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 150 mm of brick and 150 mm of glass, with an area of 6 m². The outer faces are at 23 °C and -8 °C. Find the temperature Ti at the interface between the two layers.
L1 = 150 mm, k1 = 0.72 W/(m·K)L2 = 150 mm, k2 = 0.96 W/(m·K)A = 6 m²T1 = 23 °CT3 = -8 °C
Ti = °C
2.Heat conductionHard
A wall of 100 mm concrete and 75 mm glass separates a room at 20 °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 10 m². Find the overall heat transfer coefficient U of the wall.
L1 = 100 mm, k1 = 1.4 W/(m·K)L2 = 75 mm, k2 = 0.96 W/(m·K)A = 10 m²hi = 8 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = 4 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 160 kW. The hot fluid cools from 175 to 140 °C and the cold fluid heats from 30 to 50 °C. The overall heat transfer coefficient is 300 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 160 kWU = 300 W/(m²·K)Th: 175 → 140 °CTc: 30 → 50 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a counter-flow heat exchanger both streams have the same heat capacity rate, ṁ·cp = 4 kW/K. The hot stream enters at 95 °C and the cold stream at 15 °C. The exchanger has A = 4 m² and U = 300 W/(m²·K). Find the hot outlet temperature Th,out.
Ch = Cc = 4 kW/KTh,in = 95 °CTc,in = 15 °CA = 4 m²U = 300 W/(m²·K)
Th,out = °C
3Solutions
1.Heat conduction
Answer: Ti = 5.286 °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) = (150/1000)/(0.72·6) = 0.03472 K/W
R2 = L2/(k2A) = (150/1000)/(0.96·6) = 0.02604 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 31/0.06076 = 510.2 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 23 − 510.2·0.03472 = 5.286 °C
Answer
Ti = 5.286 °C
2.Heat conduction
Answer: U = 3.179 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/(8·10) = 0.0125 K/W
R1 = L1/(k1A) = (100/1000)/(1.4·10) = 0.007143 K/W
R2 = L2/(k2A) = (75/1000)/(0.96·10) = 0.007813 K/W
Ro = 1/(hoA) = 1/(25·10) = 0.004 K/W
ΣR = 0.03146 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(10·0.03146) = 3.179 W/(m²·K)
Answer
U = 3.179 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 4.545 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 = 175 − 50 = 125 K
ΔT₂ = Th,out − Tc,in = 140 − 30 = 110 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (125 − 110)/ln(125/110) = 117.3 K
Area
A = Q̇/(U·ΔTlm) = 160000/(300·117.3) = 4.545 m²
Answer
A = 4.545 m²
4.Heat exchangers (LMTD)
Answer: Th,out = 76.54 °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
4·(95 − Th,out) = 1.2·(Th,out − 15)
Th,out = (4·95 + 1.2·15)/(4 + 1.2) = 76.54 °C
Answer
Th,out = 76.54 °C
Heat transferred: Q̇ = 73.85 kW; the cold stream leaves at 33.46 °C.
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