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
*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 supercharges to active 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
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 supercharges to active 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( /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
Pv =c
Wd=p(v2-v1)
isobaric
v/t=c
Wd=p1v1 ln(p2/p1)
isothermal
Pv=c
Wd=(p2v2-p1v1)/γ )/γ-1
Rev adiabatic/isentropic
pv =c
Wd=v(Δ =v(Δp)
isochoric
P/T=C
n
γ
Work supplied to compressor without
4-1 – Suction
:
constant pressure
1-2 – compression polytropic 2-3 – discharge
constant pressure
/n-1) + ( ) ( ) = ( /n-1) + = (1/n-1 + 1) ( ) = n/n-1 ( ) = n/n-1* ( / - 1) =› () () W=(
= n/n-1*( )
() = n/n-1*mR( )
) () () = n/n-1*mR (
Also,
√ √ , ( ) ( ) √ ( ) or √ ( )
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:
()
(since PV = C)
1-3 – polytropic
() *n/n-1 ( -1)
1-4 – adiabatic
( ) *ϒ/ϒ-1 ( -1) ()
= = / n/n-1 (
=
= IP/SP,
()
= n/n-1 (
SP-IP = Fr.P.
-1)
-1)/ ϒ/ϒ-1 (
( )
-1)
Effect of clearance of work of compression:
()
*n/n-1 (
-1) -
() ( )*n/n-1 ( -1) ()
*n/n-1 ( Also,
Also,
*n/n-1 (
-1)
is negligible
if
-1)
()
*n/n-1 (
()
* / -1 (
()
-1)
-1) -
()
* / -1 (
-1)
Also, ‘C’ = 3% to 12% of Clearance ratio ‘C’ =
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: n-1/n
Wc=n/n-1*p1v1((p2/p1)
n-1/n
-1) +n/n-1*p2p3((p3/p2)
P1V1=P2V2= Perfect intercooling to T1 n-1/n
Wc=n/n-1*p1v1((p2/p1)
n-1/n
+ (p3/p2)
-2
dwc/dP2=0 for minimum work
n-1/n
(1/p1)
n-1/n
- p3
2(n-1/n)
/p2
=0
0.5
P2=(p1p3)
n-1/2n
Wc=n/n-1*p1v1((p3/p1)
n-1/2n
Wc= (2n/n-1 v1 (p3/p1)
n-1/2n
+(p3/p1)
-1)p1
For N-stages, Similarly n-1/Nn
Wc= (Nn/n-1) *p1v1 (pN+1/p1)
-1
n-1/n
Wc= (Nn/n-1)maRT1((pL/pS)
-1)
-2
-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
1/n
=(VS+CVS)-CVS(P2/P1)
/Vs
1/n
ηvo=1+c- CVS(P2/P1)
=1-C(V1/V2-1)
N-1
T2/T1 =( V1/V2)
n-1/n
= (p2/p1)
PROBLEMS: 3
0
1) Air is to be isentropically compressed at the rate of 1 m /s from 1 bar and 20 c to 10 bar, Find the work of compressor and the volumetric effi ciency. If the clearance volume is 4% of stroke volume for all the cylinders for the following causes 1) single stage 2) two stage 3) Three stage compression Given: Free air delivered = 1.2
/s
Assume ρ=1.2 kg/
) -1)
1. Work done =ϒ/ϒ-1 mR ((
-1)
= 1.4/0.4 * 1.2 * 0.287 * 293 *( = 328.71 kw
= 1- (() -1)
-1)
= 1 – 0.04 ( = 83.28%
2. Intermediate pressure
= ϒ/ϒ-1
W=2*ϒ/ϒ-1*maRT1((P2/P1)
-1)
-1)
=2*1.4/0.4*1.2*0.287 * 293 *( =274.92 kw
=1-0.04(3.16
1/1.4
=94.9%
-1)
= √
P2=
=2.15 Bar .4/1.4
WC=3*1,4/0.4*1.2*.287*293*(2.15) =259 Kw
=1-0.04*(2.15
1/1.4
=97.09%
-1)
-1)
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 Pv compression and reversible adiabatic.Alao find the isothermal efficiency
2
Volume of air compressed per min= d /4*1*N 2
=3.14/4*0.30 *0.4*400 3
=11.31 m /min Given: d-=0.3 m l=0.4 m N=400rpm P1=1 bar T1=300k P2= 5bar 5
Isothermal compression work=P1V1ln(P2/P1)=1*10 *11.31/60*ln(5/1) = 30337.0 W. Polytropic compression work=
*n/n-1 ( 5
()
-1) .3/1.3
=1*10 *11.31/60*(1.3/1.3-1)((5/1) =36.74 kw
-1)
1.3
=constant nd
Isothermal efficiency=isotherml work/actual work)*100=30.34/36.74)*100 =82.58% Isentropic work of compression=
* ϒ/ϒ-1*((P /P ) 2
5
1
ϒ/ϒ-1
-1) .4/1.4
=1*10 *11.31/60*(1.4/1.4-1)((5/1) =38.52 kw
=36.74/38.52)*100=78.76% =actual work/adiabatic work done
-1)
