Friction (major) losses

Every metre of pipe steals a little pressure to friction. The Darcy equation puts a number on it — and shows why pipe diameter matters more than anything else.

Source: Rabie, Fluid Power Engineering, Ch. 3.

Before you start

What you need first

Reynolds number and the laminar/turbulent split (Topic 16).

What you'll be able to do

Use the Darcy equation $\Delta p=\lambda\dfrac{L}{D}\dfrac{\rho v^2}{2}$.
Get $\lambda$: $64/\mathrm{Re}$ (laminar) or the Moody chart (turbulent).
Explain why loss $\propto 1/D^4$ for laminar flow.

Start here · the loss formula

Friction loss along the pipe

$$\Delta p = \lambda\,\dfrac{L}{D}\,\dfrac{\rho v^2}{2}$$

where:
Symbol	Meaning	SI unit
$\Delta p$	pressure lost to friction	Pa
$\lambda$	friction factor (dimensionless)	—
L, D	pipe length, bore	m
$\rho$	oil density	kg/m³
v	mean velocity	m/s

The group $\tfrac{\rho v^2}{2}$ is the dynamic pressure — the pressure stored in the oil's motion. The Darcy equation simply says friction eats a fraction $\lambda\dfrac{L}{D}$ of it along the pipe.

Read it: loss grows with length, with the square of speed ($v^2$), and falls with bore. The factor $\lambda$ packages how rough/turbulent the flow is.

Getting the friction factor $\lambda$

It depends on the flow regime (Topic 16):

Laminar ($\mathrm{Re}<2300$): exact, $\ \lambda=\dfrac{64}{Re}$.
Turbulent ($\mathrm{Re}>2300$): read $\lambda$ from the Moody chart, or for smooth pipe use Blasius $\ \lambda=\dfrac{0.316}{Re^{0.25}}$.

The Moody chart is just an experimental graph of $\lambda$ against $\mathrm{Re}$. In turbulent flow $\lambda$ also depends on how rough the pipe wall is, measured as the relative roughness $\varepsilon/D$ — the average bump height $\varepsilon$ divided by the bore $D$. That is why the turbulent side of the chart is a family of curves, one per roughness: pick the curve for your pipe, then read $\lambda$ off it at your $\mathrm{Re}$. Smooth, drawn hydraulic tubing sits near the lowest curve — which is the one Blasius approximates.

Moody chart (sketch): $\lambda$ drops steeply while laminar, then levels into a band of turbulent curves — one for each pipe roughness $\varepsilon/D$ (two shown).

Diameter is king: loss $\propto 1/D^4$

For laminar flow at a fixed flow rate, put the pieces together ($\lambda=64/\mathrm{Re}$, $v=4Q/\pi D^2$) and the diameter dependence is dramatic:

$$\Delta p \;\propto\; \dfrac{1}{D^4}\quad(\text{laminar, fixed }Q)$$

Halve the bore and the friction loss goes up sixteen-fold ($2^4=16$). This is why a slightly bigger pipe is the cheapest cure for a "hot, sluggish" system — far more effective than anything else.

✏️ Try it yourself — no numbers needed

A laminar pressure line runs a bit hot from friction loss. An engineer fits a pipe of half the bore "to save weight," keeping the same flow. What happens to the friction loss, and is this a good idea?

Loss ×16. For laminar flow at fixed $Q$, $\Delta p\propto 1/D^4$; halving $D$ multiplies the loss by $2^4=16$. A bad idea: sixteen times the lost pressure means far more wasted power and heat (and the speed jumps $4\times$, risking turbulence and cavitation). To cool a sluggish line you go bigger, not smaller.

Common mistakes to avoid

Mistake	Fix
Using $\lambda=64/\mathrm{Re}$ for turbulent flow	$64/\mathrm{Re}$ is laminar only; use Moody/Blasius once $\mathrm{Re}>2300$.
Forgetting the $v^2$	Loss scales with the square of velocity.
Using $L$ or $D$ in mm	Use metres throughout; $\Delta p$ comes out in Pa.

Recap — the whole topic on one screen

$$\Delta p=\lambda\dfrac{L}{D}\dfrac{\rho v^2}{2} \qquad \lambda_{\text{lam}}=\dfrac{64}{Re}\qquad \Delta p\propto\dfrac{1}{D^4}$$

Idea	What you own now
Darcy loss	$\Delta p=\lambda(L/D)(\rho v^2/2)$
Friction factor	$64/\mathrm{Re}$ laminar; Moody/Blasius turbulent
Diameter rules	Laminar loss $\propto 1/D^4$ — a tiny bore is very costly

Next topic

Minor losses & total pressure drop

Friction is the loss along the straight pipe. Next we add the losses at the bends, tees and valves, and total everything the pump must overcome.

→ Minor losses & total pressure drop

Mistake	Fix
Using \(\lambda=64/\mathrm{Re}\) for turbulent flow	\(64/\mathrm{Re}\) is laminar only; use Moody/Blasius once \(\mathrm{Re}>2300\).
Forgetting the \(v^2\)	Loss scales with the square of velocity.
Using \(L\) or \(D\) in mm	Use metres throughout; \(\Delta p\) comes out in Pa.