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 brick and 100 mm of concrete, with an area of 10 m². The outer faces are at 22 °C and 2 °C. Find the heat flow Q̇ through the wall.
L1 = 200 mm, k1 = 0.72 W/(m·K)L2 = 100 mm, k2 = 1.4 W/(m·K)A = 10 m²T1 = 22 °CT3 = 2 °C
Q̇ = W
2.Heat conductionHard
A wall of 100 mm gypsum board and 100 mm expanded polystyrene separates a room at 21 °C from outside air at -12 °C. The convection coefficients are 10 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 6 m². Find the temperature Ts of the inside surface of the wall.
L1 = 100 mm, k1 = 0.17 W/(m·K)L2 = 100 mm, k2 = 0.035 W/(m·K)A = 6 m²hi = 10 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 250 kW. Its heat transfer area is 2.5 m². The hot fluid cools from 155 to 35 °C and the cold fluid heats from 20 to 115 °C. Find the overall heat transfer coefficient U.
Q̇ = 250 kWA = 2.5 m²Th: 155 → 35 °CTc: 20 → 115 °C
U = W/(m²·K)
4.Heat exchangers (LMTD)Hard
Hot oil (1 kg/s, cp = 2.5 kJ/(kg·K)) cools from 155 to 85 °C in a counter-flow exchanger. It heats 1.5 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 15 °C. With U = 1000 W/(m²·K), find the heat transfer area A needed.
ṁh = 1 kg/scp,h = 2.5 kJ/(kg·K)Th: 155 → 85 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 15 °CU = 1000 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 572.7 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.
Layer resistances
R1 = L1/(k1A) = (200/1000)/(0.72·10) = 0.02778 K/W
R2 = L2/(k2A) = (100/1000)/(1.4·10) = 0.007143 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 20/0.03492 = 572.7 W
Answer
Q̇ = 572.7 W
2.Heat conduction
Answer: Ts = 20.08 °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/(10·6) = 0.01667 K/W
R1 = L1/(k1A) = (100/1000)/(0.17·6) = 0.09804 K/W
R2 = L2/(k2A) = (100/1000)/(0.035·6) = 0.4762 K/W
Ro = 1/(hoA) = 1/(30·6) = 0.005556 K/W
ΣR = 0.5965 K/W
Heat loss
Q̇ = (21 − (-12))/0.5965 = 55.33 W
Ts = Troom − Q̇·Ri = 21 − 55.33·0.01667 = 20.08 °C
Answer
Ts = 20.08 °C
3.Heat exchangers (LMTD)
Answer: U = 3923 W/(m²·K)
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 = 155 − 115 = 40 K
ΔT₂ = Th,out − Tc,in = 35 − 20 = 15 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (40 − 15)/ln(40/15) = 25.49 K
Overall coefficient
U = Q̇/(A·ΔTlm) = 250000/(2.5·25.49) = 3923 W/(m²·K)
Answer
U = 3923 W/(m²·K)
4.Heat exchangers (LMTD)
Answer: A = 1.958 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) = 1·2.5·70 = 175 kW
Tc,out = 15 + 175/(1.5·4.18) = 42.91 °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 − 42.91 = 112.1 K
ΔT₂ = Th,out − Tc,in = 85 − 15 = 70 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (112.1 − 70)/ln(112.1/70) = 89.4 K
Area
A = Q̇/(U·ΔTlm) = 175000/(1000·89.4) = 1.958 m²
Answer
A = 1.958 m²
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