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% n
PV = C, n = 1.35
Solution: n-1/n
I.P = n/n-1 maR(T1((P2/P1)
n-1/n
-1)+ T2((P3/P2)
-1)
Ma = P1V1/RT 5
2
= 1*10 *3.14*0.6 *0.8*0.8*100*2/4*287*293*60 Ma = 0.7174 kg/s 0.35/1.35
I.P = 1.35/0.35 * 0.7174* 287 * (293(7 I.P = 249.9 kw B.P = I.P/ ηmech =249.9/0.85 =294 kw
0.35/1.35
-1)+313(25/7)
-1))
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 a lso 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 1
N
pressure p1 to p2 following the polytropic flow Pv =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 ϒ
following the law Pv =c. Finally the air I discharged at p3 to the receiver. Total compression work done on air is, n-1/n
Wc = n/n-1 * P1 V1 ((P2/P1)
n-1/n
-1)+n/n-1 * P2 V2 ((P3/P2)
-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
n-1/n
Wc
n-1/n
= n/n-1 P1 V1 ((P2/P1)
+ (P3/P2)
-2)
As P1 and P3 are fixed , the only variable pressure for minimum work done is P2. dW/dP2 = 0 (for minimum work) n-1/n
n/n-1 P1 V1 ((1/P1) n-1/n
(1/P1)
n-1/n
– P3
n-1/n
Or (1/P1)
-1/n
(n-1/n)P2
2(n-1/n)
/P2
n-1/n
- P3
1/n-2
(n-1/n)P2
=0
2 n-1/n
= (P3/P2 ) 2
i.e 1/P1 = P3/P2 2
P2 = P1P3 P2 =
n-1/2n
Wc
= n/n-1 P1 V1 ((P3/P1)
Wc
= 2 n/n-1 P1 V1 ((P3/P1)
n-1/2n
+ (P3/P1)
n-1/2n
-1)
-2)
) =0
For N number of stages , the pressure ratio will be P2/P1 = P3/P2 = P4/P3 = P5/P4 = . . . . . . . = P N+1/PN And minimum work done for all stages will be, Wc
n-1/Nn
= N n/n-1 P1 V1 ((P3/P1)
-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 according to law PV
1.2
= 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 compression 2. Work done and heat transferred during compression per kg of air
Solution: In fig T1 = 20 + 273 = 293, P1 = 1 bar , P2 = 10 bar 1.2
Law of compression : PV
= constant , R = 0.287 kJ/kg.k.
1.temperature at the end of compression, For compression process 1-2, we have n-1/n
T2/T1 = (P2/P1)
and compresses it
1.2-1/1.2
=(P2/P1)
= 1.468 T2 = T1 * 1.468 = 430k 2.Work done and heat transferred during compression per kg of air n-1/n
Work done W = mRT1 n/n-1 ((P2/P1)
-1) 1.2-1/1.2
= 1*0.287 * 293 * (1.2/1.2-1)(10
-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 2
Displacement volume, Vd = ᴫ/4 *D *L*N
(for single acting compressor)
2
= ᴫ/4 * (14/100) * 10/100 *1200 3
= 1.8373 m /min. 5
F.A.D = mRT1/P1 = 1.7 * 0.287 *1000 * 293/1*10 3
= 1.4295 m /min Ηvol = F.A.D/Vd * 100 = 1.4295/1.8473 * 100 = 77.38% The indicated power (I.P): n-1/n
T2/T1 = (P2/P1)
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 n-1/n
indicated power = n/n-1 mRT1 ((P2/P1)
- 1) 1.32-1/1.32
= 1.32-1/1.32 * 1.7/60 * 0.287 * 293 ((6/1) = 5.346 kw
-1)
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
3. A single stage double acting compressor has free air delivery of 14 m /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:
3
Free air delivery = 14 m / 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 bar Index of compression and expansion, n= 1.3
Clearance volume Vc = 0.05Vs Solution: Indicated power:
5
3
Mass delivered per minute, m= PV/RT = 1.013*10 *14/0.287*288*10 = 17.16 kg/min n-1/n
T2/T1 = (P2/P1)
n-1/n
T2 = T1(P2/P1)
1.3-1/1.3
= 305 * (7/0.95)
=483.5 k Indicated power = 3809.4/60 = 63.49 kw Volumetric efficiency: 1/n
V4/V3 = (P3/P4)
1/1.3
= (7/0.95) V4 = 4.65*V3
=4.65*0.05*Vs =0.233Vs
1/1.3
= (P2/P1) = 4.65
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 3
