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.

U = 129.053 KJ/m².h

Explanation:

overall heat transfer coefficient (U):

U = (mCp(ToutH2O - TinH2O))/[A((ToutAir - TinH2O)-(TinAir - ToutH2O))/(Ln(ToutAir - TinH2O)/(TinAir - ToutH2O))]

∴ A = 925 m²

H2O:

∴ mass flow H2O (m) = 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

i don't see a text, but i am gonna assume this *orbit* the curved path, usually elliptical, described by a planet, satellite, spaceship, etc., around a celestial body, as the sun.

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer