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 expanded polystyrene and 100 mm of gypsum board, with an area of 4 m². The outer faces are at 21 °C and 1 °C. Find the heat flow Q̇ through the wall.
L1 = 200 mm, k1 = 0.035 W/(m·K)L2 = 100 mm, k2 = 0.17 W/(m·K)A = 4 m²T1 = 21 °CT3 = 1 °C
Q̇ = W
2.Heat conductionHard
A wall of 250 mm glass and 100 mm wood separates a room at 22 °C from outside air at -15 °C. The convection coefficients are 6 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 15 m². Find the heat loss Q̇ through the wall.
L1 = 250 mm, k1 = 0.96 W/(m·K)L2 = 100 mm, k2 = 0.17 W/(m·K)A = 15 m²hi = 6 W/(m²·K)ho = 20 W/(m²·K)Troom = 22 °CTout = -15 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
In a condenser, steam condenses at a constant 100 °C on the outside of the tubes. Cooling water (1 kg/s, cp = 4.18 kJ/(kg·K)) flows through the tubes and heats from 10 to 30 °C. The overall heat transfer coefficient is 2000 W/(m²·K). Find the heat transfer area A needed.
Ts = 100 °Cṁ = 1 kg/scp = 4.18 kJ/(kg·K)Tc: 10 → 30 °CU = 2000 W/(m²·K)
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (0.5 kg/s, cp = 2.1 kJ/(kg·K)) cools from 120 to 80 °C in a counter-flow exchanger. It heats 3 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 30 °C. With U = 1000 W/(m²·K), find the heat transfer area A needed.
ṁh = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 120 → 80 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 30 °CU = 1000 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 12.69 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) = (200/1000)/(0.035·4) = 1.429 K/W
R2 = L2/(k2A) = (100/1000)/(0.17·4) = 0.1471 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 20/1.576 = 12.69 W
Answer
Q̇ = 12.69 W
2.Heat conduction
Answer: Q̇ = 521 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/(6·15) = 0.01111 K/W
R1 = L1/(k1A) = (250/1000)/(0.96·15) = 0.01736 K/W
R2 = L2/(k2A) = (100/1000)/(0.17·15) = 0.03922 K/W
Ro = 1/(hoA) = 1/(20·15) = 0.003333 K/W
ΣR = 0.07102 K/W
Heat loss
Q̇ = (22 − (-15))/0.07102 = 521 W
Answer
Q̇ = 521 W
3.Heat exchangers (LMTD)
Answer: A = 0.5252 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 = 1·4.18·(30 − 10) = 83.6 kW
End temperature differences
ΔT₁ = 100 − 10 = 90 K
ΔT₂ = 100 − 30 = 70 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (90 − 70)/ln(90/70) = 79.58 K
Area
A = Q̇/(U·ΔTlm) = 83600/(2000·79.58) = 0.5252 m²
Answer
A = 0.5252 m²
4.Heat exchangers (LMTD)
Answer: A = 0.6301 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) = 0.5·2.1·40 = 42 kW
Tc,out = 30 + 42/(3·4.18) = 33.35 °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 = 120 − 33.35 = 86.65 K
ΔT₂ = Th,out − Tc,in = 80 − 30 = 50 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (86.65 − 50)/ln(86.65/50) = 66.65 K
Area
A = Q̇/(U·ΔTlm) = 42000/(1000·66.65) = 0.6301 m²
Answer
A = 0.6301 m²
ExamProblems · examproblems.com/courses/heat-transfer