## Metric Problems

The unit of engine power in the SI system is the kiloWatt. Piston bore and stroke are measured in millimeters. Displacement is reported in liters. One liter is equal to 1,000 cm3 or 1,000,000 mm3. Pressure is measured in kiloPascals. The equation for engine power in kW is:

60,000 x 2

P = pressure (Pa), S = stroke (m), A = piston area (m2), N = speed (rpm).

Problem: Determine the indicated engine power for an engine in units of kW, that has a mean effective pressure of 1,050 kPa. The engine is a four-stroke cycle engine and has three cylinders. The bore (B) is 109 mm, the stroke (S) is 115 mm, and the speed is 3,000 rpm. First convert the units to base SI units

1,000 Pa

1 kPa

Next calculate the area of the piston:

A(m2) = n x — = n x --- = 0.0093313 ... or 0.00933 m2

Finally determine the power (KW):

60,000 x 2

0 rev

120,000

3380269.069

Problem: Calculate the compression ratio for the engine in the previous problem if the clearance volume is 0.063124 L.

Solution: Since we are calculating a ratio any consistent set of units can be used. Engine displacement is usually expressed in liters, therefore is this example the displacement is converted to liters because the clearance volume is in units of liters.

Start by calculating the piston displacement:

B2 1L

4.0 E - 3 = 1.0731... or 1.07 L Next calculate the compression ratio:

0.063124 L 1.133124

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