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 50 mm of glass, with an area of 4 m². The outer faces are at 21 °C and 8 °C. Find the heat flow Q̇ through the wall.
L1 = 200 mm, k1 = 0.035 W/(m·K)L2 = 50 mm, k2 = 0.96 W/(m·K)A = 4 m²T1 = 21 °CT3 = 8 °C
Q̇ = W
2.Heat conductionHard
A wall of 100 mm brick and 50 mm gypsum board separates a room at 22 °C from outside air at -14 °C. The convection coefficients are 6 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 6 m². Find the heat loss Q̇ through the wall.
L1 = 100 mm, k1 = 0.72 W/(m·K)L2 = 50 mm, k2 = 0.17 W/(m·K)A = 6 m²hi = 6 W/(m²·K)ho = 30 W/(m²·K)Troom = 22 °CTout = -14 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A shell-and-tube heat exchanger (one shell pass, two tube passes) must transfer 320 kW. The hot fluid cools from 145 to 60 °C and the cold fluid heats from 30 to 135 °C. The LMTD correction factor for this arrangement is F = 0.88, and U = 200 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 320 kWF = 0.88U = 200 W/(m²·K)Th: 145 → 60 °CTc: 30 → 135 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a parallel-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 135 to 70 °C. Water (1 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 20 °C. The inner tube has a diameter of 40 mm, and U = 500 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: 135 → 70 °Cṁc = 1 kg/sTc,in = 20 °CD = 40 mmU = 500 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Q̇ = 9.018 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) = (50/1000)/(0.96·4) = 0.01302 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 13/1.442 = 9.018 W
Answer
Q̇ = 9.018 W
2.Heat conduction
Answer: Q̇ = 341.2 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/(6·6) = 0.02778 K/W
R1 = L1/(k1A) = (100/1000)/(0.72·6) = 0.02315 K/W
R2 = L2/(k2A) = (50/1000)/(0.17·6) = 0.04902 K/W
Ro = 1/(hoA) = 1/(30·6) = 0.005556 K/W
ΣR = 0.1055 K/W
Heat loss
Q̇ = (22 − (-14))/0.1055 = 341.2 W
Answer
Q̇ = 341.2 W
3.Heat exchangers (LMTD)
Answer: A = 99.87 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 = 145 − 135 = 10 K
ΔT₂ = Th,out − Tc,in = 60 − 30 = 30 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (10 − 30)/ln(10/30) = 18.2 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 320000/(200·0.88·18.2) = 99.87 m²
Answer
A = 99.87 m²
4.Heat exchangers (LMTD)
Answer: L = 11.12 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·(135 − 70) = 48.75 kW
Tc,out = 20 + 48.75/(1·4.18) = 31.66 °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 = 135 − 20 = 115 K
ΔT₂ = Th,out − Tc,out = 70 − 31.66 = 38.34 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (115 − 38.34)/ln(115/38.34) = 69.79 K
Area and length
A = 48750/(500·69.79) = 1.397 m²
L = A/(πD) = 1.397/(π·0.04) = 11.12 m
Answer
L = 11.12 m
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