Return to Answer

The flow rate is the average velocity times the area.

If the velocity was constant, you would get a flow rate that scaled with $r^2$ (the area). But the velocity goes up for larger pipes - in fact, velocity scales with the square of the radius. And the product of these two squares gives us the 4th power relationship.

Let's break this into a few steps:

You may know that the velocity profile is a parabola - that is, if the velocity at the center is $v_0$, then the velocity at a distance $r$ from the center of a pipe with radius $R$ is given by

$$v(r) = v_0\left(1-\left(\frac{r}{R}\right)^2\right)$$

The mean velocity is exactly half the maximum velocity - you can see the proof for this in my earlier answer. So we just need to figure out how the maximum velocity scales with $r^2$. Once we know the velocity profile is quadratic, this is easy - because if we make the pipe a little bit bigger, the profile continues to follow the same parabolic shape with the same curvature.

It remains to prove for ourselves that the parabolic velocity profile is correct. This follows from the fact that the shear stress in a fluid is proportional to the viscosity times the velocity gradient. Looking at an annulus of liquid at a distance $r$ from the center, if there is a velocity gradient $\frac{dv}{dr}$ we know that the total force on the liquid inside the annulus is the pressure times the area, or $F = P\cdot A = P \cdot \pi r^2$. We also know this must equal the force due to the shear, which is the force per unit length of the annulus multiplied by the length of the circumference, so

$$\pi r^2 P = 2\pi r \frac{dv}{dr}\eta$$

It's easy to see this is a differential equation in $v$, with a parabolic solution.

$$\frac{P}{2\eta} r dr = dv\\ v = \frac{Pr^2}{2\eta} + C $$

For the boundary condition $v=0$ at $r=R$, we find that $C=-\frac{PR^2}{2\eta}$. With simple manipulation we find the same expression we had before with

$$v_0 = \frac{PR^2}{2\eta}$$.

So there you have it. Mean velocity is half the peak velocity, and peak velocity goes as the square of the radius; multiplied by the area we have the $R^4$ relationship.

The flow rate is the average velocity times the area.

If the velocity was constant, you would get a flow rate that scaled with $r^2$ (the area). But the velocity goes up for larger pipes - in fact, velocity scales with the square of the radius. And the product of these two squares gives us the 4th power relationship.

Let's break this into a few steps:

You may know that the velocity profile is a parabola - that is, if the velocity at the center is $v_0$, then the velocity at a distance $r$ from the center of a pipe with radius $R$ is given by

$$v(r) = v_0\left(1-\left(\frac{r}{R}\right)^2\right)$$

The mean velocity is exactly half the maximum velocity - you can see the proof for this in my earlier answer. So we just need to figure out how the maximum velocity scales with $r^2$. Once we know the velocity profile is quadratic, this is easy - because if we make the pipe a little bit bigger, the profile continues to follow the same parabolic shape with the same curvature.

It remains to prove for ourselves that the parabolic velocity profile is correct. This follows from the fact that the shear stress in a fluid is proportional to the viscosity times the velocity gradient. Looking at an annulus of liquid at a distance $r$ from the center, if there is a velocity gradient $\frac{dv}{dr}$ we know that the total force on the liquid inside the annulus is the pressure times the area, or $F = P\cdot A = P \cdot \pi r^2$. We also know this must equal the force due to the shear, which is the force per unit length of the annulus multiplied by the length of the circumference, so

$$\pi r^2 P = 2\pi r \frac{dv}{dr}\eta$$

It's easy to see this is a differential equation in $v$, with a parabolic solution.

$$\frac{P}{2\eta} r dr = dv\\ v = \frac{Pr^2}{4\eta} + C $$

For the boundary condition $v=0$ at $r=R$, we find that $C=-\frac{PR^2}{4\eta}$. With simple manipulation we find the same expression we had before with

$$v_0 = \frac{PR^2}{4\eta}$$.

So there you have it. Mean velocity is half the peak velocity, and peak velocity goes as the square of the radius; multiplied by the area we have the $R^4$ relationship.

The flow rate is the average velocity times the area.

If the velocity was constant, you would get a flow rate that scaled with $r^2$ (the area). But the velocity goes up for larger pipes - in fact, velocity scales with the square of the radius. And the product of these two squares gives us the 4th power relationship.

Let's break this into a few steps:

You may know that the velocity profile is a parabola - that is, if the velocity at the center is $v_0$, then the velocity at a distance $r$ from the center of a pipe with radius $R$ is given by

$$v(r) = v_0\left(1-\left(\frac{r}{R}\right)^2\right)$$

The mean velocity is exactly half the maximum velocity - you can see the proof for this in my earlier answer. So we just need to figure out how the maximum velocity scales with $r^2$. Once we know the velocity profile is quadratic, this is easy - because if we make the pipe a little bit bigger, the profile continues to follow the same parabolic shape with the same curvature.

It remains to prove for ourselves that the parabolic velocity profile is correct. This follows from the fact that the shear stress in a fluid is proportional to the viscosity times the velocity gradient. Looking at an annulus of liquid at a distance $r$ from the center, if there is a velocity gradient $\frac{dv}{dr}$ we know that the total force on the liquid inside the annulus is the pressure times the area, or $F = P\cdot A = P \cdot \pi r^2$. We also know this must equal the force due to the shear, which is the force per unit length of the annulus multiplied by the length of the circumference, so

$$\pi r^2 P = 2\pi r \frac{dv}{dr}\eta$$

It's easy to see this is a differential equation in $v$, with a parabolic solution. For the boundary condition $v=0$ at $r=R$, we find the same expression we had before.

