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 wood and 150 mm of expanded polystyrene, with an area of 4 m². The outer faces are at 19 °C and -5 °C. Find the temperature Ti at the interface between the two layers.
L1 = 100 mm, k1 = 0.17 W/(m·K)L2 = 150 mm, k2 = 0.035 W/(m·K)A = 4 m²T1 = 19 °CT3 = -5 °C
Ti = °C
2.Heat conductionHard
A wall of 50 mm concrete and 50 mm gypsum board separates a room at 22 °C from outside air at 4 °C. The convection coefficients are 10 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 4 m². Find the temperature Ts of the inside surface of the wall.
L1 = 50 mm, k1 = 1.4 W/(m·K)L2 = 50 mm, k2 = 0.17 W/(m·K)A = 4 m²hi = 10 W/(m²·K)ho = 25 W/(m²·K)Troom = 22 °CTout = 4 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A shell-and-tube heat exchanger (one shell pass, two tube passes) must transfer 720 kW. The hot fluid cools from 120 to 70 °C and the cold fluid heats from 25 to 35 °C. The LMTD correction factor for this arrangement is F = 0.88, and U = 300 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 720 kWF = 0.88U = 300 W/(m²·K)Th: 120 → 70 °CTc: 25 → 35 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1.5 kg/s, cp = 2.5 kJ/(kg·K)) cools from 100 to 55 °C in a counter-flow exchanger. It heats 2 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 25 °C. With U = 200 W/(m²·K), find the heat transfer area A needed.
ṁh = 1.5 kg/scp,h = 2.5 kJ/(kg·K)Th: 100 → 55 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 25 °CU = 200 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = 16.1 °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)/(0.17·4) = 0.1471 K/W
R2 = L2/(k2A) = (150/1000)/(0.035·4) = 1.071 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 24/1.218 = 19.7 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 19 − 19.7·0.1471 = 16.1 °C
Answer
Ti = 16.1 °C
2.Heat conduction
Answer: Ts = 18.17 °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/(10·4) = 0.025 K/W
R1 = L1/(k1A) = (50/1000)/(1.4·4) = 0.008929 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 = 0.1175 K/W
Heat loss
Q̇ = (22 − (4))/0.1175 = 153.2 W
Ts = Troom − Q̇·Ri = 22 − 153.2·0.025 = 18.17 °C
Answer
Ts = 18.17 °C
3.Heat exchangers (LMTD)
Answer: A = 43.36 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 = 120 − 35 = 85 K
ΔT₂ = Th,out − Tc,in = 70 − 25 = 45 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (85 − 45)/ln(85/45) = 62.89 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 720000/(300·0.88·62.89) = 43.36 m²
Answer
A = 43.36 m²
4.Heat exchangers (LMTD)
Answer: A = 20.5 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) = 1.5·2.5·45 = 168.8 kW
Tc,out = 25 + 168.8/(2·4.18) = 45.19 °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 = 100 − 45.19 = 54.81 K
ΔT₂ = Th,out − Tc,in = 55 − 25 = 30 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (54.81 − 30)/ln(54.81/30) = 41.17 K
Area
A = Q̇/(U·ΔTlm) = 168750/(200·41.17) = 20.5 m²
Answer
A = 20.5 m²
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