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 mineral wool and 100 mm of glass, with an area of 6 m². The outer faces are at 19 °C and -13 °C. Find the heat flow Q̇ through the wall.
L1 = 150 mm, k1 = 0.04 W/(m·K)L2 = 100 mm, k2 = 0.96 W/(m·K)A = 6 m²T1 = 19 °CT3 = -13 °C
Q̇ = W
2.Heat conductionHard
A wall of 200 mm expanded polystyrene and 50 mm concrete separates a room at 20 °C from outside air at -12 °C. The convection coefficients are 7.7 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 15 m². Find the temperature Ts of the inside surface of the wall.
L1 = 200 mm, k1 = 0.035 W/(m·K)L2 = 50 mm, k2 = 1.4 W/(m·K)A = 15 m²hi = 7.7 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = -12 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 300 kW. The hot fluid cools from 105 to 55 °C and the cold fluid heats from 30 to 70 °C. The overall heat transfer coefficient is 1500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 300 kWU = 1500 W/(m²·K)Th: 105 → 55 °CTc: 30 → 70 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1.5 kg/s, cp = 2.3 kJ/(kg·K)) cools from 110 to 70 °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 = 500 W/(m²·K), find the heat transfer area A needed.
ṁh = 1.5 kg/scp,h = 2.3 kJ/(kg·K)Th: 110 → 70 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 15 °CU = 500 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 49.82 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)/(0.04·6) = 0.625 K/W
R2 = L2/(k2A) = (100/1000)/(0.96·6) = 0.01736 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 32/0.6424 = 49.82 W
Answer
Q̇ = 49.82 W
2.Heat conduction
Answer: Ts = 19.3 °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/(7.7·15) = 0.008658 K/W
R1 = L1/(k1A) = (200/1000)/(0.035·15) = 0.381 K/W
R2 = L2/(k2A) = (50/1000)/(1.4·15) = 0.002381 K/W
Ro = 1/(hoA) = 1/(25·15) = 0.002667 K/W
ΣR = 0.3947 K/W
Heat loss
Q̇ = (20 − (-12))/0.3947 = 81.08 W
Ts = Troom − Q̇·Ri = 20 − 81.08·0.008658 = 19.3 °C
Answer
Ts = 19.3 °C
3.Heat exchangers (LMTD)
Answer: A = 6.729 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 = 105 − 70 = 35 K
ΔT₂ = Th,out − Tc,in = 55 − 30 = 25 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (35 − 25)/ln(35/25) = 29.72 K
Area
A = Q̇/(U·ΔTlm) = 300000/(1500·29.72) = 6.729 m²
Answer
A = 6.729 m²
4.Heat exchangers (LMTD)
Answer: A = 4.342 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) = 1.5·2.3·40 = 138 kW
Tc,out = 15 + 138/(1.5·4.18) = 37.01 °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 = 110 − 37.01 = 72.99 K
ΔT₂ = Th,out − Tc,in = 70 − 15 = 55 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (72.99 − 55)/ln(72.99/55) = 63.57 K
Area
A = Q̇/(U·ΔTlm) = 138000/(500·63.57) = 4.342 m²
Answer
A = 4.342 m²
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