http://www.ngvinstaller.com/ngv-compressor/ NGV Installation System Tue, 23 Feb 2010 09:49:58 +0800 http://wordpress.org/?v=2.8.4 hourly 1 http://www.ngvinstaller.com/ngv-compressor/comment-page-1/#comment-399 kirchwey Mon, 10 Nov 2008 16:38:17 +0000 http://www.ngvinstaller.com/ngv-compressor/#comment-399 Either this is isothermal compression, in which case the problem is simply about about volume ratios inversely proportional to pressure ratios, or it's a real-world situation with near-adiabatic compression, in which case you're into something that's a very complicated calculation and you haven't given enough information. In any case you haven't provided rate-of-flow information necessary to compute power. So let's go with the simple assumption and calculate what piston motions are needed to achieve the required compression ratios, and what bore ratios are needed to allow the 2nd and 3rd stages to handle the output of the previous stage. Since there are different compression ratios (4:1, 2.5:1 and 2.4:1 for the 1st, 2nd and 3rd stages respectively) and presumably this multistage pump runs from a single shaft, there have to be differences in the ratios of volumes V and heights H of the gas volume cylinders at TDC (Vt and Ht) and BDC (Vb and Hb). (The compression ratio CR = Vb / Vt = Hb / Ht.) These differences can be achieved by using a crankshaft with differing throws or by having equal throws and adjusting the connecting rod lengths or the locations of the tops of the cylinders. These must result in the compression ratios CR = Hb / Ht [Eq. 1] Let's assume each stage has the same crank throw CT. The piston stroke is 2 * CT. Then Hb - Ht = (2 * CT) [Eq. 2] and from eqs. 1 and 2, Ht = 2 * CT / (CR - 1) Assuming a 2-inch CT and solving, Ht = 4 / (CR - 1) = 1.333, 2.667 and 2.857 inches for the three stages. Thus stages 2 and 3 have Ht values that are 1.333 and 1.524 inches greater than that of stage 1. The locations of the cylinder tops and/or the connecting rod lengths must have the same differences. The bore calculations are relatively easy. Since we don't know the flow rate and shaft rpm, these will be ratios only. The input volume each stage handles is equal to the stroke times the area of the circular cross section of the cylinder. Again we'll assume a constant stroke as we did above. The 2nd cylinder handles 1 / 4 the volume of the first, so its diameter is 1 / sqrt(2) times that of the first. The 3rd cylinder handles 1 / 10 the volume of the first, so its diameter is 1 / sqrt(10) times that of the first. Either this is isothermal compression, in which case the problem is simply about about volume ratios inversely proportional to pressure ratios, or it’s a real-world situation with near-adiabatic compression, in which case you’re into something that’s a very complicated calculation and you haven’t given enough information. In any case you haven’t provided rate-of-flow information necessary to compute power. So let’s go with the simple assumption and calculate what piston motions are needed to achieve the required compression ratios, and what bore ratios are needed to allow the 2nd and 3rd stages to handle the output of the previous stage.
Since there are different compression ratios (4:1, 2.5:1 and 2.4:1 for the 1st, 2nd and 3rd stages respectively) and presumably this multistage pump runs from a single shaft, there have to be differences in the ratios of volumes V and heights H of the gas volume cylinders at TDC (Vt and Ht) and BDC (Vb and Hb). (The compression ratio CR = Vb / Vt = Hb / Ht.) These differences can be achieved by using a crankshaft with differing throws or by having equal throws and adjusting the connecting rod lengths or the locations of the tops of the cylinders. These must result in the compression ratios
CR = Hb / Ht [Eq. 1]
Let’s assume each stage has the same crank throw CT.
The piston stroke is 2 * CT. Then
Hb – Ht = (2 * CT) [Eq. 2]
and from eqs. 1 and 2,
Ht = 2 * CT / (CR – 1)
Assuming a 2-inch CT and solving, Ht = 4 / (CR – 1) = 1.333, 2.667 and 2.857 inches for the three stages. Thus stages 2 and 3 have Ht values that are 1.333 and 1.524 inches greater than that of stage 1. The locations of the cylinder tops and/or the connecting rod lengths must have the same differences.
The bore calculations are relatively easy. Since we don’t know the flow rate and shaft rpm, these will be ratios only. The input volume each stage handles is equal to the stroke times the area of the circular cross section of the cylinder. Again we’ll assume a constant stroke as we did above. The 2nd cylinder handles 1 / 4 the volume of the first, so its diameter is 1 / sqrt(2) times that of the first. The 3rd cylinder handles 1 / 10 the volume of the first, so its diameter is 1 / sqrt(10) times that of the first.

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