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 brick and 50 mm of gypsum board, with an area of 4 m². The outer faces are at 15 °C and 7 °C. Find the heat flow Q̇ through the wall.
L1 = 100 mm, k1 = 0.72 W/(m·K)L2 = 50 mm, k2 = 0.17 W/(m·K)A = 4 m²T1 = 15 °CT3 = 7 °C
Q̇ = W
2.Heat conductionHard
A wall of 150 mm gypsum board and 150 mm mineral wool separates a room at 20 °C from outside air at -3 °C. The convection coefficients are 8 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 4 m². Find the temperature Ts of the inside surface of the wall.
L1 = 150 mm, k1 = 0.17 W/(m·K)L2 = 150 mm, k2 = 0.04 W/(m·K)A = 4 m²hi = 8 W/(m²·K)ho = 30 W/(m²·K)Troom = 20 °CTout = -3 °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 80 to 60 °C and the cold fluid heats from 30 to 55 °C. Find the overall heat transfer coefficient U.
Q̇ = 250 kWA = 2.5 m²Th: 80 → 60 °CTc: 30 → 55 °C
U = W/(m²·K)
4.Heat exchangers (LMTD)Hard
Hot oil (0.5 kg/s, cp = 2.1 kJ/(kg·K)) cools from 125 to 75 °C in a counter-flow exchanger. It heats 3 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 10 °C. With U = 200 W/(m²·K), find the heat transfer area A needed.
ṁh = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 125 → 75 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 10 °CU = 200 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 73.9 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)/(0.72·4) = 0.03472 K/W
R2 = L2/(k2A) = (50/1000)/(0.17·4) = 0.07353 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 8/0.1083 = 73.9 W
Answer
Q̇ = 73.9 W
2.Heat conduction
Answer: Ts = 19.4 °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/(8·4) = 0.03125 K/W
R1 = L1/(k1A) = (150/1000)/(0.17·4) = 0.2206 K/W
R2 = L2/(k2A) = (150/1000)/(0.04·4) = 0.9375 K/W
Ro = 1/(hoA) = 1/(30·4) = 0.008333 K/W
ΣR = 1.198 K/W
Heat loss
Q̇ = (20 − (-3))/1.198 = 19.2 W
Ts = Troom − Q̇·Ri = 20 − 19.2·0.03125 = 19.4 °C
Answer
Ts = 19.4 °C
3.Heat exchangers (LMTD)
Answer: U = 3646 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 = 80 − 55 = 25 K
ΔT₂ = Th,out − Tc,in = 60 − 30 = 30 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (25 − 30)/ln(25/30) = 27.42 K
Overall coefficient
U = Q̇/(A·ΔTlm) = 250000/(2.5·27.42) = 3646 W/(m²·K)
Answer
U = 3646 W/(m²·K)
4.Heat exchangers (LMTD)
Answer: A = 3.057 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) = 0.5·2.1·50 = 52.5 kW
Tc,out = 10 + 52.5/(3·4.18) = 14.19 °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 = 125 − 14.19 = 110.8 K
ΔT₂ = Th,out − Tc,in = 75 − 10 = 65 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (110.8 − 65)/ln(110.8/65) = 85.88 K
Area
A = Q̇/(U·ΔTlm) = 52500/(200·85.88) = 3.057 m²
Answer
A = 3.057 m²
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