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 brick and 100 mm of wood, with an area of 4 m². The outer faces are at 24 °C and -11 °C. Find the temperature Ti at the interface between the two layers.
L1 = 100 mm, k1 = 0.72 W/(m·K)L2 = 100 mm, k2 = 0.17 W/(m·K)A = 4 m²T1 = 24 °CT3 = -11 °C
Ti = °C
2.Heat conductionHard
A wall of 250 mm mineral wool and 25 mm brick separates a room at 22 °C from outside air at -2 °C. The convection coefficients are 10 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 4 m². Find the temperature Ts of the inside surface of the wall.
L1 = 250 mm, k1 = 0.04 W/(m·K)L2 = 25 mm, k2 = 0.72 W/(m·K)A = 4 m²hi = 10 W/(m²·K)ho = 25 W/(m²·K)Troom = 22 °CTout = -2 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 440 kW. The hot fluid cools from 130 to 80 °C and the cold fluid heats from 30 to 60 °C. The overall heat transfer coefficient is 1000 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 440 kWU = 1000 W/(m²·K)Th: 130 → 80 °CTc: 30 → 60 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a counter-flow double-pipe heat exchanger, hot oil (0.5 kg/s, cp = 2.3 kJ/(kg·K)) flows in the inner tube and cools from 145 to 75 °C. Water (0.5 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 15 °C. The inner tube has a diameter of 20 mm, and U = 300 W/(m²·K) is based on this diameter. Find the tube length L needed.
ṁh = 0.5 kg/scp,h = 2.3 kJ/(kg·K)Th: 145 → 75 °Cṁc = 0.5 kg/sTc,in = 15 °CD = 20 mmU = 300 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Ti = 17.31 °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) = (100/1000)/(0.72·4) = 0.03472 K/W
R2 = L2/(k2A) = (100/1000)/(0.17·4) = 0.1471 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 35/0.1818 = 192.5 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 24 − 192.5·0.03472 = 17.31 °C
Answer
Ti = 17.31 °C
2.Heat conduction
Answer: Ts = 21.63 °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/(10·4) = 0.025 K/W
R1 = L1/(k1A) = (250/1000)/(0.04·4) = 1.563 K/W
R2 = L2/(k2A) = (25/1000)/(0.72·4) = 0.008681 K/W
Ro = 1/(hoA) = 1/(25·4) = 0.01 K/W
ΣR = 1.606 K/W
Heat loss
Q̇ = (22 − (-2))/1.606 = 14.94 W
Ts = Troom − Q̇·Ri = 22 − 14.94·0.025 = 21.63 °C
Answer
Ts = 21.63 °C
3.Heat exchangers (LMTD)
Answer: A = 7.402 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 counter-flow the hot inlet faces the cold outlet, and the hot outlet faces the cold inlet.
ΔT₁ = Th,in − Tc,out = 130 − 60 = 70 K
ΔT₂ = Th,out − Tc,in = 80 − 30 = 50 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (70 − 50)/ln(70/50) = 59.44 K
Area
A = Q̇/(U·ΔTlm) = 440000/(1000·59.44) = 7.402 m²
Answer
A = 7.402 m²
4.Heat exchangers (LMTD)
Answer: L = 57.22 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 = 0.5·2.3·(145 − 75) = 80.5 kW
Tc,out = 15 + 80.5/(0.5·4.18) = 53.52 °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 − 53.52 = 91.48 K
ΔT₂ = Th,out − Tc,in = 75 − 15 = 60 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (91.48 − 60)/ln(91.48/60) = 74.64 K
Area and length
A = 80500/(300·74.64) = 3.595 m²
L = A/(πD) = 3.595/(π·0.02) = 57.22 m
Answer
L = 57.22 m
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