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 150 mm of mineral wool, with an area of 10 m². The outer faces are at 23 °C and -11 °C. Find the heat flow Q̇ through the wall.
L1 = 150 mm, k1 = 1.4 W/(m·K)L2 = 150 mm, k2 = 0.04 W/(m·K)A = 10 m²T1 = 23 °CT3 = -11 °C
Q̇ = W
2.Heat conductionHard
A wall of 50 mm wood and 75 mm concrete separates a room at 20 °C from outside air at 3 °C. The convection coefficients are 10 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 10 m². Find the temperature Ts of the inside surface of the wall.
L1 = 50 mm, k1 = 0.17 W/(m·K)L2 = 75 mm, k2 = 1.4 W/(m·K)A = 10 m²hi = 10 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = 3 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A parallel-flow heat exchanger transfers 450 kW. The hot fluid cools from 150 to 130 °C and the cold fluid heats from 10 to 105 °C. The overall heat transfer coefficient is 200 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 450 kWU = 200 W/(m²·K)Th: 150 → 130 °CTc: 10 → 105 °C
A = m²
4.Heat exchangers (LMTD)Hard
Steam condenses at a constant 110 °C on the outside of the tubes of a condenser with a heat transfer area of 1 m² and U = 2500 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 = 110 °CA = 1 m²U = 2500 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 = 88.15 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)/(1.4·10) = 0.01071 K/W
R2 = L2/(k2A) = (150/1000)/(0.04·10) = 0.375 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 34/0.3857 = 88.15 W
Answer
Q̇ = 88.15 W
2.Heat conduction
Answer: Ts = 16.51 °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/(10·10) = 0.01 K/W
R1 = L1/(k1A) = (50/1000)/(0.17·10) = 0.02941 K/W
R2 = L2/(k2A) = (75/1000)/(1.4·10) = 0.005357 K/W
Ro = 1/(hoA) = 1/(25·10) = 0.004 K/W
ΣR = 0.04877 K/W
Heat loss
Q̇ = (20 − (3))/0.04877 = 348.6 W
Ts = Troom − Q̇·Ri = 20 − 348.6·0.01 = 16.51 °C
Answer
Ts = 16.51 °C
3.Heat exchangers (LMTD)
Answer: A = 33.71 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 parallel flow both inlets are at the same end, and both outlets at the other.
ΔT₁ = Th,in − Tc,in = 150 − 10 = 140 K
ΔT₂ = Th,out − Tc,out = 130 − 105 = 25 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (140 − 25)/ln(140/25) = 66.75 K
Area
A = Q̇/(U·ΔTlm) = 450000/(200·66.75) = 33.71 m²
Answer
A = 33.71 m²
4.Heat exchangers (LMTD)
Answer: Tc,out = 66.01 °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) = 2500·1/(1·4180) = 0.5981
Outlet temperature
ΔT₂ = ΔT₁·e−0.5981 = 80·0.5499 = 43.99 K
Tc,out = 110 − 43.99 = 66.01 °C
Answer
Tc,out = 66.01 °C
Heat transferred: Q̇ = 150.5 kW.
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