# Pressure Loss in Parallel Pipes

1. May 12, 2016

### cruckshank

Hi, I know this is simple but I'd like someone to clarify for me, because my lecturer wasn't clear:

Obviously when the pipes are in parallel, the head losses across them are the same. But what about the total head loss for the parallel pipes as a whole? I have 3 ideas, but I'm not sure which would be correct:

1) Treating the head losses the same way as electrical resistance, where 1/R=1/R1 + 1/R2... but with head losses instead of resistance?

3) The total head loss across all the parallel pipes is equal to that across just one of the individual pipes?

I feel like it's the third option, but I'm not entirely sure. Can you tell me which is correct (if any), and explain why please.

Thanks.

2. May 12, 2016

### Staff: Mentor

It's #1 except that head loss is a square function of velocity.

3. May 23, 2016

### insightful

But isn't head loss analogous to voltage drop? Parallel resistors have the same voltage drop.

This assumes the parallel pipes have entrances that are all in the same vessel at the same elevation to each other, and the exits are all in another vessel at the same elevation to each other.

I'd say #3.

4. May 23, 2016

### Staff: Mentor

Yes. What about what I said makes you think I'm saying something that disagrees with that?
The way they are worded, #3 could be true as well, yes.

5. May 23, 2016

### insightful

Well, you're agreeing with #1 which is treating head losses the same as electrical resistances. Is that your contention?

6. May 23, 2016

### Staff: Mentor

Yes, but maybe I'm misunderstanding what you were after there since you didn't write-out the formula for the two scenarios: each pipe section has a "resistance" and the new "resistance" of the group of pipes follows the same rules as the resistance for electricity. All must have the same head loss since the pipes are joined before and after. Adding a pipe section reduces the group resistance head loss just like it reduces the resistance and voltage drop for the wires. But it's voltage that is similar to head loss via combining the resistance formula with v=ir.

7. May 23, 2016

### insightful

This seems to say that adding a resistor reduces the voltage (rather than the resistance).

8. May 23, 2016

### Staff: Mentor

Reducing resistance in the pipe section reduces head loss in the pipe section just like reducing resistance in a set of parallel resistors (in a circuit with other resistors) reduces the voltage drop across those resistors.

9. May 23, 2016

### Staff: Mentor

As I said, you didn't give the equations you were comparing to each other, so I interpreted what you were saying, perhaps incorrectly. The two equations are:

v=ir
(voltage, amperage, resistance)
h=fQ2
(head loss, friction factor and flow rate)

"friction factor" is analogous to "resistance" and as I said in post 2, the only difference between the equations is the square function.

10. May 23, 2016

### insightful

Yeah, I think if we had a defined system it would be clear. Say two atmospheric tanks, one with a water level 10m below the other. All parallel pipes between these tanks (with submerged entrances and exits) would have a head loss of 10m regardless of how many pipes you put in.