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 brick and 50 mm of glass, with an area of 15 m². The outer faces are at 20 °C and -1 °C. Find the temperature Ti at the interface between the two layers.
L1 = 50 mm, k1 = 0.72 W/(m·K)L2 = 50 mm, k2 = 0.96 W/(m·K)A = 15 m²T1 = 20 °CT3 = -1 °C
Ti = °C
2.Heat conductionHard
A wall of 100 mm gypsum board and 100 mm mineral wool separates a room at 20 °C from outside air at -6 °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 temperature Ts of the inside surface of the wall.
L1 = 100 mm, k1 = 0.17 W/(m·K)L2 = 100 mm, k2 = 0.04 W/(m·K)A = 4 m²hi = 7.7 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = -6 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
In a condenser, steam condenses at a constant 110 °C on the outside of the tubes. Cooling water (0.5 kg/s, cp = 4.18 kJ/(kg·K)) flows through the tubes and heats from 20 to 90 °C. The overall heat transfer coefficient is 2000 W/(m²·K). Find the heat transfer area A needed.
Ts = 110 °Cṁ = 0.5 kg/scp = 4.18 kJ/(kg·K)Tc: 20 → 90 °CU = 2000 W/(m²·K)
A = m²
4.Heat exchangers (LMTD)Hard
Steam condenses at a constant 130 °C on the outside of the tubes of a condenser with a heat transfer area of 2 m² and U = 1500 W/(m²·K). Cooling water (0.5 kg/s, cp = 4.18 kJ/(kg·K)) enters the tubes at 10 °C. Find the water outlet temperature Tc,out.
Ts = 130 °CA = 2 m²U = 1500 W/(m²·K)ṁ = 0.5 kg/scp = 4.18 kJ/(kg·K)Tc,in = 10 °C
Tc,out = °C
3Solutions
1.Heat conduction
Answer: Ti = 8 °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)/(0.72·15) = 0.00463 K/W
R2 = L2/(k2A) = (50/1000)/(0.96·15) = 0.003472 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 21/0.008102 = 2592 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 20 − 2592·0.00463 = 8 °C
Answer
Ti = 8 °C
2.Heat conduction
Answer: Ts = 18.96 °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/(7.7·4) = 0.03247 K/W
R1 = L1/(k1A) = (100/1000)/(0.17·4) = 0.1471 K/W
R2 = L2/(k2A) = (100/1000)/(0.04·4) = 0.625 K/W
Ro = 1/(hoA) = 1/(25·4) = 0.01 K/W
ΣR = 0.8145 K/W
Heat loss
Q̇ = (20 − (-6))/0.8145 = 31.92 W
Ts = Troom − Q̇·Ri = 20 − 31.92·0.03247 = 18.96 °C
Answer
Ts = 18.96 °C
3.Heat exchangers (LMTD)
Answer: A = 1.572 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.
Heat duty from the water side
Q̇ = 0.5·4.18·(90 − 20) = 146.3 kW
End temperature differences
ΔT₁ = 110 − 20 = 90 K
ΔT₂ = 110 − 90 = 20 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (90 − 20)/ln(90/20) = 46.54 K
Area
A = Q̇/(U·ΔTlm) = 146300/(2000·46.54) = 1.572 m²
Answer
A = 1.572 m²
4.Heat exchangers (LMTD)
Answer: Tc,out = 101.4 °C
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.
Combine the energy balance and the rate equation
With ΔT₁ = Ts − Tc,in and ΔT₂ = Ts − Tc,out, the difference ΔT₁ − ΔT₂ equals the water temperature rise, so the rise cancels.
ṁcp(ΔT₁ − ΔT₂) = U·A·(ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂)
ln(ΔT₁/ΔT₂) = U·A/(ṁcp) = 1500·2/(0.5·4180) = 1.435
Outlet temperature
ΔT₂ = ΔT₁·e−1.435 = 120·0.238 = 28.56 K
Tc,out = 130 − 28.56 = 101.4 °C
Answer
Tc,out = 101.4 °C
Heat transferred: Q̇ = 191.1 kW.
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