Mech Sech Tech

8/15/2019 Mech Sech Tech

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(im:

• 8o study the performance of a single stage9 centrifugal compressor

• 8o get the characteristic cur es such as the 5ressure ratio ( P 2

P 1 ), isentropic

efficiency (ηs) and mechanical efficiency ηm¿ ' 3ith respect to the flo3

rate (m ) obtained9 3hile -eeping the speed of the compressor constant

(pparatus ;sed

Centrifugal compressor test rig from 5L <8 = 5(>8<E>S L8D

Location: % 11)9 ?luid Mechanics Laboratory

Compressor type: @SA+*9 from

5a$ton 5roducts

Impeller speeds: Minimum 9000

rpm9 Ma$imumA /9000 rpm

Impeller diameter : 14+ mm

Max. Pressure ratio: 1 )+:1

Inlet nozzle diameter: +/ /+ mm

Coefficient of discharge: 0 //

?igure 1 @SA+*9 from 5a$ton 5roducts 'Bith test ring





<omenclature

Mass flo3 rate m kgs

Stagnation temperature9 nlet T k

Stagnation temperature9 discharge T k Stagnation temperature9 discharge9 isentropic

compressor

T k

Specific enthalpy9 inlet h1 J kg

Specific enthalpy9 discharge h J kg

Heat loss Q J kg

(bsolute pressure9 nlet P N m2

(bsolute pressure9 discharge P N m

2

%as constant & .* for air' R J kg.k

Specific heat at constant pressure & 100+ for air' C >atio of specific heats& 1 4 for air' γ

sentropic efficiency ηs

Mechanical efficiency ηMeasuring no le diameter d mEffecti e cross section A m

2

Coefficient of discharge C (tmospheric 5ressure P N

m2

5ressure at no le discharge P N m

2

Head across no le h mm H 2 OElectrical po3er input to motor ´ w?rictional po3er lost in the dri e ´ w

5o3er input to impeller ´ w5o3er re,uired if the compressor is isentropic ´ w(ctual 5o3er for compression ´ w



8heory and Calculations:

8he air flo3 through the compressor is measured by a no le9 of S( standard form9

on the suction side of the compressor (ir9 initially at rest at atmospheric conditions9 e$pands

through the no le 8he pressure difference across the no le9 may reach *00 mm H 8he

mass flo3 rate is then gi en by:

m= 0.234 × 10 − 3( P A

√ (T 1 ))Ψ ( P 0

P A

)

( correction 3as sent hence it 3as changed to

m= 0.234 × 10 − 3

( P A

√ (T 1 ))Ψ

Bhere

Ψ =√ γ γ − 1 [( P 0

P A)2

γ −( P 0

P A)γ +1

γ ](nd

P 0 = P A− 9.81 h0

8he part of motor 3or- a ailable at the compressor shaft is obtained by -no3ing the

electrical po3er consumed by the dri e motor ;se the follo3ing empirical relation to get the

shaft 3or-:

W SH = 0.927 ( W E− 1385 )− W F Batts





ηs=W sW A

= T 2 s− T 1T 2 a− T 1

Figure 2 complete characteristics

5rocedure

G 8he speed control 3heel 3as turn cloc-3ise to set the compressor for minimum operating

speedG 8he inlet al e 3as opened and the discharge al e 3as closedG 8he stroboscope 3as s3itched on and ad usted to 11000 flashes7minG (t this condition the motor 3as started and 3ithin )0 second the discharge al e

gradually open t fully open position to allo3 air to flo3 troughG 8he speed control hand 3heel speed 3as ad usted until the e$tension of compressor shaft

appeared to rest at that moment the compressor is running at 000 rpmF 3hich is double

the stroboscope flash rateG During the test the hand 3heel 3as ad usted to ma-e the speed constantG nce a 3hole set of readings 3ere ta-en9 the e$periment 3as repeat for speeds of 49"009

)"00 and 49"00 rpms



bser ations

Table 2 shows the data for 22400

h 0 8 1 8 h 1 h B dot E

mm H C C mm Hg mm Hg Batts++0 4 4/ A" )0 10+00++0 4 4. A. / 10+00+40 4 4. A1 / 10+00+14 4 4. A 4 . 10+004"" 4 4. A44 + 10000)" 4 4/ A*" 1/ 10000

