Relative compression efficiency calculation

I like to calculate the relative compression efficiency because I do not have to know the flow of process gas. I do not have to know the driver horsepower output, the steam to the turbine, the fuel gas to the gas turbine, or the speed of the compressor. I do not have to know Z (the gas compressibility factor) or K (the ratio of the specific heats). The things I do have to know—the suction and discharge temperature and pressure—I can check with my own hands and my own tools. 

It is worth repeating that the relatively low efficiency for the appKTP crystal is due to the fact that Je (KTP) < Je (KNb03). Performing the same assessment with lithium niobate (LiNbOs) should yield up to four times the efficiency, because dg/f (LiNbOs) = 17.6 pmV. Unfortunately, insufficient power was available to measure the duration of the blue pulses from the bulk appKTP crystal. However, our calculations show that the generated blue pulses would be characterized by an uncompensated duration of 370 fs. These pulses could be compressed to around 270 fs in order to access higher peak powers. 

We now briefly discuss how thermodynamics can work for us or, better, how thermodynamics functions to solve a problem where it can help to provide the answer. We wish to illustrate this for a relatively simple problem how much work is required to compress a unit of gas per unit time (Figure 2.4) from a low to a high pressure. Figure 2.5 schematically gives the path to the answer and the structure of the solution. In fact, the same steps will have to be taken to apply thermodynamics to problems such as the calculation of the heat released from or required for a process, of the position of the chemical or phase equilibrium, or of the thermodynamic efficiency of a process. 

For well designed pumps X is normally close to zero, i.e. the pump is of rigid construction so tjE can be calculated directly from pump and fluid data. This calculation shows that the relative dead spaces df and 8dh should be minimised to achieve good volumetric efficiency. Furthermore as the slip factor should approach tjs 1 (valves should be tight and operate with no slip [15]) the volumetric efficiency can be obtained directly from the indicator diagram (Figure 9.12) and the table beneath it. The diagram shows the relation between the decompression stroke h (from 3 to 4) and the compression stroke /i2 (from 1 to 2) and the table enables Joukowsky-shock, volumetric efficiency, shock factor and piston velocity to be calculated. 

We can answer these questions using Eq. (43.1), which defines relative efficiency. The calculated numerical value of relative efficiency means nothing The equation may be used only to compare two sets of operating data. The equation is not even thermodynamically correct. But it is sufficiently correct, provided the services represented by the two sets of data are reasonably similar and the compression ratios are within 10 to 20 percent of each other. 

---