Minor losses & total pressure drop

Friction is the loss along the straight pipe; minor losses happen at the bends, tees and valves. Add them all and you know the pressure the pump must really supply.

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

Before you start

What you need first

Friction (major) loss, the Darcy equation (Topic 17).

What you'll be able to do

Use $\Delta p=\xi\,\dfrac{\rho v^2}{2}$ for a fitting.
Add up the total pressure drop.
Say when friction vs minor losses dominate.

Start here · local losses

Loss at a bend, tee or valve

Each fitting disturbs the flow and costs a pressure drop set by a loss coefficient $\xi$ (xi) times the dynamic pressure:

$$\Delta p = \xi\,\dfrac{\rho v^2}{2}$$

where:
Symbol	Meaning	SI unit
$\xi$	loss coefficient of the fitting	—
$\tfrac{\rho v^2}{2}$	dynamic pressure of the flow	Pa

The dynamic pressure $\rho v^2/2$ is the pressure the oil carries simply because it is moving — what you would recover if you brought the flow to rest. It grows with the square of speed, so faster oil carries far more of it; a fitting wastes a fraction $\xi$ of it.

Friction loss builds along the straight run; each fitting (here a 90° elbow) adds a minor loss.

Fitting	$\xi$ (Rabie, Table 3.3)
Flexible pipe connection	0.3
Pipe inlet (sharp entrance)	0.5 – 1
Pipe outlet	1
90° elbow (standard)	1.2 – 1.3
Tee junction	3.5
Screen filter	1.5 – 2.5

Valves vary widely (general data, not in Rabie's table): a fully-open gate valve $\xi\approx0.2$, an open globe valve $\xi\approx10$.

Total pressure drop the pump must supply

Add the friction loss along every run to the minor loss of every fitting — and remember the pump must also supply the load pressure at the actuator:

$$p_{\text{pump}} = p_{\text{load}} + \Delta p_{\text{friction}} + \textstyle\sum \Delta p_{\text{minor}}$$

The losses are "wasted" pressure — they turn into heat. Keeping them small (short, fat, straight, few fittings) lets the pump run at lower pressure and power, and cooler.

Friction or minor — which dominates?

Long straight runs → friction dominates (it scales with $L/D$). On a typical multi-metre line, friction can be ten-plus times the loss of a single elbow.
Short runs packed with fittings (a compact valve block) → the minor losses can be the larger share.

Always check both, but know the usual story: on a real machine's pipe runs, friction is normally the big one — which is why bore (Topic 17's $1/D^4$) matters so much.

✏️ Try it yourself — no numbers needed

A designer is told the pump pressure is "a bit high." The line is a long, thin run with just two elbows. Should they first attack the friction loss or the minor losses — and what is the single most effective change?

Attack friction first. On a long, thin run friction dominates (it scales with $L/D$ and, laminar, with $1/D^4$); two elbows are a small share. Most effective change: increase the bore — because the laminar friction loss falls as $1/D^4$, a modest bore increase slashes the loss far more than removing an elbow would.

Common mistakes to avoid

Mistake	Fix
Forgetting the load pressure	The pump supplies load plus all losses, not just losses.
Using a different $v$ for friction and minor	Same line, same $v$ in both $\rho v^2/2$ terms.
Counting one $\xi$ for a tee used as a branch and a through-run	They have different $\xi$ values — use the right one.

Recap — the whole topic on one screen

$$\Delta p_{\text{minor}}=\xi\dfrac{\rho v^2}{2} \qquad p_{\text{pump}}=p_{\text{load}}+\Delta p_{\text{friction}}+\textstyle\sum\Delta p_{\text{minor}}$$

Idea	What you own now
Minor loss	$\xi\,(\rho v^2/2)$ per fitting; sum them
Total drop	Load + friction + Σ minor = pump pressure
Who wins	Long runs → friction; compact fitting blocks → minor

Next — Theme 4

Pumps

You can now move the oil and account for what the lines cost. Theme 4 turns to the component that makes the flow in the first place — the pump.

→ Positive displacement & pump flow

Symbol	Meaning	SI unit
\(\xi\)	loss coefficient of the fitting	—
\(\tfrac{\rho v^2}{2}\)	dynamic pressure of the flow	Pa

Mistake	Fix
Forgetting the load pressure	The pump supplies load plus all losses, not just losses.
Using a different \(v\) for friction and minor	Same line, same \(v\) in both \(\rho v^2/2\) terms.
Counting one \(\xi\) for a tee used as a branch and a through-run	They have different \(\xi\) values — use the right one.

Idea	What you own now
Minor loss	\(\xi\,(\rho v^2/2)\) per fitting; sum them
Total drop	Load + friction + Σ minor = pump pressure
Who wins	Long runs → friction; compact fitting blocks → minor