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 100 mm of gypsum board, with an area of 20 m². The outer faces are at 18 °C and 2 °C. Find the heat flow Q̇ through the wall.
L1 = 250 mm, k1 = 0.04 W/(m·K)L2 = 100 mm, k2 = 0.17 W/(m·K)A = 20 m²T1 = 18 °CT3 = 2 °C
Q̇ = W
2.Heat conductionHard
A wall of 150 mm wood and 50 mm concrete separates a room at 22 °C from outside air at -5 °C. The convection coefficients are 7.7 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 15 m². Find the overall heat transfer coefficient U of the wall.
L1 = 150 mm, k1 = 0.17 W/(m·K)L2 = 50 mm, k2 = 1.4 W/(m·K)A = 15 m²hi = 7.7 W/(m²·K)ho = 30 W/(m²·K)Troom = 22 °CTout = -5 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
In a condenser, steam condenses at a constant 130 °C on the outside of the tubes. Cooling water (5 kg/s, cp = 4.18 kJ/(kg·K)) flows through the tubes and heats from 30 to 120 °C. The overall heat transfer coefficient is 2000 W/(m²·K). Find the heat transfer area A needed.
Ts = 130 °Cṁ = 5 kg/scp = 4.18 kJ/(kg·K)Tc: 30 → 120 °CU = 2000 W/(m²·K)
A = m²
4.Heat exchangers (LMTD)Hard
In a counter-flow heat exchanger both streams have the same heat capacity rate, ṁ·cp = 0.5 kW/K. The hot stream enters at 135 °C and the cold stream at 10 °C. The exchanger has A = 6 m² and U = 300 W/(m²·K). Find the hot outlet temperature Th,out.
Ch = Cc = 0.5 kW/KTh,in = 135 °CTc,in = 10 °CA = 6 m²U = 300 W/(m²·K)
Th,out = °C
3Solutions
1.Heat conduction
Answer: Q̇ = 46.8 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) = (250/1000)/(0.04·20) = 0.3125 K/W
R2 = L2/(k2A) = (100/1000)/(0.17·20) = 0.02941 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 16/0.3419 = 46.8 W
Answer
Q̇ = 46.8 W
2.Heat conduction
Answer: U = 0.9248 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/(7.7·15) = 0.008658 K/W
R1 = L1/(k1A) = (150/1000)/(0.17·15) = 0.05882 K/W
R2 = L2/(k2A) = (50/1000)/(1.4·15) = 0.002381 K/W
Ro = 1/(hoA) = 1/(30·15) = 0.002222 K/W
ΣR = 0.07208 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(15·0.07208) = 0.9248 W/(m²·K)
Answer
U = 0.9248 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 24.06 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 = 5·4.18·(120 − 30) = 1881 kW
End temperature differences
ΔT₁ = 130 − 30 = 100 K
ΔT₂ = 130 − 120 = 10 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (100 − 10)/ln(100/10) = 39.09 K
Area
A = Q̇/(U·ΔTlm) = 1.881 × 10⁶/(2000·39.09) = 24.06 m²
Answer
A = 24.06 m²
4.Heat exchangers (LMTD)
Answer: Th,out = 37.17 °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.
Constant temperature difference
The cold outlet sits opposite the hot inlet: Tc,out = Tc,in + (Th,in − Th,out), so ΔT₁ = Th,in − Tc,out = Th,out − Tc,in = ΔT₂.
ΔTlm = Th,out − Tc,in
Energy balance equals rate equation
0.5·(135 − Th,out) = 1.8·(Th,out − 10)
Th,out = (0.5·135 + 1.8·10)/(0.5 + 1.8) = 37.17 °C
Answer
Th,out = 37.17 °C
Heat transferred: Q̇ = 48.91 kW; the cold stream leaves at 107.8 °C.
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