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 gypsum board and 75 mm of glass, with an area of 8 m². The outer faces are at 21 °C and -11 °C. Find the heat flow Q̇ through the wall.
L1 = 200 mm, k1 = 0.17 W/(m·K)L2 = 75 mm, k2 = 0.96 W/(m·K)A = 8 m²T1 = 21 °CT3 = -11 °C
Q̇ = W
2.Heat conductionHard
A wall of 50 mm mineral wool and 100 mm concrete separates a room at 21 °C from outside air at -14 °C. The convection coefficients are 7.7 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 4 m². Find the heat loss Q̇ through the wall.
L1 = 50 mm, k1 = 0.04 W/(m·K)L2 = 100 mm, k2 = 1.4 W/(m·K)A = 4 m²hi = 7.7 W/(m²·K)ho = 25 W/(m²·K)Troom = 21 °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 135 to 55 °C and the cold fluid heats from 15 to 50 °C. The LMTD correction factor for this arrangement is F = 0.9, and U = 800 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 320 kWF = 0.9U = 800 W/(m²·K)Th: 135 → 55 °CTc: 15 → 50 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a counter-flow double-pipe heat exchanger, hot oil (0.3 kg/s, cp = 2.1 kJ/(kg·K)) flows in the inner tube and cools from 90 to 60 °C. Water (1 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 10 °C. The inner tube has a diameter of 25 mm, and U = 250 W/(m²·K) is based on this diameter. Find the tube length L needed.
ṁh = 0.3 kg/scp,h = 2.1 kJ/(kg·K)Th: 90 → 60 °Cṁc = 1 kg/sTc,in = 10 °CD = 25 mmU = 250 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Q̇ = 204 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)/(0.17·8) = 0.1471 K/W
R2 = L2/(k2A) = (75/1000)/(0.96·8) = 0.009766 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 32/0.1568 = 204 W
Answer
Q̇ = 204 W
2.Heat conduction
Answer: Q̇ = 93.88 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/(7.7·4) = 0.03247 K/W
R1 = L1/(k1A) = (50/1000)/(0.04·4) = 0.3125 K/W
R2 = L2/(k2A) = (100/1000)/(1.4·4) = 0.01786 K/W
Ro = 1/(hoA) = 1/(25·4) = 0.01 K/W
ΣR = 0.3728 K/W
Heat loss
Q̇ = (21 − (-14))/0.3728 = 93.88 W
Answer
Q̇ = 93.88 W
3.Heat exchangers (LMTD)
Answer: A = 7.445 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 = 135 − 50 = 85 K
ΔT₂ = Th,out − Tc,in = 55 − 15 = 40 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (85 − 40)/ln(85/40) = 59.7 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 320000/(800·0.9·59.7) = 7.445 m²
Answer
A = 7.445 m²
4.Heat exchangers (LMTD)
Answer: L = 15.56 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 = 0.3·2.1·(90 − 60) = 18.9 kW
Tc,out = 10 + 18.9/(1·4.18) = 14.52 °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 = 90 − 14.52 = 75.48 K
ΔT₂ = Th,out − Tc,in = 60 − 10 = 50 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (75.48 − 50)/ln(75.48/50) = 61.87 K
Area and length
A = 18900/(250·61.87) = 1.222 m²
L = A/(πD) = 1.222/(π·0.025) = 15.56 m
Answer
L = 15.56 m
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