In constructing a fountain, how does one calcualte
the horsepower pump motor needed to result in a
spray effect six feet in height ejecting from a one
eight inch diameter nozzle?

do you mean a one-eighth inch diameter nozzle?

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Hello,

   I'm going to treat this in the mks system of units, so as to get
the wattage required to run your fountain. I figure you need to know
this anyway, and it can be converted to horsepower at the end if
that's really necessary.

   You specify a jet of water squirting vertically from a 1/8 inch
diameter hole to a height of 6 feet. This translates to a diameter of
3.18 x 10^-3 m, and a height of (pretty nearly) 2 m. It is effectively
shooting a cylinder of water with a volume of 5.04 x 10^-6 m^3, with a
mass of 5.04 x 10^-3 kg. Now we need to determine how much initial
velocity is required to propel a body to that height, and how much
time it takes, from which we can determine the rate at which energy is
being consumed to propel the liquid.

   Neglecting air restance (and over a height of 2 meters, it's not
important anyway), the velocity aquired by a mass dropping from a
height to the ground is equal to the initial velocity required to
project it from the ground to the same height. The formula for this
is:

V = SQRT[v^2 + 2gh]

h = starting height above the ground = 2 m
g = acceleration due to gravity = 9.8 m/s^2
v = initial velocity at h = 0
V = final velocity, at the ground

   Plugging in the numbers, we get V = 6.26 m/s. This is how fast the
water must be traveling as it is ejected from the hole.

   Now for the time of ascent. The formula for this is:

t = -v = SQRT[v^2 = 2gs]/a

   Again, plugging in the same numbers from above, we get t = 0.64 s.
It'll take 64/100 of a second for ejected water to reach a height of 2
m.

   So how much water mass is ejecting each second from the nozzle?
That's [5.04 x 10^-3 kg]/[0.64 s] = 7.88 x 10^-3 kg/s. This is much
much water mass is being delivered to a height of 2 m each second.

   The kinetic energy required to propel water to this height is equal
to its potential energy when at that height. This is given simply by
the product, mgh. The pump is propelling 7.88 x 10^-3 kg/s against an
acceleration of 9.8 m/s^2 to 2 m. The kinetic energy of the mass each
second then comes to 0.15 j (joules), or in other words, the pump is
consuming at least 0.15 watts.

   Now of course, the pump will not be 100% efficient. There will be
some heat loss as it does its job. Let's say that it's 75% efficient.
This means that it must consume .20 w to propel the fountain, or a
horsepower of 2.68 x 10^-4 foot-pounds/second.

To:  qed100 from Bozobozo

Subject:Physics/hydrolics

Comment:  Thanks very much for your answer.  If you
can indulge my enlargement of the question...,
I plan to build a fountain with a 3 foot spray ring
centered in a 9' x 9' basin and would like to
have 12 nozzle spray heads in the ring, each of
which would have a 3/8 inch nozzle opening and 
spray jets 6 feet (or as you say, about 2 meters)
in height.  I need to know how many gallons per
minute would flow from these 12 jets and what
wattage (or horsepower) would be required to operate
the fountain.  The pump would be a submerged 
centripetal pump.  I apologize for not being
more specific in my original question, but I felt
I could calculate and determine these specific values
which I needed from an answer to that more "simple"
question... but I cannot!  Thanks again as your
answer was exactly what I needed, but I now realize
that I should have asked the more detailed question.

The orifice you originally specified was 1/8 inch diameter. The new
diameter is 3/8 inch. The area of the opening is given by pi*r^2, so
the ratio of the new area to the old is exactly 9/1. The mass of the
water being shot by each nozzle will be greater than my previous
estimate by a factor of 9, and so will the energy & power.

   This gives 1.8 watts per nozzle, or 2.41 x 10^-3 foot-pounds/s.
Twelve nozzles come to 21.6 watts, or  2.89 x 10^-2 foot-pounds/s.
(0.0289 foot-pounds/s)

   Each nozzle is shooting 0.071 liters/s, which is 4.26
liters/minute, or 1.12 gallons/minute, and the twelve nozzles come to
13.49 gallons/minute.

   Now just offhand, you ought to allow a very generous fudge-factor.
My calculation of the energy consumption is based only on a simplified
picture of water being ejected upwards, with an estimate of the energy
efficiency of the pump itself. If 75% efficiency turns out to be way
too high, you need to be able to adjust the ouput. Also, efficiency is
very affected by such things as how deep under the pond the recycling
inlet is (which can actually help the efficiency), and also by how
much length of pipe the water travels through between the pump & the
exit, and upon how many right turns the pipe makes, each of which
steal power. I'd conservatively guess that you should multiply all my
power figures by at least another factor of three, just to be on the
safe side, and be sure to procure a pump which can be power adjusted.

You need to look at some pump curves.  A pump curve is a graph that
shows the feet of head (how high your water will shoot) versus the
flowrate (gallons/minute).  For a centrifugal pump, the head varies
with the flowrate.  At a very low flowrate, you could shoot water
higher than you could at a higher flowrate.  Also shown on the curve
is the brake horsepower.  This is the power required at the various
flowrates.

For example, the pump curve I'm looking at right now shows some
various points - (ft. of head @ GPM).  3900ft@200GPM, 3625ft@800GPM,
2150ft@1500GPM.  Obviously this is a huge pump.

Also, you need to consider the NPSHa and NPSHr.  This is Net Positive
Suction Head available and required.  If your NPSHr is greater than
your NPSHa, your pump will cavitate (it'll sound like someone threw a
handful of gravel into your pump).  The longer it cavitates, the
faster it will break.  Your blades will dissolve away, your piping
will be eaten, and the pump will ultimately fail.

go to www.efunda.com for more about pump sizing.

It's hard to tell you how much HP you'll need without knowing the
blade design, and especially GPM.  Everything depends on GPM.  In
other words, do you want to have a micro stream of water that goes 3
ft. high, or do you want to have a massive, river of water going 3 ft.
high?

Once you choose a GPM, you'll be able to decide on a pump... and then
just size the motor to fit the brake horsepower of the pump.

AZD

I'm not sure where qed100 made a mistake, but his answers aren't
right.  Power required is sqrt(2)DA(gh)^(3/2), where D is density of
water, A is total cross sectional area of nozzles, g is acceleration
of gravity, and h is height of fountain.  For 12 3/8" diameter
nozzles, that's 36.2 watts, which is the pump power required if there
were no friction/inefficiency.  Actual electric power consumed is
probably more like 100 watts, depending on the particular pump and
geometry of the plumbing.

The water flow rate is A*sqrt(2gh), which is 0.00202 m^3/s, or 2.02
liters/sec, or 32 gallons/minute.