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The initial condition of air in an air compressor is 98 kPa and 27ºC, and it discharges air at 400 kPa. The bore and stroke are 355 mm and 381 mm, respectively, with a clearance of 5%, running at 300 rpm. Find the volume of air at suction.

Answer :

The volume of air at suction is approximately 0.1034 m³ per cycle or 31.02 m³/min.

  • Given information:

Initial pressure (P1) = 98 kPa

Initial temperature (T1) = 27°C = 300 K (converting to Kelvin)

Discharge pressure (P2) = 400 kPa

Bore (diameter) = 355 mm = 0.355 m

Stroke = 381 mm = 0.381 m

Clearance = 5%

Speed = 300 rpm

  • First, let's calculate the swept volume (Vs):

Vs = (π/4) × D² × L

where D is the bore diameter and L is the stroke length

Vs = (π/4) × (0.355)² × 0.381

Vs = 0.0377 m³

  • Now, let's calculate the clearance volume (Vc):

Vc = 5% of Vs

= 0.05 × 0.0377

= 0.00188 m³

  • The total cylinder volume (V) is:

V = Vs + Vc

= 0.0377 + 0.00188

= 0.03958 m³

  • To find the volume of air at suction, we need to use the air compression equation:

P1V1ⁿ = P2V2ⁿ

where n is the polytropic index (typically between 1.2 and 1.35 for air compressors)

Let's assume n = 1.3 (a common value for air compressors)

(P1/P2) = (V2/V1)ⁿ

(98/400) = [tex](0.03958/V1)^{1.3[/tex]

  • Solving for V1:

V1 = [tex]0.03958 \times (400/98)^{(1/1.3)[/tex]

V1 = 0.03958 × 2.6146

V1 = 0.1034 m³

  • This is the volume per cycle. To get the volume per minute:

Volume per minute = V1 × 300 rpm

= 0.1034 × 300

= 31.02 m³/min

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