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 250 mm of mineral wool and 25 mm of concrete, with an area of 15 m². The outer faces are at 23 °C and 6 °C. Find the temperature Ti at the interface between the two layers.
L1 = 250 mm, k1 = 0.04 W/(m·K)L2 = 25 mm, k2 = 1.4 W/(m·K)A = 15 m²T1 = 23 °CT3 = 6 °C
Ti = °C
2.Heat conductionHard
A wall of 100 mm wood and 150 mm brick separates a room at 20 °C from outside air at -4 °C. The convection coefficients are 8 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 12 m². Find the heat loss Q̇ through the wall.
L1 = 100 mm, k1 = 0.17 W/(m·K)L2 = 150 mm, k2 = 0.72 W/(m·K)A = 12 m²hi = 8 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = -4 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 220 kW. Its heat transfer area is 5 m². The hot fluid cools from 155 to 90 °C and the cold fluid heats from 20 to 40 °C. Find the overall heat transfer coefficient U.
Q̇ = 220 kWA = 5 m²Th: 155 → 90 °CTc: 20 → 40 °C
U = W/(m²·K)
4.Heat exchangers (LMTD)Hard
Steam condenses at a constant 120 °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 25 °C. Find the water outlet temperature Tc,out.
Ts = 120 °CA = 4 m²U = 2500 W/(m²·K)ṁ = 2 kg/scp = 4.18 kJ/(kg·K)Tc,in = 25 °C
Tc,out = °C
3Solutions
1.Heat conduction
Answer: Ti = 6.048 °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) = (250/1000)/(0.04·15) = 0.4167 K/W
R2 = L2/(k2A) = (25/1000)/(1.4·15) = 0.00119 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 17/0.4179 = 40.68 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 23 − 40.68·0.4167 = 6.048 °C
Answer
Ti = 6.048 °C
2.Heat conduction
Answer: Q̇ = 299.5 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/(8·12) = 0.01042 K/W
R1 = L1/(k1A) = (100/1000)/(0.17·12) = 0.04902 K/W
R2 = L2/(k2A) = (150/1000)/(0.72·12) = 0.01736 K/W
Ro = 1/(hoA) = 1/(25·12) = 0.003333 K/W
ΣR = 0.08013 K/W
Heat loss
Q̇ = (20 − (-4))/0.08013 = 299.5 W
Answer
Q̇ = 299.5 W
3.Heat exchangers (LMTD)
Answer: U = 485.4 W/(m²·K)
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 = 155 − 40 = 115 K
ΔT₂ = Th,out − Tc,in = 90 − 20 = 70 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (115 − 70)/ln(115/70) = 90.65 K
Overall coefficient
U = Q̇/(A·ΔTlm) = 220000/(5·90.65) = 485.4 W/(m²·K)
Answer
U = 485.4 W/(m²·K)
4.Heat exchangers (LMTD)
Answer: Tc,out = 91.28 °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 = 95·0.3023 = 28.72 K
Tc,out = 120 − 28.72 = 91.28 °C
Answer
Tc,out = 91.28 °C
Heat transferred: Q̇ = 554.1 kW.
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