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 concrete and 100 mm of mineral wool, with an area of 4 m². The outer faces are at 25 °C and -13 °C. Find the heat flow Q̇ through the wall.
L1 = 250 mm, k1 = 1.4 W/(m·K)L2 = 100 mm, k2 = 0.04 W/(m·K)A = 4 m²T1 = 25 °CT3 = -13 °C
Q̇ = W
2.Heat conductionHard
A wall of 250 mm wood and 75 mm glass separates a room at 20 °C from outside air at -3 °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 temperature Ts of the inside surface of the wall.
L1 = 250 mm, k1 = 0.17 W/(m·K)L2 = 75 mm, k2 = 0.96 W/(m·K)A = 15 m²hi = 6 W/(m²·K)ho = 20 W/(m²·K)Troom = 20 °CTout = -3 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 100 kW. The hot fluid cools from 90 to 70 °C and the cold fluid heats from 10 to 50 °C. The overall heat transfer coefficient is 200 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 100 kWU = 200 W/(m²·K)Th: 90 → 70 °CTc: 10 → 50 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a counter-flow heat exchanger both streams have the same heat capacity rate, ṁ·cp = 0.5 kW/K. The hot stream enters at 180 °C and the cold stream at 25 °C. The exchanger has A = 0.5 m² and U = 300 W/(m²·K). Find the hot outlet temperature Th,out.
Ch = Cc = 0.5 kW/KTh,in = 180 °CTc,in = 25 °CA = 0.5 m²U = 300 W/(m²·K)
Th,out = °C
3Solutions
1.Heat conduction
Answer: Q̇ = 56.75 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)/(1.4·4) = 0.04464 K/W
R2 = L2/(k2A) = (100/1000)/(0.04·4) = 0.625 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 38/0.6696 = 56.75 W
Answer
Q̇ = 56.75 W
2.Heat conduction
Answer: Ts = 17.83 °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/(6·15) = 0.01111 K/W
R1 = L1/(k1A) = (250/1000)/(0.17·15) = 0.09804 K/W
R2 = L2/(k2A) = (75/1000)/(0.96·15) = 0.005208 K/W
Ro = 1/(hoA) = 1/(20·15) = 0.003333 K/W
ΣR = 0.1177 K/W
Heat loss
Q̇ = (20 − (-3))/0.1177 = 195.4 W
Ts = Troom − Q̇·Ri = 20 − 195.4·0.01111 = 17.83 °C
Answer
Ts = 17.83 °C
3.Heat exchangers (LMTD)
Answer: A = 10.14 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 = 90 − 50 = 40 K
ΔT₂ = Th,out − Tc,in = 70 − 10 = 60 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (40 − 60)/ln(40/60) = 49.33 K
Area
A = Q̇/(U·ΔTlm) = 100000/(200·49.33) = 10.14 m²
Answer
A = 10.14 m²
4.Heat exchangers (LMTD)
Answer: Th,out = 144.2 °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
0.5·(180 − Th,out) = 0.15·(Th,out − 25)
Th,out = (0.5·180 + 0.15·25)/(0.5 + 0.15) = 144.2 °C
Answer
Th,out = 144.2 °C
Heat transferred: Q̇ = 17.88 kW; the cold stream leaves at 60.77 °C.
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