RECIPROCAL COMPRESSORS

APPLICATIONS

Pneumatic hand tools Drills Paint spraying Mining Blast furnaces Lifts rams and pneumatic conveyors

FAD:

Actual volume of air delivered by an air compressor is reduced to either NTP or STP conditions or intake conditions.

(ṁ) inlet of compressor = (ṁ)outlet of compressor

P1*V1/RT1 =P2*V2/R*T2 = Pf*Vf/TF(neglecting Vc)

Pf ,Vf , Tf are free sir conditions & Vf will be ‘FAD’.

For convenience Pf=101.325kPa

Tf=288k

If Vc is taken into account

Pf*Vf/Tf =P1(V1-V4)/T1= P2(V2-V3)/T2

P in bar M3/min ‘v’ applications

Roots 1 to 3 bar .14 to 1400 Scavenging,supercharging of ic engines

Rotary vane 1 to 8.5 150 Hinge vane compressor,expander used a integrated supercharges to active throttle(ISCAT)

Screw 1 to 3.2 3-1000 FOOD,CHEMICAL,PETROCHEMICAL,REFINING,STEEL INDUSTRIES

Single stage reciprocating

1 to 46.87 0.1 Higher pressure and low discharge

Centrifugal 11 to 3.2 60-190 Aircraft unit turbo propeller

APPLICATIONS OF VARIOUS TYPES OF COMPRESSORS

COMPRESSORS DISCHARGE PRESSURE(MPa)

DISCHARGE VOLUME(m3

/min)APPLICATIONS

Centrifugal compressor 69 170-830 Continuous duty functions as ventilation fans ,air movers ,cooling units on turbo charges and supercharges

Axial flow compressor 2 830-2300 Jet engines ,air conditioning systems in aircraft and in bleed air

Reciprocating compressor

180 780 Oil refineries ,gas pipelines ,chemical plants and refrigeration power plants

Roots blower 0.6 to 1 100 -120 Pneumatic conveying of bulk materials ,pressurized aeration of basins in sewage treatment plants ,High vacuum boosters

Rotary vane compressor 1.3 35 Vacuum pump and air motor

Screw compressor 8.3 45 Super chargers ,vacuum pumps

GOOGLE SEARCH

Applications of various compressors Discharge pressure of various compressors Discharge volume of various compressors

Air compressor Importance Industry applications Brief working principle Construction details Work supplied without clearance volume Efficiencies Work supplied with clearance volume Volumetric efficiency

Some basic formulas

WORK DONE PROCESS P,v,t relation

(p2v2-p1v1)/n-1 polytrophic Pvn=c

Wd=p(v2-v1) isobaric v/t=c

Wd=p1v1 ln(p2/p1) isothermal Pv=c

Wd=(p2v2-p1v1)/γ-1 Rev adiabatic/isentropic pvγ=c

Wd=v(Δp) isochoric P/T=C

Work supplied to compressor without V c:

4-1 – Suction constant pressure

1-2 – compression polytropic

2-3 – discharge constant pressure

W = (P2V 2−P1V 1/n-1) + P2 (V 2−V 1 )−P1(V 1−V 4)

= (P2V 2−P1V 1/n-1) + P2V 2−P1V 1

= (1/n-1 + 1) (P2V 2−P1V 1)

= n/n-1 (P2V 2−P1V 1)

= n/n-1* P1V 1 (P2V 2/P1V 1 - 1)

P1V 1n=P2V 2

n =› V 2V 1

=(P1P2

)1/n

=(P2/P1)−1 /n

w c = n/n-1*P1V 1(( P1P2 )1n (P2P1 )

−1n −1)

w c = n/n-1*mRT 1(( P2P1 )(n−1)/n

−1)

= n/n-1*mRT 1(T2T1

−1¿

T2T1

=(V 1V 2

)ϒ−1

=(P2P1

)ϒ−1 /ϒ

Also, T1T2

=(V 2V 1

)ϒ−1

=(P1P2

)ϒ−1 /ϒ

, √2√1

=(P1P2

)1/ϒ

or √1√2

=(P2P1

)1/ϒ

Comparison of work supplied among three processes

1-2 – isothermal process (PV =C)

