WIND ENERGY SYSTEMS - Cd3wd

Chapter 7—Asynchronous Loads 7–6
The pump volume capacity can also be expressed as 1.58 L/s, or 5.70 m 3 /h, or 25.1 gal/min.
We see that the SI expression for volume capacity is well under unity, while the number
of liters per second, cubic meters per hour, and gallons per minute are above unity. This
makes the non-SI units slightly easier to remember. However, we shall primarily use the SI
units, but will lapse into other units occasionally to help the reader understand those units
which have been so widely used. The conversion factors for capacity units are that 1 m 3 /sis
equal to 35.315 cubic feet per second, usually abbreviated cfs, and is also equal to 15,850.32
U. S. gallons per minute, abbreviated gal/min or gpm. One cubic meter contains 264.17 U.
S. liquid gallons and one cubic foot contains 7.4805 U. S. liquid gallons. River flows in the
United States have historically been expressed in cfs while pump capacities are more often
given in gal/min.
In power calculations we will need to express capacity in terms of mass flow rather than
volume flow. If ambient temperature water is being pumped, it is usually sufficiently accurate
to assume that one liter of water has a mass of one kilogram, or 1 m 3 has a mass of 1000
kg. We can define a mass capacity Q m kg/s as the mass flow, where Q m = 1000Q v if Q v is
expressed in the SI units of m 3 /s.
The power input to a pump is given by
P m = gQ mh
W (3)
η p
where g = 9.81 N/kg is the gravitational constant, Q m is the mass capacity of the pump
expressedinkg/s,h is the head in m, and η p is the pump mechanical efficiency. The quantity
gQ m h can be thought of as an output power
P o = gQ m h W (4)
or the energy required to raise a given mass of water a height h, divided by the time required
to do it. The mechanical efficiency includes losses in the mechanical friction between the
piston packing and the pump cylinder and also the pump rod and the water it moves through.
These losses are in addition to those included in the slip. The mechanical efficiency is usually
between 0.9 and 0.95 but can be as low as 0.5.
Example
Find the power input to the pump of the previous example if the head is 20 m and the mechanical
efficiency is 0.92.
The capacity Q m is assumed to be 1.58 kg/s. The power input is then
P m = 9.81(1.58)(20)
0.92
= 337 W
If the volume capacity is given in the English units gal/min, which we shall call Q g ,and
the head is given in feet, Eq. 3 becomes
Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001