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 wood and 50 mm of brick, with an area of 15 m². The outer faces are at 20 °C and -5 °C. Find the temperature Ti at the interface between the two layers.
L1 = 150 mm, k1 = 0.17 W/(m·K)L2 = 50 mm, k2 = 0.72 W/(m·K)A = 15 m²T1 = 20 °CT3 = -5 °C
Ti = °C
2.Heat conductionHard
A wall of 250 mm concrete and 25 mm wood separates a room at 20 °C from outside air at -8 °C. The convection coefficients are 7.7 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 15 m². Find the overall heat transfer coefficient U of the wall.
L1 = 250 mm, k1 = 1.4 W/(m·K)L2 = 25 mm, k2 = 0.17 W/(m·K)A = 15 m²hi = 7.7 W/(m²·K)ho = 30 W/(m²·K)Troom = 20 °CTout = -8 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 90 kW. The hot fluid cools from 125 to 95 °C and the cold fluid heats from 10 to 105 °C. The overall heat transfer coefficient is 1500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 90 kWU = 1500 W/(m²·K)Th: 125 → 95 °CTc: 10 → 105 °C
A = m²
4.Heat exchangers (LMTD)Hard
Steam condenses at a constant 120 °C on the outside of the tubes of a condenser with a heat transfer area of 2 m² and U = 2000 W/(m²·K). Cooling water (0.5 kg/s, cp = 4.18 kJ/(kg·K)) enters the tubes at 25 °C. Find the water outlet temperature Tc,out.
Ts = 120 °CA = 2 m²U = 2000 W/(m²·K)ṁ = 0.5 kg/scp = 4.18 kJ/(kg·K)Tc,in = 25 °C
Tc,out = °C
3Solutions
1.Heat conduction
Answer: Ti = −3.176 °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.
Layer resistances
R1 = L1/(k1A) = (150/1000)/(0.17·15) = 0.05882 K/W
R2 = L2/(k2A) = (50/1000)/(0.72·15) = 0.00463 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 25/0.06345 = 394 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 20 − 394·0.05882 = −3.176 °C
Answer
Ti = −3.176 °C
2.Heat conduction
Answer: U = 2.046 W/(m²·K)
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) = (250/1000)/(1.4·15) = 0.0119 K/W
R2 = L2/(k2A) = (25/1000)/(0.17·15) = 0.009804 K/W
Ro = 1/(hoA) = 1/(30·15) = 0.002222 K/W
ΣR = 0.03259 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(15·0.03259) = 2.046 W/(m²·K)
Answer
U = 2.046 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 1.336 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 = 125 − 105 = 20 K
ΔT₂ = Th,out − Tc,in = 95 − 10 = 85 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (20 − 85)/ln(20/85) = 44.92 K
Area
A = Q̇/(U·ΔTlm) = 90000/(1500·44.92) = 1.336 m²
Answer
A = 1.336 m²
4.Heat exchangers (LMTD)
Answer: Tc,out = 106 °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) = 2000·2/(0.5·4180) = 1.914
Outlet temperature
ΔT₂ = ΔT₁·e−1.914 = 95·0.1475 = 14.01 K
Tc,out = 120 − 14.01 = 106 °C
Answer
Tc,out = 106 °C
Heat transferred: Q̇ = 169.3 kW.
ExamProblems · examproblems.com/courses/heat-transfer