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 concrete and 25 mm of expanded polystyrene, with an area of 6 m². The outer faces are at 22 °C and 1 °C. Find the temperature Ti at the interface between the two layers.
L1 = 200 mm, k1 = 1.4 W/(m·K)L2 = 25 mm, k2 = 0.035 W/(m·K)A = 6 m²T1 = 22 °CT3 = 1 °C
Ti = °C
2.Heat conductionHard
A wall of 50 mm wood and 150 mm concrete separates a room at 22 °C from outside air at -8 °C. The convection coefficients are 6 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 20 m². Find the heat loss Q̇ through the wall.
L1 = 50 mm, k1 = 0.17 W/(m·K)L2 = 150 mm, k2 = 1.4 W/(m·K)A = 20 m²hi = 6 W/(m²·K)ho = 30 W/(m²·K)Troom = 22 °CTout = -8 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A shell-and-tube heat exchanger (one shell pass, two tube passes) must transfer 90 kW. The hot fluid cools from 140 to 100 °C and the cold fluid heats from 30 to 120 °C. The LMTD correction factor for this arrangement is F = 0.95, and U = 800 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 90 kWF = 0.95U = 800 W/(m²·K)Th: 140 → 100 °CTc: 30 → 120 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1 kg/s, cp = 2.3 kJ/(kg·K)) cools from 115 to 45 °C in a counter-flow exchanger. It heats 2 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 15 °C. With U = 800 W/(m²·K), find the heat transfer area A needed.
ṁh = 1 kg/scp,h = 2.3 kJ/(kg·K)Th: 115 → 45 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 15 °CU = 800 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = 18.5 °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)/(1.4·6) = 0.02381 K/W
R2 = L2/(k2A) = (25/1000)/(0.035·6) = 0.119 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 21/0.1429 = 147 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 22 − 147·0.02381 = 18.5 °C
Answer
Ti = 18.5 °C
2.Heat conduction
Answer: Q̇ = 997.9 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/(6·20) = 0.008333 K/W
R1 = L1/(k1A) = (50/1000)/(0.17·20) = 0.01471 K/W
R2 = L2/(k2A) = (150/1000)/(1.4·20) = 0.005357 K/W
Ro = 1/(hoA) = 1/(30·20) = 0.001667 K/W
ΣR = 0.03006 K/W
Heat loss
Q̇ = (22 − (-8))/0.03006 = 997.9 W
Answer
Q̇ = 997.9 W
3.Heat exchangers (LMTD)
Answer: A = 2.967 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 = 140 − 120 = 20 K
ΔT₂ = Th,out − Tc,in = 100 − 30 = 70 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (20 − 70)/ln(20/70) = 39.91 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 90000/(800·0.95·39.91) = 2.967 m²
Answer
A = 2.967 m²
4.Heat exchangers (LMTD)
Answer: A = 3.927 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.
Heat duty and water outlet temperature
Q̇ = ṁhcp,h(Th,in − Th,out) = 1·2.3·70 = 161 kW
Tc,out = 15 + 161/(2·4.18) = 34.26 °C
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 = 115 − 34.26 = 80.74 K
ΔT₂ = Th,out − Tc,in = 45 − 15 = 30 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (80.74 − 30)/ln(80.74/30) = 51.25 K
Area
A = Q̇/(U·ΔTlm) = 161000/(800·51.25) = 3.927 m²
Answer
A = 3.927 m²
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