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 150 mm of glass and 100 mm of concrete, with an area of 8 m². The outer faces are at 24 °C and -14 °C. Find the heat flow Q̇ through the wall.
L1 = 150 mm, k1 = 0.96 W/(m·K)L2 = 100 mm, k2 = 1.4 W/(m·K)A = 8 m²T1 = 24 °CT3 = -14 °C
Q̇ = W
2.Heat conductionHard
A wall of 50 mm expanded polystyrene and 50 mm brick separates a room at 22 °C from outside air at 4 °C. The convection coefficients are 7.7 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 6 m². Find the overall heat transfer coefficient U of the wall.
L1 = 50 mm, k1 = 0.035 W/(m·K)L2 = 50 mm, k2 = 0.72 W/(m·K)A = 6 m²hi = 7.7 W/(m²·K)ho = 20 W/(m²·K)Troom = 22 °CTout = 4 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
In a condenser, steam condenses at a constant 140 °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 15 to 90 °C. The overall heat transfer coefficient is 3500 W/(m²·K). Find the heat transfer area A needed.
Ts = 140 °Cṁ = 2 kg/scp = 4.18 kJ/(kg·K)Tc: 15 → 90 °CU = 3500 W/(m²·K)
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1 kg/s, cp = 2.1 kJ/(kg·K)) cools from 105 to 70 °C in a counter-flow exchanger. It heats 2 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 25 °C. With U = 500 W/(m²·K), find the heat transfer area A needed.
ṁh = 1 kg/scp,h = 2.1 kJ/(kg·K)Th: 105 → 70 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 25 °CU = 500 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 1335 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) = (150/1000)/(0.96·8) = 0.01953 K/W
R2 = L2/(k2A) = (100/1000)/(1.4·8) = 0.008929 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 38/0.02846 = 1335 W
Answer
Q̇ = 1335 W
2.Heat conduction
Answer: U = 0.596 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/(7.7·6) = 0.02165 K/W
R1 = L1/(k1A) = (50/1000)/(0.035·6) = 0.2381 K/W
R2 = L2/(k2A) = (50/1000)/(0.72·6) = 0.01157 K/W
Ro = 1/(hoA) = 1/(20·6) = 0.008333 K/W
ΣR = 0.2796 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(6·0.2796) = 0.596 W/(m²·K)
Answer
U = 0.596 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 2.189 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·(90 − 15) = 627 kW
End temperature differences
ΔT₁ = 140 − 15 = 125 K
ΔT₂ = 140 − 90 = 50 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (125 − 50)/ln(125/50) = 81.85 K
Area
A = Q̇/(U·ΔTlm) = 627000/(3500·81.85) = 2.189 m²
Answer
A = 2.189 m²
4.Heat exchangers (LMTD)
Answer: A = 2.574 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) = 1·2.1·35 = 73.5 kW
Tc,out = 25 + 73.5/(2·4.18) = 33.79 °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 = 105 − 33.79 = 71.21 K
ΔT₂ = Th,out − Tc,in = 70 − 25 = 45 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (71.21 − 45)/ln(71.21/45) = 57.11 K
Area
A = Q̇/(U·ΔTlm) = 73500/(500·57.11) = 2.574 m²
Answer
A = 2.574 m²
ExamProblems · examproblems.com/courses/heat-transfer