1-3 – polytropic process

1-3 – reversible adiabatic process

1-2 – isothermal process:

w cT=P1V 1 log(V 1V 2 )=mRT 1(P2P1

¿ (since PV = C)

1-3 – polytropic

w cp=P sV s*n/n-1 (( PdP s )(n−1)/n

-1)

1-4 – adiabatic

w ca=P sV s*ϒ/ϒ-1 (( PdP s )(ϒ−1)/ϒ

-1)

ηisothermal= w cT /¿ w cp = log( PdPs )/ n/n-1 (( PdP s )(n−1)/n

-1)

ηadiabatic = w cp/¿ w ca = n/n-1 (( PdP s )(n−1)/n

-1)/ ϒ/ϒ-1 (( PdP s )(ϒ−1)/ϒ

-1)

ηmechanical = IP/SP, SP-IP = Fr.P.

Effect of clearance of work of compression:

w c=P1V 1*n/n-1 (( P2P1 )(n−1)/n

-1) - P4V 4*n/n-1 (( P3P4 )(n−1)/n

-1)

w c=P1(V 1−V 4)*n/n-1 (( P2P1 )(n−1)/n

-1) if V 4 is negligible

¿ P1V 1*n/n-1 (( P2P1 )(n−1)/n

-1)

Also, w c=mRT1*n/n-1 (( P2P1 )(n−1)/n

-1)

Also, w c=P1V 1*nc/nc-1 (( P2P1 )(nc−1)/nc

-1) - P4V 4*ne/ne-1 (( P3P4 )(ne−1)/ne

-1)

Clearance ratio ‘C’ = V c /V s

Also, ‘C’ = 3% to 12% of V s

Thus the effect of clearance is to reduce the volume of air actually sucked in working cycle

Two stage air compressor with inter cooling and without clearance volume:

Wc=n/n-1*p1v1((p2/p1)n-1/n-1) +n/n-1*p2p3((p3/p2)n-1/n-1)

P1V1=P2V2= Perfect intercooling to T1

Wc=n/n-1*p1v1((p2/p1)n-1/n + (p3/p2)n-1/n-2

dwc/dP2=0 for minimum work

(1/p1)n-1/n- p3n-1/n/p2

2(n-1/n)=0

P2=(p1p3)0.5

Wc=n/n-1*p1v1((p3/p1)n-1/2n+(p3/p1)n-1/2n-2

Wc= (2n/n-1 v1 (p3/p1)n-1/2n-1)p1

For N-stages,

Similarly

Wc= (Nn/n-1) *p1v1 (pN+1/p1)n-1/Nn-1

Wc= (Nn/n-1)maRT1((pL/pS)n-1/n-1)

VOLUMETRIC EFFICIENCY

η= Actual volume of air intake per cycle/volume of Air which could theoretically fill the swept volume under optimum condition

ηvol=Va/VS=V1-V4/VS

=(VS+CVS)-CVS(P2/P1)1/n/Vs

ηvo=1+c- CVS(P2/P1)1/n

=1-C(V1/V2-1)

T2/T1 =( V1/V2)N-1= (p2/p1)n-1/n

PROBLEMS:

1) Air is to be isentropically compressed at the rate of 1 m3/s from 1 bar and 200 c to 10 bar, Find the work of compressor and the volumetric efficiency. If the clearance volume is 4% of stroke volume for all the cylinders for the following causes1) single stage2) two stage3) Three stage compression

Given:

Free air delivered = 1.2 m3/s

vc=0.04 vs

Assume ρ=1.2 kg/ m3

1. Work done =ϒ/ϒ-1 mRT 1((P2/P1¿¿ϒϒ−1-1)

= 1.4/0.4 * 1.2 * 0.287 * 293 *(100.4 /1.4-1)

= 328.71 kw

ηvol = 1- V c /V s ¿-1)

