# Question about flow rate in the source flow

• tracker890 Source h
In summary, the conversation discusses the incorrect factor of ##\frac 1 r## in the third line from the end of the equation for ##d\varphi## and clarifies the correct equations for the gradient in cylindrical coordinates and the differential of a function of three variables. It also mentions the dimensions of the gradient and differential and the incorrect dimensions in the second term of the incorrect equation.
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.)

Last edited:
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.

TSny said:
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?

tracker890 Source h said:
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##?

haruspex said:
Isn't ##{\textbf{e}}_\theta=rd\theta##?
Different：
##{\textbf{e}}_\theta## is vector
##rd\theta## is scalar

tracker890 Source h said:
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 said:
$$d\psi=\frac{\partial\psi}{\partial r}dr+\frac{\partial\psi}{\partial\theta}d\theta$$
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 eθ with dθ instead of with rdθ. Check the dimensionality of each.

tracker890 Source h said:
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 $$d\mathrm\psi=\frac{\partial\mathrm\psi}{\partial r}dr+\frac{\partial\mathrm\psi}{\partial\mathrm\theta}d\theta+\frac{\partial\mathrm\psi}{\partial z}dz$$

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

tracker890 Source h said:
$$d\psi=\frac{\partial\psi}{\partial r}dr+\frac{\partial\psi}{\partial\theta}d\theta+\frac{\partial\psi}{\partial z}dz$$
$$\because r\;and\;z=constant\;\;\;\therefore d\psi=\frac{\partial\psi}{\partial\theta}d\theta$$
Yes

tracker890 Source h

## 1. What is flow rate?

Flow rate is the measure of how much fluid (liquid or gas) is flowing through a particular point in a given amount of time. It is typically measured in volume per unit time, such as liters per second or cubic feet per minute.

## 2. How is flow rate calculated?

The flow rate can be calculated by dividing the volume of fluid that passes through a specific point by the amount of time it takes to pass through that point. This can be represented by the equation: flow rate = volume/time.

## 3. What factors affect flow rate?

The flow rate can be affected by a number of factors, including the size and shape of the source flow, the viscosity of the fluid, the pressure and temperature of the fluid, and any obstructions or restrictions in the flow path.

## 4. Why is flow rate important?

Flow rate is important in many scientific and engineering applications, such as in the design of plumbing systems, the study of fluid dynamics, and the measurement of air or water pollution. It is also used in medical settings to monitor the flow of fluids in the body.

## 5. How can flow rate be measured?

Flow rate can be measured using a variety of methods, including flow meters, pressure gauges, and timed volume measurements. The method chosen will depend on the type of fluid being measured and the accuracy required for the specific application.

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