂ = Th,out − Tc,out = 70 − 26.96 = 43.04 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (85 − 43.04)/ln(85/43.04) = 61.66 K
Area and length
A = 75000/(150·61.66) = 8.109 m²
L = A/(πD) = 8.109/(π·0.032) = 80.66 m
Answer
L = 80.66 m
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