# Why friction losses depend on velocity squared?

This is the reason that suction pipes of centrifugal pumps are most of the times much wider than the discharge pipes and that restriction of blood flow in arteries is always an alarming event.

Does anybody know any mathematical proof or explanation why friction losses depend on fluid velocity squared?

• It is only a crude approximation. For laminar flow, the frictional loss is proportional to velocity to the first power. For turbulent flow in smooth pipes, the power is between 1 and 2. For turbulent flow in rough pipes, the power approaches 2 at high Reynolds numbers. – Chet Miller Jun 18 at 18:39

I believe the dependence on fluid velocity squared is only true for well developed turbulent flow. Consider the Darcy-Weisbach equation for flow through pipes:

$$\Delta h_f=f \frac{L}{D}\frac{V^2}{2g}\tag{1}$$

For laminar flow through circular pipes:

$$f=\frac{64}{\text{Re}}$$

With $$\text{Re}$$ the dimensionless Reynolds number: $$\text{Re}=\frac{VD}{\nu}$$

Insertion into $$(1)$$ reduces the exponent of $$V$$ to $$1$$.

According to this source the exponent $$2$$ has been arrived by observation:

By observation, the head loss is roughly proportional to the square of the flow rate in most engineering flows (fully developed, turbulent pipe flow). This observation leads to the Darcy-Weisbach equation for head loss due to friction

The DW equation can also be derived with the Buckingham $$\pi$$ theorem by means of two dimensionless groups:

$$\Big(\frac{L}{D}\Big)\text{ and }\Big(\frac{V^2}{2g\Delta h_f}\Big)$$

So that:

$$\Big(\frac{L}{D}\Big)\times \Big(\frac{V^2}{2g\Delta h_f}\Big)=\text{constant}$$

which can be reworked to $$(1)$$.

Hope this helps.

• Thanks for the upvote. Edit made. – Gert Jun 17 at 22:19