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 200 mm of mineral wool and 100 mm of gypsum board, with an area of 10 m². The outer faces are at 22 °C and 8 °C. Find the temperature Ti at the interface between the two layers.
L1 = 200 mm, k1 = 0.04 W/(m·K)L2 = 100 mm, k2 = 0.17 W/(m·K)A = 10 m²T1 = 22 °CT3 = 8 °C
Ti = °C
2.Heat conductionHard
A wall of 150 mm wood and 50 mm gypsum board separates a room at 21 °C from outside air at -8 °C. The convection coefficients are 7.7 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 4 m². Find the heat loss Q̇ through the wall.
L1 = 150 mm, k1 = 0.17 W/(m·K)L2 = 50 mm, k2 = 0.17 W/(m·K)A = 4 m²hi = 7.7 W/(m²·K)ho = 25 W/(m²·K)Troom = 21 °CTout = -8 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A shell-and-tube heat exchanger (one shell pass, two tube passes) must transfer 710 kW. The hot fluid cools from 85 to 45 °C and the cold fluid heats from 30 to 55 °C. The LMTD correction factor for this arrangement is F = 0.88, and U = 500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 710 kWF = 0.88U = 500 W/(m²·K)Th: 85 → 45 °CTc: 30 → 55 °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 165 °C and the cold stream at 25 °C. The exchanger has A = 1 m² and U = 800 W/(m²·K). Find the hot outlet temperature Th,out.
Ch = Cc = 0.5 kW/KTh,in = 165 °CTc,in = 25 °CA = 1 m²U = 800 W/(m²·K)
Th,out = °C
3Solutions
1.Heat conduction
Answer: Ti = 9.474 °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) = (200/1000)/(0.04·10) = 0.5 K/W
R2 = L2/(k2A) = (100/1000)/(0.17·10) = 0.05882 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 14/0.5588 = 25.05 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 22 − 25.05·0.5 = 9.474 °C
Answer
Ti = 9.474 °C
2.Heat conduction
Answer: Q̇ = 86.16 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.
Resistances in series
Ri = 1/(hiA) = 1/(7.7·4) = 0.03247 K/W
R1 = L1/(k1A) = (150/1000)/(0.17·4) = 0.2206 K/W
R2 = L2/(k2A) = (50/1000)/(0.17·4) = 0.07353 K/W
Ro = 1/(hoA) = 1/(25·4) = 0.01 K/W
ΣR = 0.3366 K/W
Heat loss
Q̇ = (21 − (-8))/0.3366 = 86.16 W
Answer
Q̇ = 86.16 W
3.Heat exchangers (LMTD)
Answer: A = 74.57 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 (counter-flow)
In counter-flow the hot inlet faces the cold outlet, and the hot outlet faces the cold inlet.
ΔT₁ = Th,in − Tc,out = 85 − 55 = 30 K
ΔT₂ = Th,out − Tc,in = 45 − 30 = 15 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (30 − 15)/ln(30/15) = 21.64 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 710000/(500·0.88·21.64) = 74.57 m²
Answer
A = 74.57 m²
4.Heat exchangers (LMTD)
Answer: Th,out = 78.85 °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·(165 − Th,out) = 0.8·(Th,out − 25)
Th,out = (0.5·165 + 0.8·25)/(0.5 + 0.8) = 78.85 °C
Answer
Th,out = 78.85 °C
Heat transferred: Q̇ = 43.08 kW; the cold stream leaves at 111.2 °C.
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