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 200 mm of concrete and 75 mm of wood, with an area of 6 m². The outer faces are at 22 °C and -5 °C. Find the temperature Ti at the interface between the two layers.
L1 = 200 mm, k1 = 1.4 W/(m·K)L2 = 75 mm, k2 = 0.17 W/(m·K)A = 6 m²T1 = 22 °CT3 = -5 °C
Ti = °C
2.Heat conductionHard
A wall of 100 mm expanded polystyrene and 150 mm mineral wool separates a room at 20 °C from outside air at -3 °C. The convection coefficients are 6 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 4 m². Find the overall heat transfer coefficient U of the wall.
L1 = 100 mm, k1 = 0.035 W/(m·K)L2 = 150 mm, k2 = 0.04 W/(m·K)A = 4 m²hi = 6 W/(m²·K)ho = 20 W/(m²·K)Troom = 20 °CTout = -3 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 170 kW. The hot fluid cools from 165 to 150 °C and the cold fluid heats from 30 to 150 °C. The overall heat transfer coefficient is 200 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 170 kWU = 200 W/(m²·K)Th: 165 → 150 °CTc: 30 → 150 °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 2 m² and U = 2000 W/(m²·K). Cooling water (1.5 kg/s, cp = 4.18 kJ/(kg·K)) enters the tubes at 25 °C. Find the water outlet temperature Tc,out.
Ts = 110 °CA = 2 m²U = 2000 W/(m²·K)ṁ = 1.5 kg/scp = 4.18 kJ/(kg·K)Tc,in = 25 °C
Tc,out = °C
3Solutions
1.Heat conduction
Answer: Ti = 15.4 °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) = (200/1000)/(1.4·6) = 0.02381 K/W
R2 = L2/(k2A) = (75/1000)/(0.17·6) = 0.07353 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 27/0.09734 = 277.4 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 22 − 277.4·0.02381 = 15.4 °C
Answer
Ti = 15.4 °C
2.Heat conduction
Answer: U = 0.1465 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/(6·4) = 0.04167 K/W
R1 = L1/(k1A) = (100/1000)/(0.035·4) = 0.7143 K/W
R2 = L2/(k2A) = (150/1000)/(0.04·4) = 0.9375 K/W
Ro = 1/(hoA) = 1/(20·4) = 0.0125 K/W
ΣR = 1.706 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(4·1.706) = 0.1465 W/(m²·K)
Answer
U = 0.1465 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 16.83 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 = 165 − 150 = 15 K
ΔT₂ = Th,out − Tc,in = 150 − 30 = 120 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (15 − 120)/ln(15/120) = 50.49 K
Area
A = Q̇/(U·ΔTlm) = 170000/(200·50.49) = 16.83 m²
Answer
A = 16.83 m²
4.Heat exchangers (LMTD)
Answer: Tc,out = 65.09 °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/(1.5·4180) = 0.638
Outlet temperature
ΔT₂ = ΔT₁·e−0.638 = 85·0.5284 = 44.91 K
Tc,out = 110 − 44.91 = 65.09 °C
Answer
Tc,out = 65.09 °C
Heat transferred: Q̇ = 251.4 kW.
ExamProblems · examproblems.com/courses/heat-transfer