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 200 mm of gypsum board and 100 mm of expanded polystyrene, with an area of 15 m². The outer faces are at 17 °C and -1 °C. Find the heat flow Q̇ through the wall.
L1 = 200 mm, k1 = 0.17 W/(m·K)L2 = 100 mm, k2 = 0.035 W/(m·K)A = 15 m²T1 = 17 °CT3 = -1 °C
Q̇ = W
2.Heat conductionHard
A wall of 100 mm glass and 75 mm expanded polystyrene separates a room at 22 °C from outside air at -13 °C. The convection coefficients are 7.7 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 6 m². Find the temperature Ts of the inside surface of the wall.
L1 = 100 mm, k1 = 0.96 W/(m·K)L2 = 75 mm, k2 = 0.035 W/(m·K)A = 6 m²hi = 7.7 W/(m²·K)ho = 25 W/(m²·K)Troom = 22 °CTout = -13 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 360 kW. The hot fluid cools from 130 to 80 °C and the cold fluid heats from 20 to 95 °C. The overall heat transfer coefficient is 1000 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 360 kWU = 1000 W/(m²·K)Th: 130 → 80 °CTc: 20 → 95 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (0.5 kg/s, cp = 2.5 kJ/(kg·K)) cools from 150 to 120 °C in a counter-flow exchanger. It heats 0.5 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 15 °C. With U = 300 W/(m²·K), find the heat transfer area A needed.
ṁh = 0.5 kg/scp,h = 2.5 kJ/(kg·K)Th: 150 → 120 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 15 °CU = 300 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 66.94 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) = (200/1000)/(0.17·15) = 0.07843 K/W
R2 = L2/(k2A) = (100/1000)/(0.035·15) = 0.1905 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 18/0.2689 = 66.94 W
Answer
Q̇ = 66.94 W
2.Heat conduction
Answer: Ts = 20.12 °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/(7.7·6) = 0.02165 K/W
R1 = L1/(k1A) = (100/1000)/(0.96·6) = 0.01736 K/W
R2 = L2/(k2A) = (75/1000)/(0.035·6) = 0.3571 K/W
Ro = 1/(hoA) = 1/(25·6) = 0.006667 K/W
ΣR = 0.4028 K/W
Heat loss
Q̇ = (22 − (-13))/0.4028 = 86.89 W
Ts = Troom − Q̇·Ri = 22 − 86.89·0.02165 = 20.12 °C
Answer
Ts = 20.12 °C
3.Heat exchangers (LMTD)
Answer: A = 7.762 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 = 130 − 95 = 35 K
ΔT₂ = Th,out − Tc,in = 80 − 20 = 60 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (35 − 60)/ln(35/60) = 46.38 K
Area
A = Q̇/(U·ΔTlm) = 360000/(1000·46.38) = 7.762 m²
Answer
A = 7.762 m²
4.Heat exchangers (LMTD)
Answer: A = 1.127 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 and water outlet temperature
Q̇ = ṁhcp,h(Th,in − Th,out) = 0.5·2.5·30 = 37.5 kW
Tc,out = 15 + 37.5/(0.5·4.18) = 32.94 °C
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 = 150 − 32.94 = 117.1 K
ΔT₂ = Th,out − Tc,in = 120 − 15 = 105 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (117.1 − 105)/ln(117.1/105) = 110.9 K
Area
A = Q̇/(U·ΔTlm) = 37500/(300·110.9) = 1.127 m²
Answer
A = 1.127 m²
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