WATER SUPPLY, PLUMBING, ETC., FOR COUNTRY HOMES. 
29 
Example 2. It is desired to find the necessary fall from the spring 
to the ram in order to supply the storage tank with 2 gallons per 
minute, when the drive water supply in the spring is 10 gallons per 
minute and one height to which the water is to be pumped is 40 feet. 
2X40 
Substituting in the equation H= —^ 
10 
= 8 feet. Add one-third 
to allow for friction, making the total fall necessary 10.66 feet. 
Example 3. It is desired to find how much water will be delivered 
into the storage tank if the drive water supply is 6 gallons per minute, 
the fall is 10 feet, and the height to which the water is to be pumped 
is 40 feet. Substituting m the 
QxH 
equation q = -^ — : 
40 
= 1.5 
and the fall is 15 feet. Substituting in the equation Ji 
gallons per minute. Deduct one-third of this result to allow for 
friction, making the quantity delivered per minute 1 gallon. 
Example 4. It is desired to find how high 1 gallon per minute 
can be pumped if the drive water supply is 4 gallons per minute 
QxH _ 
9 
4 X 15 
— r — = 60 feet. Deduct one-third to allow for friction, making the 
result 40 feet. 
The above computations are only approximate, but should give a 
good general idea of the operation of a ram. 
The following table gives commercial estimates of the quantities of 
water delivered in 24 hours under certain conditions: 
Capacity of hydraulic rams. 
Power 
head 
in 
feet. 
Pumping head in feet— 
4 
10 
15 
20 
30 
140 
50 
60 
70 
80 
90 
100 
120 
1,0 
160 
180 
200 
2 
3 
4 
5 
6 
7 
8 
9 
10 
12 
14 
16 
18 
20 
22 
24 
26 
28 
30 
540 
192 
301 
432 
540 
128 
192 
256 
345 
432 
505 
96 
144 
192 
240 
302 
378 
432 
485 
540 
64 
96 
128 
160 
192 
235 
270 
300 
360 
430 
505 
43 
72 
96 
120 
144 
168 
192 
216 
1252 
301 
353 
432 
486 
540 
29 
58 
77 
96 
115 
134 
154 
173 
192 
230 
270 
323 
390 
430 
475 
520 
24 
43 
64 
80 
96 
112 
128 
144 
160 
192 
224 
257 
303 
336 
370 
405 
470 
505 
540 
37 
55 
69 
82 
96 
110 
124 
137 
165 
192 
220 
247 
288 
303 
346 
375 
430 
465 
27 
43 
60 
72 
84 
96 
108 
120 
144 
168 
192 
216 
240 
264 
288 
328 
354 
405 
24 
38 
53 
64 
75 
86 
96 
107 
128 
150 
171 
192 
214 
235 
256 
278 
300 
336 
29 
43 
57 
67 
77 
86 
96 
115 
135 
154 
173 
192 
212 
230 
250 
269 
288 
24 
30 
43 
50 
64 
72 
80 
96 
112 
128 
144 
160 
176 
192 
208 
224 
240 
26 
31 
36 
55 
62 
68 
82 
96 
110 
124 
137 
151 
164 
178 
192 
206 
27 
31 
43 
54 
60 
72 
84 
96 
108 
120 
132 
144 
156 
168 
180 
24 
28 
38 
43 
53 
64 
75 
85 
96 
107 
118 
128 
139 
149 
160 
"25*" 
29 
39 
43 : 
57 
67 
86 ! 
96 
105 
115 
125 
134 
144 
1 Multiply factor opposite "power head" and under "pumping head'' by the number of gallons per 
minute used by the engine and the result will be the number of gallons delivered per day. Example : With 
a supply of 6 gallons per minute, 10 feet fall, 40 feet elevation, No. 10 or 15 engine will deliver 1,512 gallons 
per day; 6X252=1,512. 
This table will give only approximate quantities since the results will vary with the length of delivery 
pipe. Due consideration of pipe friction will give more correct results. 
The efficiency developed is governed by the ratio of fall to pumping head, being greatest for a ratio of 1 to 
2\ or 1 to 3, and the ram will not usually work well when the ratio is over 1 to 25, friction in the delivery 
pipe beine duly considered. 