" 4 +0 A11 1 *+0011 4 +) A1)) " "000) 4 +* A144 1 4+00

Table 3 shows the data for 23600 !"

h 0 8 1 8 h 1 h B dot Emm H C C mm Hg mm Hg Batts

+/) 4 4/ A1) / 11+00"01 4 +0 A* ) 1 000+.+ ) + +0 A1 )1 1 000+*0 4 +0 A 0 40 1 000+14 ) + +0 A+ / 11+0041/ ) + +1 A*. 10+00

* ) + + A1 14 /0004. ) + ++ A14. * *000

Table 4 shows the data for 24600 !"

h 0 8 1 8 h 1 h B dot Emm H C C mm Hg mm Hg Batts

"4/ ) +1 A" )4 14+00")4 ) + +1 A/ )4 14000")0 ) + +1 + A1 )4 14+00"10 ) + + A " ) 14+00+4" ) + + + A+) . 1)00044" ) + +) A.. 4 1 000)10 ) + +4 A1)0 1" 100001"4 ) + +" + A1"0 . *+00) ) + " A1*0 +000



>esults

0 0+ 0 1 0 1+ 0 0 + 0 ) 0 )+14

)4

+4

*4

/4

114

1)4

1+4

1*4

1/4

14

0 /+

1

1 0+

1 1

1 1+

1

1 +

@ariation of Efficiencies and 5ressure >atio & 400'

eta M 5olynomial &eta M' eta S

5olynomial &eta S' 5ressure >atio 5olynomial &5ressure >atio'

Mass flo3 rate &-g7s '

Efficiency !Mecanical9 sentropic# &I' 5ressure >atio

Figure 3 how mechanical and isentropic efficiencies var# with mass flow rate for 22400 which was the closet to 22000

0 0+ 0 1 0 1+ 0 0 + 0 ) 0 )+14

1/

4

/

)4

)/

44

4/

+4

+/

"4

0

+0

100

1+0

00

+0

@ariation of Efficiencies 3ith Mass ?lo3 >ate & 400'

eta S 5olynomial &eta S' eta M 5olynomial &eta M'

Mass ?lo3 >rate &-g7s'

sentropic Efficiency &I' Mchanical Efficiency &I'

Figure 4 how the isentropic and mechanical efficienc# with mass flow rate with impeller speed of 22400



0 0+ 0 1 0 1+ 0 0 + 0 ) 0 )+14

4

)4

44

+4

"4

*4

0

0

40

"0

.0

100

1 0

140

@ariation of Efficiencies 3ith Mass ?lo3 >ate & )"00'


Mass ?lo3 >ate &-g7s'


Figure $ how the isentropic and mechanical efficienc# with mass flow rate with impeller speed of 23600

0 0+ 0 1 0 1+ 0 0 + 0 ) 0 )+14

4

)4

44

+4

"4

*4

0

+0

100

1+0

00

+0

@ariation of Efficiencies 3ith Mass ?lo3 >ate & 4"00'




Figure 6 how the isentropic and mechanical efficienc# with mass flow rate with impeller speed of 24600



0 0+ 0 1 0 1+ 0 0 + 0 ) 0 )+14

4

)4

44

+4

"4

*4

@ariation of 0sentropic Efficiencies 3ith Mass ?lo3 >ate for Different 0mpeller Speeds

)"00 5olynomial & )"00' 4"00 5olynomial & 4"00'

400 5olynomial & 400'


sentropic Efficiencies &I'

Figure % how the isentropic efficiencies with mass flow rate with &ifferent impeller speed

0 0+ 0 1 0 1+ 0 0 + 0 ) 0 )+1

1 0+

1 1

1 1+

1

1 +

1 )

1 )+

Variation of Presure Ratio with The Mass flow Rate

400 5olynomial & 400' )"00 5olynomial & )"00' 4"00

5olynomial & 4"00'

