D

"It is customary to select diameters of large plungers in full-inch or 2-inch sizes; this requires adjustment of the area.

Standard Procedure

8. Assume a reasonable approximate level of hoop stresses sh along the inner surface of the cylinder barrel. The cylinder diameter expansion will be

where E = modulus of elasticity assumed to be 30 X 106 lb/in2 (207 GPa)

9. On the basis of c0and sh, determine the volume of liquid required to compensate for compression of liquid Vcl as well as for expansion of cylinder Vec, During planishing, the pressures rise to only about 10% of rated; therefore, the compressibility of the liquid and expansion of the cylinder may be neglected.

30,000

0.045 in

30,000

0.023 in

: : 58.4 = 0.058 cm where dem is the main cylinder diameter expansion and des is the side cylinder expansion.

9. Compression of the liquid takes place along the entire length of the cylinder barrel. Assuming the length of the barrel to be approximately S + ^ S, or 48 + 25 = 72 in (122 + 61 = 183 cm): V-! = 72 : : 2420 : : 0.015 = 2614 in3

(183 : : 15,600 : : 0.015 = 42,800 cm3) Omitting quantities small in the second order.

= 72 :: 1.57 :: 0.001 : !>■ o||:i : :Z>2(183 :: 1.57 :: 0001 : !>■ = 0.287 :: D2) For the center cylinder:

0.113 : : 452 = 229 in3 ( 0.2 87 : : 1142 = 3730 cm3) For the side cylinders:

= V^-3 = 0.113 : : 232 = 60 in3 (0.287 : : 58.42 = 980 cm3) = 229 + (2 : : 60) = 349 in3 [3730 + (2 : : 980) = 5700 cm3] Total additional volume of liquid required. 2614 + 349 = 2963 in3per stroke (42,800 + 5700 = 48,500 cm3)

Standard Procedure

10. Determine the total amount of liquid V required per cogging stroke.

11. Check whether specified speeds are compatible with the required number of cogging strokes. If not, increase the specified speeds (or reduce the number of strokes).

12. Determine the liquid requirements for the pull-back stroke (Figures 6 and 7). In general, the pull-back cylinders have an area equal to 10% of the area of the main cylinders and twice as long a stroke as the pressing stroke; in general, the required volume is therefore, with sufficient accuracy:

10. V = 9680 + 2963 = 12,643 in3 (156,000 + 48,500 = 204,500 cm3) This value shows what additional burden can be imposed on the power plant by unnecessarily generous dimensioning of the total stroke.

Pressing time: § = 2 s Fast return time: § = 1.33 s Valving time (3 switches) = 0.45 s Total cycle time = 4.45 s

This time is too long to allow 15 cogging strokes per minute. Increase fast advance and fast return to 480 in/min = 8 in/s (20 cm/s) and pressing to 180 in/min = 3 in/s (8 cm/s). Then

Fast advance time: § = 0.5 s Pressing time: § = 1.33 s Fast return time: § = 1 s Valving time = 0.45 s

12. 12,643 X 0.2 = 2528 in3 (204,500 X 0.2 = 40,900 cm3)

Standard Procedure

13. Determine the pumping requirements for the pump station with or without an accumulator. Without an accumulator:

231 x tp V 1000 x tt where V is the volume in gallons (cubic meters) required for one pressing stroke and tp is the pressing time in seconds. With an accumulator:

' 231 x tc V 106 x tc where Vtc is the total volume in gallons (cubic meters) of pressurized liquid required during one cycle and tc is the cycle time in seconds. 14. Determine size of the accumulator required to store the accumulated vol-p ume Vs of liquid:

Vs = gpm (acc) X - VR ^md/h X - Vj where VR is the volume required for the return stroke in gallons (cubic meters).

Air volume required, based on 10% pressure fluctuation for isothermic condition, is 10VS. For adiabatic or, more often, polytropic conditions and 10% pressure fluctuation, the air volume is

Although n = 1.4 is the exponent for adiabatic compression, n = 1.3 is considered satisfactory, even for severe and demanding conditions, which require full utilization of press stroke and a fast cycle. Based on n = 1.3,

13. Without an accumulator: 12,643 : : 60

2469 gpm

204,500

553 m3/h

It is evident that the use of an accumulator will result in significant savings and in a substantial reduction of the peak load.

V^ (isometric) = 1 - : : 32.9 = 329 gal = 10 : : 4.4 = 44 ft3 (10 : : 0.125 = 1.5 m3)

End of Days Apocalypse

End of Days Apocalypse

This work on 2012 will attempt to note them allfrom the concepts andinvolvement by the authors of the Bible and its interpreters and theprophecies depicted in both the Hopi petroglyphs and the Mayan calendarto the prophetic uttering of such psychics, mediums, and prophets asNostradamus, Madame Blavatsky, Edgar Cayce, and Jean Dixon.

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