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 50 mm of mineral wool and 75 mm of expanded polystyrene, with an area of 6 m². The outer faces are at 21 °C and 4 °C. Find the heat flow Q̇ through the wall.
L1 = 50 mm, k1 = 0.04 W/(m·K)L2 = 75 mm, k2 = 0.035 W/(m·K)A = 6 m²T1 = 21 °CT3 = 4 °C
Q̇ = W
2.Heat conductionHard
A wall of 50 mm brick and 50 mm gypsum board separates a room at 22 °C from outside air at -6 °C. The convection coefficients are 6 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 12 m². Find the temperature Ts of the inside surface of the wall.
L1 = 50 mm, k1 = 0.72 W/(m·K)L2 = 50 mm, k2 = 0.17 W/(m·K)A = 12 m²hi = 6 W/(m²·K)ho = 25 W/(m²·K)Troom = 22 °CTout = -6 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
In a condenser, steam condenses at a constant 105 °C on the outside of the tubes. Cooling water (1.5 kg/s, cp = 4.18 kJ/(kg·K)) flows through the tubes and heats from 10 to 80 °C. The overall heat transfer coefficient is 2000 W/(m²·K). Find the heat transfer area A needed.
Ts = 105 °Cṁ = 1.5 kg/scp = 4.18 kJ/(kg·K)Tc: 10 → 80 °CU = 2000 W/(m²·K)
A = m²
4.Heat exchangers (LMTD)Hard
Steam condenses at a constant 100 °C on the outside of the tubes of a condenser with a heat transfer area of 1.5 m² and U = 1000 W/(m²·K). Cooling water (0.5 kg/s, cp = 4.18 kJ/(kg·K)) enters the tubes at 30 °C. Find the water outlet temperature Tc,out.
Ts = 100 °CA = 1.5 m²U = 1000 W/(m²·K)ṁ = 0.5 kg/scp = 4.18 kJ/(kg·K)Tc,in = 30 °C
Tc,out = °C
3Solutions
1.Heat conduction
Answer: Q̇ = 30.06 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) = (50/1000)/(0.04·6) = 0.2083 K/W
R2 = L2/(k2A) = (75/1000)/(0.035·6) = 0.3571 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 17/0.5655 = 30.06 W
Answer
Q̇ = 30.06 W
2.Heat conduction
Answer: Ts = 13.82 °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/(6·12) = 0.01389 K/W
R1 = L1/(k1A) = (50/1000)/(0.72·12) = 0.005787 K/W
R2 = L2/(k2A) = (50/1000)/(0.17·12) = 0.02451 K/W
Ro = 1/(hoA) = 1/(25·12) = 0.003333 K/W
ΣR = 0.04752 K/W
Heat loss
Q̇ = (22 − (-6))/0.04752 = 589.2 W
Ts = Troom − Q̇·Ri = 22 − 589.2·0.01389 = 13.82 °C
Answer
Ts = 13.82 °C
3.Heat exchangers (LMTD)
Answer: A = 4.185 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 = 1.5·4.18·(80 − 10) = 438.9 kW
End temperature differences
ΔT₁ = 105 − 10 = 95 K
ΔT₂ = 105 − 80 = 25 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (95 − 25)/ln(95/25) = 52.43 K
Area
A = Q̇/(U·ΔTlm) = 438900/(2000·52.43) = 4.185 m²
Answer
A = 4.185 m²
4.Heat exchangers (LMTD)
Answer: Tc,out = 65.85 °C
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.
Combine the energy balance and the rate equation
With ΔT₁ = Ts − Tc,in and ΔT₂ = Ts − Tc,out, the difference ΔT₁ − ΔT₂ equals the water temperature rise, so the rise cancels.
ṁcp(ΔT₁ − ΔT₂) = U·A·(ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂)
ln(ΔT₁/ΔT₂) = U·A/(ṁcp) = 1000·1.5/(0.5·4180) = 0.7177
Outlet temperature
ΔT₂ = ΔT₁·e−0.7177 = 70·0.4879 = 34.15 K
Tc,out = 100 − 34.15 = 65.85 °C
Answer
Tc,out = 65.85 °C
Heat transferred: Q̇ = 74.92 kW.
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