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 100 mm of mineral wool and 100 mm of expanded polystyrene, with an area of 6 m². The outer faces are at 24 °C and 5 °C. Find the heat flow Q̇ through the wall.
L1 = 100 mm, k1 = 0.04 W/(m·K)L2 = 100 mm, k2 = 0.035 W/(m·K)A = 6 m²T1 = 24 °CT3 = 5 °C
Q̇ = W
2.Heat conductionHard
A wall of 250 mm expanded polystyrene and 50 mm wood separates a room at 22 °C from outside air at 3 °C. The convection coefficients are 7.7 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 12 m². Find the heat loss Q̇ through the wall.
L1 = 250 mm, k1 = 0.035 W/(m·K)L2 = 50 mm, k2 = 0.17 W/(m·K)A = 12 m²hi = 7.7 W/(m²·K)ho = 20 W/(m²·K)Troom = 22 °CTout = 3 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A shell-and-tube heat exchanger (one shell pass, two tube passes) must transfer 110 kW. The hot fluid cools from 120 to 80 °C and the cold fluid heats from 10 to 80 °C. The LMTD correction factor for this arrangement is F = 0.8, and U = 800 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 110 kWF = 0.8U = 800 W/(m²·K)Th: 120 → 80 °CTc: 10 → 80 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a counter-flow double-pipe heat exchanger, hot oil (1 kg/s, cp = 2.3 kJ/(kg·K)) flows in the inner tube and cools from 130 to 75 °C. Water (0.5 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 10 °C. The inner tube has a diameter of 50 mm, and U = 500 W/(m²·K) is based on this diameter. Find the tube length L needed.
ṁh = 1 kg/scp,h = 2.3 kJ/(kg·K)Th: 130 → 75 °Cṁc = 0.5 kg/sTc,in = 10 °CD = 50 mmU = 500 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Q̇ = 21.28 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.
Layer resistances
R1 = L1/(k1A) = (100/1000)/(0.04·6) = 0.4167 K/W
R2 = L2/(k2A) = (100/1000)/(0.035·6) = 0.4762 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 19/0.8929 = 21.28 W
Answer
Q̇ = 21.28 W
2.Heat conduction
Answer: Q̇ = 29.93 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/(7.7·12) = 0.01082 K/W
R1 = L1/(k1A) = (250/1000)/(0.035·12) = 0.5952 K/W
R2 = L2/(k2A) = (50/1000)/(0.17·12) = 0.02451 K/W
Ro = 1/(hoA) = 1/(20·12) = 0.004167 K/W
ΣR = 0.6347 K/W
Heat loss
Q̇ = (22 − (3))/0.6347 = 29.93 W
Answer
Q̇ = 29.93 W
3.Heat exchangers (LMTD)
Answer: A = 3.206 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 (counter-flow)
In counter-flow the hot inlet faces the cold outlet, and the hot outlet faces the cold inlet.
ΔT₁ = Th,in − Tc,out = 120 − 80 = 40 K
ΔT₂ = Th,out − Tc,in = 80 − 10 = 70 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (40 − 70)/ln(40/70) = 53.61 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 110000/(800·0.8·53.61) = 3.206 m²
Answer
A = 3.206 m²
4.Heat exchangers (LMTD)
Answer: L = 25.9 m
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 = 1·2.3·(130 − 75) = 126.5 kW
Tc,out = 10 + 126.5/(0.5·4.18) = 70.53 °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 = 130 − 70.53 = 59.47 K
ΔT₂ = Th,out − Tc,in = 75 − 10 = 65 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (59.47 − 65)/ln(59.47/65) = 62.2 K
Area and length
A = 126500/(500·62.2) = 4.068 m²
L = A/(πD) = 4.068/(π·0.05) = 25.9 m
Answer
L = 25.9 m
ExamProblems · examproblems.com/courses/heat-transfer