## Power

then.

vSubstituting PjQj for wRTj from Equation 2.7,

For two stages, the above equation can be expanded to add the interstage conditions for the second stage. Note the subscript i is added to the second set of terms to reflect the second-stage inlet.

For a first trial at sizing or for estimates, the Equation 3.13 can be differentiated and solved for Pi? with the result, fwn*p2

This expression can be changed to

lUJi

Substituting the term r for the pressure ratio, the following results r„, - r.

Equation 3.16 can be generalized for optimum work division by dividing the pressure ratio into a set of balanced values,

The values for pressure ratio in a practical case must include allowance for pressure drop in the interstage piping. In the sizing procedure used by manufacturers, certain adjustments must be made to the ideal for incremental cylinder sizes and allowable rod loading. Efficiency is represented by r|cyl.

To assist the engineer in making estimates, the curve in Figure 3-6 gives values of efficiency plotted against pressure ratios. The values on the curve include a 95% mechanical efficiency and a valve velocity of 3,000 feet per minute. Table 3-1 and Table 3-2 are included to permit a correction to be made to the compressor horsepower for specific gravity and low inlet pressure. They are included to help illustrate the influence of these factors to the power required. The application of these factors to

 00* - 89» -| B8* - 87* - as« - 84* - 83* "J 82* - / 81* - / 80* - / 79* - y 78* - / 77* - / 78* - / 75* -J / 74* - / 73* 72* - 71* — 70* - • •1 ■ 1"1 1" " 1 T— T-T" T-T" ' ' ' '

PRESSURE RATIO

PRESSURE RATIO

Figure 3-6. Reciprocating compressor efficiencies plotted against pressure ratio with a valve velocity of 3,000 fpm and a mechanical efficiency of 95%.

Table 3-1

Efficiency Multiplier for Specific Gravity

Table 3-1

Efficiency Multiplier for Specific Gravity

 SG fp 1.5 1.3 1.0 0.8 0.6 2.0 0.99 1.0 1.0 1.0 1.01 1.75 0.97 0.99 1.0 1.01 1.02 r 1.5 0.94 0.97 1.0 1.02 1.04

Source: Modified courtesy of the Gas Processors Suppliers Association

Source: Modified courtesy of the Gas Processors Suppliers Association

Table 3-2

Efficiency Multiplier for Low Pressure

Table 3-2

Efficiency Multiplier for Low Pressure

 Pressure Psia rP 10 14.7 20 40 60 80 100 150 3.0 .990 1.00 1.00 1.00 1.00 1.00 1.00 1.00 2.5 .980 .985 .990 .995 1.00 1.00 1.00 1.00 2.0 .960 .965 .970 .980 .990 1.00 1.00 1.00 1.5 .890 .900 .920 .940 .960 .980 .990 1.00

Source. Modified courtesy of the Gas Processors Association and Ingersoll-Rand.

Source. Modified courtesy of the Gas Processors Association and Ingersoll-Rand.

efficiency value is arbitrary. While it is recognized that the efficiency is not necessarily the element affected, the desire is to modify the power required per the criteria in the tables.

The efficiency correction accomplishes this. These corrections become more significant at the lower pressure ratios.

### Valve Loss

The efficiency values are affected by several losses: ring slippage, packing leakage, and valve losses. Valve losses are generally the most significant and are made of several components such as channel loss, loss in the valve opening, and leakage. Also, because of inertia and imperfect damping properties of the gas, the valve may have transient losses due to bounce. The manufacturer, therefore, modifies the valve lift to suit the gas specified. For example, an air compressor might be furnished with a lift of .100 inch. The same compressor being furnished for a low molecular service such as a hydrogen-rich gas, might use a lift of .032 inches. The problem with the higher lift is that hydrogen lacks the damping properties of air and, as a result, the valve would experience excessive bounce. The effect on the compressor would be loss in efficiency and higher valve maintenance.

The valve porting influences volumetric efficiency by contributing to the minimum clearance volume. If the porting must be enlarged to reducc the flow loss, it is done at the expense of minimum clearance volume

This is just one example of the many compromises the engineer is faced with while designing the compressor. The subject of valve design is involved and complex. For individuals wishing to obtain more information on the subject, references discussing additional aspects of valves are included at the end of the chapter [2, 3, 4, 5, 6 |.

To calculate the valve velocity for evaluation purposes, use the following equation. This equation is based on the equation given in API 618.

where v = average gas velocity, fpm Pd = piston displacement per cylinder, ft3/min A = total inlet or discharge valve area per cylinder, in.2

To calculate Pd, use Equation 3.1 for single-acting cylinders, Equation 3.2 for double-acting cylinders without a tail rod, and Equation 3.3 for double-acting cylinders with a tail rod.

The area, A, is the product of actual lift and the valve opening periphery and is the total for all inlet or discharge valves in a cylinder. The lift is a compressor vendor-furnished number.

Example 3-1

Calculate the suction capacity, horsepower, discharge temperature, and piston speed for the following single-stage double acting compressor.

Bore: 6 inches Stroke: 12 inches Speed: 300 rpm Rod diameter: 2lA inches Clearance: 12% Gas: C02

Inlet pPressure: 1,720 psia Discharge pressure: 3,440 psia Inlet temperature: 115°F

Calculate the piston displacement using Equation 3.2 and dividing by 1,728 in.3 per ft3 to convert the output to cfm.

1,728x4

Step 1. Calculate volumetric efficiency using Equations 3.5 and 3.6. To complete the calculation for volumetric efficiency, the compressibilities are needed to evaluate the f term of Equation 3.6. Using Equations 2.11 and 2.12 for the inlet conditions,

From the generalized compressibility charts (see Appendix B),

Step 2. At this point the discharge temperature must be calculated to arrive at a value for the discharge compressibility.

3.440

t1 = 2 14.7°F discharge temperature Calculate the discharge compressibility:

3.440

From ihc generalized compressibility charts, Z2 - 0.575

From Equation 3.6, calculate f:

Calculate the volumetric efficiency using Equation 3.5. Use .05 for L because of the high differential pressure:

Ev = .93 volumetric efficiency

Now calculate suction capacity using Equation 3.7:

Qt = 100.1 cfm suction capacity

Step 3. Piston speed is calculated using Equation 3.8 converting the stroke to feet by dividing the equation by 12 inches per foot:

PS = 600 fpm piston speed

Step 4. Calculate the power required. Refer to Figure 3-6 and select the efficiency at a pressure ratio of 2.0. The value from the curve is 79%. Equation 3.18 is used to calculate power. The constants 144 in2ft2 and 33,000 ft-lbs/min/hp have been used to correct the equation for the unit from the example.

144x1,720x100.1 1.3

Wcvj = 714.8 hp cylinder horsepower

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