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The initial condition of air in an air compressor is 98 kPa and 27ºC, and it discharges at 500 kPa. The bore and stroke are 360 mm and 380 mm, respectively, with a percent clearance of 7%, running at 300 rpm.

Find the mass flow of air at suction.

A. 0.1886 kg/s
B. 0.11 kg/s
C. 0.33 kg/s
D. 0.44 kg/s

Answer :

The mass flow of air at suction is approximately 0.238 kg/s, so none of the options are correct.

Step 1: Calculate the Bore and Stroke Volumes

Bore (D) = 360 mm = 0.36 m

Stroke (L) = 380 mm = 0.38 m

Clearance Volume (Vc) = (Percent Clearance / 100) × Cylinder Volume

The cylinder volume (V) can be calculated using the formula:

V = π/4 × D² × L

V = π/4 × (0.36)² × 0.38

V = π/4 × 0.1296 × 0.38

V = π/4 × 0.049728 m³

V ≈ 0.03907 m³

Now calculate the clearance volume:

Vc = (7 / 100) × 0.03907

Vc = 0.002735 m³

Step 2: Calculate the Total Volume per Cycle

Total Volume (Vt) = V + Vc

Vt = 0.03907 + 0.002735

Vt ≈ 0.041805 m³

Step 3: Calculate the Theoretical Airflow Rate

RPM = 300

Cycles per second = RPM / 60

Cycles per second = 300 / 60

Cycles per second = 5

Airflow Rate (Q) = Vt × Cycles per second

Q = 0.041805 × 5

Q ≈ 0.209025 m³/s

Step 4: Calculate the Density of Air at Suction Conditions

Suction Pressure (P1) = 98 kPa

Suction Temperature (T1) = 27ºC

Suction Temperature (T1) = 27 + 273.15

Suction Temperature (T1) = 300.15 K

Use the Ideal Gas Law to find density:

ρ = P1 / (R × T1)

Where R for air = 287 J/(kg·K)

ρ = 98,000 / (287 × 300.15)

ρ ≈ 1.139 kg/m³

Step 5: Calculate the Mass Flow Rate

Mass Flow Rate (ṁ) = ρ × Q

ṁ = 1.139 × 0.209025

ṁ ≈ 0.238 kg/s

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