Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

- Se prod thermo_chapter_2_compressor by VJTI Production 3818 views
- Thermodynamics Problems Chapter 1 by VJTI Production 14156 views
- Se prod thermo_examples_mst by VJTI Production 4191 views
- Thermodynamics Lecture 1 by VJTI Production 25291 views
- Thermodynamics Chapter 3- Heat Tran... by VJTI Production 31144 views
- Se prod thermo_chapter_4_i.c.engines by VJTI Production 3969 views

No Downloads

Total views

4,077

On SlideShare

0

From Embeds

0

Number of Embeds

1

Shares

0

Downloads

334

Comments

0

Likes

3

No embeds

No notes for slide

- 1. <ul><li>Compressor </li></ul><ul><li>Examples </li></ul>ME0223 SEM-IV Applied Thermodynamics & Heat Engines Applied Thermodynamics & Heat Engines S.Y. B. Tech. ME0223 SEM - IV Production Engineering
- 2. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 1 A single-stage reciprocating compressor takes 1m 3 of air per minute at 1.013 bar and 15 0 C and delivers it at 7 bar. Assuming that the law of compression is PV 1.35 = constant, and the clearance is negligible, calculate the indicated power. Mass of air delivered per min. : Delivery Temperature : Indicated Work : … .ANS
- 3. Example 2 If the compressor in last example is driven at 300 rpm and is a single acting, single-cylinder machine, calculate the cylinder bore required, assuming a stoke to bore ration of 1.5:1.Calculate the power of the motor required to drive the compressor if the mechanical efficiency of the compressor is 85 % and that of the motor transmission is 90%. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Volume dealt with per min. at inlet = 1 m 3 /min Hence, Vol. drawn in per cycle = 1/300 = 0.00333 m3/cycle = Cylinder Volume Thus, cylinder bore, D = 0.1414 m or 141.4 mm … .ANS Power input to the compressor = … .ANS
- 4. ME0223 SEM-IV Applied Thermodynamics & Heat Engines <ul><li>An air compressor takes in air at 1 bar and 20 ºC and compresses it according to law PV 1.2 = C. It is then delivered to a receiver at a constant pressure of 10 bar. R = 0.287 kJ/kg.K. Determine : </li></ul><ul><li>Temperature at the end of compression. </li></ul><ul><li>Work done and heat transferred during compression per kg of air. </li></ul>Example 3 P 1 P 2 V 1 V 2 3 2 4 1 Delivery Temperature : Work done : … .ANS
- 5. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 3….contd P 1 P 2 V 1 V 2 3 2 4 1 Heat Transferred : … .ANS
- 6. Example 4 Following data relate to a performance test of a single-acting 14 cm X 10 cm reciprocating compressor: Suction Pressure =1 bar Suction temperature = 20 0 C Discharge Pressure = 6 bar Discharge temperature = 180 0 C Speed of compressor = 1200 rpm Shaft Power = 6.25 kW Mass of air delivered =1.7 kg/min Calculate the following: 1. The actual volumetric efficiency. 4. The indicated power. 2. The isothermal efficiency. 5. The mechanical efficiency. 3. The overall isothermal efficiency. ME0223 SEM-IV Applied Thermodynamics & Heat Engines P 1 P 2 V 1 V 2 3 2 4 1 … .for single-acting compressor
- 7. Example 4….contd ME0223 SEM-IV Applied Thermodynamics & Heat Engines P 1 P 2 V 1 V 2 3 2 4 1 … .ANS Indicated Power : … .ANS
- 8. Example 4….contd ME0223 SEM-IV Applied Thermodynamics & Heat Engines P 1 P 2 V 1 V 2 3 2 4 1 Isothermal Power : … .ANS … .ANS … .ANS
- 9. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 5 A single-stage double-acing air compressor is required to deliver 14 m 3 of air per min measured at 1.013 bar and 15 O C. the delivery pressure is 7 bar and the speed 300 rpm. Take the clearance volume as 5 % of swept volume with the compression and expansion index of n = 1.3. Calculate: 1. Swept volume of the cylinder 2. Indicated power 3. Delivery temperature. Swept Vol. V s = V 1 – V 3 = V 1 – V c = V 1 – 0.05 V s Vol. induced per cycle = ( V 1 – V 4 ) P 1 =1.013 bar P 2 =7 bar V 1 V 4 6 2 5 1 3 4 V 3 Swept Volume, V 1 -V 3 =V s V 3 =V c = 0.05 V s 288 K V 1 = 1.05 V s
