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 150 mm of expanded polystyrene, with an area of 20 m². The outer faces are at 23 °C and 8 °C. Find the temperature Ti at the interface between the two layers.
L1 = 200 mm, k1 = 0.17 W/(m·K)L2 = 150 mm, k2 = 0.035 W/(m·K)A = 20 m²T1 = 23 °CT3 = 8 °C
Ti = °C
2.Heat conductionHard
A wall of 250 mm concrete and 150 mm glass separates a room at 22 °C from outside air at -9 °C. The convection coefficients are 8 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 15 m². Find the temperature Ts of the inside surface of the wall.
L1 = 250 mm, k1 = 1.4 W/(m·K)L2 = 150 mm, k2 = 0.96 W/(m·K)A = 15 m²hi = 8 W/(m²·K)ho = 20 W/(m²·K)Troom = 22 °CTout = -9 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
In a condenser, steam condenses at a constant 100 °C on the outside of the tubes. Cooling water (2 kg/s, cp = 4.18 kJ/(kg·K)) flows through the tubes and heats from 30 to 80 °C. The overall heat transfer coefficient is 3500 W/(m²·K). Find the heat transfer area A needed.
Ts = 100 °Cṁ = 2 kg/scp = 4.18 kJ/(kg·K)Tc: 30 → 80 °CU = 3500 W/(m²·K)
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1.5 kg/s, cp = 2.5 kJ/(kg·K)) cools from 135 to 100 °C in a counter-flow exchanger. It heats 2 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 20 °C. With U = 800 W/(m²·K), find the heat transfer area A needed.
ṁh = 1.5 kg/scp,h = 2.5 kJ/(kg·K)Th: 135 → 100 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 20 °CU = 800 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = 19.77 °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)/(0.17·20) = 0.05882 K/W
R2 = L2/(k2A) = (150/1000)/(0.035·20) = 0.2143 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 15/0.2731 = 54.92 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 23 − 54.92·0.05882 = 19.77 °C
Answer
Ti = 19.77 °C
2.Heat conduction
Answer: Ts = 14.4 °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.
Resistances in series
Ri = 1/(hiA) = 1/(8·15) = 0.008333 K/W
R1 = L1/(k1A) = (250/1000)/(1.4·15) = 0.0119 K/W
R2 = L2/(k2A) = (150/1000)/(0.96·15) = 0.01042 K/W
Ro = 1/(hoA) = 1/(20·15) = 0.003333 K/W
ΣR = 0.03399 K/W
Heat loss
Q̇ = (22 − (-9))/0.03399 = 912.1 W
Ts = Troom − Q̇·Ri = 22 − 912.1·0.008333 = 14.4 °C
Answer
Ts = 14.4 °C
3.Heat exchangers (LMTD)
Answer: A = 2.992 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.
Heat duty from the water side
Q̇ = 2·4.18·(80 − 30) = 418 kW
End temperature differences
ΔT₁ = 100 − 30 = 70 K
ΔT₂ = 100 − 80 = 20 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (70 − 20)/ln(70/20) = 39.91 K
Area
A = Q̇/(U·ΔTlm) = 418000/(3500·39.91) = 2.992 m²
Answer
A = 2.992 m²
4.Heat exchangers (LMTD)
Answer: A = 1.837 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.
Heat duty and water outlet temperature
Q̇ = ṁhcp,h(Th,in − Th,out) = 1.5·2.5·35 = 131.3 kW
Tc,out = 20 + 131.3/(2·4.18) = 35.7 °C
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 = 135 − 35.7 = 99.3 K
ΔT₂ = Th,out − Tc,in = 100 − 20 = 80 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (99.3 − 80)/ln(99.3/80) = 89.3 K
Area
A = Q̇/(U·ΔTlm) = 131250/(800·89.3) = 1.837 m²
Answer
A = 1.837 m²
ExamProblems · examproblems.com/courses/heat-transfer