After combining FM Global's orifice equation[1] (which is a modified form
of Pitot's 1732 equation )
with the Bernoulli equation (presented in ~ 1738)
(apparently those two guys were doing a lot of studying too, Steve,
back in the day)
one rearranges and solves for P, pitot through substitution of velocity.
P,pitot = (P1 – P2)* (2/ρ)*( π/{4*a*C})^2 eq (1a)
While not a perfect reflection of reality (no friction or turbulence
considered), if we assume
P1 = P,static [Pa]
P2 = P,residual [Pa]
C is the hydrant discharge coefficient that FM states varies from 0.6 ->
0.9 [2]
and the constant term (2/ρ)*( π/{4*a*C})^2 is 1.24 and is
dimensionless
found by using a density, ρ, for sweet water, C=0.9 and 'a' = 0.035
which is converted from FM's units of {bar,L,min} to SI.
The equation (1a) suggests that, yup, there are conditions where P, pitot
will be larger than P,resdual, even with flat ground.
Those conditions become easier to create when the hydrant discharge
coefficient gets smaller (when there is more non-uniform water discharge) or
when the pressure drop is larger (created either by larger water
discharge or a very small cross-sectional gravity-source of water or a pump
with a really steep performance curve).
The small diameter water source is impractical with money constraints,
and not many listed fire pumps have extremely steep performance-curves.
Equation (1a) suggests:
a). with an inset hydrant outlet (C = 0.6) and a P,static of 5 bar (75
psi), the pitot pressure matches the residual pressure IF the residual
pressure drop is about 27%.
b). with a smooth well-rounded hydrant outlet (C=0.9) and the same
P,static, the pitot pressure matches the residual pressure if the residual
pressure drop is 45%.
I have seldom got anywhere near pressure drops that large, it would take a
large opening or openings.
ICYMI, pipe diameter falls out of the set of independent variables (it
doesn't influence the results)... at least in theory.
NFPA 291 recommends a 25% pressure drop, and AWWA recommends at least 10
psi.
Even at the 25% pressure drop recommended by NFPA 291, theory suggests we
will not see P,pitot closely approach P,residual unless the hydrant
discharge is very uneven.
The remaining question I have is... "why NFPA 291 does not openly endorse
P,residual and P,pitot at the same hydrant?". It could be another case of,
"we have always done it
that way." Given the previously mentioned seldom fully-accounted for
parameters in water flow testing, it seems that the turbulence issue at the
proposed single-measurement
hydrant would be an issue of minor consequence. Australia tests pitot and
residual pressure at the same hydrant, but maybe that is because their
water curls in the other direction.
[1]. FM Global Data Sheet 3-0, pp. 47, equation 1
[2]. FM Global, op. cit., pp. 52, Table 4, row 3
Scot Deal
Excelsior Risk & Fire Engineering
gms: +420 606 872 129 (GMT + 1)
On Thu, Jan 3, 2019 at 12:15 AM Steve Leyton <[email protected]>
wrote:
> I will leave the science part to Scot and Cecil; my college thesis was
> titled, “Our Friend the Beaver”. And that was in architecture school …
>
>
>
> You’re right about the basic theorem, hence my question/assumption about
> the large diameter main feeding the test hydrants. It’s not uncommon to
> have extremely generous flows with low static and residual pressures if the
> main is 12”, 16” and above. Bernoulli indeed …
>
>
>
> SML
>
>
>
>
>
> *From:* Sprinklerforum [mailto:
> [email protected]] *On Behalf Of *Skyler
> Bilbo
> *Sent:* Wednesday, January 02, 2019 11:20 AM
> *To:* [email protected]
> *Subject:* Re: Pitot Pressure Above Residual Pressure
>
>
>
> Steve,
>
>
>
> This was actually very helpful. I was thinking of it wrong. Our pitots
> measure velocity pressure. The gauge on the test hydrant is measuring
> normal pressure inside of the pipe, or hydrant. I think I have it sorted,
> but feel free to correct me. A better explanation is below.
>
>
>
> -The normal pressure is the pressure acting on the walls of the pipe, and
> is what is typically measured with our regular gauges.
>
> -The velocity pressure is the pressure acting on anything that is
> perpendicular to the direction of flow, like one of our pitot gauges (it
> would be the pressure you would feel pushing you if you tried to stand in
> front of a flowing hydrant)
>
> -The total pressure is both of these things combined.
>
>
>
> Velocity pressure goes up as you increase the velocity of the water, which
> can be accomplished by going from a large pipe to a small one (like going
> from an 8" water main to a 2-1/2" connection on a fire hydrant; 1,000 GPM
> in an 8" main travels at about 5.96 ft/sec, which equals a velocity
> pressure of 0.24 psi; 1000 GPM comes out of a 2-1/2" hydrant at about 65
> ft/sec *that's why it shoots out so far* with a velocity pressure of about
> 28.8 psi, which is a pitot pressure of about 35.5 psi, if the opening
> coefficient is 0.9). This velocity pressure is dependent on the velocity
> of the water.
>
>
>
> I was wrong in my original thinking. Hopefully my explanation is useful
> to others.
>
>
>
> I don't think the pitot reading should/could ever be larger than the
> static pressure, however (assuming elevation is the same, no additional
> water supplies kick on, and no negative gauge pressure possible), due to
> conservation of energy. The static pressure is the total pressure when no
> water is flowing, and no matter how much water is flowing after that, no
> combination of velocity pressure or normal pressure could ever exceed this
> total pressure.
>
>
>
>
>
> Thanks guys,
>
> Skyler Bilbo
>
>
>
>
>
> On Wed, Jan 2, 2019 at 11:33 AM Steve Leyton <[email protected]>
> wrote:
>
> Pitot measures velocity pressure, residual is atmospheric pressure.
> There’s not a fixed correlation between the two values – I’m guessing that
> the main supplying the test hydrants is a very large diameter one?
>
>
>
> Steve Leyton
>
>
>
>
>
> _______________________________________________
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>
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>
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