Question
Download Solution PDFA horizontal jet of water with a velocity of 10 m/s and cross-sectional area of 10 mm2 strikes a flat plate held normal to the flow direction. The density of water is 1000 kg/m3. The total force on the plate due to the jets is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Force Exerted by Jet on a Flat Plate Normal to Jet:
Force on the plate due to impact, F = ρa(v)2
where, ρ = density, a = Area of jet, v = velocity of jet, u = Plate velocity
Given: v = 10 m/s, a = 10 mm2 = 10 × 10-6 m2 , ρ = 1000 kg/m3.
F = ρa(v)2
F = 1000 × 10 × 10-6 × 100
F = 1 N
Additional Information
Force Exerted by Jet on Moving Flat Plate Normal to Jet:
- Force on the plate due to impact, F = ρa(v - u)2
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