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 100 mm of gypsum board and 150 mm of concrete, with an area of 12 m². The outer faces are at 17 °C and 6 °C. Find the heat flow Q̇ through the wall.
L1 = 100 mm, k1 = 0.17 W/(m·K)L2 = 150 mm, k2 = 1.4 W/(m·K)A = 12 m²T1 = 17 °CT3 = 6 °C
Q̇ = W
2.Heat conductionHard
A wall of 150 mm glass and 100 mm mineral wool separates a room at 21 °C from outside air at 0 °C. The convection coefficients are 6 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 12 m². Find the temperature Ts of the inside surface of the wall.
L1 = 150 mm, k1 = 0.96 W/(m·K)L2 = 100 mm, k2 = 0.04 W/(m·K)A = 12 m²hi = 6 W/(m²·K)ho = 20 W/(m²·K)Troom = 21 °CTout = 0 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A shell-and-tube heat exchanger (one shell pass, two tube passes) must transfer 150 kW. The hot fluid cools from 80 to 40 °C and the cold fluid heats from 20 to 50 °C. The LMTD correction factor for this arrangement is F = 0.85, and U = 800 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 150 kWF = 0.85U = 800 W/(m²·K)Th: 80 → 40 °CTc: 20 → 50 °C
A = m²
4.Heat exchangers (LMTD)Hard
Steam condenses at a constant 120 °C on the outside of the tubes of a condenser with a heat transfer area of 1 m² and U = 3000 W/(m²·K). Cooling water (1.5 kg/s, cp = 4.18 kJ/(kg·K)) enters the tubes at 15 °C. Find the water outlet temperature Tc,out.
Ts = 120 °CA = 1 m²U = 3000 W/(m²·K)ṁ = 1.5 kg/scp = 4.18 kJ/(kg·K)Tc,in = 15 °C
Tc,out = °C
3Solutions
1.Heat conduction
Answer: Q̇ = 189.8 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) = (100/1000)/(0.17·12) = 0.04902 K/W
R2 = L2/(k2A) = (150/1000)/(1.4·12) = 0.008929 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 11/0.05795 = 189.8 W
Answer
Q̇ = 189.8 W
2.Heat conduction
Answer: Ts = 19.78 °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/(6·12) = 0.01389 K/W
R1 = L1/(k1A) = (150/1000)/(0.96·12) = 0.01302 K/W
R2 = L2/(k2A) = (100/1000)/(0.04·12) = 0.2083 K/W
Ro = 1/(hoA) = 1/(20·12) = 0.004167 K/W
ΣR = 0.2394 K/W
Heat loss
Q̇ = (21 − (0))/0.2394 = 87.72 W
Ts = Troom − Q̇·Ri = 21 − 87.72·0.01389 = 19.78 °C
Answer
Ts = 19.78 °C
3.Heat exchangers (LMTD)
Answer: A = 8.944 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 = 80 − 50 = 30 K
ΔT₂ = Th,out − Tc,in = 40 − 20 = 20 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (30 − 20)/ln(30/20) = 24.66 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 150000/(800·0.85·24.66) = 8.944 m²
Answer
A = 8.944 m²
4.Heat exchangers (LMTD)
Answer: Tc,out = 54.93 °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.
Combine the energy balance and the rate equation
With ΔT₁ = Ts − Tc,in and ΔT₂ = Ts − Tc,out, the difference ΔT₁ − ΔT₂ equals the water temperature rise, so the rise cancels.
ṁcp(ΔT₁ − ΔT₂) = U·A·(ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂)
ln(ΔT₁/ΔT₂) = U·A/(ṁcp) = 3000·1/(1.5·4180) = 0.4785
Outlet temperature
ΔT₂ = ΔT₁·e−0.4785 = 105·0.6197 = 65.07 K
Tc,out = 120 − 65.07 = 54.93 °C
Answer
Tc,out = 54.93 °C
Heat transferred: Q̇ = 250.3 kW.
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