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 concrete and 50 mm of glass, with an area of 20 m². The outer faces are at 19 °C and -15 °C. Find the temperature Ti at the interface between the two layers.
L1 = 100 mm, k1 = 1.4 W/(m·K)L2 = 50 mm, k2 = 0.96 W/(m·K)A = 20 m²T1 = 19 °CT3 = -15 °C
Ti = °C
2.Heat conductionHard
A wall of 50 mm concrete and 25 mm wood separates a room at 21 °C from outside air at -13 °C. The convection coefficients are 10 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 20 m². Find the overall heat transfer coefficient U of the wall.
L1 = 50 mm, k1 = 1.4 W/(m·K)L2 = 25 mm, k2 = 0.17 W/(m·K)A = 20 m²hi = 10 W/(m²·K)ho = 25 W/(m²·K)Troom = 21 °CTout = -13 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 440 kW. The hot fluid cools from 145 to 105 °C and the cold fluid heats from 25 to 75 °C. The overall heat transfer coefficient is 800 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 440 kWU = 800 W/(m²·K)Th: 145 → 105 °CTc: 25 → 75 °C
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 110 °C and the cold stream at 20 °C. The exchanger has A = 5 m² and U = 300 W/(m²·K). Find the hot outlet temperature Th,out.
Ch = Cc = 0.5 kW/KTh,in = 110 °CTc,in = 20 °CA = 5 m²U = 300 W/(m²·K)
Th,out = °C
3Solutions
1.Heat conduction
Answer: Ti = −0.6627 °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)/(1.4·20) = 0.003571 K/W
R2 = L2/(k2A) = (50/1000)/(0.96·20) = 0.002604 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 34/0.006176 = 5506 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 19 − 5506·0.003571 = −0.6627 °C
Answer
Ti = −0.6627 °C
2.Heat conduction
Answer: U = 3.098 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/(10·20) = 0.005 K/W
R1 = L1/(k1A) = (50/1000)/(1.4·20) = 0.001786 K/W
R2 = L2/(k2A) = (25/1000)/(0.17·20) = 0.007353 K/W
Ro = 1/(hoA) = 1/(25·20) = 0.002 K/W
ΣR = 0.01614 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(20·0.01614) = 3.098 W/(m²·K)
Answer
U = 3.098 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 7.344 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 = 145 − 75 = 70 K
ΔT₂ = Th,out − Tc,in = 105 − 25 = 80 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (70 − 80)/ln(70/80) = 74.89 K
Area
A = Q̇/(U·ΔTlm) = 440000/(800·74.89) = 7.344 m²
Answer
A = 7.344 m²
4.Heat exchangers (LMTD)
Answer: Th,out = 42.5 °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·(110 − Th,out) = 1.5·(Th,out − 20)
Th,out = (0.5·110 + 1.5·20)/(0.5 + 1.5) = 42.5 °C
Answer
Th,out = 42.5 °C
Heat transferred: Q̇ = 33.75 kW; the cold stream leaves at 87.5 °C.
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