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 100 mm of brick and 25 mm of wood, with an area of 12 m². The outer faces are at 25 °C and -8 °C. Find the heat flow Q̇ through the wall.
L1 = 100 mm, k1 = 0.72 W/(m·K)L2 = 25 mm, k2 = 0.17 W/(m·K)A = 12 m²T1 = 25 °CT3 = -8 °C
Q̇ = W
2.Heat conductionHard
A wall of 50 mm expanded polystyrene and 100 mm wood separates a room at 20 °C from outside air at 1 °C. The convection coefficients are 10 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 8 m². Find the overall heat transfer coefficient U of the wall.
L1 = 50 mm, k1 = 0.035 W/(m·K)L2 = 100 mm, k2 = 0.17 W/(m·K)A = 8 m²hi = 10 W/(m²·K)ho = 20 W/(m²·K)Troom = 20 °CTout = 1 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 360 kW. The hot fluid cools from 85 to 65 °C and the cold fluid heats from 10 to 40 °C. The overall heat transfer coefficient is 1500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 360 kWU = 1500 W/(m²·K)Th: 85 → 65 °CTc: 10 → 40 °C
A = m²
4.Heat exchangers (LMTD)Hard
Steam condenses at a constant 110 °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 (3 kg/s, cp = 4.18 kJ/(kg·K)) enters the tubes at 25 °C. Find the water outlet temperature Tc,out.
Ts = 110 °CA = 4 m²U = 2500 W/(m²·K)ṁ = 3 kg/scp = 4.18 kJ/(kg·K)Tc,in = 25 °C
Tc,out = °C
3Solutions
1.Heat conduction
Answer: Q̇ = 1385 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) = (100/1000)/(0.72·12) = 0.01157 K/W
R2 = L2/(k2A) = (25/1000)/(0.17·12) = 0.01225 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 33/0.02383 = 1385 W
Answer
Q̇ = 1385 W
2.Heat conduction
Answer: U = 0.4615 W/(m²·K)
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/(10·8) = 0.0125 K/W
R1 = L1/(k1A) = (50/1000)/(0.035·8) = 0.1786 K/W
R2 = L2/(k2A) = (100/1000)/(0.17·8) = 0.07353 K/W
Ro = 1/(hoA) = 1/(20·8) = 0.00625 K/W
ΣR = 0.2709 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(8·0.2709) = 0.4615 W/(m²·K)
Answer
U = 0.4615 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 4.816 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 = 85 − 40 = 45 K
ΔT₂ = Th,out − Tc,in = 65 − 10 = 55 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (45 − 55)/ln(45/55) = 49.83 K
Area
A = Q̇/(U·ΔTlm) = 360000/(1500·49.83) = 4.816 m²
Answer
A = 4.816 m²
4.Heat exchangers (LMTD)
Answer: Tc,out = 71.71 °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/(3·4180) = 0.7974
Outlet temperature
ΔT₂ = ΔT₁·e−0.7974 = 85·0.4505 = 38.29 K
Tc,out = 110 − 38.29 = 71.71 °C
Answer
Tc,out = 71.71 °C
Heat transferred: Q̇ = 585.7 kW.
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