Summary+Compressors - Lecture notes 3 PDF

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Chapter 12 - Reciprocating Gas Compressor Re a dpa ge3 8 1–3 8 3 , 3 8 5–3 8 7 , 3 9 2–3 9 4  Make sure that you understand how a typical reciprocating compressor works.  Explain the terms: i n d i c a t e dp o we r / wo r k , s h a f t p o we r / wo r k , i s o t h e r ma l wo r k , i n p u t p o we r ,

c o n d i t i o no f mi n i mu mwo r k , me c h a n i c a l e ffic i e n c y , i s o t h e r ma l e ffic i e n c y , v o l u me t r i ce ffi 

– as they apply to a reciprocating gas compressor. Use equations in the formula sheet to calculate indicated power (NB!NB! A l wa y sc h o o s e

t h ee q u a t i o nt h eu s e st h eg i v e nv a l u e s ) General Information and Terms: Stroke – Distance the piston moves = L Bore – inside diameter of the cylinder = d Stroke volume (Di s t a n c et h ep i s t o nmo v e sXc y l i n d e rv o l u me ) = Vs = d2L Volume of the cylinder = Va Clearance volume (Unusabl ev o l u mei nt h ec y l i n d e rb e c a u s eo f v a l v e so r o t h e rme c h a n i c a l p a r t s ; v o l u meo f a i r a l r e a d yi nt h e c y l i n d e r wh e ni n t a k es t a r t s ) = Vc (Vol u meo f a i rr e ma i n i n gi nt h ec y l i n d e r a f t e rt h ea i ri se x p e l l e d ) =Vd No clearance:

Vc = Vd = 0 and

Va = Vs

V d =V c With clearance:

Vc = %Vs and

[] P2

1 n

P1

and Va = Vc + Vs

V˙ =V a −V d Volume induced (vol u meo f a i rmo v e dwh e nt h ec y l i n d e rmo v e so u t a n db a c ko n c e ) = N ×acting V˙ =V a −V d N ×1 Single acting – the cylinder is filled and emptied once V˙ =V a −V d Double acting – the cylinder if filled and emptied twice N ×2

Free air delivery (FAD) – inlet condition used to design the compressor; (TFAD = Tinlet design ; PFAD = Pinlet design) If TFAD = T1 ; PFAD = P1 then If TFAD ≠ T1 ; PFAD ≠ P1 then

Volumetric efficiency =

FAD/cycle , V = ( V a −V d )

FAD/ cycle , V = ( V a −V d)

{( ) }

V c P2 1n m V V a−V d ηV = = = =1− −1 ms V s Vs V s P1

Mass flow through the compressor is constant -

min=

T FAD×P1 T 1×P FAD

; ms =

P1V s RT 1

and m=

P1 V P2 V P V =mFAD= FAD =mout = RT 1 RT 2 RT FAD

P 1V RT 1

No Clearance ; Single Acting Noc l e a r a n c e;Si n gl ea c t i ng o u t l e t

T h ec y l i n d e rfi l l sf r o m V=0 m3 t o V a=V s a t P1( i n l e t p r e s s u r e )

i nl e t Co mp r e s s i o na n dEx p a n s i o n

V˙ V induced= =V a =V s N

T2 n

p V=c o n s t a n t;

T1

=

[] P2

n−1 n

P1

;

Noc l e a r a n c e;Si n gl ea c t i ng

P1 P2

=

T h eg a si nt h ec y l i n d e ri sc o mp r e s s e d t oP2( d e l i v e r yp r e s s u r e )

[ ] V2

n

V1

T h ec y l i n d e ri se mp t i e df r o m V=Vd t o V=0 m3 a t P2( d e l i v e r yp r e s s u r e )

out l e t

i n l e t

With Clearance ; Single Acting Wi t hc l e a r a nc e;Si ngl ea c t i ng o u t l e t T h ec y l i n d e rfi l l sf r o m V=VC=%Vsm3 t o V=( v o l u meo f t h ec y l i n d e r )Va=( 1 + %) Vs a t P1( i n l e t p r e s s u r e )

i nl e t Co mp r e s s i o na n dEx p a n s i o n

V induced=

( )

P2 V˙ =V a −V d =(1+%)V s +%V s P1 N ×1 T2

n

p V=c o n s t a n t;