= 1 – 0.04 (101 /1.4-1)

= 83.28%

2. Intermediate pressure

P2= √P1P4

W=2*ϒ/ϒ-1*maRT1((P2/P1)ϒ/ϒ-1-1)

=2*1.4/0.4*1.2*0.287 * 293 *(3.160.4 /1.4-1)

=274.92 kw

ηvol=1-0.04(3.161/1.4-1)

=94.9%

P2=3√P1P2

=3√10

=2.15 Bar

WC=3*1,4/0.4*1.2*.287*293*(2.15).4/1.4-1)

=259 Kw

ηvol¿

¿=1-0.04*(2.151/1.4-1)

=97.09%

2.A single stage single acting reciprocating air compressor with 0.3 m bore and 0.4 m stroke runs at 400 rpm.the suction pressure is 1 bar at 300k and the elivery preure is 5 bar.Find the power required to run it if compression is thermal,compression follows the Pv1.3=constant nd compression and reversible adiabatic.Alao find the isothermal efficiency

Volume of air compressed per min=ᴫd2/4*1*N

=3.14/4*0.302*0.4*400

=11.31 m3/min

Given:

d-=0.3 m

l=0.4 m

N=400rpm

P1=1 bar

T1=300k

P2= 5bar

Isothermal compression work=P1V1ln(P2/P1)=1*105*11.31/60*ln(5/1)

= 30337.0 W.

Polytropic compression work=P1V 1*n/n-1 (( P2P1 )(n−1)/n

-1)

=1*105*11.31/60*(1.3/1.3-1)((5/1).3/1.3-1)

=36.74 kw

Isothermal efficiency=isotherml work/actual work)*100=30.34/36.74)*100

=82.58%

Isentropic work of compression=P1V 1* ϒ/ϒ-1*((P2/P1)ϒ/ϒ-1-1)

=1*105*11.31/60*(1.4/1.4-1)((5/1).4/1.4-1)

=38.52 kw

ηadiabatic¿

¿=36.74/38.52)*100=78.76%

ηadiabatic¿

¿=actual work/adiabatic work done

3. A double acting two stage compressor delivers air at 25 bar. The pressure and temperature of air at the beginning of compression in L.P. cylinder are 1 bar and 20 degree Celsius . the temperature of air coming out from an intercooler between two stages is 40 degree Celsius and the pressure is 7 bar. The diameter and stroke of L.P cylinder are 60 cm and 80 cm and N = 100 rpm. The ηvol = 80%. Find the power of the electric motor required to drive the compressor assuming ηmech = 85%

Given:

Pd = 25 bar, P1 = 1 bar , T1 = 20+273 = 293 k, T3 = 313 k , P2 = 7 bar

D = 60 cm, l = 80 cm ,N = 100 rpm , ηvol = 80%, ηmech = 85%

PVn = C, n = 1.35

Solution:

I.P = n/n-1 maR(T1((P2/P1)n-1/n -1)+ T2((P3/P2)n-1/n -1)

Ma = P1V1/RT

= 1*105*3.14*0.62*0.8*0.8*100*2/4*287*293*60

Ma = 0.7174 kg/s

I.P = 1.35/0.35 * 0.7174* 287 * (293(70.35/1.35-1)+313(25/7)0.35/1.35-1))

I.P = 249.9 kw

B.P = I.P/ ηmech

=249.9/0.85

=294 kw

MULTI STAGE RECIPROCATING AIR COMPRESSOR

Compression of air to high pressure in one stage has many disadvantages,

Less volumetric efficiency Higher work of compression Leakage past the piston Ineffective cooling of the air Robust construction of the cylinder(to withstand delivery pressure)

Hence, multistage compression is necessary for efficient compression with higher volumetric efficiency.

In multistage compression, the air can be cooled perfectly at intermediate pressures resulting in same power required to drive the compressor, than compared to a single stage compressor of same delivery pressure and flow rate. The mechanical balance of the machine is also better in multistage compression due to phasing out of operation in stages.

