__Answer to Article #57 quiz__: "Pump NPSH and Developed
Head"

The
available NPSHA is a net (total) positive suction head, over vapor pressure of
a pumped liquid. (A more detailed background on this can be found in a topic
"How does pump suction limit the flow?").

We will
first determine a total suction head (Hs), which is equal to static water level
above the pump centerline plus the pressure (in units of head) at the liquid
surface, minus the losses at the suction side of a pump.

80o
water has specific gravity of 1.0, and atmospheric pressure, converted to feet
of head is equal to 14.7 psia x 2.31 / 1.0 = 34.0
feet.

Then,
Hs = 30+34-10 = 54 feet - a total suction head, in front of a pump.

Vapor
pressure of 80 ^{0}F water is 0.5 psia, or
0.5 x 2.31 / 1.0 = 1.2 feet.

NPSHA
can now be calculated as:

NPSHA = Hs - Hvapor
= 54 - 1.2 = 52.8 feet.

Suction
gage shows static pressure, which can be found by subtracting velocity head and
a gage elevation (the higher the gage is positioned, the low its static
pressure reading would be).

Velocity
is equal to flow divided by the pipe net open area (with coefficient 0.321, if
flow is in gpm, and area in square inches). Making a simplifying
assumption that the net pipe open area is based on its nominal diameter, we
get:

A_{SUCTION} = 3.14 x 3^{2}
/ 4 = 7.1 in^{2}, and

V_{SUCTION} = 200 x 0.321 /
7.1 = 9.1 ft/sec, and

Velocity
head (dynamic) is h_{vel.suction} = 9.12 /
64.4 = 1.3 feet

This
now allows us to calculate the gage pressure in absolute units:

H_suction_gage (abs) = 54 - 1.4 - 4 = 48.6 feet, or, subtracting the atmospheric
component:

H_suction_gage (gage) = 48.6 - 34 = 14.6 feet, which would translate into a psi(gage)
units as:

P_suction_gage = 14.6 x 1.0 / 2.31 = 6.3 psig - this is what we would see on the
gage dial.

Next, let's
find a discharge head, which is (similar to suction) a sum of static pressure,
velocity head, and correction for the gage elevation.

Discharge
gage reads 100 psig, which is equivalent to 100+14.7 = 114.7 psig, and, in
units of feet is: 114.7 x 2.31 / 1.0 = 265.0 feet.

Static
pressure at the pump centerline is then: 265 + 20 = 285 feet.

Discharge
pipe flow area is 3.14 x 2^{2} / 4 = 3.1 in^{2}, and

Vdisch
= 200 x 0.321 / 3.1 = 20.4 ft/sec, producing a velocity head of

hvel suction = 20.42 / 64.4 = 6.5 feet.

Total
discharge head is then Hd =
285 + 6.5 = 291.5 feet.

Pump
developed head is a difference between the discharge and suction heads:

H=Hd - Hs

H =
291.5 - 54 = 237.5 feet, which is the equivalent to saying that the pump
generates 237.5 x 1.0 / 2.31 = 102.8 psi differential
pressure.

Note
that simply taking a difference between a discharge a suction gage readings
would give us 100 - 6.3 = 93.7 psi, i.e. almost a 9%
error.

Usually
the elevation between the gages is not as significant as is this example, and
the flow velocities are usually lower (larger pipe), but, in general, it is
important to remember to account for all components of the pump head, in
calculations.