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 250 mm of mineral wool and 25 mm of brick, with an area of 6 m². The outer faces are at 16 °C and 4 °C. Find the temperature Ti at the interface between the two layers.
L1 = 250 mm, k1 = 0.04 W/(m·K)L2 = 25 mm, k2 = 0.72 W/(m·K)A = 6 m²T1 = 16 °CT3 = 4 °C
Ti = °C
2.Heat conductionHard
A wall of 250 mm gypsum board and 100 mm wood separates a room at 22 °C from outside air at 2 °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 heat loss Q̇ through the wall.
L1 = 250 mm, k1 = 0.17 W/(m·K)L2 = 100 mm, k2 = 0.17 W/(m·K)A = 8 m²hi = 8 W/(m²·K)ho = 25 W/(m²·K)Troom = 22 °CTout = 2 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A shell-and-tube heat exchanger (one shell pass, two tube passes) must transfer 170 kW. The hot fluid cools from 115 to 70 °C and the cold fluid heats from 25 to 60 °C. The LMTD correction factor for this arrangement is F = 0.85, and U = 300 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 170 kWF = 0.85U = 300 W/(m²·K)Th: 115 → 70 °CTc: 25 → 60 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1 kg/s, cp = 2.1 kJ/(kg·K)) cools from 95 to 75 °C in a counter-flow exchanger. It heats 1.5 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 30 °C. With U = 800 W/(m²·K), find the heat transfer area A needed.
ṁh = 1 kg/scp,h = 2.1 kJ/(kg·K)Th: 95 → 75 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 30 °CU = 800 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = 4.066 °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) = (250/1000)/(0.04·6) = 1.042 K/W
R2 = L2/(k2A) = (25/1000)/(0.72·6) = 0.005787 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 12/1.047 = 11.46 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 16 − 11.46·1.042 = 4.066 °C
Answer
Ti = 4.066 °C
2.Heat conduction
Answer: Q̇ = 71.95 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/(8·8) = 0.01563 K/W
R1 = L1/(k1A) = (250/1000)/(0.17·8) = 0.1838 K/W
R2 = L2/(k2A) = (100/1000)/(0.17·8) = 0.07353 K/W
Ro = 1/(hoA) = 1/(25·8) = 0.005 K/W
ΣR = 0.278 K/W
Heat loss
Q̇ = (22 − (2))/0.278 = 71.95 W
Answer
Q̇ = 71.95 W
3.Heat exchangers (LMTD)
Answer: A = 13.38 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 = 115 − 60 = 55 K
ΔT₂ = Th,out − Tc,in = 70 − 25 = 45 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (55 − 45)/ln(55/45) = 49.83 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 170000/(300·0.85·49.83) = 13.38 m²
Answer
A = 13.38 m²
4.Heat exchangers (LMTD)
Answer: A = 1.022 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) = 1·2.1·20 = 42 kW
Tc,out = 30 + 42/(1.5·4.18) = 36.7 °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 = 95 − 36.7 = 58.3 K
ΔT₂ = Th,out − Tc,in = 75 − 30 = 45 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (58.3 − 45)/ln(58.3/45) = 51.36 K
Area
A = Q̇/(U·ΔTlm) = 42000/(800·51.36) = 1.022 m²
Answer
A = 1.022 m²
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