required to run the compressor if it has to deliver 30 m of air per hour measured at inlet conditions. Given: Pressure of intake air (L.P. cylinder) , P 1 = 103 KPa Temperature of intake air, T1 = 27 + 293 = 300 k Pressure of air entering H.P. cylinder , P 2 = 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 3
Volume of air delivered = 30 m /h
Solution: Power required to run the compressor: 5
Mass of air compressed , m = (103*10 )*30/(0.287*1000*300) 35.89 kg/h For the compression process 1- 2’ , we have ’
ϒ-1/ϒ
T2 /T1 = (P2/P1)
0.4/1.4
= (700/103) =1.7289
T2’ = 300 * 1.7289 = 518.7 k Similarly for the compression process 2-3 , we have ϒ-1/ϒ
T3/T2=(P3/P2)
1.4-1/1.4
=(4000/700)
=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 0
3
at 1 bar and 15 c of free air,is 16 m /min.Apressure and temperature of air inside the cylinder 0
duing suction are 0.96 bar and 30 c 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: 3
0
F.A.D=16 m /min(measured at 1 Bar and 15 c) 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 5
M=pv/RT=1*10 *16/287*288=19.36 Kg/min
To fin ‘T2’ using the relation, n-1/n
T2/T1=(P2/P1)
1.3-1/1.3
T2=T1(P2/P1)
=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 .
2
F .A.D=(3.14*D L*2N*ηvd….for double acting air compressor------(1) To find ηvol proceed as follows 1/N
V4/V3=(P3/P4)
1/3
=(P2/P1)
1/3
=(6/0.96)
=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 2
16 = ᴫ/4 * D *0.3 * 2* 500 *0.799 1/2
D = (16*4 / ᴫ * 0.2 * 500 * 0.799) = 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 2
2
Pi = Ps / Pd = Ps * 9Ps = 9*Ps Pi = 3Ps
i.e. Pi/Ps = 3 n-1/n
Now using the equation , T i/Ts = (Pi/Ps) 1.3-1/1.3
T i = Ts * 3
1.3-1/1.3
=3
1.3-1/1.3
= 288*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 5
V1 -V4 = mRTs/Ps = 0.015 * 287 *288/ 1.013 * 10 3
= 0.0122 m / cycle.
1/n
ηvol = V1 – V4 / Vs = 1+k –k*(Pi/Ps) = 1+0.05 – 0.05 * 3
1/1.3
= 0.934 3
Vs = V1 – V4/ηvol = 0.0122/0.934 = 0.0131 m / cycle 3
Swept volume of L.P. cylinder V s(L.P.)= 0.0131 m
o
For the high pressure stage , a mass of 0.015 kg/ cycle is drawn in at 15 c and a pressure of Pi = 3*1.013 = 3.039 bar 5
3
Volume drawn in = 0.015 * 287 * 288 / 3.039*10 = 0.00408 m /cycle 1/n
ηvol = 1+k-k(Pd/Pi)
and since Vc/Vs = k is the same as for the low pressure stage and also P d/Pi = Pi/Ps then ηvol is 0.934 as above swept volume of H.P. stage , 3
Vs(H.P.) = 0.00408 / 0.934 = 0.004367 m
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, 3
Vs(H.P.) = VL.P/3 = 0.0131/3 = 0.000437 m
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 boiler 2. 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
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Equivalent evaporation = total heat required to evaporate feed water at 100 c 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 gas Cpg = specific heat capacity of the flue gas Tg = temperature of the flue gas leaving chimney Ta = 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 bar o
T = temperature of boiler = 100 c
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 = m c * cv
mc = mass of the carbon 5) 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 o
40 c. find the equivalent evaporation and boiler efficiency.
Given: Mass of the coal mf = 300kg
c.v. of the coal = 30000kj/kg pressure = 12bar o
temperature = 40 c dryness fraction x = 0.98 to find: 1. Equivalent evaporation 2. 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
