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 50 mm of expanded polystyrene, with an area of 8 m². The outer faces are at 20 °C and -12 °C. Find the heat flow Q̇ through the wall.
L1 = 150 mm, k1 = 0.72 W/(m·K)L2 = 50 mm, k2 = 0.035 W/(m·K)A = 8 m²T1 = 20 °CT3 = -12 °C
Q̇ = W
2.Heat conductionHard
A wall of 150 mm expanded polystyrene and 100 mm mineral wool separates a room at 20 °C from outside air at -11 °C. The convection coefficients are 6 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 4 m². Find the overall heat transfer coefficient U of the wall.
L1 = 150 mm, k1 = 0.035 W/(m·K)L2 = 100 mm, k2 = 0.04 W/(m·K)A = 4 m²hi = 6 W/(m²·K)ho = 30 W/(m²·K)Troom = 20 °CTout = -11 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A parallel-flow heat exchanger transfers 320 kW. The hot fluid cools from 95 to 70 °C and the cold fluid heats from 20 to 35 °C. The overall heat transfer coefficient is 800 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 320 kWU = 800 W/(m²·K)Th: 95 → 70 °CTc: 20 → 35 °C
A = m²
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 2.5 m² and U = 3000 W/(m²·K). Cooling water (1.5 kg/s, cp = 4.18 kJ/(kg·K)) enters the tubes at 15 °C. Find the water outlet temperature Tc,out.
Ts = 120 °CA = 2.5 m²U = 3000 W/(m²·K)ṁ = 1.5 kg/scp = 4.18 kJ/(kg·K)Tc,in = 15 °C
Tc,out = °C
3Solutions
1.Heat conduction
Answer: Q̇ = 156.4 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) = (150/1000)/(0.72·8) = 0.02604 K/W
R2 = L2/(k2A) = (50/1000)/(0.035·8) = 0.1786 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 32/0.2046 = 156.4 W
Answer
Q̇ = 156.4 W
2.Heat conduction
Answer: U = 0.1431 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/(6·4) = 0.04167 K/W
R1 = L1/(k1A) = (150/1000)/(0.035·4) = 1.071 K/W
R2 = L2/(k2A) = (100/1000)/(0.04·4) = 0.625 K/W
Ro = 1/(hoA) = 1/(30·4) = 0.008333 K/W
ΣR = 1.746 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(4·1.746) = 0.1431 W/(m²·K)
Answer
U = 0.1431 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 7.621 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 parallel flow both inlets are at the same end, and both outlets at the other.
ΔT₁ = Th,in − Tc,in = 95 − 20 = 75 K
ΔT₂ = Th,out − Tc,out = 70 − 35 = 35 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (75 − 35)/ln(75/35) = 52.48 K
Area
A = Q̇/(U·ΔTlm) = 320000/(800·52.48) = 7.621 m²
Answer
A = 7.621 m²
4.Heat exchangers (LMTD)
Answer: Tc,out = 88.25 °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) = 3000·2.5/(1.5·4180) = 1.196
Outlet temperature
ΔT₂ = ΔT₁·e−1.196 = 105·0.3023 = 31.75 K
Tc,out = 120 − 31.75 = 88.25 °C
Answer
Tc,out = 88.25 °C
Heat transferred: Q̇ = 459.3 kW.
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