# Homework Help: Understanding pipe ''head loss''

1. Jan 3, 2017

### axe34

1. The problem statement, all variables and given/known data
I am trying to understand a problem that I made for myself (although this is based on other problems I've encountered).

We have a tank (reservoir) 2.9534 m in height and a head loss of 2.88 m from beginning to end; the loss is due to friction etc. There is discharge to atmosphere; see diagram 1 below

The velocity of water at the exit is found via bernoulli's equations:
V1^2/2g + Z1 + p1/rho.g - losses = V2^2/2g + Z2 + p2/rho.g

With respect to a streamline from the top of the reservoir to the exit of the pipe, I get V2 as 1.2 m/s (with V1 = 0, Z1 = 2.9534, losses = 2.88, Z2 = 0 and p1 and p2 = atmospheric pressure).

With respect to diagram 2, I always thought that head loss at any point was 2.9534 - H. Say I move this piezometer to the right hand side with H = 2.88 m then I get the same velocity as before using bernoulli equations. Why? This isn't the definition of head loss?

2. Relevant equations
As above

3. The attempt at a solution
As above

2. Jan 9, 2017

### BvU

Hi axe,
since you didn't get a reply from others, I am taking this up again; see if we can make some progress.

But I want to deal with this understanding part separately and treat it as a steady-state situation. Because:

I think part of our problem in understanding all this comes from the mixup between a steady-state treatment and the dynamic treatment .
Steady state is already pretty involved; generally dynamics are considered in a later stage and for special circumstances (surge, water hammer, process control and such). And even then we often find the dynamic situation is approached as a sequence of steady states.

Let's start witrh the simplest possible mechanical energy balance $E_{\rm pot} + E_{\rm kin} = {\rm Constant}\ \Rightarrow\ {dE\over dt} = 0$ which often appears as a simplified form of the Bernoulli equation. However, there are considerable losses due to friction etc.

From the top of the reservoir to the outflow you get $\rho \,\Phi_V \, g\;{\rm height} - {\rm losses} = \rho\, \Phi_v {1\over 2}\, v^2$ ($\Phi_V$ is the volume flow} and with $v = 1.2$ m/s the kinetic energy of the outflow is pretty small compared to the available head.

In fact the $v$ at the outflow is the result of a balance between head and head loss as also described in your section 2: $\Delta H = H - H_f$ for steady state becomes $H - H_f = 0$ (no more acceleration). $\Delta H$ is NOT a head loss but the remaining head that drives the acceleration. The total head loss is $H_f$. Very confusing due to the dynamic consideration.

Head loss is not a characteristic at a point, but the difference in head between two points. So I would change
into "The head loss wrt the reservoir surface is .. " And as you can see here in the third picture, the piezometer doesn't show the head but instead shows head minus $v^2 \over 2g$.

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Now in the steady state situation you have (see 2.3) and (2.4) $\Delta H = 0\ \Rightarrow\ {\rm height} = 4 f {l\over d} {v^2\over 2g}$ where $f$ depends on $v$ (If the flow is Laminar: $f =\displaystyle { 16\mu\over \rho v d}$ ; for the 16 see here)

3. Jan 11, 2017

### CivilSigma

The simplest way to look at head loss is to analyse what is happening in direction of water flow. And no, the head loss is not constant throughout the entire system.

First of all , head loss is defined as the frictional head loss and local head-loss (energy lost due to fittings etc..).

$$H_f = h_l + h_f= k_f \cdot \frac{v^2}{2g} + \frac{\lambda \cdot L}{D} \cdot \frac{v^2}{2g}$$

In the equation $$h_f$$ is the local energy loss and $$k_f$$ is the local losses coefficient, for example k=1 for sudden expansion from the pipe to the reservoir. $$h_f$$ is the frictional head losses and as you can see it the equation it is a function of the length of the piping system.

Also, when we look at the system, you "read" it in direction of water flow and depending on where we do our analysis, we will have different frictional losses.

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For figure #2, you really can't perform the calculations because you don't know the head loss between the segments (you simply assumed it is 0 which is inconsistent with the original question that frictional losses exist). In reality and for sake of consistency with the question, you would of had a LARGER velocity calculated

Last edited: Jan 11, 2017