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 250 mm of concrete and 150 mm of glass, with an area of 12 m². The outer faces are at 23 °C and 2 °C. Find the heat flow Q̇ through the wall.
L1 = 250 mm, k1 = 1.4 W/(m·K)L2 = 150 mm, k2 = 0.96 W/(m·K)A = 12 m²T1 = 23 °CT3 = 2 °C
Q̇ = W
2.Heat conductionHard
A wall of 250 mm expanded polystyrene and 25 mm concrete separates a room at 22 °C from outside air at -10 °C. The convection coefficients are 10 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 6 m². Find the temperature Ts of the inside surface of the wall.
L1 = 250 mm, k1 = 0.035 W/(m·K)L2 = 25 mm, k2 = 1.4 W/(m·K)A = 6 m²hi = 10 W/(m²·K)ho = 30 W/(m²·K)Troom = 22 °CTout = -10 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 460 kW. Its heat transfer area is 6 m². The hot fluid cools from 95 to 45 °C and the cold fluid heats from 15 to 80 °C. Find the overall heat transfer coefficient U.
Q̇ = 460 kWA = 6 m²Th: 95 → 45 °CTc: 15 → 80 °C
U = W/(m²·K)
4.Heat exchangers (LMTD)Hard
Steam condenses at a constant 105 °C on the outside of the tubes of a condenser with a heat transfer area of 3 m² and U = 1500 W/(m²·K). Cooling water (1 kg/s, cp = 4.18 kJ/(kg·K)) enters the tubes at 30 °C. Find the water outlet temperature Tc,out.
Ts = 105 °CA = 3 m²U = 1500 W/(m²·K)ṁ = 1 kg/scp = 4.18 kJ/(kg·K)Tc,in = 30 °C
Tc,out = °C
3Solutions
1.Heat conduction
Answer: Q̇ = 752.6 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) = (250/1000)/(1.4·12) = 0.01488 K/W
R2 = L2/(k2A) = (150/1000)/(0.96·12) = 0.01302 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 21/0.0279 = 752.6 W
Answer
Q̇ = 752.6 W
2.Heat conduction
Answer: Ts = 21.56 °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/(10·6) = 0.01667 K/W
R1 = L1/(k1A) = (250/1000)/(0.035·6) = 1.19 K/W
R2 = L2/(k2A) = (25/1000)/(1.4·6) = 0.002976 K/W
Ro = 1/(hoA) = 1/(30·6) = 0.005556 K/W
ΣR = 1.216 K/W
Heat loss
Q̇ = (22 − (-10))/1.216 = 26.32 W
Ts = Troom − Q̇·Ri = 22 − 26.32·0.01667 = 21.56 °C
Answer
Ts = 21.56 °C
3.Heat exchangers (LMTD)
Answer: U = 3543 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 = 95 − 80 = 15 K
ΔT₂ = Th,out − Tc,in = 45 − 15 = 30 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (15 − 30)/ln(15/30) = 21.64 K
Overall coefficient
U = Q̇/(A·ΔTlm) = 460000/(6·21.64) = 3543 W/(m²·K)
Answer
U = 3543 W/(m²·K)
4.Heat exchangers (LMTD)
Answer: Tc,out = 79.44 °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) = 1500·3/(1·4180) = 1.077
Outlet temperature
ΔT₂ = ΔT₁·e−1.077 = 75·0.3408 = 25.56 K
Tc,out = 105 − 25.56 = 79.44 °C
Answer
Tc,out = 79.44 °C
Heat transferred: Q̇ = 206.7 kW.
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