*Biosystems Engineering - Food Science Vegetable oil at 25℃ (density = 910 kgm3, viscosity 0.084 Pa...
*Biosystems Engineering - Food Science
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Vegetable oil at 25℃ (density = 910 kgm3, viscosity 0.084 Pa s) s flowing with a volumetric average velocity of 0.85 m/s through 2 inch smooth (Ɛ = 0) steel tubing (ID = 4.749 cm)
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What is the Reynolds number for this flow situation?
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What is the friction factor?
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What is the pressure drop over the 120 m section of pipe?
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Homework Answers
Reynolds number = density * velocity * internal diameter / viscosity (flow will through this inner section)
910*0.85*0.04749/0.084 = 437.3 Ans
since Reynold number (Re) < 2300 so flow is laminar
now friction factor for laminar flow(f) =64/Re = 0.14635 Ans
now pressure drop for 120 m section
head loss = f*length * velocity2/d*2g = 0.14635*120*0.852/0.04749*19.62 = 13.618 m
so pressure drop = head loss * g* density = 13.618*9.81*910 = 121569.6306 Pa or 121.569 Kpa Ans
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*Biosystems Engineering - Food Science
Vegetable oil at 25℃ (density = 910 kgm3, viscosity
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