TWO STAGE AIR COMPRESSOR WITH INTERCOOLER

The arrangement of a two stage compressor is given in the figure below. Air is sucked into the low pressure cylinder and compressed into intermediate pressure. This compound air at higher temperature from l.p cylinder flows into a air-cooler. In the inter cooler the air is cooled to its initial temperature, In the high pressure cylinder, the air is then compressed to the final delivery tube.

The compressed air from the H.P cylinder is also passed through an after cooler sometimes to send cooled air to the reservoir n avoid the reservoir to the store hot air.

The inter cooler and after coolers are simple heat exchangers. The coolant may be water or any other fluid which passes through the tubes secured between the two end plates and the air circulates over the through a system of baffles. Baffles are provided to ensure intimate contact between the tubes and the air in long path.

The indicator diagram for low pressure and high pressure cylinders. Suction to the low pressure stage is represented by 7-1 and this is followed by a compression process from pressure p1 to p2

1 following the polytropic flow PvN=c.

Thus the air is then discharged from the low pressure cylinder to the inter cooler where it is cooled at constant pressure condition 2, to the initial temperature T .From the inter cooler the air is sucked into the high pressure cylinder followed by a compression process from p2-p3 following the law Pvϒ=c. Finally the air I discharged at p3 to the receiver.

Total compression work done on air is,

Wc = n/n-1 * P1 V1 ((P2/P1)n-1/n-1)+n/n-1 * P2 V2 ((P3/P2)n-1/n-1)

assuming the same value for the index of compression n and assuming that the intercooler is perfect ,that is the air is cooled in the inter cooler to original temperature, T1 and there is no pressure drop in the intercooler we have,

P1V1=P2V2 (isothermal condition)

P2’=P1

Wc = n/n-1 P1 V1 ((P2/P1)n-1/n + (P3/P2)n-1/n-2)

As P1 and P3 are fixed , the only variable pressure for minimum work done is P2.

dW/dP2 = 0 (for minimum work)

n/n-1 P1 V1 ((1/P1)n-1/n (n-1/n)P2-1/n- P3

n-1/n(n-1/n)P21/n-2) =0

(1/P1)n-1/n – P3n-1/n/P2

2(n-1/n) = 0

Or (1/P1)n-1/n = (P3/P22)n-1/n

i.e 1/P1 = P3/P22

P22 = P1P3

P2 = √P1P3

Wc = n/n-1 P1 V1 ((P3/P1)n-1/2n + (P3/P1)n-1/2n-2)

Wc = 2 n/n-1 P1 V1 ((P3/P1)n-1/2n-1)

For N number of stages , the pressure ratio will be

P2/P1 = P3/P2 = P4/P3 = P5/P4 = . . . . . . . = PN+1/PN

And minimum work done for all stages will be,

Wc = N n/n-1 P1 V1 ((P3/P1)n-1/Nn-1)

The following assumptions are used for the minimum work required to compress the air , in multi stage compressor .

1. The air is cooled to the initial temperature after each stage of compression.2. The pressure ratio in each stage is same .3. Work done in all stages is equal if the same index of compression is used.

Problems:

1. An air compressor takes in air at 1 bar and 20 degree Celsius and compresses it according to law PV1.2 = constant. It is then delivered to a receiver at a constant pressure of 10 bar . R = 0.287 kJ/kg.k.

Determine :

1. Temperature at the end of compression2. Work done and heat transferred during compression per kg of air

Solution:

In fig

T1 = 20 + 273 = 293, P1 = 1 bar , P2 = 10 bar

Law of compression : PV1.2 = constant , R = 0.287 kJ/kg.k.

