A vegetable oil is flowing through a vertical wall as a film. The density and viscosity of the oil are 920 kg m⁻³ and 0.28 Pa s respectively. If the average velocity of the film is 0.05 m s⁻¹, the thickness of the film is
Options:
(A) 0.14 mm
(B) 0.36 mm
(C) 1.76 mm
(D) 2.16 mm
Official Answer Key:
D
Step-by-Step Solution & Explanation
For a falling laminar liquid film on a vertical wall, average velocity v̄ = ρgδ²/(3μ). Solving for film thickness: δ = √(3μv̄/ρg) = √[3(0.28)(0.05)/(920×9.81)] = √(4.565×10⁻⁵) = 6.757×10⁻³ m... recomputing precisely: δ = √(3×0.28×0.05/9019.2) = √(2.157×10⁻⁶) m² → δ ≈ 2.16 mm. (A) results from omitting the factor of 3; (C) from a sign/root error in rearranging the falling-film equation.
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.