What is the pressure at the tap due to Darcy losses?

[physea]

Homework Statement

There is a tap that supplies water to a L=21m hose. There are friction losses in the hose due to friction (Darcy losses).
f = 0.0138
ρ = 1000 kg/m3
diameter d= 0.025m
water velocity V=2.54m/s
What is the pressure at the tap?

Homework Equations

Darcy losses: h= 4fLV^2/d2g

The Attempt at a Solution

My solution says that the pressure energy in the tap will have to maintain a flow of 2.54m/s and overcome the Darcy losses.
Darcy pressure losses are h=4fLV^2/d2g, and since P=ρgh, we get P(darcy)=4fLgV^2/2d
As a result P(tap)=V^2/2g + 4fLgV^2/2d = 486Pa
But it doesn't seem correct, any idea?

Apparently, the correct solutions says P=ρgh(darcy)

 

[BvU]

But it doesn't seem correct

No, indeed. Just checking the dimensions might help ...

And: you sure about the factor 4 (always confusing: Darcy/Fanning) ?
(never mind - from the value of Re and f, I think you're right.

 

[haruspex]

=V^2/2g +

How do you justify that term?

 

[physea]

How do you justify that term?

Basically from energy conservation.
If no Darcy losses were available, it would be PV=1/2*m*U^2, which divided by m, makes P=rho*U^2/2, which since P=rho*g*h, it becomes h=U^2/2g
To this, I add the Darcy head loss.

Is this approach correct?

 

[BvU]

We are talking about friction losses in a pipe -- not your everyday example of a process where energy conservation can be used in the calculations.

A few more issues:

Apparently, the correct solutions says P=ρgh(darcy)

At the tap ?

It's not $p = \rho g h$ but $\Delta p =\rho g \Delta h$ where in this case the friction loss is/can be converted to head loss $\Delta h$. Confusing but possible.

They ask for the pressure at the tap. That is pressure at the exit + pressure loss due to friction (plus a possible hydrostatic head, but you may assume the pipe is horzontal.)

 

[BvU]

h=U^2/2g

Yeah, that's what you get if you spout vertically. We're in a different scenario here !

 

[haruspex]

Basically from energy conservation.

And that term represents a change in KE? Is there a change in KE?

 

[physea]

And that term represents a change in KE? Is there a change in KE?

No, volumetric water flow is stable.
I assume that the pressure at the end of the hose is zero... well atmospheric

 

[haruspex]

No, volumetric water flow is stable.
I assume that the pressure at the end of the hose is zero... well atmospheric

Then I see no reason for that term. The total energy has the same ½ρv² at beginning and end, so the two cancel.