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 gypsum board and 75 mm of brick, with an area of 6 m². The outer faces are at 15 °C and -15 °C. Find the heat flow Q̇ through the wall.
L1 = 200 mm, k1 = 0.17 W/(m·K)L2 = 75 mm, k2 = 0.72 W/(m·K)A = 6 m²T1 = 15 °CT3 = -15 °C
Q̇ = W
2.Heat conductionHard
A wall of 50 mm concrete and 50 mm expanded polystyrene separates a room at 22 °C from outside air at -14 °C. The convection coefficients are 6 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 10 m². Find the temperature Ts of the inside surface of the wall.
L1 = 50 mm, k1 = 1.4 W/(m·K)L2 = 50 mm, k2 = 0.035 W/(m·K)A = 10 m²hi = 6 W/(m²·K)ho = 30 W/(m²·K)Troom = 22 °CTout = -14 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 320 kW. The hot fluid cools from 95 to 80 °C and the cold fluid heats from 25 to 55 °C. The overall heat transfer coefficient is 200 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 320 kWU = 200 W/(m²·K)Th: 95 → 80 °CTc: 25 → 55 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a counter-flow double-pipe heat exchanger, hot oil (0.3 kg/s, cp = 2.5 kJ/(kg·K)) flows in the inner tube and cools from 90 to 60 °C. Water (1 kg/s, cp = 4.18 kJ/(kg·K)) in the annulus enters at 25 °C. The inner tube has a diameter of 50 mm, and U = 200 W/(m²·K) is based on this diameter. Find the tube length L needed.
ṁh = 0.3 kg/scp,h = 2.5 kJ/(kg·K)Th: 90 → 60 °Cṁc = 1 kg/sTc,in = 25 °CD = 50 mmU = 200 W/(m²·K)
L = m
3Solutions
1.Heat conduction
Answer: Q̇ = 140.6 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)/(0.17·6) = 0.1961 K/W
R2 = L2/(k2A) = (75/1000)/(0.72·6) = 0.01736 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 30/0.2134 = 140.6 W
Answer
Q̇ = 140.6 W
2.Heat conduction
Answer: Ts = 18.39 °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/(6·10) = 0.01667 K/W
R1 = L1/(k1A) = (50/1000)/(1.4·10) = 0.003571 K/W
R2 = L2/(k2A) = (50/1000)/(0.035·10) = 0.1429 K/W
Ro = 1/(hoA) = 1/(30·10) = 0.003333 K/W
ΣR = 0.1664 K/W
Heat loss
Q̇ = (22 − (-14))/0.1664 = 216.3 W
Ts = Troom − Q̇·Ri = 22 − 216.3·0.01667 = 18.39 °C
Answer
Ts = 18.39 °C
3.Heat exchangers (LMTD)
Answer: A = 33.97 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 = 95 − 55 = 40 K
ΔT₂ = Th,out − Tc,in = 80 − 25 = 55 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (40 − 55)/ln(40/55) = 47.1 K
Area
A = Q̇/(U·ΔTlm) = 320000/(200·47.1) = 33.97 m²
Answer
A = 33.97 m²
4.Heat exchangers (LMTD)
Answer: L = 15.5 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 = 0.3·2.5·(90 − 60) = 22.5 kW
Tc,out = 25 + 22.5/(1·4.18) = 30.38 °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 = 90 − 30.38 = 59.62 K
ΔT₂ = Th,out − Tc,in = 60 − 25 = 35 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (59.62 − 35)/ln(59.62/35) = 46.22 K
Area and length
A = 22500/(200·46.22) = 2.434 m²
L = A/(πD) = 2.434/(π·0.05) = 15.5 m
Answer
L = 15.5 m
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