SSC JE Civil Engineering 2018 Set 1 Solved Paper

© examsiri.com

Question : 139 of 200

Marks: +1, -0

The velocity potential which follow the equation of continuity is ______.

$x ^2y$
$x ^2 -y^2$
cos x
$x^ 2 + y^2$

Solution:

Equation of continuity :-

{∂p}/{∂t}+ {∂(ρu)}/{∂x}+{∂(ρv)}/{∂y}+{∂(ρw)}/{∂x}=0

According to given conditions :-

{∂ u }/{∂x}+{∂ v }/{∂y}=0

u={∂ ϕ }/{∂x},v={∂ ϕ }/{∂y}

Now, ⇒

{∂^2 ϕ }/{∂x^2}+{∂ ^2ϕ }/{∂y^2}=0

Case 1:-
If

ϕ = x^2y

⇒

2 ≠ y + 0 ≠ 0

Case 2:-
If

ϕ = (x^2– y^2)

⇒ +2 – 2 = 0

© examsiri.com

Go to Question:

Prev Question

Next Question