Question
Download Solution PDFFor the following stream function calculate velocity at a point (1, 2)
1) ψ = 4xy
2) ψ = 3x2y - y3
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
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,
1) ψ = 4xy
We know that:
\({\rm{u}} = - \frac{{\partial {\rm{\psi}}}}{{\partial {\rm{y}}}} = - 4{\rm{x}}\)
And
\({\rm{v}} = + \frac{{\partial {\rm{\psi}}}}{{\partial {\rm{x}}}} = 4{\rm{y}}\)
Velocity, = ui + vj
\(\vec v\) = (-4x) i + (4y) j
Velocity at point (1, 2) i.e. at x = 1 and y = 2 is:
\(\vec v\) = - 4i + 8j
Or \(\left| {{\rm{\vec v}}} \right| = {\rm{v}} = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( 8 \right)}^2}} = \sqrt{80} ~{\rm{m}}/{\rm{s}}\)
Given that,
2) ψ = 3x2y - y3
\({\rm{u}} = - \frac{{\partial {\rm{\psi}}}}{{\partial {\rm{y}}}} = -3x^2+3y^2\)
And
\({\rm{v}} = + \frac{{\partial {\rm{\psi}}}}{{\partial {\rm{x}}}} = 6x{\rm{y}}\)
Velocity, = ui + vj
v = (-3x2 + 3y2)i + (6xy)j
Velocity at point (1, 2) i.e. at x = 1 and y = 2 is:
v = -9i + 12j
\(\left| {{\rm{\vec v}}} \right| = {\rm{v}} = \sqrt {{{\left( { - 9} \right)}^2} + {{\left( 12 \right)}^2}} =15 ~{\rm{m}}/{\rm{s}}\)
Last updated on Jul 1, 2025
-> JKSSB Junior Engineer recruitment exam date 2025 for Civil and Electrical Engineering has been rescheduled on its official website.
-> JKSSB JE exam will be conducted on 31st August (Civil), and on 24th August 2025 (Electrical).
-> JKSSB JE application form correction facility has been started. Candidates can make corrections in the JKSSB recruitment 2025 form from June 23 to 27.
-> JKSSB JE recruitment 2025 notification has been released for Civil Engineering.
-> A total of 508 vacancies has been announced for JKSSB JE Civil Engineering recruitment 2025.
-> JKSSB JE Online Application form will be activated from 18th May 2025 to 16th June 2025
-> Candidates who are preparing for the exam can access the JKSSB JE syllabus PDF from official website of JKSSB.
-> The candidates can check the JKSSB JE Previous Year Papers to understand the difficulty level of the exam.
-> Candidates also attempt the JKSSB JE Mock Test which gives you an experience of the actual exam.