RECIPROCATING COMPRESSORS
There are various compressor designs:
Rotary vane; Centrifugal & Axial flow (typically used on gas
turbines); Lobe (Roots blowers), and Reciprocating.
The main advantages of the reciprocating compressor are that
it can achieve high pressure ratios (but at comparatively low
mass flow rates) and is relatively cheap.
It is a piston and cylinder device with (automatic) spring
controlled inlet and exhaust valves. Delivery is usually to a
receiver. The receiver is effectively a store of energy used to
drive (eg) compressed air tools.
Inlet
Delivery
BDC
TDC
¬ Swept vol. ®
Clearance vol.
Receiver
1
Reciprocating compressors usually compress air but are also
used in refrigeration where they compress a superheated
vapour (to which the gas laws strictly do not apply).
In order to be practical there is a clearance between the piston
crown and the top of the cylinder. Air 'trapped' in this clearance
volume is never delivered, it expands as the piston moves
back and limits the volume of fresh air which can be induced
to a value less than the swept volume.
The induced volume flow is an important purchasing
parameter. It is called the "Free Air Delivery" (FAD), and it
measures the capacity of a compressor in terms of the air flow
it can handle. It is normally measured at standard sea level
(SSL) atmospheric conditions and allows the capacities (size)
of compressors to be compared.
N.B. The induced mass per cycle must equal the delivered
mass per cycle (continuity!), although the induced and
delivered volumes will be different.
2
Cycle Analysis
The cycle may be analysed as two non-flow (compression and
expansion) processes and two flow processes (delivery and
induction)
PROCESS
GROSS WORK
1®2
Compression
p2V2 - p1V1
n-1
2®3
Delivery
p2(V2-V3)
3®4
Expansion
4®1
Induction
p4V4 - p3V3
n-1
p1(V4-V1)
Note that we assume polytropic compression and expansion.
This is because some degree of cooling is usually attempted
for reasons we shall see later.
If no cooling were attempted n becomes g.
On p-V co-ordinates:
pressure (kPa)
1000
900
800
700
3
2
600
500
400
300
200
100
0
4
0
0.2
1
0.4
0.6
0.8
Volume (litres)
1
1.2
3
The work per cycle is given by: å gross work
p4V4 - p3V3+ p1(V4-V1)
work per cycle = p2V2 - p1V1 + p2(V2-V3) +
n-1
n-1
p4V4 - p1V1 + p1(V4-V1)+ p2V2 - p3V3 + p2(V2-V3)
n-1
n-1
but p1=p4 & p2=p3
=
work per cycle =
p1(V4-V1)
p2(V2-V3) p2(V2-V3)
+
p
1(V4-V1) +
+
n-1
n-1
= p1(V4-V1) {1+ 1 } + p2(V2-V3){1+ 1 }
n-1
n-1
but
\
mass delivered = mass induced
p2(V2-V3)
=
RT2
p1(V1-V4)
RT1
p2(V2-V3) = p1(V1-V4) T2
T1
n
work per cycle = p1(V1-V4) { n-1 } [ T2 -1]
T1
for a polytropic process :
T2 = (
T1
n-1
p2 ) n
p1
=
n-1
n
rp
Noting that (V1-V4) is the induced volume (Vind), and p1 is the
inlet pressure (pin) we may re-arrange and write:
n-1
work per cycle = n pin Vind { rp n -1}
n-1
NB Power required = work per cycle x cycles per sec
4
Volumetric Efficiency
We have already noted that the induced volume is less than
the swept volume. To enable this effect to be evaluated we
define volumetric efficiency (hvol) as:
hvol =
Induced volume
Swept volume
= V1-V4
Vs
but
n
p3V3
=
n
p4V4
\
V4 = V3 rp
1
n
V3 is the clearance volume (Vc), and V1 = Vc + Vs
hvol = Vc + Vs Vs
\
hvol
1
Vc rp n
1
V
c
=1( rp n - 1 )
Vs
The reference conditions (p & T) at which the volumetric
efficiency is measured should always be quoted (it would
normally be SSL conditions).
