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 250 mm of glass and 75 mm of expanded polystyrene, with an area of 8 m². The outer faces are at 19 °C and -1 °C. Find the heat flow Q̇ through the wall.
L1 = 250 mm, k1 = 0.96 W/(m·K)L2 = 75 mm, k2 = 0.035 W/(m·K)A = 8 m²T1 = 19 °CT3 = -1 °C
Q̇ = W
2.Heat conductionHard
A wall of 250 mm expanded polystyrene and 25 mm gypsum board separates a room at 21 °C from outside air at -11 °C. The convection coefficients are 10 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 4 m². Find the heat loss Q̇ through the wall.
L1 = 250 mm, k1 = 0.035 W/(m·K)L2 = 25 mm, k2 = 0.17 W/(m·K)A = 4 m²hi = 10 W/(m²·K)ho = 20 W/(m²·K)Troom = 21 °CTout = -11 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 220 kW. Its heat transfer area is 6 m². The hot fluid cools from 120 to 70 °C and the cold fluid heats from 15 to 35 °C. Find the overall heat transfer coefficient U.
Q̇ = 220 kWA = 6 m²Th: 120 → 70 °CTc: 15 → 35 °C
U = W/(m²·K)
4.Heat exchangers (LMTD)Hard
In a counter-flow heat exchanger both streams have the same heat capacity rate, ṁ·cp = 1.5 kW/K. The hot stream enters at 80 °C and the cold stream at 10 °C. The exchanger has A = 2 m² and U = 800 W/(m²·K). Find the hot outlet temperature Th,out.
Ch = Cc = 1.5 kW/KTh,in = 80 °CTc,in = 10 °CA = 2 m²U = 800 W/(m²·K)
Th,out = °C
3Solutions
1.Heat conduction
Answer: Q̇ = 66.58 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) = (250/1000)/(0.96·8) = 0.03255 K/W
R2 = L2/(k2A) = (75/1000)/(0.035·8) = 0.2679 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 20/0.3004 = 66.58 W
Answer
Q̇ = 66.58 W
2.Heat conduction
Answer: Q̇ = 17.2 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/(10·4) = 0.025 K/W
R1 = L1/(k1A) = (250/1000)/(0.035·4) = 1.786 K/W
R2 = L2/(k2A) = (25/1000)/(0.17·4) = 0.03676 K/W
Ro = 1/(hoA) = 1/(20·4) = 0.0125 K/W
ΣR = 1.86 K/W
Heat loss
Q̇ = (21 − (-11))/1.86 = 17.2 W
Answer
Q̇ = 17.2 W
3.Heat exchangers (LMTD)
Answer: U = 532.1 W/(m²·K)
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 = 120 − 35 = 85 K
ΔT₂ = Th,out − Tc,in = 70 − 15 = 55 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (85 − 55)/ln(85/55) = 68.92 K
Overall coefficient
U = Q̇/(A·ΔTlm) = 220000/(6·68.92) = 532.1 W/(m²·K)
Answer
U = 532.1 W/(m²·K)
4.Heat exchangers (LMTD)
Answer: Th,out = 43.87 °C
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.
Constant temperature difference
The cold outlet sits opposite the hot inlet: Tc,out = Tc,in + (Th,in − Th,out), so ΔT₁ = Th,in − Tc,out = Th,out − Tc,in = ΔT₂.
ΔTlm = Th,out − Tc,in
Energy balance equals rate equation
1.5·(80 − Th,out) = 1.6·(Th,out − 10)
Th,out = (1.5·80 + 1.6·10)/(1.5 + 1.6) = 43.87 °C
Answer
Th,out = 43.87 °C
Heat transferred: Q̇ = 54.19 kW; the cold stream leaves at 46.13 °C.
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