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 mineral wool and 150 mm of brick, with an area of 20 m². The outer faces are at 24 °C and 1 °C. Find the temperature Ti at the interface between the two layers.
L1 = 100 mm, k1 = 0.04 W/(m·K)L2 = 150 mm, k2 = 0.72 W/(m·K)A = 20 m²T1 = 24 °CT3 = 1 °C
Ti = °C
2.Heat conductionHard
A wall of 50 mm expanded polystyrene and 75 mm brick separates a room at 20 °C from outside air at -8 °C. The convection coefficients are 8 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 = 50 mm, k1 = 0.035 W/(m·K)L2 = 75 mm, k2 = 0.72 W/(m·K)A = 15 m²hi = 8 W/(m²·K)ho = 30 W/(m²·K)Troom = 20 °CTout = -8 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 290 kW. The hot fluid cools from 95 to 40 °C and the cold fluid heats from 10 to 80 °C. The overall heat transfer coefficient is 1000 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 290 kWU = 1000 W/(m²·K)Th: 95 → 40 °CTc: 10 → 80 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a counter-flow heat exchanger both streams have the same heat capacity rate, ṁ·cp = 3 kW/K. The hot stream enters at 145 °C and the cold stream at 10 °C. The exchanger has A = 4 m² and U = 500 W/(m²·K). Find the hot outlet temperature Th,out.
Ch = Cc = 3 kW/KTh,in = 145 °CTc,in = 10 °CA = 4 m²U = 500 W/(m²·K)
Th,out = °C
3Solutions
1.Heat conduction
Answer: Ti = 2.769 °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) = (100/1000)/(0.04·20) = 0.125 K/W
R2 = L2/(k2A) = (150/1000)/(0.72·20) = 0.01042 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 23/0.1354 = 169.8 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 24 − 169.8·0.125 = 2.769 °C
Answer
Ti = 2.769 °C
2.Heat conduction
Answer: U = 0.5913 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/(8·15) = 0.008333 K/W
R1 = L1/(k1A) = (50/1000)/(0.035·15) = 0.09524 K/W
R2 = L2/(k2A) = (75/1000)/(0.72·15) = 0.006944 K/W
Ro = 1/(hoA) = 1/(30·15) = 0.002222 K/W
ΣR = 0.1127 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(15·0.1127) = 0.5913 W/(m²·K)
Answer
U = 0.5913 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 13.4 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 = 95 − 80 = 15 K
ΔT₂ = Th,out − Tc,in = 40 − 10 = 30 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (15 − 30)/ln(15/30) = 21.64 K
Area
A = Q̇/(U·ΔTlm) = 290000/(1000·21.64) = 13.4 m²
Answer
A = 13.4 m²
4.Heat exchangers (LMTD)
Answer: Th,out = 91 °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
3·(145 − Th,out) = 2·(Th,out − 10)
Th,out = (3·145 + 2·10)/(3 + 2) = 91 °C
Answer
Th,out = 91 °C
Heat transferred: Q̇ = 162 kW; the cold stream leaves at 64 °C.
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