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7.6. THEORY AND CALCULATIONS OF GAS COMPRESSION 155
TABLE 7.9. Five Rotary Compressors for a Common Service
Helical Spiral Straight Sliding Liquid
Screw Axial Lobes Vanes Liner
Suction loss 8, 9.35 1.32 0.89 0.90 1.40
Discharge ioss Oe 7.35 1.04 0.70 0.70 1.10
Intrinsic cow. B 1.185 1.023 1.016 1.016 1.025
Adiabatic eff. qad 85.6 97.7 98.5 98.5 97.9
Slippage W, (%) 28.5 16.6 11.8 11.8 3.0
Slip e%. qs (%) 71.5 83.4 88.2 88.2 97.0
Thermal eff. qt (%) 89.2 93.7 95.8 95.5 42.5
Volumetric eff, E,,, 58.0 85.7 89.1 89.9 96.6
Displacement (cfrn) 14,700 1 1,550 1 1,220 11,120 10,370
Rotor dia. (in.) 26.6 25.2 27.0 65.0 45.5
Commercial size, d x L 25x25 22x33 22x33 46~92~ 43 x 4klb
Speed (rpm) 3,500 1,250 593 284 378
Motor (HP) 1,700 800 750 750 1.400
Service factor 1.09 1.11 1.10 1.12 1.10
Discharge temp "F 309 270 262 263 120
"Twin 32.5 x 65 or triplet 26.5 x 33 (667 rpm) are more realistic.
bTwwin 32 X 32 (613 rpm) alternate where L = d.
(Evans, 1979).
is the ratio of heat capacities at constant pressure and constant In multistage centrifugal compression it is justifiable to take the
volume and average of the inlet and outlet compressibilities so that the work
becomes
C" =R - Cp. (7.22)
(k-1)lk
-
__
A related expression of some utility is Ws=H2-Hl= (kkl)(Z1~Z2)RTl[(~) -I]. (7.26)
TZ/T1 = (Pz/P1)(k-1)l'C. (7.23)
When friction is present, the problem is handled with empirical
Since k ordinarily is a fairly strong function of the temperature, a efficiency factors. The isentropic compression efficiency is defined as
suitable average value must be used in Eq. (7.20) and related ones.
Under adiabatic conditions the flow work may be written as isentropic work or enthalpy change
(7.27)
9 " = actually required work or enthalpy change '
W = Sr, - Hl V dP. (7.24)
=
Accordingly,
Upon substitution of Eq. (7.20) into Eq. ('7.24) and integration, the
isentropic work becomes W = AH = Ws/qs = (AH)s/qs. (7.28)
When no other information is available about the process gas, it is
justifiable to find the temperature rise from
(7.25)
so that
TABLE 7.10. Specifications of Liquid Liner Compressors
T2= Tl(l + (1/"/P1)'"~"'k- 1 I. (7.30)
Compressor Pressure Capacity Motor Speed
(size) (psi) (cuft/rnin) (HP) (rpm) A case with variable heat capacity is worked out in Example
5 1020 40 7.5.
K-5 10 990 60 For mixtures, the heat capacity to use is the sum of the mol
15 870 100 ('.
75
20 650
621 26 7; 3500
1251 35 120 40 1750
1256 440 100 1750 EXAMPLE 7.4
Gas Compression, Isentropic and True Final Temperatures
621 23 10 3500 With k = 1.4, P2/Pl = 3 and qs = 0.71; the final temperatures are
1251 k0 110 50 1750 (Tz)s = 1.369T1 and Tz = 1.519T1 with Eqs. (7.24) and (7.31).
1256 41 0 150 1750
(Nash Enginleering Co.).