Question about flow rate in the source flow

[tracker890 Source h]

Homework Statement:

Determine flow rate per unite width in the source flow

Relevant Equations:

flow rate equation

Please help me to understand what wrong with method 2.
ref.Flowrate Between Streamlines
(Thank you for your time and consideration.)

 

[TSny]

In the third line from the end where you have an equation for $d\varphi$, the last term should not have a factor of $\frac 1 r$. Note how the factor of $\frac 1 r$ makes the term have the wrong dimensions.

 

[tracker890 Source h]

In the third line from the end where you have an equation for $d\varphi$, the last term should not have a factor of $\frac 1 r$. Note how the factor of $\frac 1 r$ makes the term have the wrong dimensions.

But about the gradent in cylindrical coordinates is $\nabla\mathrm\psi=\frac{\partial\mathrm\psi}{\partial r}{\textbf{e}}_r+\frac1r\frac{\partial\mathrm\psi}{\partial\mathrm\theta}{\textbf{e}}_\theta+\frac{\partial\mathrm\psi}{\partial z}{\textbf{e}}_z$, so $d\mathrm\psi=\frac{\partial\mathrm\psi}{\partial r}dr+\frac1r\frac{\partial\mathrm\psi}{\partial\mathrm\theta}d\theta+\frac{\partial\mathrm\psi}{\partial z}dz$.
Is the above thinking wrong?

 

[haruspex]

But about the gradent in cylindrical coordinates is $\nabla\mathrm\psi=\frac{\partial\mathrm\psi}{\partial r}{\textbf{e}}_r+\frac1r\frac{\partial\mathrm\psi}{\partial\mathrm\theta}{\textbf{e}}_\theta+\frac{\partial\mathrm\psi}{\partial z}{\textbf{e}}_z$, so $d\mathrm\psi=\frac{\partial\mathrm\psi}{\partial r}dr+\frac1r\frac{\partial\mathrm\psi}{\partial\mathrm\theta}d\theta+\frac{\partial\mathrm\psi}{\partial z}dz$.
Is the above thinking wrong?

Isn't ${\textbf{e}}_\theta=rd\theta$?

 

[tracker890 Source h]

Isn't ${\textbf{e}}_\theta=rd\theta$?

Different：
${\textbf{e}}_\theta$ is vector
$rd\theta$ is scalar

 

[haruspex]

Different：
${\textbf{e}}_\theta$ is vector
$rd\theta$ is scalar

Yes, I forgot to make the dθ bold. But the point is you need a factor r.

 

[tracker890 Source h]

$$d\psi=\frac{\partial\psi}{\partial r}dr+\frac{\partial\psi}{\partial\theta}d\theta$$
reference as follows：
he Gradient in Polar Coordinates and other Orthogonal Coordinate Systems

 

[haruspex]

$$d\psi=\frac{\partial\psi}{\partial r}dr+\frac{\partial\psi}{\partial\theta}d\theta$$
reference as follows：
he Gradient in Polar Coordinates and other Orthogonal Coordinate Systems

Yes, for the case of constant z; so?
Compare that with what you wrote in post #3. You have a 1/r factor on the second term which should not be there. It appears in your version because you wrongly replaced e_θ with dθ instead of with rdθ. Check the dimensionality of each.

 

[TSny]

But about the gradent in cylindrical coordinates is $\nabla\mathrm\psi=\frac{\partial\mathrm\psi}{\partial r}{\textbf{e}}_r+\frac1r\frac{\partial\mathrm\psi}{\partial\mathrm\theta}{\textbf{e}}_\theta+\frac{\partial\mathrm\psi}{\partial z}{\textbf{e}}_z$, so $d\mathrm\psi=\frac{\partial\mathrm\psi}{\partial r}dr+\frac1r\frac{\partial\mathrm\psi}{\partial\mathrm\theta}d\theta+\frac{\partial\mathrm\psi}{\partial z}dz$.
Is the above thinking wrong?

