I am reading "Fundamental of Aerodynamics " by J.D.Anderson Fifth edition, At pages 177 and 178 (section 2.14 Stream funciton), the book said: given 2D steady flow, from: $$\frac{dy}{dx}=\frac{v}{u}$$ with u and v are known function of x and y, then we obtain f(x,y) = c by intergration.

If we have steamline 1 is f(x,y) = c1 and streamline 2 is f(x,y) = c2, then why is the mass flow between two streamlines equal (c2 - c1)? The book said that but there is no derivation of that.

As far as I can do: enter image description here


1 Answer 1


Short answer: Because that's how we've defined $c_1$ and $c_2$.

Long answer: Read 2.14 carefully. As you've said, we have $f(x,y)=c$ via integration. Anderson then chooses $\bar{\psi}$ as $f$ (a convention) to get $\bar{\psi}=c_1$ and $\bar{\psi}=c_2$ for the two streamlines. However, we know that $c_1$ and $c_2$ are just arbitrary constants of integration that we can set to whatever we want. Anderson just happens to know that later down the road it will be helpful to

define the numerical value of $\bar{\psi}$ such that the difference $\Delta\bar{\psi}$ [between the two streamlines] is equal to the mass flow between the two streamlines.


$\Delta\bar{\psi}$ equaling mass flow (per unit depth) between streamlines is natural. For a steady flow, the mass flow inside a given streamtube is constant along the tube; the mass flow across any cross section of the tube is the same. Since $\Delta\bar{\psi}$ is equal to this mass flow, then $\Delta\bar{\psi}$ itself is constant for a given streamtube.

  • $\begingroup$ OMG, I just found out that it's just defined ;)) I also just know that: with that definition, we can obtain the velocity field from such stream fuctions. Thanks $\endgroup$
    – Dat
    Jul 9, 2017 at 3:30

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