So there you have it. Mean velocity is half the peak velocity, and peak velocity goes as the square of the radius; multiplied by the area we have the $r^4$ relationship.

The flow rate is the average velocity times the area.

If the velocity was constant, you would get a flow rate that scaled with $r^2$ (the area). But the velocity goes up for larger pipes - in fact, velocity scales with the square of the radius. And the product of these two squares gives us the 4th power relationship.

Let's break this into a few steps:

You may know that the velocity profile is a parabola - that is, if the velocity at the center is $v_0$, then the velocity at a distance $r$ from the center of a pipe with radius $R$ is given by

$$v(r) = v_0\left(1-\left(\frac{r}{R}\right)^2\right)$$

The mean velocity is exactly half the maximum velocity - you can see the proof for this in my earlier answer. So we just need to figure out how the maximum velocity scales with $r^2$. Once we know the velocity profile is quadratic, this is easy - because if we make the pipe a little bit bigger, the profile continues to follow the same parabolic shape with the same curvature.

It remains to prove for ourselves that the parabolic velocity profile is correct. This follows from the fact that the shear stress in a fluid is proportional to the viscosity times the velocity gradient. Looking at an annulus of liquid at a distance $r$ from the center, if there is a velocity gradient $\frac{dv}{dr}$ we know that the total force on the liquid inside the annulus is the pressure times the area, or $F = P\cdot A = P \cdot \pi r^2$. We also know this must equal the force due to the shear, which is the force per unit length of the annulus multiplied by the length of the circumference, so

$$\pi r^2 P = 2\pi r \frac{dv}{dr}\eta$$

It's easy to see this is a differential equation in $v$, with a parabolic solution.

$$\frac{P}{2\eta} r dr = dv\\ v = \frac{Pr^2}{2\eta} + C $$

For the boundary condition $v=0$ at $r=R$, we find that $C=-\frac{PR^2}{2\eta}$. With simple manipulation we find the same expression we had before with

$$v_0 = \frac{PR^2}{2\eta}$$.

So there you have it. Mean velocity is half the peak velocity, and peak velocity goes as the square of the radius; multiplied by the area we have the $R^4$ relationship.

The flow rate is the average velocity times the area.

If the velocity was constant, you would get a flow rate that scaled with $r^2$ (the area). But the velocity goes up for larger pipes - in fact, velocity scales with the square of the radius. And the product of these two squares gives us the 4th power relationship.

Let's break this into a few steps:

You may know that the velocity profile is a parabola - that is, if the velocity at the center is $v_0$, then the velocity at a distance $r$ from the center of a pipe with radius $R$ is given by

$$v(r) = v_0\left(1-\left(\frac{r}{R}\right)^2\right)$$

The mean velocity is exactly half the maximum velocity - you can see the proof for this in my earlier answer. So we just need to figure out how the maximum velocity scales with $r^2$. Once we know the velocity profile is quadratic, this is easy - because if we make the pipe a little bit bigger, the profile continues to follow the same parabolic shape with the same curvature.

It remains to prove for ourselves that the parabolic velocity profile is correct. This follows from the fact that the shear stress in a fluid is proportional to the viscosity times the velocity gradient. Looking at an annulus of liquid at a distance $r$ from the center, if there is a velocity gradient $\frac{dv}{dr}$ we know that the total force on the liquid inside the annulus is the pressure times the area, or $F = P\cdot A = P \cdot \pi r^2$. We also know this must equal the force due to the shear, which is the force per unit length of the annulus multiplied by the length of the circumference, so

$$\pi r^2 P = 2\pi r \frac{dv}{dr}\eta$$

It's easy to see this is a differential equation in $v$, with a parabolic solution. For the boundary condition $v=0$ at $r=R$, we find the same expression we had before.

So there you have it. Mean velocity is half the peak velocity, and peak velocity goes as the square of the radius; multiplied by the area we have the $r^4$ relationship.

The flow rate is the average velocity times the area.

If the velocity was constant, you would get a flow rate that scaled with $r^2$ (the area). But the velocity goes up for larger pipes - in fact, velocity scales with the square of the radius. And the product of these two squares gives us the 4th power relationship.

Let's break this into a few steps:

You may know that the velocity profile is a parabola - that is, if the velocity at the center is $v_0$, then the velocity at a distance $r$ from the center of a pipe with radius $R$ is given by

$$v(r) = v_0\left(1-\left(\frac{r}{R}\right)^2\right)$$

The mean velocity is exactly half the maximum velocity - you can see the proof for this in my earlier answer. So we just need to figure out how the maximum velocity scales with $r^2$. Once we know the velocity profile is quadratic, this is easy - because if we make the pipe a little bit bigger, the profile continues to follow the same parabolic shape with the same curvature.

It remains to prove for ourselves that the parabolic velocity profile is correct. This follows from the fact that the shear stress in a fluid is proportional to the viscosity times the velocity gradient. Looking at an annulus of liquid at a distance $r$ from the center, if there is a velocity gradient $\frac{dv}{dr}$ we know that the total force on the liquid inside the annulus is the pressure times the area, or $F = P\cdot A = P \cdot \pi r^2$. We also know this must equal the force due to the shear, which is the force per unit length of the annulus multiplied by the length of the circumference, so

$$\pi r^2 P = 2\pi r \frac{dv}{dr}\eta$$

It's easy to see this is a differential equation in $v$, with a parabolic solution. For the boundary condition $v=0$ at $r=R$, we find the same expression we had before.

So there you have it. Mean velocity is half the peak velocity, and peak velocity goes as the square of the radius; multiplied by the area we have the $r^4$ relationship.