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 concrete and 25 mm of gypsum board, with an area of 4 m². The outer faces are at 18 °C and -13 °C. Find the heat flow Q̇ through the wall.
L1 = 150 mm, k1 = 1.4 W/(m·K)L2 = 25 mm, k2 = 0.17 W/(m·K)A = 4 m²T1 = 18 °CT3 = -13 °C
Q̇ = W
2.Heat conductionHard
A wall of 50 mm mineral wool and 75 mm gypsum board separates a room at 21 °C from outside air at 5 °C. The convection coefficients are 8 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 = 50 mm, k1 = 0.04 W/(m·K)L2 = 75 mm, k2 = 0.17 W/(m·K)A = 15 m²hi = 8 W/(m²·K)ho = 25 W/(m²·K)Troom = 21 °CTout = 5 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 20 kW. The hot fluid cools from 95 to 55 °C and the cold fluid heats from 20 to 40 °C. The overall heat transfer coefficient is 500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 20 kWU = 500 W/(m²·K)Th: 95 → 55 °CTc: 20 → 40 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (0.5 kg/s, cp = 2.1 kJ/(kg·K)) cools from 100 to 35 °C in a counter-flow exchanger. It heats 0.5 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 15 °C. With U = 1500 W/(m²·K), find the heat transfer area A needed.
ṁh = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 100 → 35 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 15 °CU = 1500 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 487.8 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)/(1.4·4) = 0.02679 K/W
R2 = L2/(k2A) = (25/1000)/(0.17·4) = 0.03676 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 31/0.06355 = 487.8 W
Answer
Q̇ = 487.8 W
2.Heat conduction
Answer: Ts = 19.92 °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/(8·15) = 0.008333 K/W
R1 = L1/(k1A) = (50/1000)/(0.04·15) = 0.08333 K/W
R2 = L2/(k2A) = (75/1000)/(0.17·15) = 0.02941 K/W
Ro = 1/(hoA) = 1/(25·15) = 0.002667 K/W
ΣR = 0.1237 K/W
Heat loss
Q̇ = (21 − (5))/0.1237 = 129.3 W
Ts = Troom − Q̇·Ri = 21 − 129.3·0.008333 = 19.92 °C
Answer
Ts = 19.92 °C
3.Heat exchangers (LMTD)
Answer: A = 0.904 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 = 95 − 40 = 55 K
ΔT₂ = Th,out − Tc,in = 55 − 20 = 35 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (55 − 35)/ln(55/35) = 44.25 K
Area
A = Q̇/(U·ΔTlm) = 20000/(500·44.25) = 0.904 m²
Answer
A = 0.904 m²
4.Heat exchangers (LMTD)
Answer: A = 1.353 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) = 0.5·2.1·65 = 68.25 kW
Tc,out = 15 + 68.25/(0.5·4.18) = 47.66 °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 = 100 − 47.66 = 52.34 K
ΔT₂ = Th,out − Tc,in = 35 − 15 = 20 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (52.34 − 20)/ln(52.34/20) = 33.62 K
Area
A = Q̇/(U·ΔTlm) = 68250/(1500·33.62) = 1.353 m²
Answer
A = 1.353 m²
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