1.temperature at the end of compression,

For compression process 1-2, we have

T2/T1 = (P2/P1)n-1/n =(P2/P1)1.2-1/1.2

= 1.468

T2 = T1 * 1.468 = 430k

2.Work done and heat transferred during compression per kg of air

Work done W = mRT1 n/n-1 ((P2/P1)n-1/n -1)

= 1*0.287 * 293 * (1.2/1.2-1)(101.2-1/1.2 -1)

= 236.13 kJ/kg of air

Heat transferred Q = W +ΔU

= P1V1 – P2V2/n-1 + Cv(T2-T1)

=mR(T1-T2)/n-1 + Cv(T2- T1)

= (T2-T1)(Cv-R/n-1)

= (430 – 293 )(0.78 – 0.287 / 1.2 -1)

Q = -98.23 kJ/kg

Negative sign indicates heat rejection

2.following data relate to a performance test of a single acting 14cm * 10cm reciprocating compressor

Suction pressure = 1 bar

Suction temperature = 20°C

Discharge pressure = 6 bar

Discharge temperature = 180°C

Speed of compressor = 1200 rpm

Shaft power = 6.25 kw

Mass of air delivered = 1.7 kg/min

Calculate the following,

The actual volumetric efficiency, the indicated power, the isothermal efficiency, the mechanical efficiency, the overall isothermal efficiency.

Solution:

Given: P1 = 1 bar

T1 = 20+273 = 293k

P2 = 6 bar

T2 = 180+273 = 453k.

N= = 1200 rpm

Pshaft = 6.25 kw

Ma = 1.7 kg/min

Displacement volume, Vd = ᴫ/4 *D2*L*N (for single acting compressor)

= ᴫ/4 * (14/100)2 * 10/100 *1200

= 1.8373 m3/min.

F.A.D = mRT1/P1 = 1.7 * 0.287 *1000 * 293/1*105

= 1.4295 m3/min

Ηvol = F.A.D/Vd * 100 = 1.4295/1.8473 * 100

= 77.38%

The indicated power (I.P):

T2/T1 = (P2/P1)n-1/n or n-1/n = ln(T2/T1)/ln(P2/P1)

1/n = 1- ln(453/293)/ln(6/1)

n= 1.32

hence , index of compression , n = 1.32

indicated power = n/n-1 mRT1 ((P2/P1)n-1/n - 1)

= 1.32-1/1.32 * 1.7/60 * 0.287 * 293 ((6/1)1.32-1/1.32 -1)

= 5.346 kw

The isothermal efficiency, ηiso:

Isothermal power = mRT1 *ln(P2/P1)

= 1.7/60 *0.287*293 ln(6/1)

= 4.269 kw

ηiso = 4.269/5.346 *100

= 79.85%

The mechanical efficiency, ηmech:

ηmech =indicated power / shaft power * 100

= 5.346/6.25 * 100

=85.5%

The overall isothermal efficiency, ηoverall:

ηoverall = isothermal power / shaft power * 100

= 4.269 / 6.25 * 100

= 68.3%

3. A single stage double acting compressor has free air delivery of 14 m3/min measured at 1.013 bar and 15°C. the pressure and temperature in the cylinder during induction are 0.95 bar and 32° C. the delivery pressure is 7 bar and index of compression and expansion , n= 1.3. the clearance volume is 5% of the swept volume. Calculate Indicated power required,Volumetric efficiency.

Given:

Free air delivery = 14 m3 / min (measured at 1.013 bar and 15° C)Induction pressure, P1 = 0.95 bar,Induction temperature, T1 = 32+273 = 305 k

Delivery pressure , P2 = 7 barIndex of compression and expansion, n= 1.3

Clearance volume Vc = 0.05Vs

Solution:

Indicated power:

Mass delivered per minute, m= PV/RT = 1.013*105*14/0.287*288*103

= 17.16 kg/min

T2/T1 = (P2/P1)n-1/n

T2 = T1(P2/P1)n-1/n = 305 * (7/0.95)1.3-1/1.3

=483.5 k

Indicated power = 3809.4/60 = 63.49 kw

Volumetric efficiency:

V4/V3 = (P3/P4)1/n = (P2/P1)1/1.3

= (7/0.95)1/1.3 = 4.65

V4 = 4.65*V3

=4.65*0.05*Vs

=0.233Vs

V1 – V4 = V1 – 0.233Vs = 1.05Vs – 0.233Vs = 0.817Vs

m = PV/RT = P1(V1 – V4)/RT1

F.A.D/cycle, v. = (V1- V4)TP1/T1P

(where P1 and T1 are the suction conditions)

