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 gypsum board and 100 mm of concrete, with an area of 12 m². The outer faces are at 22 °C and 4 °C. Find the heat flow Q̇ through the wall.
L1 = 150 mm, k1 = 0.17 W/(m·K)L2 = 100 mm, k2 = 1.4 W/(m·K)A = 12 m²T1 = 22 °CT3 = 4 °C
Q̇ = W
2.Heat conductionHard
A wall of 150 mm glass and 75 mm concrete separates a room at 22 °C from outside air at 1 °C. The convection coefficients are 6 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 = 0.96 W/(m·K)L2 = 75 mm, k2 = 1.4 W/(m·K)A = 4 m²hi = 6 W/(m²·K)ho = 25 W/(m²·K)Troom = 22 °CTout = 1 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A shell-and-tube heat exchanger (one shell pass, two tube passes) must transfer 800 kW. The hot fluid cools from 80 to 30 °C and the cold fluid heats from 10 to 70 °C. The LMTD correction factor for this arrangement is F = 0.8, and U = 300 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 800 kWF = 0.8U = 300 W/(m²·K)Th: 80 → 30 °CTc: 10 → 70 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (2 kg/s, cp = 2.1 kJ/(kg·K)) cools from 80 to 60 °C in a counter-flow exchanger. It heats 0.5 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 20 °C. With U = 500 W/(m²·K), find the heat transfer area A needed.
ṁh = 2 kg/scp,h = 2.1 kJ/(kg·K)Th: 80 → 60 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 20 °CU = 500 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 226.5 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.
Layer resistances
R1 = L1/(k1A) = (150/1000)/(0.17·12) = 0.07353 K/W
R2 = L2/(k2A) = (100/1000)/(1.4·12) = 0.005952 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 18/0.07948 = 226.5 W
Answer
Q̇ = 226.5 W
2.Heat conduction
Answer: Q̇ = 201.7 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/(6·4) = 0.04167 K/W
R1 = L1/(k1A) = (150/1000)/(0.96·4) = 0.03906 K/W
R2 = L2/(k2A) = (75/1000)/(1.4·4) = 0.01339 K/W
Ro = 1/(hoA) = 1/(25·4) = 0.01 K/W
ΣR = 0.1041 K/W
Heat loss
Q̇ = (22 − (1))/0.1041 = 201.7 W
Answer
Q̇ = 201.7 W
3.Heat exchangers (LMTD)
Answer: A = 231 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 = 80 − 70 = 10 K
ΔT₂ = Th,out − Tc,in = 30 − 10 = 20 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (10 − 20)/ln(10/20) = 14.43 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 800000/(300·0.8·14.43) = 231 m²
Answer
A = 231 m²
4.Heat exchangers (LMTD)
Answer: A = 5.847 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) = 2·2.1·20 = 84 kW
Tc,out = 20 + 84/(0.5·4.18) = 60.19 °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 = 80 − 60.19 = 19.81 K
ΔT₂ = Th,out − Tc,in = 60 − 20 = 40 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (19.81 − 40)/ln(19.81/40) = 28.73 K
Area
A = Q̇/(U·ΔTlm) = 84000/(500·28.73) = 5.847 m²
Answer
A = 5.847 m²
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