BMEE303L
Unit III
Positive Displacement Compressors
Dr. Feroskhan M
Associate Professor, SMEC
Compressors – Introduction
• Used to raise the pressure of gases
• Applications
– refrigeration/air conditioning
– industrial processes
– pneumatic tools
– vehicles (super/turbo charging, air assisted braking, injection etc.)
• Classification
– Positive displacement: Pressure raised by confining the gas to a small
volume. This is achieved by an actuator (e.g. piston). Unsteady
(pulsating) flow. A reservoir is used to remove fluctuations.
– Rotodynamic: A rotating element (e.g. impeller) imparts KE to the gas
stream. It is then decelerated in a diverging passage to raise the
pressure. Steady flow device. Can be centrifugal/axial flow type
Reciprocating compressor is the most common
positive displacement compressor
Construction and Working
• Construction is similar to that of an IC engine – piston reciprocates within
the cylinder. Has connecting rod, crankshaft etc.
• Operation is similar to a non-firing engine – suction occurs during
downstroke, compression + delivery in upstroke – one crank revolution
• Valves are opened and closed by the Dp between manifold and cylinder
• Piston can be single acting or double acting (compression on both sides)
Double Acting Compressor
Delivery happens in both directions of piston motion
Rotary Positive Displacement Compressors
Screw compressor
Lobe compressor
Vane compressor
Scroll compressor
Operating Cycle
• Clearance volume neglected. V1 – Va = displacement volume
• a-1 : Suction stroke, air drawn in at ambient pressure
• 1-2 : Suction valve closes at 1, Compression (pVn = constant)
• 2-b : Delivery valve opens at 2, delivery at high pressure
• Work done per cycle = p dv = - v dp = area a-1-2-b-a
• Value is negative as work is consumed
p2V2 p1V1
W p2V2 p1V1
n 1
n
p2V2 p1V1
n 1
n (BDC)
mRT1 T2 / T1 1
(TDC)
n 1
T1 = intake temperature,
n n 1 n n 1
W mRT1 r p n 1 p1V1 r p n 1 rp = pressure ratio,
n 1 n 1 m = mass per cycle
Nature of Compression Process
• Compression work for different processes :
1
Isentropic :
mRT1 rp 1
1
n 1
mRT1 rp n 1
n
Polytropic :
n 1
Isothermal : mRT1 lnrp
• Power consumption (P = work/cycle * cycles per second)
P = WN/60 for single acting compressors
P = 2WN/60 for double acting compressors
• Work is minimum when compression is isothermal (process 1-2’ in
previous slide)
• In actual case, compression is an irreversible process (involves friction),
results in an increase in entropy, consumes more work
Efficiency Indices
• Isothermal efficiency compares the actual work with isothermal work
Isothermal work
T
Actual work
• Isentropic efficiency compares actual work to isentropic work
Isentropic work
s
Actual work
• Volumetric efficiency is the ratio of actual volume of gas taken in per
cycle to the ideal volume (displacement/swept/stroke volume)
Vactual m actual
v
Vd mideal
Effect of Clearance Volume
• Clearance volume is usually < 5% of
displacement volume
• This is quantified by the clearance factor,
e = Vc/Vd
• At end of delivery, some high pressure gas
remains in the clearance volume (pt 4)
• During suction stroke, this gas expands to
pt 5 before fresh air starts coming in
• In effect, the clearance volume hampers the intake of gas
• Effective intake volume (inducted volume) = V2 – V5
• It can be shown that the specific work input (work per unit mass) is
n 1
RT1 rp n 1
unaffected due to clearance volume n
w
n 1
1 e rp n 1
V V 1
• Volumetric efficiency is given by: v 2 5
V2 V1
Example - Single Stage Compressor
• A single cylinder, single acting air compressor (30 cm bore, 40 cm stroke)
operating at 100 rpm takes in air at 1 bar and 20 oC and compresses it to a
pressure of 5 bar. The compression process follows the law pV1.2 = const.
Calculate a) the mean effective pressure, b) power consumption, and c)
isothermal efficiency neglecting clearance volume. e) Find the volumetric
efficiency when the clearance factor is 4%.
• Given: p1 = 1 x 105 Pa ; p2 = 5 x 105 Pa ; rp = 5; T1 = 293 K; n = 1.2
a) Volume at end of suction (for no clearance) = V1 = p (d2/4) L = 0.028 m3
For the given polytropic process, work input per cycle = W = 5.22 kJ
Mean effective pressure = W/V1 = 1.85 bar
b) Power consumption = WN/60 = 8.7 kW
c) For isothermal process, cyclic work, WT = p1V1ln(rp) = 4.55 kJ
Isothermal efficiency, T = WT/W = 87.2%
d) For e = 0.04, volumetric efficiency, vol = 88.7%
Multi-staging
• Multi-staging is used when the values of rp are very high (above 5)
• Deviation between polytropic and isothermal increases with rp
• Final temperature is also high in polytropic compression: affects
component life, lubrication
• High rp also lowers volumetric efficiency
• Solution: when rp is very high compress in stages with intermediate cooling
• Principle: Polytropic compression + isobaric cooling isothermal
• Compression isothermal for infinite stages
• Up to 4 stages may be used
• Each stage has an rp of up to 5
• More stages: lower work consumption , higher v
• For k stages with intercooling (same initial T) and
each handling same mass,
n 1 n 1 n 1
mRT1 rp1 n rp 2 n rpk n k
n
W
n 1
• Work is least when rp1 = rp2 = …. rpk same work in all stages
Example - Multi-stage Compressor
• A two stage air compressor with ideal intercooler pressure and perfect
intercooling (what does this mean?) compresses air from 1 bar to 16 bar at
the rate of 5 m3/min. Mechanical efficiency of the compressor is 80%. Assume
polytropic compression and expansion (n=1.3). Calculate: a) the power
consumption, b) volumetric efficiency when the clearance volume of the LP
cylinder is 3% of its stroke volume, and c) power consumption for an
equivalent single stage compressor.
• Given: p1 = 1 x 105 Pa ; p3 = 16 x 105 Pa ; V1N/60 = 5/60 m3/s ; e = 0.03
a) With ideal intercooler pressure, rp1 = rp2 = √16 = 4
Total indicated power = 2* power of stage 1 = 27.23 kW
Actual power consumption = Pi/mech = 34.04 kW
b) For e = 0.03, volumetric efficiency of LP cylinder, vol = 94.3%
c) For equivalent single stage compressor (rp = 16), indicated power = 32.36 kW
Power consumption = Pi/mech = 40.45 kW
Thank You !
