Page 160 -

..................................... Special Applications

8.2.5.2 Example 2 - Approximate pump performance on stock
when water performance is known.

Find the performance for the Model 3175, 4 x 6-18 at 1180 rpm from
curve on Figure 8.12 above, with an impeller diameter suitable for the
rating of 600 gpm at 5% Kraft stock at 100 feet Total Head from
Example 1.
1. Find the water rating correction factor and approximate water
ratings as outlined in Example 1.
Kp = 0.87
Qw = 690 gpm
Hw = 115 ft.
NOTE: If the approximate water rating for 3.5% stock or over,
exceeds the best efficiency capacity, use a larger pump.
Refer to Figure 8.12. The capacity at the best efficiency point
.
(b.e.p.) for an impeller diameter slightly smaller than required for
the water rating, is 880 gpm. Multiply 880 gpm times the
correction factor to find the pump capacity at b.e.p, on stock.

880 x 0.87 = 766 gpm

Find the ratio of the stock capacity at the rating (600 gpm) to the
.
stock capacity at b.e.p. (766).

600
QP= 7~ x 8 = 78%

For capacities not at the b.e.p., the percent consistency is
,
subtracted from Qp found in Step 3 above.

Qp = 78- 5 = 73%

Find the correction factor for 73% capacity Qp from the chart, Kp
,
= 0.905
Divide the rated stock Head and Capacity by the new correction
factor to estimate the final water rating.

600
Qw - 0.905 = 663gpm

lOO
Hw - = 110 ft. Total Head
0.905

.......................................................................................................................................................... - .......................... 139 m