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AIR AND GAS COMPRESSOR

AIR AND GAS COMPRESSOR

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- 1. Compressor is a machine used to compressed air or gas to a final pressure exceeding 241.25 KPa gage. TYPES OF COMPRESSORS Centrifugal Compressors - For low pressure and high capacity application Rotary Compressors - For medium pressure and low capacity application Reciprocating Compressors - For high pressure and low capacity application USES OF COMPRESSED AIR Operation of small engines Pneumatic tools Air hoists Industrial cleaning by air blast Tire inflation Paint Spraying Air lifting of liquids Manufacture of plastics and other industrial products To supply air in mine tunnels Other specialized industrial applications ANALYSIS OF CENTRIFUGAL AND ROTARY TYPE W 2 1 Q Assuming: KE = 0; and PE = 0 Q = h + KE + PE + W (Steady state - steady flow equation) Q = h + W For a compressor, work is done on the system; thus -W = h - Q; Let -W = W (compressor work) A) Isentropic Compression (PVk = C) P 2 W = -VdP PVk = c dP V 1 V For Isentopic compression' Q = 0 W = h ; W = -VdP W = mCp(T2-T1) k 1 kmRT P k 1 2 W 1 KW k - 1 P 1 Q=0 PV1 = mRT1 P,V, and T relationship k 1 k P 2 2 T P 1 1 T k 1 V 1 V 2
- 2. where m - mass flow rate of the gas in kg/sec W - work in KW P - pressure in KPa T - temperature in K R - gas constant in KJ/kg-K V - volume flow rate in m3/sec B) Polytropic Compression (PVn = C) P 2 W = -VdP PVn =C dP V 1 V For Polytropic compression, Q 0 W = h - Q ; W = -VdP h = mCp(T2 - T1) Q = mCn(T2 - T1) n1 nmRT P n 1 2 W 1 KW n - 1 P 1 k n C Cv n 1 n P 2 2 T P 1 1 T n1 n V 1 V 2 n1 C) Isothermal Compression (PV = C) P 2 W = -VdP PV = C dP V 1 V W = -Q P W P V ln 2 1 1 P 1 P1V1 = mRT1
- 3. valves cylinder piston piston-rod d HE P P2 P1 CE L 2 3 4 CVD D V1’ 1 V VD HE - Head end CE - Crank end L - length of stroke, m VD - Displacement volume, m3/sec D - diameter of bore, m d - diameter of piston rod, m A) For Isentropic Compression and RE-expansion process (PVK = C),no heat is removed from the gas. W = h ; W = -VdP W = mCp(T2-T1) k 1 kmRT1 P2 k W 1 KW k -1 P1 Q=0 PV1' = mRT1 P,V, and T relationship k 1 k 1 P k V1 2 V T1 P1 2 where: V1' - volume flow rate measured at intake, m3/sec m - mass flow rate corresponding V1', kg/sec B) For Polytropic Compression and Re-expansion process (PVn = C), some amount of heat is removed from the gas. W = h - Q ; W = -VdP h = mCp(T2 - T1) Q = mCn(T2 - T1) T2
- 4. n 1 nmRT1 P2 n W 1 KW n -1 P1 P1V1' = mRT1 k n Cn Cv KJ/kg-K 1n n 1 n 1 P n V1 2 V T1 P1 2 C) For Isothermal compression and re-expansion process (PV = C), an amount of heat equivalent to the compression work is removed from the gas. W = -Q P W P1 V1'ln 2 P1 P1V1' = mRT1 T2 PERCENT CLEARANCE Clearance Volume C Displacement Volume V C 3 x 100% VD For compressor design, values of percent clearance C ranges from 3 to 10 %. V3 = CVD where: V3 - clearance volume VOLUMETRIC EFFICIENCY v = Volume flow rate at intake x 100% Displacement Volume V ηv 1' x 100% VD A) For Isentropic Compression (PVk= C) 1/k P2 ηv 1 C C x 100% P 1 B) For Polytropic Compression (PVn = C) 1/n P2 ηv 1 C C x 100% P 1 C) For Isothermal Compression (PV = C) P ηv 1 C C 2 x 100% P 1
- 5. DISPLACEMENT VOLUME A) For single acting VD = LD2Nn' m3/sec 4(60) B) For Double acting without considering the volume of piston rod VD = 2LD2Nn' m3/sec 4(60) C) For Double acting considering volume of piston rod VD = LNn'[2D2 - d2] m3/sec 4(60) where: L - length of stroke, m D - diameter of bore, m d - diameter of piston rod, m n' - no. of cylinders ACTUAL VOLUMETRIC EFFICIENCY V ηva a x 100 % VD Va - actual volume of air or gas drawn in MEAN EFFECTIVE PRESSURE W Pm KPa VD W in KJ, KJ/kg, KW VD in m3, m3/kg, m3/sec PISTON SPEED PS = 2LN m/min PS = 2LN m/sec 60 EFFICIENCY A) COMPRESSION EFFICIENCY cn = Ideal Work x 100% Indicated Work B) MECHANICAL EFFICIENCY m = Indicated Work x 100% Brake Work C) COMPRESSOR EFFICIENCY c = cn = m = Ideal Work x 100% Brake work MULTISTAGE COMPRESSION Multi staging is simply the compression of air or gas in two or more cylinders in place of a single cylinder compressor. It is used in reciprocating compressors when pressure of 300 KPa and above are desired, in order to: Save power Limit the gas discharge temperature Limit the pressure differential per cylinder Prevent vaporization of lubricating oil and to prevent its ignition if the temperature becomes too high. It is common practice for multi-staging to cool the air or gas between stages of compression in an intercooler, and it is this cooling that affects considerable saving in power.
