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Charra [1.4K]
3 years ago
15

Water at a rate of 45,500 kg/h is heated from 80 to 150°C in a heat exchanger having two shell passes and eight tube passes with

a total surface area of 925 m2 . Hot exhaust gases having approximately the same thermophysical properties as air enter at 350°C and exit at 175°C. Determine the overall heat transfer coefficient.
Chemistry
1 answer:
Vlad [161]3 years ago
4 0

Answer:

U = 129.053 KJ/m².h

Explanation:

overall heat transfer coefficient (U):

U = (<em>m</em>Cp(ToutH2O - TinH2O))/[A((ToutAir - TinH2O)-(TinAir - ToutH2O))/(Ln(ToutAir - TinH2O)/(TinAir - ToutH2O))]

∴ A = 925 m²

H2O:

∴ mass flow H2O (<em>m</em>) = 45500 Kg/h

∴ Cp H2O = 4186 J/Kg°C

∴ Tin = 80°C

∴ Tout = 150°C

Air:

∴ Tin = 350°C

∴ Tout = 175°C

⇒ U = ((45500Kg/h)(4186J/Kg°C)(150-80))/(925m²)[((175-80)-(350-150))Ln((175-80)/(350-150))]

⇒ U = (1.333 E10 J/h)/(925m²)[(-150)Ln(0.475)]

⇒ U = 129052.73 J/m².h = 129.053 KJ/m².h

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C = –134830.08 / –321050.7

C = 0.42 J/gºC

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