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 gypsum board and 75 mm of wood, with an area of 10 m². The outer faces are at 19 °C and -8 °C. Find the heat flow Q̇ through the wall.
L1 = 200 mm, k1 = 0.17 W/(m·K)L2 = 75 mm, k2 = 0.17 W/(m·K)A = 10 m²T1 = 19 °CT3 = -8 °C
Q̇ = W
2.Heat conductionHard
A wall of 200 mm brick and 75 mm expanded polystyrene separates a room at 22 °C from outside air at -8 °C. The convection coefficients are 10 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 10 m². Find the overall heat transfer coefficient U of the wall.
L1 = 200 mm, k1 = 0.72 W/(m·K)L2 = 75 mm, k2 = 0.035 W/(m·K)A = 10 m²hi = 10 W/(m²·K)ho = 30 W/(m²·K)Troom = 22 °CTout = -8 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
In a condenser, steam condenses at a constant 130 °C on the outside of the tubes. Cooling water (0.5 kg/s, cp = 4.18 kJ/(kg·K)) flows through the tubes and heats from 20 to 70 °C. The overall heat transfer coefficient is 2000 W/(m²·K). Find the heat transfer area A needed.
Ts = 130 °Cṁ = 0.5 kg/scp = 4.18 kJ/(kg·K)Tc: 20 → 70 °CU = 2000 W/(m²·K)
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1.5 kg/s, cp = 2.5 kJ/(kg·K)) cools from 80 to 45 °C in a counter-flow exchanger. It heats 1.5 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 10 °C. With U = 800 W/(m²·K), find the heat transfer area A needed.
ṁh = 1.5 kg/scp,h = 2.5 kJ/(kg·K)Th: 80 → 45 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 10 °CU = 800 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 166.9 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) = (200/1000)/(0.17·10) = 0.1176 K/W
R2 = L2/(k2A) = (75/1000)/(0.17·10) = 0.04412 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 27/0.1618 = 166.9 W
Answer
Q̇ = 166.9 W
2.Heat conduction
Answer: U = 0.3915 W/(m²·K)
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/(10·10) = 0.01 K/W
R1 = L1/(k1A) = (200/1000)/(0.72·10) = 0.02778 K/W
R2 = L2/(k2A) = (75/1000)/(0.035·10) = 0.2143 K/W
Ro = 1/(hoA) = 1/(30·10) = 0.003333 K/W
ΣR = 0.2554 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(10·0.2554) = 0.3915 W/(m²·K)
Answer
U = 0.3915 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 0.6334 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 from the water side
Q̇ = 0.5·4.18·(70 − 20) = 104.5 kW
End temperature differences
ΔT₁ = 130 − 20 = 110 K
ΔT₂ = 130 − 70 = 60 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (110 − 60)/ln(110/60) = 82.49 K
Area
A = Q̇/(U·ΔTlm) = 104500/(2000·82.49) = 0.6334 m²
Answer
A = 0.6334 m²
4.Heat exchangers (LMTD)
Answer: A = 3.94 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) = 1.5·2.5·35 = 131.3 kW
Tc,out = 10 + 131.3/(1.5·4.18) = 30.93 °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 − 30.93 = 49.07 K
ΔT₂ = Th,out − Tc,in = 45 − 10 = 35 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (49.07 − 35)/ln(49.07/35) = 41.64 K
Area
A = Q̇/(U·ΔTlm) = 131250/(800·41.64) = 3.94 m²
Answer
A = 3.94 m²
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