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 expanded polystyrene and 100 mm of wood, with an area of 4 m². The outer faces are at 15 °C and 6 °C. Find the temperature Ti at the interface between the two layers.
L1 = 50 mm, k1 = 0.035 W/(m·K)L2 = 100 mm, k2 = 0.17 W/(m·K)A = 4 m²T1 = 15 °CT3 = 6 °C
Ti = °C
2.Heat conductionHard
A wall of 50 mm mineral wool and 25 mm gypsum board separates a room at 22 °C from outside air at -14 °C. The convection coefficients are 8 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 20 m². Find the temperature Ts of the inside surface of the wall.
L1 = 50 mm, k1 = 0.04 W/(m·K)L2 = 25 mm, k2 = 0.17 W/(m·K)A = 20 m²hi = 8 W/(m²·K)ho = 30 W/(m²·K)Troom = 22 °CTout = -14 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 40 kW. The hot fluid cools from 170 to 130 °C and the cold fluid heats from 15 to 110 °C. The overall heat transfer coefficient is 500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 40 kWU = 500 W/(m²·K)Th: 170 → 130 °CTc: 15 → 110 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a counter-flow double-pipe heat exchanger, hot oil (0.5 kg/s, cp = 2.1 kJ/(kg·K)) flows in the inner tube and cools from 110 to 55 °C. Water (0.8 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 25 °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 = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 110 → 55 °Cṁc = 0.8 kg/sTc,in = 25 °CD = 25 mmU = 250 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Ti = 8.625 °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.
Layer resistances
R1 = L1/(k1A) = (50/1000)/(0.035·4) = 0.3571 K/W
R2 = L2/(k2A) = (100/1000)/(0.17·4) = 0.1471 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 9/0.5042 = 17.85 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 15 − 17.85·0.3571 = 8.625 °C
Answer
Ti = 8.625 °C
2.Heat conduction
Answer: Ts = 19.11 °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/(8·20) = 0.00625 K/W
R1 = L1/(k1A) = (50/1000)/(0.04·20) = 0.0625 K/W
R2 = L2/(k2A) = (25/1000)/(0.17·20) = 0.007353 K/W
Ro = 1/(hoA) = 1/(30·20) = 0.001667 K/W
ΣR = 0.07777 K/W
Heat loss
Q̇ = (22 − (-14))/0.07777 = 462.9 W
Ts = Troom − Q̇·Ri = 22 − 462.9·0.00625 = 19.11 °C
Answer
Ts = 19.11 °C
3.Heat exchangers (LMTD)
Answer: A = 0.9463 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 = 170 − 110 = 60 K
ΔT₂ = Th,out − Tc,in = 130 − 15 = 115 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (60 − 115)/ln(60/115) = 84.54 K
Area
A = Q̇/(U·ΔTlm) = 40000/(500·84.54) = 0.9463 m²
Answer
A = 0.9463 m²
4.Heat exchangers (LMTD)
Answer: L = 63.48 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 = 0.5·2.1·(110 − 55) = 57.75 kW
Tc,out = 25 + 57.75/(0.8·4.18) = 42.27 °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 − 42.27 = 67.73 K
ΔT₂ = Th,out − Tc,in = 55 − 25 = 30 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (67.73 − 30)/ln(67.73/30) = 46.33 K
Area and length
A = 57750/(250·46.33) = 4.986 m²
L = A/(πD) = 4.986/(π·0.025) = 63.48 m
Answer
L = 63.48 m
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