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 150 mm of glass and 150 mm of wood, with an area of 8 m². The outer faces are at 17 °C and -15 °C. Find the heat flow Q̇ through the wall.
L1 = 150 mm, k1 = 0.96 W/(m·K)L2 = 150 mm, k2 = 0.17 W/(m·K)A = 8 m²T1 = 17 °CT3 = -15 °C
Q̇ = W
2.Heat conductionHard
A wall of 200 mm glass and 100 mm concrete separates a room at 22 °C from outside air at -1 °C. The convection coefficients are 8 W/(m²·K) inside and 30 W/(m²·K) outside, and the wall area is 12 m². Find the heat loss Q̇ through the wall.
L1 = 200 mm, k1 = 0.96 W/(m·K)L2 = 100 mm, k2 = 1.4 W/(m·K)A = 12 m²hi = 8 W/(m²·K)ho = 30 W/(m²·K)Troom = 22 °CTout = -1 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A shell-and-tube heat exchanger (one shell pass, two tube passes) must transfer 250 kW. The hot fluid cools from 170 to 120 °C and the cold fluid heats from 25 to 60 °C. The LMTD correction factor for this arrangement is F = 0.88, and U = 200 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 250 kWF = 0.88U = 200 W/(m²·K)Th: 170 → 120 °CTc: 25 → 60 °C
A = m²
4.Heat exchangers (LMTD)Hard
In a counter-flow heat exchanger both streams have the same heat capacity rate, ṁ·cp = 3 kW/K. The hot stream enters at 135 °C and the cold stream at 25 °C. The exchanger has A = 1.5 m² and U = 800 W/(m²·K). Find the hot outlet temperature Th,out.
Ch = Cc = 3 kW/KTh,in = 135 °CTc,in = 25 °CA = 1.5 m²U = 800 W/(m²·K)
Th,out = °C
3Solutions
1.Heat conduction
Answer: Q̇ = 246.5 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) = (150/1000)/(0.96·8) = 0.01953 K/W
R2 = L2/(k2A) = (150/1000)/(0.17·8) = 0.1103 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 32/0.1298 = 246.5 W
Answer
Q̇ = 246.5 W
2.Heat conduction
Answer: Q̇ = 630 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/(8·12) = 0.01042 K/W
R1 = L1/(k1A) = (200/1000)/(0.96·12) = 0.01736 K/W
R2 = L2/(k2A) = (100/1000)/(1.4·12) = 0.005952 K/W
Ro = 1/(hoA) = 1/(30·12) = 0.002778 K/W
ΣR = 0.03651 K/W
Heat loss
Q̇ = (22 − (-1))/0.03651 = 630 W
Answer
Q̇ = 630 W
3.Heat exchangers (LMTD)
Answer: A = 13.88 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 (counter-flow)
In counter-flow the hot inlet faces the cold outlet, and the hot outlet faces the cold inlet.
ΔT₁ = Th,in − Tc,out = 170 − 60 = 110 K
ΔT₂ = Th,out − Tc,in = 120 − 25 = 95 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (110 − 95)/ln(110/95) = 102.3 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 250000/(200·0.88·102.3) = 13.88 m²
Answer
A = 13.88 m²
4.Heat exchangers (LMTD)
Answer: Th,out = 103.6 °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
3·(135 − Th,out) = 1.2·(Th,out − 25)
Th,out = (3·135 + 1.2·25)/(3 + 1.2) = 103.6 °C
Answer
Th,out = 103.6 °C
Heat transferred: Q̇ = 94.29 kW; the cold stream leaves at 56.43 °C.
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