V= 0.817Vs*288*0.95/305*1.013

= 0.723Vs

Volumetric efficiency, ηvol = V/Vs = 0.723 Vs/Vs

= 72.3%

4. Air at 103 kpa and 27° C is drawn into reciprocating , two stage L.P cylinder of a air compressor and is isentropically compressed to 700 kpa. The air is then cooled at constant pressure 37° C in an inter cooler and is then again compressed isentropically to 4 MPa in the H.P cylinder and is delivered at this pressure . determine the power required to run the compressor if it has to deliver 30 m3 of air per hour measured at inlet conditions.

Given:

Pressure of intake air (L.P. cylinder) , P1 = 103 KPa

Temperature of intake air, T1 = 27 + 293 = 300 k

Pressure of air entering H.P. cylinder , P2 = 700 KPa

Temperature of air entering H.P. cylinder, T2 = 37 + 273 = 310 k

Pressure of air after compression in H.P cylinder , P3 = 4 MPa

Volume of air delivered = 30 m3/h

Solution:

Power required to run the compressor:

Mass of air compressed , m = (103*105)*30/(0.287*1000*300)

35.89 kg/h

For the compression process 1- 2’ , we have

T2’/T1 = (P2/P1)ϒ-1/ϒ

= (700/103)0.4/1.4

=1.7289

T2’ = 300 * 1.7289

= 518.7 k

Similarly for the compression process 2-3 , we have

T3/T2=(P3/P2)ϒ-1/ϒ=(4000/700)1.4-1/1.4

=1.645

T3= 310* 1.645=510.1 K

Work reqiuredto run the compressor,

W=ϒ/ϒ-1*(m*R(T2’-T1)+m*R(T3-T2))

= ϒ/ϒ-1*(m*R((T2’-T1)+(T3-T5)))

=(1.4/1.4-1)*35.89/3600*0.287((518.7-300)+(510.7-310))

=4.194 K

24.8 C) A free air delivered by a single srage double acting reciprocating compressor,measured at 1 bar and 150 c of free air,is 16 m3/min.Apressure and temperature of air inside the cylinder duing suction are 0.96 bar and 30 0c respectively and delivery pressure is 6 bar.The compressor has a clearance of 4 % of the swept volume and the mean piston speed is limitd to 300 m/min.Determine

1) The power input to the compressor if mechanical eddiciency is 90% and the compression efficiency is 85%.

2) Stroke and bore if the compressor runs at 500 rpm.

Take index of compression and expansion as 1.4

Given:

F.A.D=16 m3/min(measured at 1 Bar and 150c)

P1=0.96 bar

T1=303 K

n= 1.3,V3=VC=0.04 VS

P2=6 Bar

1)Piston input to compressor

Mass flow rate of compressor

M=pv/RT=1*105*16/287*288=19.36 Kg/min

To fin ‘T2’ using the relation,

T2/T1=(P2/P1)n-1/n

T2=T1(P2/P1)1.3-1/1.3=462.4 K

Power input to the compressor=(n/n-1 m*R(T2-T1))*1/ηmech*ηcomp

=((1.3-1/1.3)* 19.36*.287*(462.4-303)*1/0.9*0.5

=5016.9 kJ/min=83.6 kj/s

2)stroke (L) and bore (D)

Piston speed =2LN

300 =2*L*500

L= 0.3 m or 300 mm

F..A.D=(3.14*D2L*2N*ηvd….for double acting air compressor------(1)

