Mechanical Engineering - Hydraulic Machines - Discussion
Discussion Forum : Hydraulic Machines - Section 1 (Q.No. 45)
45.
The force exerted by a jet of water (in a direction normal to flow) impinging on a fixed plate inclined at an angle θ with the jet is
Discussion:
8 comments Page 1 of 1.
Vinay BEL said:
5 years ago
Given one is the correct answer! I agree.
Vihan said:
5 years ago
Let say Normal force to the plate is fn.
So force coponent in the direction normal to the force is fnsinθ.
Fn = w/g(vvcosθ).
So w/gv(vvcosθsinθ).
W/2g(vvsin2θ).
Where sin2θ=2sinθcosθ.
So force coponent in the direction normal to the force is fnsinθ.
Fn = w/g(vvcosθ).
So w/gv(vvcosθsinθ).
W/2g(vvsin2θ).
Where sin2θ=2sinθcosθ.
Lutuna said:
7 years ago
Here we can use a formula SIN a * COS a=SIN2a/2.
Shad said:
8 years ago
Good @Vasi.
Vasi said:
9 years ago
It is normal to flow, Not normal to plate. Hence, Given one is correct answer!
Jwala said:
9 years ago
Answer is correct it is calculated based on momentum equations.
Rishav Shaw said:
9 years ago
In the third and fourth option, it will be (sinθ) ^2 instead of sin 2*θ.
Harry said:
9 years ago
How it is wav^2 x sin^2 theta / 2g? Explain.
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