- 6. A) 2 - Stage Compression without pressure drop in the intercooler 1 suction 2 Qx 3 4 discharge Intercooler 1stStage 2nd Stage For an ideal multistage compression, with perfect inter-cooling and minimum work, the cylinder were properly designed so that: the work at each stage are equal the air in the intercooler is cooled back to the initial temperature no pressure drop occurs in the intercooler the pressure at each stage are equal W1 = W2 ; T1 = T3 ; P2 = P3 = Px where: W1 - work of the LP cylinder (1st stage) W2 - work of the HP cylinder (2nd stage) Px = ideal intercooler pressure, optimum pressure Assuming polytropic compression and expansion processes: P P4 Px P1 5 4 6 7 3 2 8 W1 = W2 T1 = T3 P2 = P3 = P x W = W1 + W2 Px - the ideal intercooler pressure or optimum pressure Work 1st Stage: n 1 nmRT P n 1 2 W 1 KW 1 n - 1 P 1 Work 2nd Stage: n 1 nmRT P n 3 4 W 1 KW 2 n - 1 P 3 PVn = C 1
- 7. Pressure Ratio: P2 P4 P1 P3 but P2 = P3 = Px Px P4 P1 Px then Px P1P4 Since W1 =W2, the total work W is; n1 2nmRT1 P2 n W 1 P n -1 1 KW substituting Px to W, it follows that n1 2nmRT1 P4 2n W 1 KW n -1 P1 By performing an energy balance on the inter-cooler Qx = mCP(T3 - T2) T-S Diagram: T P4 T2 = T4 T1 = T3 Px P1 Qx 4 2 3 1 S B) 2 stage compressor with pressure drop in the intercooler For 2 stage compression with pressure drop in the intercooler, P2 P3.The air in the intercooler may or may not be cooled to the initial temperature, and the work at each stage may or may not be equal, thus the work W = W1 + W2 Work 1st Stage: n 1 nmRT1 P2 n W1 1 KW n -1 P1 Work 2nd Stage: n 1 nmRT3 P4 n W2 1 KW n -1 P3
- 8. The total work W is; W = W1 + W2 The pressure, P2 P3, but the 1st stage may compress the air or gas to the optimum intercooler pressure P x, but a pressure drop will occur in the inter-cooler. P 5 4 P4 P2 P3 7 2 6 3 P1 8 1 V Heat Rejected in the inter-cooler Qx = mCp(T3 - T2) C. Three-Stage compressor without pressure drop in the intercooler Qx Qy suction 1 2 3 4 LP Intercooler 1st Stage 5 6 discharge HP Inercooler 2nd stage 3rd stage Considering Polytropic compression and expansion processes and with perfect inter-cooling; Work of 1st stage cylinder: n 1 nmRT1 P2 n W1 1 KW n -1 P1 Work of the 2nd stage cylinder: n 1 nmRT3 P4 n W2 1 KW n -1 P3 Work of the 3rd stage cylinder: n 1 nmRT5 P6 n W3 1 KW n - 1 P5 For perfect inter-cooling: W1 = W2 = W3 T1 = T3 = T5
- 9. and P2 P1 But Thus P4 P3 P6 P5 P2 = P3 = Px (Ideal LP Intercooler pressure) P4 = P5 = Py (Ideal HP Intercooler pressure) Px P1 Py Px P6 Py By expressing Px and Py in terms of P1 & P6: 2 Px 3 P1 P6 Py 3 P1P6 2 The total compressor work is equal to: W = W1 + W2 + W3 but: W1 = W2 = W3 ;therefore W = 3W1 n 1 3nmRT1 P2 n W 1 KW n -1 P1 then substituting Px andPy then simplify, the result is: n 1 3nmRT1 P6 3n W 1 KW n -1 P1 For multistage compression with minimum work and perfect inter-cooling and no pressure drop in the inter-coolers between stages, the following conditions apply: 1. the work at each stage are equal 2. the pressure ratio between stages are equal 3. the air temperature in the inter-coolers are cooled to the original temperature T1 4. the total work W is equal to
- 10. n 1 SnmRT1 P2S Sn W 1 KW n -1 P1 where S - number of stages Example An ideal 3-stage air compressor with intercoolers handles air at the rate of 2 kg/min. The suction pressure is 101 Kpa, suction temperature is 21C, delivery pressure is 5000 KPa. Assuming perfect inter-cooling and minimum work, calculate total power required if compressor efficiency is 60% and both compression and expansion processes are PVn=C, where n = 1.2. (21 KW) Given m 2 kg / min P1 101 KPa T1 21 273 294 K P6 5000KPa ec 0.60 PV n C n 1 .2 3nmRT1 W n -1 P n 13n 6 1 P1 KW 1. 2 1 3(1.2)(2)(0.287)(294) 5000 3(1.2) W 1 60(1.2 - 1) 101 W 20.41 KW

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