Page 145 - Petroleum Production Engineering, A Computer-Assisted Approach

Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 140 3.1.2007 8:54pm Compositor Name: SJoearun

11/140 EQUIPMENT DESIGN AND SELECTION
n h io
1
The compression ratio in each stage should be less than e v ¼ 0:96 1 « r s 1 , (11:63)
k
six to increase compression efficiency. The equation to
calculate stage-compression ratio is where « is the clearance ratio defined as the clearance
1=n s volume at the end of the piston stroke divided by the
P dis
r s ¼ , (11:56) entire volume of the chamber (volume contacted by the gas
P in in the cylinder). In addition, there is a mechanical efficiency
where P dis , P in , and n s are final discharge pressure, inlet e m of the compressor and its prime mover. This results in
pressure, and number of stages, respectively. two separate expressions for calculating the required HP t
For a two-stage compression, the compression ratio for for reciprocating compressors and rotary compressors. The
each stage should be required minimum input prime mover motor to practically
r ffiffiffiffiffiffiffiffi operate the compressor (either reciprocating or rotary) is
P dis
r s ¼ : (11:57) HP t
P in HP in ¼ , (11:64)
e v e m
Using Eq. (11.50), we can write the total power require-
ment for the two-stage compressor as where e v 0:80 0:99 and e m 0:80 to 0:95 for recipro-
" # cating compressors, and e v ¼ 1:0 and e m 0:70 to 0:75 for
k P dis1 k 1 rotary compressors.
k
P total ¼ w t RT in1 1
k 1 P in1 Equation (11.64) stands for the input power required
" # by the compressor, which is the minimum power to be
k P dis2 k 1 provided by the prime mover. The prime movers usually
k
þ w t RT in2 1 : (11:58) have fixed power HP p under normal operating conditions.
k 1 P in2
The usable prime mover power ratio is
The ideal intercooler will cool the gas flow stage one to HP in
stage two to the temperature entering the compressor. PR ¼ : (11:65)
HP p
Thus, we have T in1 ¼ T in2 . Also, the pressure P in2 ¼ P dis1 .
Equation (11.58) may be written as If the prime mover is not fully loaded by the compressor, its
" # rotary speed increases and fuel consumption thus increases.
k P dis1 k 1 Figure 11.7 shows fuel consumption curves for prime movers
k
P total ¼ w t RT in1 1
k 1 P in1 using gasoline, propane/butane, and diesel as fuel. Figure 11.8
" # presents fuel consumption curve for prime movers using nat-
k P dis2 k 1 ural gas as fuel. It is also important to know that the prime
k
þ w t RT in1 1 : (11:59) mover power drops with surface location elevation (Fig. 11.9).
k 1 P dis1
We can find the value of P dis1 that will minimize the power ExampleProblem11.2 Considerathree-stagereciprocating
required, P total . We take the derivative of Eq. (11.59) with compressor that is rated at q ¼ 900 scfm and a maximum
respect to P dis1 and set this equal to zero and solve for pressure capability of p max ¼ 240 psig (standard conditions
P dis1 . This gives at sea level). The diesel prime mover is a diesel motor
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (naturally aspirated) rated at 300 horsepower (at sea-level
P dis1 ¼ P in1 P dis2 , conditions). The reciprocating compressor has a clearance
which proves Eq. (11.57). ratio of « ¼ 0:06 and e m 0:90. Determine the gallons/hr
For the two-stage compressor, Eq. (11.59) can be of fuel consumption if the working backpressure is 150 psig,
rewritten as anddofor
" #
k P dis2 k 1 1. operating at sea level
2k
P total ¼ 2 w t RT 1 1 : (11:60) 2. operating at 6,000 ft.
k 1 P in1
Solution
The ideal intercooling does not extend to the gas exiting
the compressor. Gas exiting the compressor is governed 1. Operating at sea level:
by Eq. (11.41). Usually there is an adjustable after-cooler r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r ffiffiffiffiffiffiffiffiffiffiffi
on a compressor that allows the operators to control r ffiffiffiffiffiffiffi 3 150 þ 14:7 3 164:7
p dis
the temperature of the exiting flow of gas. For greater r s ¼ 3 ¼ ¼ ¼ 2:24
p in 14:7 14:7
number of stages, Eq. (11.60) can be written in field
units as n h 1 io
" # e v ¼ 0:96 1 0:06 (2:24) 1:4 1 ¼ 0:9151
k 1
n s p 1 q 1 k p 2 n s k
HP t ¼ 1 (11:61)
229:2 (k 1) p 1 Required theoretical power to compress the gas:
"
or 0:4 #
" # 14:7(900) 1:4 164:7 3(1:4)
k 1 HP t ¼ (3) 1 ¼ 156:8hp
k n s k
181:79n s p b T 1 Q MM p 2 229:2 0:4 14:7
HP t ¼ 1 : (11:62)
T b (k 1) p 1
Required input power to the compressor:
In the above, p 1 (psia) is the intake pressure of the gas and
p 2 (psia) is the outlet pressure of the compressor after the HP r ¼ HP t ¼ 156:8 ¼ 190:3hp
final stage, q 1 is the actual cfm of gas into the compressor, e m e v 0:90(0:9151)
HP t is the theoretical horsepower needed to compress the
gas. This HP t value has to be matched with a prime mover Since the available power from the prime mover is 300 hp,
motor. The proceeding equations have been coded in the which is greater than HP r , the prime mover is okay. The
spreadsheet ReciprocatingCompressorPower.xls for quick power ratio is
calculations. PR ¼ 190:3 ¼ 0:634 or 63:4%:
Reciprocating compressors have a clearance at the end 300:0
of the piston. This clearance produces a volumetric effi- From Fig. 11.7, fuel usage is approximately 0.56 lb/hp-hr.
ciency e v . The relation is given by The weight of fuel requirement is, therefore,