Mass low Rate !"g#s$

Pressure Ratio

Figure ' this shows how the pressure ratio var# with the mass flow rate for different .!."(s

Surge line



Discussion and conclusion:

%iscussion

n this e$periment the performance of a single stage centrifugal compressor

3as studied and the characteristic cur es 3ere plotted ?e3 issues accrued during the

e$periment 8he most successful cur e 3as the pressure ration bet3een the outlet pressure

and the inlet pressure 3ere as mass flo3 rate the 5ressure ratio increased until 1 + -g7s after

that the pressure ratio started dropping (nother thing can be seen form the as the speed

increased as the pressure ratio increased 8his a typical beha er can for the pressure ratio &see

?igure ' 8he isentropic efficiency 3as increasing until around 1 + -g7s the dropped this is

also a typical beha er and if the 3as a source of error to be mention is the use of

¿1.4 , C p= 1005 9a better 3ay to do this 3ould ha e been to e aluated at each temperature

or finding the enthalpies for each case 8he mechanical efficiency had huge noticeable error

since in general it canJt be higher than the isentropic efficiency and should be less than 100I

and in most of the reading this 3as not the case hence this error could be traced bac- to t3o

sources 3hich are the po3er of the shaft or the actual po3er and since the actual po3er 3as

can produce that big of an error then the issue most li-ely is for the shaft po3er 3here an

empirical formula 3as used to calculate this formula should be reAderi ed

?or clarification the efficiencies 3ere put in different graphs classified by speed and

the t3o different y a$is 3ere used since the mechanical efficiencies could ha e ruined the

graph by not allo3ing the reader to see the isentropic efficiency clearly

Conclusion

n conclusion the e$periment should be repeated after testing and calibrating the

e,uipment (nd erify the empirical formula and the loo-up table gi en 8he Data from the

other group could not ha e been used since there 3ere fe3 readings and did not co er the

needed speeds



References

)entrifugal compressor & 01+9 + 1+' >etrie ed from 5etroBi-i:

http:77petro3i-i org7Centrifugal compressor

Cherno 9 M & 01)' &*+,- /F T*+T +T & F/ )* T ,F - )/"! *++/ .

Lappeenranta: L(55EE<>(<8( ;< @E>S 8K ? 8ECH< L %K

%hanbariannaeeni9 ( 9 = %ha anfarihashemi 9 % & 01 9 March' 5rotecting ( Centrifugal

Compressor ?rom Surge !ipeline nd -as ournal







&ample calculation:

?or 400 sample calculation 3as done

h 0 8 1 8 h 1 h B dot Emm H C C mm Hg mm Hg Batts++0 4 4/ A" )0 10+00

P 0 = P A− 9.81 ∗550 = 101325 − 9.81 ∗550 = 95930 Pa

P 1 =− 6∗133.322 + P A= 100525 Pa

P 2 = 30 ∗133.322 + P A= 105325 Pa

T 1 = 24 +273 = 297 K

T 2 = 49 +273 = 322 K

m= 0.234 ∗10 − 3∗( P A

√ T 1 )∗

Ra!"#0 a=( P 0

P A)= 95930

101325= 0.9468

=√( γ γ − 1∗($a!"#0 a

(2

γ )− $a!"#0 a

γ +1

γ ))= 0.2241 %#$ γ = 1.4

m= 0.234 ∗10 − 3∗( P A

√ T 1 )∗ = 0.3083



nterpolation of W F = 1640

W Sh= 0.927 ∗( W E− 1380 )− ( W F )= 6815

W A= m∗1005 ∗

(T 2

a− T 1

)= 7746

%#$T 2 s }

$a!"#21 =( P2

P1 )= 1.048

a 21 = γ − 1

γ

T 2 s= T 1∗$a!"#21

a 21

W S= m∗1005 ∗(T 2 s− T 1 )= 1234

η M =( W A∗100

W SH )= 113.7

Bhich can be true

Hence the error could be traced bac- to W SH being t3o lo3 or W A being too big

f 3e assume that the sensors ha e no issues then the empirical formula to calculate

W SH is 3rong

ηS= W S∗100 1

W A= 15.94

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