T1

=

[] P2 P1

n−1 n

;

P1 P2

=

[ ] V2 V1

n

1 n

Wi t hc l e a r a nc e;Si ngl ea c t i ng T h eg a si nt h ec y l i n d e ri sc o mp r e s s e dt o P2( d e l i v e r yp r e s s u r e ) T h ec y l i n d e ri se mp t i e df r o m V=( v o l u meo f t h ec y l i n d e r )Va=( 1 + %) Vs t o

out l e t

V d =V C V=

i n l e t

[] P2

1 n

P1

a t P2( d e l i v e r yp r e s s u r e )

With Clearance ; Double Acting Wi t hc l e a r a nc e;Doub l ea c t i ng

out l e t

o u t l e t

i nl e t

i n l e t

T h ec y l i n d e rfi l l sf r o m V=VC=%Vsm3 t o V=( v o l u meo f t h ec y l i n d e r )Va=( 1 + %) Vs a t P1( i n l e t p r e s s u r e )

Co mp r e s s i o na n dEx p a n s i o n

( )

P2 V˙ V induced= =V a −V d =(1+%)V s +%V s P1 N ×2 T2 p Vn=c o n s t a n t;

out l e t

T1

=

[] P2 P1

n−1 n

;

P1 P2

=

1 n

[] V2

n

V1

Wi t hc l e a r a nc e;Si ngl ea c t i ng T h eg a si nt h ec y l i n d e ri sc o mp r e s s e dt o P2( d e l i v e r yp r e s s u r e ) T h ec y l i n d e ri se mp t i e df r o m V=( v o l u meo f t h ec y l i n d e r )Va=( 1 + %) Vs t o

o u t l e t

V d =V C V=

i n l e t

i nl e t

[] P2

1 n

P1

a t P2( d e l i v e r yp r e s s u r e )

Example 148kg/min Nitrogen comes into this system at 20 0C and 100kPa and is delivered at 500kPa. The compressor used is a single stage, double acting reciprocating compressor, working at a speed of 500rpm. The clearance volume is 5% of the swept volume and the compressor has a stroke to bore ratio of 1.2 / 1. The compression and re-expansion index is n = 1.5. (Nitrogen = 1.389, RNitrogen = 0.297kJ/kg.K and CP Nitrogen = 1.040kJ/kg.K.)

Calculate the following: (a) (b) (c) (d) (e)

The swept volume The actual length of the cylinder in m The Indicated power The isothermal efficiency The volumetric efficiency

(a)

[ ]

1

[ ]

P2 n 500 V c =0 .05 V s ; V a =1. 05 V s ; V d =V c =0 . 05 V s P1 100

1 1. 5 =0 .146 V

s

mRT 1 148×0 . 297×( 20 + 273) m3 =128 . 791 = 100 min P1 128. 791 V = V induced= =0. 129 m3 N ×2 500×2 V induced=V a −V d =1 . 05V s −0 .146 V s=0. 904 V s =0 .129 V s =0 . 143 m3 V=

(b)

V s =πd 2 L ; L=1 . 2 d 0 .143=πd2 ( 1 . 2 d) d=0 . 336 m ; L =1. 2× 0. 336 = 0 . 403 m ( stroke length ) V a =1 .05 V s= 1 . 05× 0 . 143=0. 150 m V a =πd 2 L 0 .150=π (0 .336 )2 L L=0 . 423 m ( length of the cylinder ) (c)

{( ) }

P2 n IP= ×P1 V P1 n−1

n−1 n

{( ) }

128.791 500 1.5 −1 = ×100× 100 60 0.5

0. 5 1. 5 −1

=457.193kW

(d)

W isothermal=mRT ×ln

ηiso=

W isothermal IP

P 2 148 500 = =345 . 469 kW ×0 .297×( 20 + 273 )×ln P1 60 100

( )

( )

345 . 469 = =0. 756 457 . 193

(e) ηv =

V induced 0 .129 = =0 . 902 Vs 0 .143

Tutorial Study Examples 12.1 – 12.5 pg 384, 385 – 386, 387 – 388, 390 – 392, 394 – 396 Do Problems 12.1 – 12.4 and 12.7 – 12.8 pg 416 – 417...