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 25 mm of concrete, with an area of 8 m². The outer faces are at 25 °C and 5 °C. Find the temperature Ti at the interface between the two layers.
L1 = 150 mm, k1 = 0.96 W/(m·K)L2 = 25 mm, k2 = 1.4 W/(m·K)A = 8 m²T1 = 25 °CT3 = 5 °C
Ti = °C
2.Heat conductionHard
A wall of 200 mm concrete and 100 mm expanded polystyrene separates a room at 22 °C from outside air at -5 °C. The convection coefficients are 8 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 6 m². Find the overall heat transfer coefficient U of the wall.
L1 = 200 mm, k1 = 1.4 W/(m·K)L2 = 100 mm, k2 = 0.035 W/(m·K)A = 6 m²hi = 8 W/(m²·K)ho = 25 W/(m²·K)Troom = 22 °CTout = -5 °C
U = W/(m²·K)
3.Heat exchangers (LMTD)Medium
In a condenser, steam condenses at a constant 130 °C on the outside of the tubes. Cooling water (4 kg/s, cp = 4.18 kJ/(kg·K)) flows through the tubes and heats from 25 to 45 °C. The overall heat transfer coefficient is 3000 W/(m²·K). Find the heat transfer area A needed.
Ts = 130 °Cṁ = 4 kg/scp = 4.18 kJ/(kg·K)Tc: 25 → 45 °CU = 3000 W/(m²·K)
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 105 °C and the cold stream at 20 °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 = 105 °CTc,in = 20 °CA = 1.5 m²U = 800 W/(m²·K)
Th,out = °C
3Solutions
1.Heat conduction
Answer: Ti = 7.051 °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.
Layer resistances
R1 = L1/(k1A) = (150/1000)/(0.96·8) = 0.01953 K/W
R2 = L2/(k2A) = (25/1000)/(1.4·8) = 0.002232 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 20/0.02176 = 919 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 25 − 919·0.01953 = 7.051 °C
Answer
Ti = 7.051 °C
2.Heat conduction
Answer: U = 0.316 W/(m²·K)
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/(8·6) = 0.02083 K/W
R1 = L1/(k1A) = (200/1000)/(1.4·6) = 0.02381 K/W
R2 = L2/(k2A) = (100/1000)/(0.035·6) = 0.4762 K/W
Ro = 1/(hoA) = 1/(25·6) = 0.006667 K/W
ΣR = 0.5275 K/W
Overall coefficient
U = 1/(A·ΣR) = 1/(6·0.5275) = 0.316 W/(m²·K)
Answer
U = 0.316 W/(m²·K)
3.Heat exchangers (LMTD)
Answer: A = 1.178 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.
Heat duty from the water side
Q̇ = 4·4.18·(45 − 25) = 334.4 kW
End temperature differences
ΔT₁ = 130 − 25 = 105 K
ΔT₂ = 130 − 45 = 85 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (105 − 85)/ln(105/85) = 94.65 K
Area
A = Q̇/(U·ΔTlm) = 334400/(3000·94.65) = 1.178 m²
Answer
A = 1.178 m²
4.Heat exchangers (LMTD)
Answer: Th,out = 67.22 °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·(105 − Th,out) = 1.2·(Th,out − 20)
Th,out = (1.5·105 + 1.2·20)/(1.5 + 1.2) = 67.22 °C
Answer
Th,out = 67.22 °C
Heat transferred: Q̇ = 56.67 kW; the cold stream leaves at 57.78 °C.
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