## Example

To keep from complicating the example with real world considerations, a few simplifying assumptions will be made. In all cases, the compressor will be considered to be 100% efficient. Intercooling will be perfect, that is, no pressure drop will be considered and the cooler return gas will be the same temperature to the first stage of the compressor.

Gas: nitrogen Molecular weight: 29 Ratio of specific heats: 1.4 Inlet pressure: 20 psia Outlet pressure: 180 psia Inlet temperature: 80°F Weight flow: 100 lb/min

Calculate the theoretical power for each case: (1) no intercooling, (2) one intercooler, (3) two intercoolers, (4) isothermal compression.

Because of the assumption that efficiency is 100%, Equations 2.70 and 2.73 yield the same results.

R ■= 1.545/29 R = 53.3 ft lb/lb °R Ti = 80 + 460 T] = 540°R k/(k - 1) = 1.4/.4 k/(k - I ) = 3.5 (k - i)/k = .4/1.4 (k- l)/k = .286

W, = 1.0 x 53.3 x 540 x 3.5(9.286 - 1) Ha = 88,107 ft lb/lb (no intercooler)

Before proceeding with the power calculation, the head for each cooled case will be calculated. In the idealized case, the most efficient division of work for minimum power is achieved by taking the nth root of the pressure ratio, where n is the number of uncooled sections or compression stages in the parlance of the process engineer. For one cooler, n = 2.

Ha = 100737(3-286- 1) Ha = 37,189 ft lb/lb (one cooler) Ha = 100,737(2.08 286 - 1) Ha = 23,473 ft lb/lb (two coolers)

Calculate the power using Equation 2.77 and setting rja = 1.0 and mechanical losses = 0.

100 X 87,983

Multiply the head by n for the cooled cases.

100x23,473

Using Equation 2.84 to establish the theoretical limit of isothermal compression,

Taking the horsepower values due to cooling, comparing them to the uncooled case, and converting to percentage,

225.4

x 100

267.0

= 84.4% of the uncooled hp, one cooler

213 3 267.0

267.0

- 71.8% of the uncooled hp, isothermal case

It can be seen by comparing the percentages that the benefit of cooling diminishes as each cooler is added. This is particularly noticeable in light of the comparatively small horsepower reduction brought about by isothermal compression since this represents the effect of an infinite number of coolers. The first step was a 15.6% decrease in power, while the second cooler only reduced it another 4.5 percentage points. Even the addition of an infinite number of coolers (isothermal case) added just 12.6 percentage points, a decrease less than the percentage achieved with ihe first cooler. While the economic impact must be evaluated in each case, this illustration does demonstrate that intercooling does save horsepower. in a practical evaluation, some of the idealized values used in the illustration must be replaced with anticipated actual values, such as real efficiency for the compressor, pressure drops in the coolers and piping, and the true outlet temperature expected from the coolers, based on cooling medium temperature. Because the point was made about the decrease in outlet temperature, these values will be calculated to make the exam pie complete. Note that with temperature as with horsepower, the first increment of cooling yields the largest return. The Equation 2.65 is used to make the calculation.

t2 = 540(3 )-28® - 460 t2 = 279°F (one cooler) t2 = 540(2.08)-286 - 460 t2 = 205°F (two coolers)

The equations presented in this chapter should have general application to most compressors, particularly the ones to be discussed in the following chapters. As each compressor is covered, additional equations will be introduced.

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