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 glass and 50 mm of mineral wool, with an area of 4 m². The outer faces are at 21 °C and 1 °C. Find the heat flow Q̇ through the wall.
L1 = 100 mm, k1 = 0.96 W/(m·K)L2 = 50 mm, k2 = 0.04 W/(m·K)A = 4 m²T1 = 21 °CT3 = 1 °C
Q̇ = W
2.Heat conductionHard
A wall of 150 mm brick and 150 mm mineral wool separates a room at 21 °C from outside air at -12 °C. The convection coefficients are 8 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 20 m². Find the temperature Ts of the inside surface of the wall.
L1 = 150 mm, k1 = 0.72 W/(m·K)L2 = 150 mm, k2 = 0.04 W/(m·K)A = 20 m²hi = 8 W/(m²·K)ho = 30 W/(m²·K)Troom = 21 °CTout = -12 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 90 kW. The hot fluid cools from 175 to 55 °C and the cold fluid heats from 25 to 155 °C. The overall heat transfer coefficient is 500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 90 kWU = 500 W/(m²·K)Th: 175 → 55 °CTc: 25 → 155 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1.5 kg/s, cp = 2.3 kJ/(kg·K)) cools from 155 to 85 °C in a counter-flow exchanger. It heats 3 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 25 °C. With U = 800 W/(m²·K), find the heat transfer area A needed.
ṁh = 1.5 kg/scp,h = 2.3 kJ/(kg·K)Th: 155 → 85 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 25 °CU = 800 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 59.08 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) = (100/1000)/(0.96·4) = 0.02604 K/W
R2 = L2/(k2A) = (50/1000)/(0.04·4) = 0.3125 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 20/0.3385 = 59.08 W
Answer
Q̇ = 59.08 W
2.Heat conduction
Answer: Ts = 20 °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/(8·20) = 0.00625 K/W
R1 = L1/(k1A) = (150/1000)/(0.72·20) = 0.01042 K/W
R2 = L2/(k2A) = (150/1000)/(0.04·20) = 0.1875 K/W
Ro = 1/(hoA) = 1/(30·20) = 0.001667 K/W
ΣR = 0.2058 K/W
Heat loss
Q̇ = (21 − (-12))/0.2058 = 160.3 W
Ts = Troom − Q̇·Ri = 21 − 160.3·0.00625 = 20 °C
Answer
Ts = 20 °C
3.Heat exchangers (LMTD)
Answer: A = 7.298 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 − 155 = 20 K
ΔT₂ = Th,out − Tc,in = 55 − 25 = 30 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (20 − 30)/ln(20/30) = 24.66 K
Area
A = Q̇/(U·ΔTlm) = 90000/(500·24.66) = 7.298 m²
Answer
A = 7.298 m²
4.Heat exchangers (LMTD)
Answer: A = 3.646 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) = 1.5·2.3·70 = 241.5 kW
Tc,out = 25 + 241.5/(3·4.18) = 44.26 °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 = 155 − 44.26 = 110.7 K
ΔT₂ = Th,out − Tc,in = 85 − 25 = 60 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (110.7 − 60)/ln(110.7/60) = 82.8 K
Area
A = Q̇/(U·ΔTlm) = 241500/(800·82.8) = 3.646 m²
Answer
A = 3.646 m²
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