15.8-4. Heat Transfer in a Finned Tube Exchanger. Air at an average temperature

Question

15.8-4. Heat Transfer in a Finned Tube Exchanger. Air at an average temperature of 50°C is being heated by flowing outside a steel tube (k 45.1 W/m K) having an inside diameter of 35 mm and a wall thickness of 3 mm. The outside of the tube is covered with 16 longitudinal steel fins with a length L 13 mm and a thickness of t1.0 mm. Condensing steam inside the tube at 120°C has a coefficient of 7000 W/m2 K. The outside coefficient of the air has been estimated as 30 W/m2 K. Neeng fouling factors and using a tube 1.0 m long, calculate the overall heat-transfer coefficient U based on the inside area A

Explanation / Answer

Inside heat transfer coefficient hi = 7000 W/m2.K

Outside heat transfer coefficient ho = 30 W/m2.K

Area of inside pipe Ai = 3.14 x Di x L

= 3.14 x 35 mm x 1m/1000mm x 1 m

= 0.1099 m2

Dia of outside tube = 35 + 2*3 = 41 mm = 0.041 m

Area of steel tube As = 3.14 x L x DLM

= 3.14 x 1 m x (Do - Di)/ln(Do/Di)

= 3.14 x 1m x (0.041 - 0.035) / ln(0.041/0.035)

= 0.119 m2

Aeb = (3.14*Do*L) - 16*thickness of fins*L

= (3.14 x 0.041 x 1m) - (16 x 1mm x 1m/1000mm x 1 m)

= 0.11274 m2

Area of fins Af = 16 x (thickness x 1m + 2 x Length of fins x 1 m)

= 16 x (0.001 m x 1m + 2 x 0.013 m x 1 m)

= 0.432 m2

Fin efficiency n = tanh (mL) / mL

m = (hoP/kA)^0.5

= { [30 x (2*1 + 2*0.001)] / [45.1 x 0.001 x 1] }^0.5

= 36.533

n = tanh (36.533*0.013) / (36.533*0.013)

= tanh (0.474929) / (0.474929)

= 0.44217/0.474929

= 0.9310

Thermal resistance due to inside convection

R1 = 1/(hi x Ai)

= 1/(0.1099 x 7000) = 0.0012998 K/W

Thermal resistance due to steel tube conduction

R2 = (Do - Di) / (k x As)

= (0.041 - 0.035) / (45.1 x 0.119)

= 0.001118 K/W

Thermal resistance due to outside convection

R3 = 1/[ho (Aeb + n*Af)]

= 1/ [30 x (0.11274 + 0.9310*0.432)]

= 0.064733 K/W

Total resistance R = R1 + R2 + R3

= 0.0012998 + 0.001118 + 0.064733

= 0.0671508 K/W

Overall heat transfer coefficient

Ui = 1/(Ai x R)

= 1/(0.1099 x 0.0671508)

= 135.50 W/m2-K

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