tag:blogger.com,1999:blog-4315911668447344547.post1760704703938515210..comments2016-06-23T22:28:43.533-07:00Comments on Hydraulic Answers: Impulse Test Flow RateJDKhttp://www.blogger.com/profile/16796281957964741569noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-4315911668447344547.post-62115320441939678202009-09-07T05:27:05.625-07:002009-09-07T05:27:05.625-07:00pnachtwey makes excellent observations about the p...pnachtwey makes excellent observations about the pressure requirements in this application. His insights must be taken into account. If the pressure in the cylinder must go down to 0 PSI, oil flow will be very challenging. The "Delta P" of the system must always be taken into account.<br /><br />However, Valliappan was looking for pump flow to perform this test. I think it is important to distinguish between pump flow and flow to and from the cylinder (System Flow). To accomplish the 8 Hz sine wave, the max theoretical system flow (max. slope of the sine wave) calculates to approx. 60 GPM as pnatchwey points out. Delta P, Pump Flow, Accumulators, limitations etc. must be considered.JDKhttp://www.blogger.com/profile/16796281957964741569noreply@blogger.comtag:blogger.com,1999:blog-4315911668447344547.post-196028429743118292009-09-06T18:30:00.127-07:002009-09-06T18:30:00.127-07:00The calculations are for an ideal system. 60gpm p...The calculations are for an ideal system. 60gpm probably will not work. Also, it is extermely difficult to get the pressure down to 0 psi because the oil simply will not exit that quickly. Likewise it is hard to get the pressure up to 410bar if the system pressure is only 410bar. A pressure difference beyond the required extremes of the pressure sinewave is required. If a the pressure sine wave must truly go from 0 to 410 bar then more research is required. The calculations so far have not taken into account the equaiton Q=K*sqrt(delta P).<br /><br />Also, there should be accumulators with servo valves. Pumps cannot possible respond fast enough or control accurately enough. A pump doesn't need to supply the peak flow if energy can be store in the accumulator during the low flow times. <br /><br />The motion controller is important but the knowledge required to do the prelimary design is even more important. We have lots of experience at this.<br />http://www.qualitymag.com/CDA/Articles/Web_Exclusive/BNP_GUID_9-5-2006_A_10000000000000580356<br />This application is also doing sinusoidal testing but for pipes.<br />I have more testing examples.pnachtweyhttp://www.blogger.com/profile/12912750159539760736noreply@blogger.comtag:blogger.com,1999:blog-4315911668447344547.post-15782254022210516872009-09-06T14:45:27.664-07:002009-09-06T14:45:27.664-07:00Thank you pnachtwey.
The slope in pressure increas...Thank you pnachtwey.<br />The slope in pressure increase as the sine wave crosses the horizontal axis (its maximum value) would result in the maximum needed flow rate of approximately 60 GPM. <br />After further investigation of the sine wave, I agree with this number. Again, thank you pnatchtwey for the help. Looks like a Delta Motion Controller should be used here.JDKhttp://www.blogger.com/profile/16796281957964741569noreply@blogger.comtag:blogger.com,1999:blog-4315911668447344547.post-91454129669151930292009-09-05T21:20:00.114-07:002009-09-05T21:20:00.114-07:00A good hydraulic motion controller that is used in...A good hydraulic motion controller that is used in testing applications can do this easily provided the hydraulic system can actual respond that fast.<br />http://www.deltamotion.com/products/motion/rmc70/index.php<br /><br />I compute a higher flow rate.<br />The amplitude of the sine wave is (410*bar/2) convert to psi = 2973 psi. At 8Hz the pressure rate, dp, will be amplitude*2*PI*Hz=149414 psi/sec. Since dp=B*dv/Vol and we know we must rearrange do compute dv. dv=dp*Vol/B<br />I assumed B, the bulk modulus of oil is 200000 psi. Vol is 5.026l*61.025in^3/l. The result is 229 in^3/s which is a little less than one gallon per second or 60 GPM.pnachtweyhttp://www.blogger.com/profile/12912750159539760736noreply@blogger.comtag:blogger.com,1999:blog-4315911668447344547.post-26835664488795223562009-09-03T18:15:38.041-07:002009-09-03T18:15:38.041-07:00Using .56%/1000 PSI (added Volume)
Pressurizing 5....Using .56%/1000 PSI (added Volume)<br />Pressurizing 5.026 Liters in 60 ms to 410 Bar = a flow rate of 173 Liters/Min is required. However using an 8 HZ Sine wave steady flow during 60 ms is not a good assumption. Added oil vs pressure build up is linear. Therefore the slope as the sine wave crosses the horizontal axis will give the approximate maximum flow. A pump with a larger flow than this slope dictates and an accumulator may be a viable option here.JDKhttp://www.blogger.com/profile/16796281957964741569noreply@blogger.com