Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
754 views
in General by (108k points)
closed by
The stream function for a two dimensional flow is given by ψ = 2xy. The velocity at (2, 2) is:
1. 4√2
2. 4
3. 2√2
4. √2

1 Answer

0 votes
by (111k points)
selected by
 
Best answer
Correct Answer - Option 1 : 4√2

Concept:

Stream Function:

It is defined as the scalar function of space and time, such that its partial derivative with respect to any direction gives the velocity component at right angles to that direction.

It is denoted by ψ and defined only for two-dimensional flow.

\(v = \frac{{\partial \psi }}{{\partial x}};u = - \frac{{\partial \psi }}{{\partial y}}\)

Properties of Stream function:

  • If stream function exists, it is a possible case of fluid flow which may be rotational or irrotational
  • If the stream function satisfies the Laplace equation i.e. \(\frac{{{\partial ^2}\psi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\psi }}{{\partial {y^2}}} = 0\), it is a case of irrotational flow

Calculation:

Let u and v are horizontal and vertical components of fluid velocity.

Given that, ψ = 2xy

We know that:

\({\rm{u}} = - \frac{{\partial {\rm{\psi}}}}{{\partial {\rm{y}}}} = - 2{\rm{x}}\)

And

\({\rm{v}} = + \frac{{\partial {\rm{\psi}}}}{{\partial {\rm{x}}}} = 2{\rm{y}}\)

Velocity, = ui + vj

\(\vec v\) = (-2x) i + (2y) j

Velocity at point (2, 2) i.e. at x = 2 and y = 2 is:

\(\vec v\) = - 4i + 4j

Or \(\left| {{\rm{\vec v}}} \right| = {\rm{v}} = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( 4 \right)}^2}} = 4\sqrt2\ {\rm{m}}/{\rm{s}}\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...