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					5

Compressors

5.1 Introduction · definition: “a compressor is a device that produces flow and/or pressure in a fluid by the expenditure of work”. · · a compressor operates between a low pressure source and a higher pressure delivery. the compression process is, of course, irreversible. This arises mostly from a number of fluid dynamic effects, including departures from adiabatic behaviour, friction losses and, in some cases, shock waves. The final delivery temperature and enthalpy are therefore higher than these ideal values because of these irreversibilities:

5.2 Types of compressors · there are two basic forms of compressor: 1. reciprocating: usually characterised by intermittent delivery, large pressure ratio, small mass flow. 2. rotary: usually characterised by continuous delivery, small to moderate pressure ratio, larger mass flow 5.3 Reciprocating compressors · a reciprocating compressor can be analysed in 2 ways: 1. Open, steady flow system, taking in a fluid at a low pressure, delivery at a higher pressure. 2. A closed piston and cylinder system containing the working fluid, with the process repeated to obtain the flow (this is not quite a complete model because we must include a mass transfer). · small capacity compressors are almost all reciprocating designs.

5.3.1 Zero clearance volume reciprocating compressors · the cycle is composed of 3 distinct processes: 1. inlet (mass transferred into volume) – inlet valve open 2. compression – both valves closed 3. delivery (mass transferred out of volume) – delivery valve open · schematically, this can be represented by:

· ·

where V 1 = swept volume and V 2 = clearance volume = 0 (i.e cylinder head touches top of casing) on a PV diagram, this is represented as:

·

in terms of each process: 1. inlet: cylinder head moves down from V 2 (=0) to V 1 , drawing in fluid. 2. compression: cylinder head moves up and, since both valves are closed, fluid is compressed.

3. delivery: whilst cylinder head is moving up, delivery valve is opened and fluid leaves the cavity. · · · considering the work done per cycle: W = Wab + Wbc + Wda { { {
<0 <0 >0

note that for a compression, work is done on the fluid and is therefore taken as negative. for a polytropic process between points 1 and 2: W12 = ò pdV
1 2

= cò

2

1

dV Q pV n = c Vn
2

c é 1 ù = (1 - n) ê V n -1 ú1 ë û = · ·

1 ( p 2V 2 - p 1V 1 ) (1 - n)

if process is isobaric (n=0): W 12= p(V 2-V 1) the work per cycle is then:
W = Wab + Wbc + Wda = = ( p 2Vb - p 1Va ) + p 2 (Vc - Vb ) + p 1 (Va - Vd ) { { (1 - n) =0 =0 ( p 2Vb - p 1Va ) - p 2Vb + p 1Va (1 - n)

é1 - (1 - n) ù = ( p 2Vb - p 1Va ) ê ë 1- n ú û n = ( p 2Vb - p 1Va ) 1- n

·

for an ideal gas: p 1Va = m a RT 1

& p 2Vb = m b RT 2
· and the mass delivered m = m a = m b . Thus, the overall work per cycle becomes n W= mR(T 2 -T 1 ) 1- n for a polytropic process with an ideal gas:

·

T 2 æ p2 ö ÷ =ç T 1 ç p1 ÷ è ø

n -1 n

· thus, the work per cycle can be expressed as
n -1 é ù æ p2 ö n n ê W= mRT1 ç - 1ú êè p 1 ÷ ú (1 - n) ø ê ú ë û

· and the power input is:
n -1 é ù æ p2 ö n & = n mRT êç & 1 W - 1ú êè p 1 ÷ ú (1 - n) ø ê ú ë û

5.3.2 Compressors with clearance volume · real compressors must have valves to pass flow into and out of the cylinder. The result is a ‘clearance volume’ always being present above the piston.

· ·

the clearance volume can be as much as 20% of the swept volume, although it can be significantly smaller with more sophisticated designs. compression ratio (based on volumes): swept volume rv = clearance volume on a PV diagram:

·

·

the cycle consists of 4 processes (cf. 3 processes in zero clearance volume case): 1. 2. 3. 4. inlet a®b inlet valve open, mass transferred in compression b®c both valves closed delivery c®d delivery valve open, mass transferred out expansion d®a both valves closed

· · ·

note that, as before, this is not a true ‘closed’ thermodynamic cycle because the mass changes. also note that a mass of gas which corresponds to the clearance volume (V d ) , at delivery pressure and temperature, remains in the cylinder. considering the work done per cycle (see PV diagram):

W = area ( abcd ) = area (1bc2 ) - area (1ad2 ) 14 3 14 3 24 24
zero clearance volume process zero clearance volume process

·

using the expressions for work derived earlier for zero clearance volumes: n n W= m b R(T 2 -T 1 ) m a R (T 2 -T 1 ) 1- n (1 - n) n (m b -m a ) R (T 2 -T 1 ) = (1 - n) since:

·

mass delivered = (mass in cylinder at c) - (mass in cylinder at d) =m c -m d = m b -m a Q m b = mc & m a = m d

· it follows that
W=
· · where m = mass delivered note: 1. that this expression is the same as that for a compressor without clearance, delivering the same amount of gas. 2. note this derivation assumed that the work of compressing the clearance volume gas is returned when the gas is expanded i.e. n COMP = n EXP . 5.3.3 Volumetric efficiency · the volumetric efficiency relates the actual to the ideal flow rates: hVOL = ·

n mR (T 2 -T 1 ) (1 - n)

mass of delivered gas mass of gas in the swept volume at inlet P&T

since by convention the gas compressor flow rates are commonly expressed using inlet conditions: volume of air delivered at inlet P&T hVOL = swept volume it therefore follows that the clearance volume has the effect of reducing the volumetric efficiency, which increases the size of the plant for a given delivery volume. the effect of clearance volume on n VOL can be explained graphically:

