A single-stage, three-cylinder piston air compressor with a diameter of 180 mm and a stroke length of 2500 mm per cylinder runs at rotational of 240 rpm. The intake pressure is 101,325 kPa and the is 20 deg * C Air is delivered into pasand the index of compression (n) = 1, 2 The clearance volumeris 5% he stroke volume. Take the characteristic gas constant as 0,289 kJ/kg.K.

Calculate each of the following:

4.1.1 Stroke volume in m³

4.1.2

Mass of air in kg/min

4.1.3

Volumetric efficiency

1 answer

Let's go through each of the calculations step by step.

4.1.1 Stroke Volume in m³

The stroke volume (V_s) for a single cylinder is calculated using the formula:

\[ V_s = \pi \left(\frac{d}{2}\right)^2 \cdot L \]

Where:

  • \( d \) = diameter of the cylinder = 180 mm = 0.18 m (convert mm to m)
  • \( L \) = stroke length = 2500 mm = 2.5 m (convert mm to m)

First, calculate the piston area:

\[ \text{Area} = \pi \left(\frac{0.18}{2}\right)^2 = \pi \times (0.09)^2 = \pi \times 0.0081 \approx 0.025446 \text{ m}^2 \]

Now calculate the stroke volume for one cylinder:

\[ V_s = \text{Area} \times L = 0.025446 , \text{m}^2 \times 2.5 , \text{m} \approx 0.063615 , \text{m}^3 \]

Since there are three cylinders, the total stroke volume (V_t) is:

\[ V_t = 3 \times V_s = 3 \times 0.063615 \approx 0.190845 , \text{m}^3 \]

4.1.2 Mass of Air in kg/min

First, we need to calculate the mass flow rate of air using the ideal gas equation. However, we first need to find the effective volume flow rate and then the mass flow rate.

The stroke volume per minute gives the volume of air delivered:

  • Number of strokes per minute (for one cylinder) = 240
  • Total strokes per minute (for three cylinders) = \( 240 \times 3 = 720 \text{ strokes/min} \)

Now, calculate the volume flow rate:

\[ \text{Volume flow rate} = V_t \times \text{strokes/min} = 0.063615 , \text{m}^3 \times 240 = 15.274 , \text{m}^3/\text{min} \]

Now we need to calculate the density of air (\(\rho\)) at the given conditions. Use the ideal gas law:

\[ \rho = \frac{P}{R \cdot T} \]

Where:

  • \( P = 101325 , \text{Pa} \)
  • \( R = 0.289 , \text{kJ/kg.K} = 289 , \text{J/kg.K} \)
  • \( T = 20 , \text{°C} = 293.15 , \text{K} \) (convert °C to Kelvin)

Calculate the density:

\[ \rho = \frac{101325}{289 \cdot 293.15} \approx \frac{101325}{84709.35} \approx 1.195 , \text{kg/m}^3 \]

Now we can calculate the mass flow rate:

\[ \text{Mass flow rate} = \text{Volume flow rate} \times \rho = 15.274 \times 1.195 \approx 18.271 , \text{kg/min} \]

4.1.3 Volumetric Efficiency

Volumetric efficiency (\(\eta_v\)) is given by:

\[ \eta_v = \frac{\text{Actual volume flow rate}}{\text{Theoretical volume flow rate}} \times 100% \]

The theoretical volume flow rate (\(Q_t\)) can be calculated based on the volume of the air in the cylinder:

Using the clearance volume \(V_c\)

\[ V_c = 0.05 \times V_s = 0.05 \times 0.063615 \approx 0.00318075 , \text{m}^3 \]

The total volume for one cycle (considering clearance volume) for one cylinder would be:

\[ V_t + V_c = V_s + V_c = 0.063615 + 0.00318075 \approx 0.06679575 , \text{m}^3 \]

Considering 3 cylinders:

\[ Q_t = 3 \times (V_s + V_c) \times \text{strokes/min} = 3 \times 0.06679575 \times 240 = 48.084192 , \text{m}^3/\text{min} \]

Now calculate volumetric efficiency

\[ \eta_v = \frac{15.274}{48.084192} \times 100% \approx 31.8 % \]

Final Answers

  • 4.1.1 Stroke Volume (m³): \(0.190845 , \text{m}^3\)
  • 4.1.2 Mass of Air (kg/min): \(18.271 , \text{kg/min}\)
  • 4.1.3 Volumetric Efficiency: \(31.8%\)