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 100 mm of mineral wool, with an area of 6 m². The outer faces are at 25 °C and -1 °C. Find the temperature Ti at the interface between the two layers.
L1 = 100 mm, k1 = 1.4 W/(m·K)L2 = 100 mm, k2 = 0.04 W/(m·K)A = 6 m²T1 = 25 °CT3 = -1 °C
Ti = °C
2.Heat conductionHard
A wall of 200 mm mineral wool and 150 mm gypsum board separates a room at 22 °C from outside air at 4 °C. The convection coefficients are 8 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 8 m². Find the temperature Ts of the inside surface of the wall.
L1 = 200 mm, k1 = 0.04 W/(m·K)L2 = 150 mm, k2 = 0.17 W/(m·K)A = 8 m²hi = 8 W/(m²·K)ho = 25 W/(m²·K)Troom = 22 °CTout = 4 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A parallel-flow heat exchanger transfers 340 kW. The hot fluid cools from 110 to 65 °C and the cold fluid heats from 20 to 45 °C. The overall heat transfer coefficient is 1500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 340 kWU = 1500 W/(m²·K)Th: 110 → 65 °CTc: 20 → 45 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a counter-flow heat exchanger both streams have the same heat capacity rate, ṁ·cp = 2 kW/K. The hot stream enters at 85 °C and the cold stream at 10 °C. The exchanger has A = 5 m² and U = 300 W/(m²·K). Find the hot outlet temperature Th,out.
Ch = Cc = 2 kW/KTh,in = 85 °CTc,in = 10 °CA = 5 m²U = 300 W/(m²·K)
Th,out = °C
3Solutions
1.Heat conduction
Answer: Ti = 24.28 °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)/(1.4·6) = 0.0119 K/W
R2 = L2/(k2A) = (100/1000)/(0.04·6) = 0.4167 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 26/0.4286 = 60.67 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 25 − 60.67·0.0119 = 24.28 °C
Answer
Ti = 24.28 °C
2.Heat conduction
Answer: Ts = 21.63 °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.
Resistances in series
Ri = 1/(hiA) = 1/(8·8) = 0.01563 K/W
R1 = L1/(k1A) = (200/1000)/(0.04·8) = 0.625 K/W
R2 = L2/(k2A) = (150/1000)/(0.17·8) = 0.1103 K/W
Ro = 1/(hoA) = 1/(25·8) = 0.005 K/W
ΣR = 0.7559 K/W
Heat loss
Q̇ = (22 − (4))/0.7559 = 23.81 W
Ts = Troom − Q̇·Ri = 22 − 23.81·0.01563 = 21.63 °C
Answer
Ts = 21.63 °C
3.Heat exchangers (LMTD)
Answer: A = 4.87 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 = 110 − 20 = 90 K
ΔT₂ = Th,out − Tc,out = 65 − 45 = 20 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (90 − 20)/ln(90/20) = 46.54 K
Area
A = Q̇/(U·ΔTlm) = 340000/(1500·46.54) = 4.87 m²
Answer
A = 4.87 m²
4.Heat exchangers (LMTD)
Answer: Th,out = 52.86 °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
2·(85 − Th,out) = 1.5·(Th,out − 10)
Th,out = (2·85 + 1.5·10)/(2 + 1.5) = 52.86 °C
Answer
Th,out = 52.86 °C
Heat transferred: Q̇ = 64.29 kW; the cold stream leaves at 42.14 °C.
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