·

·

·

an increase in the clearance volume (Vd ® Vd ') has a proportionally greater impact on (Va ® Va ') than on change in swept volume (Vsw ® Vsw ') . Thus: hVOL ' < hVOL

·

deriving an expression for n VOL : (V - V ) @ P 1 , T 1 hVOL = c d VSW

= = =

Vb - Va VSW VSW - (Va - VCL ) VSW VSW - VCL (V a / V CL -1) VSW
1/ n 1/ n - 1ù V û Q a = V a = æ p2 ö ç ÷ VCL V d è p 1 ø

VSW - VCL é( p 2 / p 1 ) ë = VSW
· therefore:

hVOL

V = 1 - CL V SW

éæ p ö1/ n ù êç 2 ÷ - 1ú êè p 1 ø ú ë û

5.3.4 Mean effective pressure · the mean effective pressure ( pM or MEP ) is the pressure that would produce the same work as the actual cycle of the same swept volume:

· thus:

W = pM VSW & & W = pM VSW N

5.3.5 Effect of pressure ratio · for a given swept volume, hVOL decreases as the pressure ratio increases. This can be seen in the PV diagram:

·

it follows that multistage compression should be considered for high pressure ratio applications in order to achieve reasonable n VOL :

·

note the decreasing cylinder volume for each subsequent stage.

5.3.6 Intercooling · the delivery temperature of compressed gas can be high, even for moderate delivery pressures. Cooling between stages can therefore be desirable.

· ·

for complete intercooling, the gas leaving the first compression stage is cooled back to T 1 before further compression in the second stage. note: 1. intercooling reduces the work of compression 2. the more stages, the greater the work saving (in the limit, many stage intercooled compression approaches an isothermal process).

·

calculating the ( i.e. Ti = T 1 ) :

optimal

intermediate

pressure for

complete

intercooling

W = Wab + Wcd
n -1 n -1 é ù é ù n n n êæ p i ö - 1ú + n mRT êæ p 2 ö - 1ú = mRT1 ç ÷ 1 ç êè p i ÷ ú êè p 1 ø ú 1- n 1- n ø ê ú ê ú ë û ë û

· minimum work output occurs at:
n -1 é n -1 1- 2 n ù n dW n ê n - 1 æ 1 ö . p i -1/ n - n - 1 p 2 n . p i n ú =0= mRT1 ê n ç p1 ÷ ú dp i 1- n n è ø ê ú ë û

· simplifying further:
æ 1 ö ç ÷ è p1 ø

n -1 n

p i -1/ n = p 2

n -1 n

.p i

1- 2 n n

p i n n = ( p 1. p 2 ) 1 24 4 3
pi
2 ( n-1) n

(2 n -1) 1 -

n -1 n

· thus:
Þ p i 2 = p 1. p 2 or

pi p2 = p1 p i

·

i.e. each stage has the same pressure ratio and work output.

5.4 Rotary compressors · rotary (or rotating) compressors take numerous forms · for example, the ‘roots blower’ is used for low pressure ratio, large volume flow rate applications e.g. supercharging, powder transfer:

· ·

commonly have 2 or 3 lobes, geared in phase. the ‘vane compressor’ features vanes that retreat into rotor hub as rotor rotates:

·

and is used for high volume flow rates with low pressure ratios

· the ‘screw compressor’ delivers moderately high pressures and higher mass flow
rates:

5.4.1 Centrifugal compressors

· · ·

high rotational speed high exit velocity becomes static pressure in the diffuser. used in turbochargers, small gas turbines.

5.4.2 Axial flow compressors

· note that only the rotor blades are shown in the above picture – the stators are
connected to the (removed) casing · · · multistage blading with the pressure rise distributed between the stationary (stator) and rotating (rotor) blades. high rotational speeds, but lower than centrifugal used in large gas turbines (including aircraft) and industrial compressors.

5.4.3 Steady flow analysis of rotating compressors · heat transfer is negligible in most cases, meaning that irreversibility is mainly the result of boundary layer losses and shock losses:

·

& assuming that Q » 0 & Dn 2 / 2 » 0 : æ ö 2 & - W = mD ç h + n + gz ÷ & & Q ç 2 {÷ { ~0 ÷ ç ~0 è ø thus: & & & -W = mDh = mC p (T2 '- T 1 ) æT ' ö & = mC pT1 ç 2 - 1÷ èT1 ø

·

·

T ' æ p 'ö For an ideal gas: 2 = ç 2 ÷ T 1 è p1 ø

g -1 g

. It follows that:
g -1

é ù g êæ p 2 ö ú & & Þ -W = mC pT1 êç ÷ - 1ú êè p 1 ø ú ë û
· remembering that a real compressor is not isotropic: h '- h 1 T 2 '- T 1 hc = 2 = if C p = const. h 2 -h 1 T 2 -T 1 thus: 1 T 2 -T 1 = (T 2 '- T 1 ) hc
g -1 é ù & mC pT 1 êæ p 2 ö g ú & = & -W ç ÷ - 1ú h c êè p 1 ø ê ú ë û of course, the performance of a real compressor is complicated by other effects. We can appreciate that the pressure rise is dependent on the rotor speed, the ambient conditions and the impedance of the pumping circuit.

·

·

·

these effects can be seen in a ‘compressor map’, which relates the pressure rise & to the rotor speed (N) and the massflow (m ) :


				
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posted:7/19/2009
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