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 250 mm of concrete and 75 mm of expanded polystyrene, with an area of 10 m². The outer faces are at 16 °C and 5 °C. Find the heat flow Q̇ through the wall.
L1 = 250 mm, k1 = 1.4 W/(m·K)L2 = 75 mm, k2 = 0.035 W/(m·K)A = 10 m²T1 = 16 °CT3 = 5 °C
Q̇ = W
2.Heat conductionHard
A wall of 200 mm mineral wool and 100 mm brick separates a room at 22 °C from outside air at -12 °C. The convection coefficients are 8 W/(m²·K) inside and 20 W/(m²·K) outside, and the wall area is 6 m². Find the temperature Ts of the inside surface of the wall.
L1 = 200 mm, k1 = 0.04 W/(m·K)L2 = 100 mm, k2 = 0.72 W/(m·K)A = 6 m²hi = 8 W/(m²·K)ho = 20 W/(m²·K)Troom = 22 °CTout = -12 °C
Ts = °C
3.Heat exchangers (LMTD)Medium
A shell-and-tube heat exchanger (one shell pass, two tube passes) must transfer 800 kW. The hot fluid cools from 95 to 50 °C and the cold fluid heats from 10 to 70 °C. The LMTD correction factor for this arrangement is F = 0.92, and U = 300 W/(m²·K). Find the heat transfer area A needed.
Q̇ = 800 kWF = 0.92U = 300 W/(m²·K)Th: 95 → 50 °CTc: 10 → 70 °C
A = m²
4.Heat exchangers (LMTD)Hard
Hot oil (0.5 kg/s, cp = 2.3 kJ/(kg·K)) cools from 175 to 105 °C in a counter-flow exchanger. It heats 1.5 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 25 °C. With U = 500 W/(m²·K), find the heat transfer area A needed.
ṁh = 0.5 kg/scp,h = 2.3 kJ/(kg·K)Th: 175 → 105 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)Tc,in = 25 °CU = 500 W/(m²·K)
A = m²
3Solutions
1.Heat conduction
Answer: Q̇ = 47.38 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) = (250/1000)/(1.4·10) = 0.01786 K/W
R2 = L2/(k2A) = (75/1000)/(0.035·10) = 0.2143 K/W
Heat flow
Q̇ = (T1 − T3)/(R1 + R2) = 11/0.2321 = 47.38 W
Answer
Q̇ = 47.38 W
2.Heat conduction
Answer: Ts = 21.2 °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.
Resistances in series
Ri = 1/(hiA) = 1/(8·6) = 0.02083 K/W
R1 = L1/(k1A) = (200/1000)/(0.04·6) = 0.8333 K/W
R2 = L2/(k2A) = (100/1000)/(0.72·6) = 0.02315 K/W
Ro = 1/(hoA) = 1/(20·6) = 0.008333 K/W
ΣR = 0.8856 K/W
Heat loss
Q̇ = (22 − (-12))/0.8856 = 38.39 W
Ts = Troom − Q̇·Ri = 22 − 38.39·0.02083 = 21.2 °C
Answer
Ts = 21.2 °C
3.Heat exchangers (LMTD)
Answer: A = 90.82 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 = 95 − 70 = 25 K
ΔT₂ = Th,out − Tc,in = 50 − 10 = 40 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (25 − 40)/ln(25/40) = 31.91 K
Area
A = Q̇/(U·F·ΔTlm,cf) = 800000/(300·0.92·31.91) = 90.82 m²
Answer
A = 90.82 m²
4.Heat exchangers (LMTD)
Answer: A = 1.519 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) = 0.5·2.3·70 = 80.5 kW
Tc,out = 25 + 80.5/(1.5·4.18) = 37.84 °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 = 175 − 37.84 = 137.2 K
ΔT₂ = Th,out − Tc,in = 105 − 25 = 80 K
Log-mean temperature difference
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (137.2 − 80)/ln(137.2/80) = 106 K
Area
A = Q̇/(U·ΔTlm) = 80500/(500·106) = 1.519 m²
Answer
A = 1.519 m²
ExamProblems · examproblems.com/courses/heat-transfer