To find ηvol proceed as follows

V4/V3=(P3/P4)1/N=(P2/P1)1/3=(6/0.96)1/3=4.094

V4=4.094*V3=4.094*0.04VS=0.1637 VS

V1-V4= V1-0.1637 VS=1.04 VS-0.1637 VS=0.8763 VS

m= pv/RT =P1(V1-V4)/RT1

V=(V1-V4) T/T1*P1/P

Where P1 and T1 are suction conditions

V= 0.8763 Vs * 288/303* 0.96

= 0.799Vs

ηvol = V/Vs = 0.799Vs/Vs = 0.799

substituting the values in (1), we get

16 = ᴫ/4 * D2 *0.3 * 2* 500 *0.799

D = (16*4 / ᴫ * 0.2 * 500 * 0.799)1/2

= 0.29m or 290 mm

24.22) in a single acting two stage reciprocating compressor 4.5 kg of air per min are compressed from 1.013 bar

Solution:

Amount of air compressed m = 4.5 kg/min

Suction conditions, Ps = 1.013 bar, Ts = 15 + 273 = 288 k

Pressure ratio Pd/Ps = 9

Also Pi/Ps = Pd/Pi ……….

Compression expansion index , n = 1.3

Clearance volume in each stage = 5% swept volume

Speed of the compressor , N = 300 rpm

1) Indicted power :

Pi/Ps = Pd/Pi

Pi2 = Ps / Pd = Ps * 9Ps = 9*Ps

2

Pi = 3Ps i.e. Pi/Ps = 3

Now using the equation , Ti/Ts = (Pi/Ps)n-1/n = 31.3-1/1.3

Ti = Ts * 31.3-1/1.3 = 288*31.3-1/1.3

= 371 k

Now as n,m and temperature difference are the same for both stages the the work done in each stage is the same .

Total work required per min = 2* n/n-1*mR* (Ti – Ts)

= 2 *1.3/1.3-1 * 4.5 *0.287 (371 – 288)

= 929 kj/min

Indicated power = 929/60 = 1.8 kw

2) The cylinder swept volume requied:

The mass indicated per cycle m=4.5/300 0.015 kg/cycle

The mass is passed through each stage in turn

For the L.P. pressure cylinder

V1 -V4 = mRTs/Ps = 0.015 * 287 *288/ 1.013 * 105

= 0.0122 m3 / cycle.

ηvol = V1 – V4 / Vs = 1+k –k*(Pi/Ps)1/n

= 1+0.05 – 0.05 * 31/1.3

= 0.934

Vs = V1 – V4/ηvol = 0.0122/0.934 = 0.0131 m3/ cycle

Swept volume of L.P. cylinder Vs(L.P.)= 0.0131 m3

For the high pressure stage , a mass of 0.015 kg/ cycle is drawn in at 15 o c and a pressure of Pi = 3*1.013 = 3.039 bar

Volume drawn in = 0.015 * 287 * 288 / 3.039*105 = 0.00408 m3/cycle

ηvol = 1+k-k(Pd/Pi)1/n

and since Vc/Vs = k is the same as for the low pressure stage and also Pd/Pi = Pi/Ps then ηvol is 0.934 as above

swept volume of H.P. stage ,

Vs(H.P.) = 0.00408 / 0.934 = 0.004367 m3

it may be noted that the clearance ratio k = Vc/Vs is the same in each cylinder , and the suction temperatures are the same since intercooling is complete , therefore the swept volumes are in the ratio of the suction pressures,

Vs(H.P.) = VL.P/3 = 0.0131/3 = 0.000437 m3

BOILER:

Boilers are also known as steam generators. In boilers , the heat energy produced during the combustion of fuel is transferred to

water for generating steam

Classification:

1. Fire tube boiler2. Water tube boiler

Fire tube boiler:

The hot gases produced during fuel combustion passes through the tubes which are surrounded by water

Eg: Cochran, lancashire boilers.

Water tube boiler:

Water flows through the tube which are surrounded by hot flue gases.

Eg: bab cock and Wilcox, laMont boilers.

La Mont boiler:

Type : water tube boiler

Working:

The water is circulated by afeed pump. The water first passes through the economizer . economizer is used to heat the feed water before the gasses passes through the chimney. The feed water then reaches the drum . from the storage drum, the circulation pump takes the water to the evaporater.from the evaporator , the steam gets separated and then forced through superheater.