[The concept of hvol applies also to reciprocating engines.]
hvol
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Vc = 0.05
Vs
n = 1.27
0
10
20
30
40
rp
50
5
Volumetric Efficiency referred to SSL conditions.
In testing a compressor, the measured induced volume flow
will be that of the actual test inlet conditions.
It is unlikely that these inlet conditions will be SSL.
We therefore need to refer our results to SSL conditions.
(Measured)
.
Inlet >> Vi
Ti
SSL
Ts
Ps
Pi
The mass flow of gas must be the same both at SSL (s)
conditions and at Inlet (i) conditions.
.
.
ms = mi
.
.
p
i
V
i
psVs
=
RTi
RTs
.
Vs
=
pi Ts
ps Ti
.
Vi
.
. sides by V
dividing both
swept
hvol(SSL) =
pi Ts
ps Ti
hvol(inlet)
6
Compressor Efficiency
If we plot the specific work (kJ/kg delivered) against the
polytropic index n we obtain:
n-1
w = n RTin { rp n -1}
n-1
250
rp=8
200
w
kJ/kg
150
rp=4
100
50
0
1.1
1.2
Polytropic
n
1.3
1.4
Adiabatic
compression
Isothermal
compression
1
It is clear that the closer the compression is to isothermal the
less work is required.
The savings are greater at higher pressure ratios
(eg above 18% @ rp=4 and 27% @ rp=8)
It is also apparent that isothermal compression represents
the ideal minimum work input for a compression process
(where cooling is feasible).
7
We can therefore define compressor efficiency as:
hiso =
Isothermal work per cycle
Actual work per cycle
If we recalculate the work input assuming isothermal
compression [ W12 = p1V1 ln(p1/p2) etc] it is found that:
hiso =
ln rp
n-1
n
{
r
p n -1}
n-1
Note that this efficiency is known as the isothermal efficiency.
The degree of cooling possible during a single stage
compression process tends to be limited. It improves at low
speeds but this limits compressor capacity.
One way of improving efficiency, especially at higher
compression ratios and speeds, is to go to multistage
compression with cooling of the gas between each stage.
8
Multistage compression
To avoid unacceptable reductions in compressor capacity
(RPM and volumetric efficiency) and to minimise power input
with high compression ratios, multistaging with inter-cooling is
used.
The number of stages will normally be between two and four.
intercoolers
> out
in >
Stage 1
Stage 2
Stage 3
Each stage may be treated as a separate compressor,
however, with multistaging, all will normally rotate at the same
speed.
The volumetric efficiency of the compressor as a whole is
determined by the first stage.
If the inter-cooling is such that the inlet temperature to the
following stage(s) is the same as the inlet temperature to the
first stage we have ideal (or perfect) inter-cooling.
Since the work input to any stage is dependent on the
pressure ratio across it, it should be possible to minimise the
total work input by the correct choice of compression ratio
across each stage.
9
Optimum stage pressure ratio
Assume we have two stages of compression with ideal
intercooling and the same index of compression (and
expansion) 'n' in each stage.
Total Work per cycle
n-1
n-1
= n ép1in V1ind { rp1 n -1} + p2in V2ind { rp2 n -1}ù
n-1 ë
û
Since the mass induced by the first stage must be equal to
the mass induced by the second stage:
p1in V1ind
p2in V2ind
=
R Tin
R Tin
n-1
n-1
n -1} ù
\ Total Work per cycle = n p1in V1ind é{ rp1 n -1} + { rp2
n-1
ë
û
If p1 is the inlet pressure and p2 the final delivery pressure,
let pi = the inter-stage pressure:
then
rp1 = pi
p1
rp2 = p2
pi
&
If we substitute the above in the expression for Total Work and
differentiate wrt pi, we can find pi for minimum Total Work.
½
whence
pi = [p1 p2]
or
rp1 = rp2
½
p
2
=( )
p1
We could extend the same method to N stages with the result
that, for minimum work input, the pressure ratio across each
stage must be the same and equal to the Nth root of the
overall pressure ratio.
rp(opt) = rp(overall)
1
N
10