# Thermofluids, bernoullis equation head loss

1. Apr 1, 2015

### TheRedDevil18

Hi, I just have a few questions on bernoullis equation dealing with power and head loss

If the equation for power is p = mass flow rate * g * Hloss, Then is g only used when lets say a pipe is at an elevation ?, because in my book I have an example where g is not used to calculate the power and I assumed it was because the pipe was level. Also what exactly is Hloss and why is it sometimes measured in metres and sometimes in kilowatts ?, doesn't it have to do with friction so it should be in joules

2. Apr 1, 2015

### Alex.malh

Hello,

Can you tell me which units you use in your equation?
Normally its:
P [W] = Q [m³/s] x dP [N/m²]
[W] = [Nm/s]

Hloss is cannot be in watt. Or in m or in Pa.
If its in watt it is probably recalculated from pressure or head loss to power.

3. Apr 1, 2015

### TheRedDevil18

p(W) = mass flow rate(kg/s) * g(m/s2) * Hloss(m)

How does that give you watts ?

4. Apr 1, 2015

### Alex.malh

Yes, that is watt.

[kg] x [m/s²] = [N] = m x a = F
you can rewrite your equation as:
1/s x (kg x m/s²) x m
=1/s x N x m
=Nm/s
=W

5. Apr 1, 2015

### Staff: Mentor

$\frac{kgm}{s^2}$ = N
N x m / sec = watts

Last edited: Apr 1, 2015
6. Apr 1, 2015

### Alex.malh

Don't you mean N instead of Pa?

7. Apr 1, 2015

### Staff: Mentor

Yes. Sorry. I'll go back and edit it. Thanks for catching that.

Chet

8. Apr 1, 2015

### Staff: Mentor

The head loss is often used as a surrogate for the pressure loss. The head loss (in meters) is equal to the pressure loss in Pa divided by the (fluid density (kg/m^3) times g (m/s^2). So H = ΔP/(ρg). You equation for the power is equivalent to the pressure loss times the volumetric flow rate of fluid.

Chet

9. Apr 1, 2015

### TheRedDevil18

Ok, thanks. In an example from my book the equation for power was given as p = mass flow rate*Hloss, without the g but the units for Hloss was given in J/kg so multiplying that by the mass flow rate would give J/s = power. So basically these equations are just based on what units you are given ?

Also in benoullis equation which is

p1/pg + vel1^2/2g + z1 = p2/pg + vel2^2/2g + z2 + Hloss

You say that p/pg = Hloss, then why is it added again on the right hand side of the equation ?

Last edited: Apr 1, 2015
10. Apr 1, 2015

### Staff: Mentor

If the g is omitted, then there is either a mistake or they are using Imperial units.
This depends on the context, which hasn't been provided. However, in this equation, it looks like they are using Hloss to represent the pressure drop resulting from "hydrodynamic frictional drag."

Chet

11. Apr 1, 2015

### TheRedDevil18

So is p/pg just a different type of Hloss ? ,because if it where the same then why would they add it twice

12. Apr 1, 2015

### Staff: Mentor

The pressures p1 and p2 appearing in this equation are the measured pressures at two locations along a flow channel. The pressure change between these two locations is determined by the change in elevation, the change in kinetic energy, and the frictional pressure loss.

Chet