Boiler accessories:

1. Economizer:

Economizer transfers the excess heat in the flue gasses to the feed water. Economizer helps to improve the boiler’s efficiency.

2. Super heater: Super heater is used to increase the temperature of saturate steam without raising the pressure. It is placed in the path of flue gasses.

3. Safety valves: Safety valves are used to prevent the explosion of the boiler due to high internal pressure. The function of safety valve is to release the pressure when it exceeds the danger level.

4. Steam stop valve: Steam stop valve is used to

To control the flow rate of steam from boiler To shut off steam flow completely if required

5. Blow off cock:

The function of blow off cock is

To empty the boiler To remove mud and scale at the bottom of the boiler

6. Feed check valve: Feed check valve is used to regulate the supply of water pumped into the boiler

7. Fusible plug: Fusible plug is used to put off the fire in boiler when the water level in the boiler falls below unsafe limit.

Performance of boiler:

Boiler performance is measured by its evaporative capacity

Equivalent evaporation = total heat required to evaporate feed water at 1000c at normal atm/2257

E is always more than 1 for all boilers

E=mw(h-hw)/2257

mw=mass of feed water evaporated/kg of coal

h=steam enthalpy

hw=enthalpy of feed water

latent heat=2257kj/kg

(h-hw)/2257=evaporation factor

Boiler efficiency=heat used to produce steam/heat generated in the furnace

η=ms(h-hw)/mf*cv

ms = mass of the steam

mf = mass of the fuel

cv = calorific value of the fuel kj/kg

heat balance:

1) Heat loss in the dry flue gases = mg * cpg (Tg – Ta)Mg = mass of the dry flue gasCpg = specific heat capacity of the flue gas Tg = temperature of the flue gas leaving chimneyTa = temperature of the air

2) Heat lost to conversion of water present in the fuel to the steam = mw(hs –hw) = mw(hg + cp(Ts-T) –hw) =mw(2676+ cp(Ts-100) –hw)Ts = temperature of the superheated steam Hg = 2676kj/kg at 1.013 barT = temperature of boiler = 100oc

3) Heat lost to the water formed by the combustion of hydrogen /kg of the fuel into steam=9H2(2676+ cp(Ts-100) –hw)

4) Heat lost to the unburnt carbon = mc * cv

mc = mass of the carbon5) Heat lost to incomplete combustion of carbon to carbon monoxide due to inefficient air

supply = mco (combustion heat of co2 – combustion heat of co)mco = mass of the carbon monoxide

6) Heat lost to radiation = heat supplied – (heat to steam + all heat losses)

Problem:

In the boiler 300kg of coal with a calorific value of 30000kj/kg is used to evaporate 1800kg of water to steam at 12 bar. Dryness fraction of steam is 0.98. and the feed water temperature is 40oc. find the equivalent evaporation and boiler efficiency.

Given:

Mass of the coal mf = 300kg

c.v. of the coal = 30000kj/kg

pressure = 12bar

temperature = 40o c

dryness fraction x = 0.98

to find:

1. Equivalent evaporation2. Efficiency of the boiler

Solution:

From the steam table

At 12 bar , hf = 798.4 kj/kg

Hfg =1984.3 kj/kg

Hg = 2782.7 kj/kg

h=hf+x*hfg

=798.4 + 0.98*1984.3

h=2745.3kj/kg

(h=enthalpy of steam)

Hw = cp.Tw

=4.18 *40

=167.2 kj/kg

hw=enthalpy of water

mass of water evaporated per kg of coal

mw = ms/mf = 1800/300 = 6 kg/kg of coal

equivalent evaporation:

E = mw *(h- hw)/2257

= 6(2745.3 – 167.2)/2257

E = 6.85 kg/kg of coal

Efficiency of the boiler:

η= ms(h – hw)/mf * c

= 1800(2745.3 – 167.2)/300*30000

= 0.5156

=51.6%

Solution: 1) equivalent evaporation = 6.85 kg/kg of coal

2)boiler efficiency =s 51.6%