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 150 mm of brick and 100 mm of gypsum board, with an area of 6 m². The outer faces are at 20 °C and -8 °C. Find the temperature Ti at the interface between the two layers.
L1 = 150 mm, k1 = 0.72 W/(m·K)L2 = 100 mm, k2 = 0.17 W/(m·K)A = 6 m²T1 = 20 °CT3 = -8 °C
Ti = °C
2.Heat conductionHard
A wall of 150 mm gypsum board and 100 mm mineral wool separates a room at 22 °C from outside air at -3 °C. The convection coefficients are 6 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 8 m². Find the heat loss Q̇ through the wall.
L1 = 150 mm, k1 = 0.17 W/(m·K)L2 = 100 mm, k2 = 0.04 W/(m·K)A = 8 m²hi = 6 W/(m²·K)ho = 30 W/(m²·K)Troom = 22 °CTout = -3 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
In a condenser, steam condenses at a constant 105 °C on the outside of the tubes. Cooling water (1.5 kg/s, cp = 4.18 kJ/(kg·K)) flows through the tubes and heats from 30 to 65 °C. The overall heat transfer coefficient is 3000 W/(m²·K). Find the heat transfer area A needed.
Ts = 105 °Cṁ = 1.5 kg/scp = 4.18 kJ/(kg·K)Tc: 30 → 65 °CU = 3000 W/(m²·K)
A = m²
4.Heat exchangers (LMTD)Hard
Steam condenses at a constant 105 °C on the outside of the tubes of a condenser with a heat transfer area of 4 m² and U = 2500 W/(m²·K). Cooling water (2 kg/s, cp = 4.18 kJ/(kg·K)) enters the tubes at 20 °C. Find the water outlet temperature Tc,out.
Ts = 105 °CA = 4 m²U = 2500 W/(m²·K)ṁ = 2 kg/scp = 4.18 kJ/(kg·K)Tc,in = 20 °C
Tc,out = °C
3Solutions
1.Heat conduction
Answer: Ti = 12.68 °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) = (150/1000)/(0.72·6) = 0.03472 K/W
R2 = L2/(k2A) = (100/1000)/(0.17·6) = 0.09804 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 28/0.1328 = 210.9 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 20 − 210.9·0.03472 = 12.68 °C
Answer
Ti = 12.68 °C
2.Heat conduction
Answer: Q̇ = 55.83 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/(6·8) = 0.02083 K/W
R1 = L1/(k1A) = (150/1000)/(0.17·8) = 0.1103 K/W
R2 = L2/(k2A) = (100/1000)/(0.04·8) = 0.3125 K/W
Ro = 1/(hoA) = 1/(30·8) = 0.004167 K/W
ΣR = 0.4478 K/W
Heat loss
Q̇ = (22 − (-3))/0.4478 = 55.83 W
Answer
Q̇ = 55.83 W
3.Heat exchangers (LMTD)
Answer: A = 1.314 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 = 1.5·4.18·(65 − 30) = 219.4 kW
End temperature differences
ΔT₁ = 105 − 30 = 75 K
ΔT₂ = 105 − 65 = 40 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (75 − 40)/ln(75/40) = 55.68 K
Area
A = Q̇/(U·ΔTlm) = 219450/(3000·55.68) = 1.314 m²
Answer
A = 1.314 m²
4.Heat exchangers (LMTD)
Answer: Tc,out = 79.3 °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) = 2500·4/(2·4180) = 1.196
Outlet temperature
ΔT₂ = ΔT₁·e−1.196 = 85·0.3023 = 25.7 K
Tc,out = 105 − 25.7 = 79.3 °C
Answer
Tc,out = 79.3 °C
Heat transferred: Q̇ = 495.8 kW.
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