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 50 mm of expanded polystyrene, with an area of 20 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 = 1.4 W/(m·K)L2 = 50 mm, k2 = 0.035 W/(m·K)A = 20 m²T1 = 25 °CT3 = -14 °C
Ti = °C
2.Heat conductionHard
A wall of 150 mm brick and 100 mm gypsum board separates a room at 22 °C from outside air at -12 °C. The convection coefficients are 10 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 15 m². Find the heat loss Q̇ through the wall.
L1 = 150 mm, k1 = 0.72 W/(m·K)L2 = 100 mm, k2 = 0.17 W/(m·K)A = 15 m²hi = 10 W/(m²·K)ho = 25 W/(m²·K)Troom = 22 °CTout = -12 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A parallel-flow heat exchanger transfers 70 kW. The hot fluid cools from 120 to 80 °C and the cold fluid heats from 25 to 50 °C. The overall heat transfer coefficient is 1500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 70 kWU = 1500 W/(m²·K)Th: 120 → 80 °CTc: 25 → 50 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1 kg/s, cp = 2.5 kJ/(kg·K)) cools from 145 to 85 °C in a counter-flow exchanger. It heats 1 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 20 °C. With U = 300 W/(m²·K), find the heat transfer area A needed.
ṁh = 1 kg/scp,h = 2.5 kJ/(kg·K)Th: 145 → 85 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 20 °CU = 300 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = 21.45 °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)/(1.4·20) = 0.007143 K/W
R2 = L2/(k2A) = (50/1000)/(0.035·20) = 0.07143 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 39/0.07857 = 496.4 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 25 − 496.4·0.007143 = 21.45 °C
Answer
Ti = 21.45 °C
2.Heat conduction
Answer: Q̇ = 544.5 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.
Resistances in series
Ri = 1/(hiA) = 1/(10·15) = 0.006667 K/W
R1 = L1/(k1A) = (150/1000)/(0.72·15) = 0.01389 K/W
R2 = L2/(k2A) = (100/1000)/(0.17·15) = 0.03922 K/W
Ro = 1/(hoA) = 1/(25·15) = 0.002667 K/W
ΣR = 0.06244 K/W
Heat loss
Q̇ = (22 − (-12))/0.06244 = 544.5 W
Answer
Q̇ = 544.5 W
3.Heat exchangers (LMTD)
Answer: A = 0.8276 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 = 120 − 25 = 95 K
ΔT₂ = Th,out − Tc,out = 80 − 50 = 30 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (95 − 30)/ln(95/30) = 56.39 K
Area
A = Q̇/(U·ΔTlm) = 70000/(1500·56.39) = 0.8276 m²
Answer
A = 0.8276 m²
4.Heat exchangers (LMTD)
Answer: A = 6.542 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) = 1·2.5·60 = 150 kW
Tc,out = 20 + 150/(1·4.18) = 55.89 °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 = 145 − 55.89 = 89.11 K
ΔT₂ = Th,out − Tc,in = 85 − 20 = 65 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (89.11 − 65)/ln(89.11/65) = 76.42 K
Area
A = Q̇/(U·ΔTlm) = 150000/(300·76.42) = 6.542 m²
Answer
A = 6.542 m²
ExamProblems · examproblems.com/courses/heat-transfer