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 concrete and 100 mm of gypsum board, with an area of 10 m². The outer faces are at 22 °C and -10 °C. Find the heat flow Q̇ through the wall.
L1 = 150 mm, k1 = 1.4 W/(m·K)L2 = 100 mm, k2 = 0.17 W/(m·K)A = 10 m²T1 = 22 °CT3 = -10 °C
Q̇ = W
2.Heat conductionHard
A wall of 150 mm concrete and 100 mm brick separates a room at 22 °C from outside air at -14 °C. The convection coefficients are 7.7 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 10 m². Find the heat loss Q̇ through the wall.
L1 = 150 mm, k1 = 1.4 W/(m·K)L2 = 100 mm, k2 = 0.72 W/(m·K)A = 10 m²hi = 7.7 W/(m²·K)ho = 20 W/(m²·K)Troom = 22 °CTout = -14 °C
Q̇ = W
3.Heat exchangers (LMTD)Medium
A counter-flow heat exchanger transfers 120 kW. The hot fluid cools from 170 to 60 °C and the cold fluid heats from 25 to 80 °C. The overall heat transfer coefficient is 1500 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 120 kWU = 1500 W/(m²·K)Th: 170 → 60 °CTc: 25 → 80 °C
A = m²
4.Heat exchangers (LMTD)Hard
Steam condenses at a constant 130 °C on the outside of the tubes of a condenser with a heat transfer area of 8 m² and U = 1000 W/(m²·K). Cooling water (1.5 kg/s, cp = 4.18 kJ/(kg·K)) enters the tubes at 30 °C. Find the water outlet temperature Tc,out.
Ts = 130 °CA = 8 m²U = 1000 W/(m²·K)ṁ = 1.5 kg/scp = 4.18 kJ/(kg·K)Tc,in = 30 °C
Tc,out = °C
3Solutions
1.Heat conduction
Answer: Q̇ = 460.2 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)/(1.4·10) = 0.01071 K/W
R2 = L2/(k2A) = (100/1000)/(0.17·10) = 0.05882 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 32/0.06954 = 460.2 W
Answer
Q̇ = 460.2 W
2.Heat conduction
Answer: Q̇ = 845.3 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/(7.7·10) = 0.01299 K/W
R1 = L1/(k1A) = (150/1000)/(1.4·10) = 0.01071 K/W
R2 = L2/(k2A) = (100/1000)/(0.72·10) = 0.01389 K/W
Ro = 1/(hoA) = 1/(20·10) = 0.005 K/W
ΣR = 0.04259 K/W
Heat loss
Q̇ = (22 − (-14))/0.04259 = 845.3 W
Answer
Q̇ = 845.3 W
3.Heat exchangers (LMTD)
Answer: A = 1.374 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 = 170 − 80 = 90 K
ΔT₂ = Th,out − Tc,in = 60 − 25 = 35 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (90 − 35)/ln(90/35) = 58.23 K
Area
A = Q̇/(U·ΔTlm) = 120000/(1500·58.23) = 1.374 m²
Answer
A = 1.374 m²
4.Heat exchangers (LMTD)
Answer: Tc,out = 102.1 °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) = 1000·8/(1.5·4180) = 1.276
Outlet temperature
ΔT₂ = ΔT₁·e−1.276 = 100·0.2792 = 27.92 K
Tc,out = 130 − 27.92 = 102.1 °C
Answer
Tc,out = 102.1 °C
Heat transferred: Q̇ = 452 kW.
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