Integration of the Stream Function & Velocity Potential

In summary, the conversation discusses how to go from a partial derivative of Ψ with respect to θ equaling m, to Ψ being equal to mθ plus a term f1(r). This term is necessary because Ψ is assumed to be a function of both θ and r. Similarly, the conversation talks about how to go from a partial derivative of Ψ with respect to r equaling 0, to Ψ being equal to a constant c1 plus a term g1(θ). Again, this additional term is needed because Ψ is assumed to be a function of both θ and r. The context of the problem usually provides information about what variables any functions are in terms of.
  • #1
influx
164
2
767bc5.png


How did they go from ∂Ψ/dθ = m to Ψ = mθ + f1(r) ?

I mean I understand how they got mθ but can someone explain the f1(r) term?

Similarly, how did they go from -∂Ψ/dr = 0 to Ψ = c1 + g1(θ) ?

I understand how they got c1 but why the additional term?

I can't provide further working out as I think this is supposed to be pretty simple, I'm just missing something..

Thanks
 
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  • #2
Moved this thread out of the HW sections. Note that if you post in the HW sections, you must include the homework template.
influx said:
767bc5.png


How did they go from ∂Ψ/dθ = m to Ψ = mθ + f1(r) ?

I mean I understand how they got mθ but can someone explain the f1(r) term?
Ψ is assumed to be a function of both θ and r, so when you integrate ∂Ψ/dθ with respect to θ, you have to include a "constant" of integration that is a function of r alone. As a check, take the partial derivative with respect to θ of the right side, and you should get what they have on the left side.
influx said:
Similarly, how did they go from -∂Ψ/dr = 0 to Ψ = c1 + g1(θ) ?

I understand how they got c1 but why the additional term?
See above, with the difference that now the integration is with respect to r.
influx said:
I can't provide further working out as I think this is supposed to be pretty simple, I'm just missing something..

Thanks
 
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  • #3
Mark44 said:
Moved this thread out of the HW sections. Note that if you post in the HW sections, you must include the homework template.
Ψ is assumed to be a function of both θ and r, so when you integrate ∂Ψ/dθ with respect to θ, you have to include a "constant" of integration that is a function of r alone. As a check, take the partial derivative with respect to θ of the right side, and you should get what they have on the left side.
See above, with the difference that now the integration is with respect to r.
That makes perfect sense. One question though, in this case we know that Ψ is a function of both θ and r but what if we were not aware of this? So say we were told to integrate ∂Ψ/dθ = m without being informed that Ψ is a function of both θ and r?

Also for:

cea3a9.md.png


How do we know that Ψ is a function of both x and y?

Thanks
 
  • #4
influx said:
That makes perfect sense. One question though, in this case we know that Ψ is a function of both θ and r but what if we were not aware of this? So say we were told to integrate ∂Ψ/dθ = m without being informed that Ψ is a function of both θ and r?
Since you are asked to find the partial derivative ∂Ψ/∂θ, you can be certain that Ψ is a function of θ and some other variable, most likely r, as the problem is evidently working with polar coordinates. The context of the problem will usually provide information about what variables any functions are in terms of.
influx said:
Also for:

cea3a9.md.png


How do we know that Ψ is a function of both x and y?
The fact that the flows are shown in an x-y coordinate system is a strong clue.
 
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  • #5
Mark44 said:
Since you are asked to find the partial derivative ∂Ψ/∂θ, you can be certain that Ψ is a function of θ and some other variable, most likely r, as the problem is evidently working with polar coordinates. The context of the problem will usually provide information about what variables any functions are in terms of.
The fact that the flows are shown in an x-y coordinate system is a strong clue.
Thank you very much. Makes perfect sense! :)
 

1. What is the purpose of integrating the stream function and velocity potential?

The integration of the stream function and velocity potential is used to determine the velocity and direction of a fluid flow in a two-dimensional, irrotational flow field. It allows for a more complete understanding of the flow behavior and can be used to solve various fluid mechanics problems.

2. How is the stream function related to the velocity potential?

The stream function and velocity potential are mathematically related through the Cauchy-Riemann equations. Specifically, the partial derivatives of the stream function with respect to the x and y coordinates are equal to the negative partial derivatives of the velocity potential with respect to the y and x coordinates, respectively.

3. What is the equation for the stream function?

The equation for the stream function is given by ψ = constant, where ψ is the stream function and the constant is determined by the boundary conditions of the flow field. In polar coordinates, the equation becomes ψ = f(r, θ), where r is the radial distance and θ is the angular coordinate.

4. How is the stream function used to determine the velocity components?

The stream function is used in conjunction with the continuity equation to determine the velocity components in a two-dimensional, irrotational flow field. The x- and y-components of velocity can be calculated using the partial derivatives of the stream function with respect to the y and x coordinates, respectively.

5. What are some common applications of integrating the stream function and velocity potential?

Some common applications of integrating the stream function and velocity potential include determining the flow characteristics around objects, such as airfoils and ships, and analyzing the behavior of fluids in pipes and channels. It is also used in the study of potential flow, which is a simplified model for fluid flow that is often used in aerodynamics and hydrodynamics.

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