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 mineral wool and 25 mm of expanded polystyrene, with an area of 4 m². The outer faces are at 22 °C and -8 °C. Find the temperature Ti at the interface between the two layers.
L1 = 150 mm, k1 = 0.04 W/(m·K)L2 = 25 mm, k2 = 0.035 W/(m·K)A = 4 m²T1 = 22 °CT3 = -8 °C
Ti = °C
2.Heat conductionHard
A wall of 150 mm glass and 75 mm expanded polystyrene separates a room at 20 °C from outside air at -12 °C. The convection coefficients are 6 W/(m²·K) inside and 25 W/(m²·K) outside, and the wall area is 8 m². Find the temperature Ts of the inside surface of the wall.
L1 = 150 mm, k1 = 0.96 W/(m·K)L2 = 75 mm, k2 = 0.035 W/(m·K)A = 8 m²hi = 6 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = -12 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A shell-and-tube heat exchanger (one shell pass, two tube passes) must transfer 320 kW. The hot fluid cools from 160 to 85 °C and the cold fluid heats from 20 to 65 °C. The LMTD correction factor for this arrangement is F = 0.9, and U = 500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 320 kWF = 0.9U = 500 W/(m²·K)Th: 160 → 85 °CTc: 20 → 65 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (1 kg/s, cp = 2.1 kJ/(kg·K)) cools from 80 to 55 °C in a counter-flow exchanger. It heats 2 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 15 °C. With U = 200 W/(m²·K), find the heat transfer area A needed.
ṁh = 1 kg/scp,h = 2.1 kJ/(kg·K)Th: 80 → 55 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 15 °CU = 200 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Ti = −3.2 °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)/(0.04·4) = 0.9375 K/W
R2 = L2/(k2A) = (25/1000)/(0.035·4) = 0.1786 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 30/1.116 = 26.88 W
Interface temperature
The temperature drops across the first layer by Q̇·R₁.
Ti = T1 − Q̇·R1 = 22 − 26.88·0.9375 = −3.2 °C
Answer
Ti = −3.2 °C
2.Heat conduction
Answer: Ts = 17.87 °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/(6·8) = 0.02083 K/W
R1 = L1/(k1A) = (150/1000)/(0.96·8) = 0.01953 K/W
R2 = L2/(k2A) = (75/1000)/(0.035·8) = 0.2679 K/W
Ro = 1/(hoA) = 1/(25·8) = 0.005 K/W
ΣR = 0.3132 K/W
Heat loss
Q̇ = (20 − (-12))/0.3132 = 102.2 W
Ts = Troom − Q̇·Ri = 20 − 102.2·0.02083 = 17.87 °C
Answer
Ts = 17.87 °C
3.Heat exchangers (LMTD)
Answer: A = 8.995 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 = 160 − 65 = 95 K
ΔT₂ = Th,out − Tc,in = 85 − 20 = 65 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (95 − 65)/ln(95/65) = 79.05 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 320000/(500·0.9·79.05) = 8.995 m²
Answer
A = 8.995 m²
4.Heat exchangers (LMTD)
Answer: A = 5.383 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) = 1·2.1·25 = 52.5 kW
Tc,out = 15 + 52.5/(2·4.18) = 21.28 °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 = 80 − 21.28 = 58.72 K
ΔT₂ = Th,out − Tc,in = 55 − 15 = 40 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (58.72 − 40)/ln(58.72/40) = 48.76 K
Area
A = Q̇/(U·ΔTlm) = 52500/(200·48.76) = 5.383 m²
Answer
A = 5.383 m²
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