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 50 mm of wood and 100 mm of mineral wool, with an area of 6 m². The outer faces are at 16 °C and -5 °C. Find the heat flow Q̇ through the wall.
L1 = 50 mm, k1 = 0.17 W/(m·K)L2 = 100 mm, k2 = 0.04 W/(m·K)A = 6 m²T1 = 16 °CT3 = -5 °C
Q̇ = W
2.Heat conductionHard
A wall of 150 mm mineral wool and 75 mm concrete separates a room at 21 °C from outside air at 5 °C. The convection coefficients are 6 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 4 m². Find the temperature Ts of the inside surface of the wall.
L1 = 150 mm, k1 = 0.04 W/(m·K)L2 = 75 mm, k2 = 1.4 W/(m·K)A = 4 m²hi = 6 W/(m²·K)ho = 20 W/(m²·K)Troom = 21 °CTout = 5 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 150 kW. The hot fluid cools from 110 to 80 °C and the cold fluid heats from 20 to 80 °C. The overall heat transfer coefficient is 1500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 150 kWU = 1500 W/(m²·K)Th: 110 → 80 °CTc: 20 → 80 °C
A = m²
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 95 °C and the cold stream at 30 °C. The exchanger has A = 1.5 m² and U = 800 W/(m²·K). Find the hot outlet temperature Th,out.
Ch = Cc = 1.5 kW/KTh,in = 95 °CTc,in = 30 °CA = 1.5 m²U = 800 W/(m²·K)
Th,out = °C
3Solutions
1.Heat conduction
Answer: Q̇ = 45.09 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) = (50/1000)/(0.17·6) = 0.04902 K/W
R2 = L2/(k2A) = (100/1000)/(0.04·6) = 0.4167 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 21/0.4657 = 45.09 W
Answer
Q̇ = 45.09 W
2.Heat conduction
Answer: Ts = 20.34 °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/(6·4) = 0.04167 K/W
R1 = L1/(k1A) = (150/1000)/(0.04·4) = 0.9375 K/W
R2 = L2/(k2A) = (75/1000)/(1.4·4) = 0.01339 K/W
Ro = 1/(hoA) = 1/(20·4) = 0.0125 K/W
ΣR = 1.005 K/W
Heat loss
Q̇ = (21 − (5))/1.005 = 15.92 W
Ts = Troom − Q̇·Ri = 21 − 15.92·0.04167 = 20.34 °C
Answer
Ts = 20.34 °C
3.Heat exchangers (LMTD)
Answer: A = 2.31 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 = 110 − 80 = 30 K
ΔT₂ = Th,out − Tc,in = 80 − 20 = 60 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (30 − 60)/ln(30/60) = 43.28 K
Area
A = Q̇/(U·ΔTlm) = 150000/(1500·43.28) = 2.31 m²
Answer
A = 2.31 m²
4.Heat exchangers (LMTD)
Answer: Th,out = 66.11 °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·(95 − Th,out) = 1.2·(Th,out − 30)
Th,out = (1.5·95 + 1.2·30)/(1.5 + 1.2) = 66.11 °C
Answer
Th,out = 66.11 °C
Heat transferred: Q̇ = 43.33 kW; the cold stream leaves at 58.89 °C.
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