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 wood and 25 mm of brick, with an area of 10 m². The outer faces are at 25 °C and -14 °C. Find the temperature Ti at the interface between the two layers.
L1 = 200 mm, k1 = 0.17 W/(m·K)L2 = 25 mm, k2 = 0.72 W/(m·K)A = 10 m²T1 = 25 °CT3 = -14 °C
Ti = °C
2.Heat conductionHard
A wall of 100 mm glass and 25 mm gypsum board separates a room at 20 °C from outside air at 1 °C. The convection coefficients are 7.7 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 = 100 mm, k1 = 0.96 W/(m·K)L2 = 25 mm, k2 = 0.17 W/(m·K)A = 10 m²hi = 7.7 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = 1 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 370 kW. Its heat transfer area is 2.5 m². The hot fluid cools from 130 to 95 °C and the cold fluid heats from 15 to 80 °C. Find the overall heat transfer coefficient U.
Q̇ = 370 kWA = 2.5 m²Th: 130 → 95 °CTc: 15 → 80 °C
U = W/(m²·K)
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 4 m² and U = 3000 W/(m²·K). Cooling water (1.5 kg/s, cp = 4.18 kJ/(kg·K)) enters the tubes at 20 °C. Find the water outlet temperature Tc,out.
Ts = 100 °CA = 4 m²U = 3000 W/(m²·K)ṁ = 1.5 kg/scp = 4.18 kJ/(kg·K)Tc,in = 20 °C
Tc,out = °C
3Solutions
1.Heat conduction
Answer: Ti = −12.88 °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)/(0.17·10) = 0.1176 K/W
R2 = L2/(k2A) = (25/1000)/(0.72·10) = 0.003472 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 39/0.1211 = 322 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 25 − 322·0.1176 = −12.88 °C
Answer
Ti = −12.88 °C
2.Heat conduction
Answer: Ts = 14.14 °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/(7.7·10) = 0.01299 K/W
R1 = L1/(k1A) = (100/1000)/(0.96·10) = 0.01042 K/W
R2 = L2/(k2A) = (25/1000)/(0.17·10) = 0.01471 K/W
Ro = 1/(hoA) = 1/(25·10) = 0.004 K/W
ΣR = 0.04211 K/W
Heat loss
Q̇ = (20 − (1))/0.04211 = 451.2 W
Ts = Troom − Q̇·Ri = 20 − 451.2·0.01299 = 14.14 °C
Answer
Ts = 14.14 °C
3.Heat exchangers (LMTD)
Answer: U = 2319 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 = 130 − 80 = 50 K
ΔT₂ = Th,out − Tc,in = 95 − 15 = 80 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (50 − 80)/ln(50/80) = 63.83 K
Overall coefficient
U = Q̇/(A·ΔTlm) = 370000/(2.5·63.83) = 2319 W/(m²·K)
Answer
U = 2319 W/(m²·K)
4.Heat exchangers (LMTD)
Answer: Tc,out = 88.2 °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) = 3000·4/(1.5·4180) = 1.914
Outlet temperature
ΔT₂ = ΔT₁·e−1.914 = 80·0.1475 = 11.8 K
Tc,out = 100 − 11.8 = 88.2 °C
Answer
Tc,out = 88.2 °C
Heat transferred: Q̇ = 427.6 kW.
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