-----------------
$\nabla\mathrm\psi=\frac{\partial\mathrm\psi}{\partial r}{\textbf{e}}_r+\frac1r\frac{\partial\mathrm\psi}{\partial\mathrm\theta}{\textbf{e}}_\theta+\frac{\partial\mathrm\psi}{\partial z}{\textbf{e}}_z$

This equation is correct. It expresses the gradient of $\psi$. The gradient of $\psi$ is a vector quantity. The dimensions of the gradient of $\psi$ are the dimensions of $\psi$ divided by distance. You can check that each term on the right of the equation has these dimensions.
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$d\mathrm\psi=\frac{\partial\mathrm\psi}{\partial r}dr+\frac1r\frac{\partial\mathrm\psi}{\partial\mathrm\theta}d\theta+\frac{\partial\mathrm\psi}{\partial z}dz$

This equation is not correct. $d \psi$ is the differential of psi. This is a scalar quantity that represents a small change in $\psi$ when the variables $r, \theta$ and $z$ are varied by $dr$, $d\theta$, and $dz$ respectively. The dimensions of $d \psi$ are the same as the dimensions of $\psi$. Each term on the right side of the equation should have dimensions of $\psi$. But the second term on the right side of your equation has dimensions of $\psi$ divided by distance ($r$).

For an arbitrary function $f$ of three variables $u, v, w$, the differential of $f(u, v, w)$ is $$df = \frac{\partial f}{\partial u} du + \frac{\partial f}{\partial v}dv + \frac{\partial f}{\partial w}dw $$ So $d \psi$ is $$d\mathrm\psi=\frac{\partial\mathrm\psi}{\partial r}dr+\frac{\partial\mathrm\psi}{\partial\mathrm\theta}d\theta+\frac{\partial\mathrm\psi}{\partial z}dz$$

 

[tracker890 Source h]

-----------------
$\nabla\mathrm\psi=\frac{\partial\mathrm\psi}{\partial r}{\textbf{e}}_r+\frac1r\frac{\partial\mathrm\psi}{\partial\mathrm\theta}{\textbf{e}}_\theta+\frac{\partial\mathrm\psi}{\partial z}{\textbf{e}}_z$

This equation is correct. It expresses the gradient of $\psi$. The gradient of $\psi$ is a vector quantity. The dimensions of the gradient of $\psi$ are the dimensions of $\psi$ divided by distance. You can check that each term on the right of the equation has these dimensions.
------------------

$d\mathrm\psi=\frac{\partial\mathrm\psi}{\partial r}dr+\frac1r\frac{\partial\mathrm\psi}{\partial\mathrm\theta}d\theta+\frac{\partial\mathrm\psi}{\partial z}dz$

This equation is not correct. $d \psi$ is the differential of psi. This is a scalar quantity that represents a small change in $\psi$ when the variables $r, \theta$ and $z$ are varied by $dr$, $d\theta$, and $dz$ respectively. The dimensions of $d \psi$ are the same as the dimensions of $\psi$. Each term on the right side of the equation should have dimensions of $\psi$. But the second term on the right side of your equation has dimensions of $\psi$ divided by distance ($r$).

For an arbitrary function $f$ of three variables $u, v, w$, the differential of $f(u, v, w)$ is $$df = \frac{\partial f}{\partial u} du + \frac{\partial f}{\partial v}dv + \frac{\partial f}{\partial w}dw $$ So $d \psi$ is $$d\mathrm\psi=\frac{\partial\mathrm\psi}{\partial r}dr+\frac{\partial\mathrm\psi}{\partial\mathrm\theta}d\theta+\frac{\partial\mathrm\psi}{\partial z}dz$$

Thank you for your clear answer, so $d\psi$ is
$$d\psi=\frac{\partial\psi}{\partial r}dr+\frac{\partial\psi}{\partial\theta}d\theta+\frac{\partial\psi}{\partial z}dz$$
$$ \because r\;and\;z=constant\;\;\;\therefore d\psi=\frac{\partial\psi}{\partial\theta}d\theta $$

 

[TSny]

$$d\psi=\frac{\partial\psi}{\partial r}dr+\frac{\partial\psi}{\partial\theta}d\theta+\frac{\partial\psi}{\partial z}dz$$
$$ \because r\;and\;z=constant\;\;\;\therefore d\psi=\frac{\partial\psi}{\partial\theta}d\theta $$

Yes