- 10. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 5….contd V 4 = 4.423 V 3 = 4.423 X 1.05 V s = 0.221 V s (V 1 – V 4 ) = 1.05 V s – 0.221 V s = 0.0233 m 3 … .ANS Swept Volume, Delivery Temperature : … .ANS Indicated Power : … .ANS
- 11. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 6 A four-cylinder double-acting compressor is required to compress 30 m 3 / min of air at 1 bar and 27 0 C to a pressure of 16 bar. Determine the size of motor required and cylinder dimensions if the following data is given : Speed of compressor = 320 rpm. Clearance volume = 4 % Stroke : Bore ratio = 1.2 Mechanical Efficiency = 82 % Index = 1.32 Assume no pressure change in suction valves and air gets heated by 12 0 C during suction stroke. Net Work done : … .ANS Motor Power : … .ANS
- 12. Example 6….contd ME0223 SEM-IV Applied Thermodynamics & Heat Engines Vol. Efficiency : Swept Vol. per cylinder : … .ANS … .ANS
- 13. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 7 Air at 103 kPa and 27 0 C is drawn in L.P. cylinder of a two-stage air compressor and is isentropically compressed to 700 kPa. The air is then cooled at constant pressure at 37 0 C in an intercooler and is then again compressed isentropically to 4 MPa in the H.P. cylinder, and is deliverer at this pressure. Determine the power required to run the compressor if it has to deliver 30 m 3 of air per hour measured at inlet conditions. P 1 = 103 kPa P 3 = 4 MPa 3 2 1 P 2 = 700 kPa L.P. H.P. Volume Mass of air delivered per min. : Temperature for compression process 1-2’ : Temperature for compression process 2-3 :
- 14. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 7….contd Work required to run the compressor : … .ANS
- 15. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 8 A trial on a two-stage single-acting reciprocating compressor gave the following data : Free air delivered =6 m 3 /min Delivery Pr. = 40 bar Atm. Pr. and Temp. = 1 bar 27 ºC Speed = 400 rpm Intermediate Pr. = 6 bar Temp. at inlet to 2 nd stage = 27 ºC Law of Compression = PV 1.3 = C Mechanical Efficiency = 80 % Stroke of L.P. = Stroke of H.P. = Diameter of L.P. P 1 = 1 bar P 3 = 40 bar 3 2 1 P 2 = 6 bar L.P. H.P. Volume Assuming Vol. Efficiency of 100 % , … .ANS Swept Vol. of H.P. cylinder = Vol. of air at 6 bar, 27 ºC … .ANS
- 16. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 8….contd Indicated Work : … .ANS
- 17. ME0223 SEM-IV Applied Thermodynamics & Heat Engines <ul><li>A two-stage single-acting reciprocating air compressor takes in air at the rate of 0.2 m 3 /sec. The intake pressure and temperature are 0.1 MPa and 16 ºC. The air is compressed to a final pressure of 0.7 MPa. The intermediate pressure is ideal and intercooling is perfect. The compression index in both stages is 1.25 and the compressor runs at 600 rpm. Neglecting clearance, determine : </li></ul><ul><li>Intermediate Pr. 2. Total Vol. of each cylinder. </li></ul><ul><li>3. Power required to drive compressor 4. Rate of heat rejection in intercooler </li></ul>Example 9 Intermediate Pr., P 2 : … .ANS Vol. of cylinder 1, V s1 : … .ANS Vol. of cylinder 2, V s2 : … .ANS
- 18. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 9….contd Power Required : … .ANS Mass of air handled : Delivery Temperature : Heat rejected in Intercooler : … .ANS
- 19. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 10 A two-stage air compressor with complete intercooling delivers air to the mains at a pressure of 30 bar, the suction conditions being 1 bar and 15 0 C. if both cylinders have same stroke, find the ratio of cylinder diameter for the efficiency of compression to be maximum. Assume index of compression to be 1.3. P 1 P 3 3 2 1 P 2 L.P. H.P. Volume 2’ 5 6 4 For max. efficiency :
- 20. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 10 Delivery Temperature : For Const. Pr. Process 2 – 2’ : Thus, we get : Hence : … .ANS

No public clipboards found for this slide

Be the first to comment