7.1 Define the following terms with regard to an internal combustion engine: 7.1.1 Work done 7.1.2 Clearance volume 7.2 FIGURE 7.2 below shows a diagram of a four-stroke, spark-ignition (S.I.) engine - NSC Mechanical Technology Automotive - Question 7 - 2022 - Paper 1

Clearance volume is the space above the piston when it is at Top Dead Center (TDC) in an internal combustion engine. It is critical in determining the engine's compression ratio, allowing for proper air-fuel mixture intake and combustion efficiency.

The diagram in FIGURE 7.2 represents an indicator diagram or a pressure-volume diagram for a four-stroke spark-ignition (S.I.) engine.

The unit of mean effective pressure is kPa (kilopascals) or can also be expressed as N/m².

To calculate the swept volume (SV), use the formula:

$SV = \frac{\pi D^2}{4} \times L$

where:

• $D$ = Diameter of the bore in cm = 7 cm,
• $L$ = Stroke length in cm = 6.5 cm.

Substituting the values:

$SV = \frac{\pi (7)^2}{4} \times 6.5 = 250.13 \: cm^3$

To find the original clearance volume (CV), the relationship is given by:

$CV = \frac{SV}{CR - 1}$

where $CR$ (Compression Ratio) = 9. Therefore:

$CV = \frac{250.13}{9 - 1} = 31.27 \, cm^3$

Assuming the original clearance volume remains unchanged:

Given:

• New bore diameter = 7.2 cm
• New compression ratio = 10 Using the volume formula again:

$SV = CV \times (CR - 1)$

Now we calculate the new swept volume:

$SV = 31.27 \times (10 - 1) = 281.42 \: cm^3$

Now using the swept volume to determine the new stroke length:

$L = \frac{SV \times 4}{\pi D^2}$

Substituting for $D$:

$L = \frac{281.42 \times 4}{\pi (7.2)^2} = 6.91 \: cm = 69.12 \, mm$

Indicated power (IP) can be calculated using:

$IP = \frac{MAP \times A \times N}{2}$

• $MAP$ = Mean effective pressure = 1250 kPa = 1250 x 10³ N/m²
• $A$ = Area = $\frac{\pi D^2}{4} = \frac{\pi (0.08)^2}{4} = 5.03 \times 10^{-3} \,m^2$
• $N$ = (2500 r/min) converting to firing strokes per second: $N = \frac{2500}{60} = 41.67 \, strokes/sec$

Substituting the values:

$IP = (1250 \times 10^{3}) \times 5.03 \times 10^{-3} \times 41.67 \approx 52.39 \, kW$

Torque (T) can be calculated using:

$T = \frac{BP}{2\pi N}$

Where:

• $BP$ = Brake power = 46.08 kW = 46.08 \times 10^3 W
• $N$ = 2500 r/min = \frac{2500}{60} = 41.67 , r/s

Substituting into the equation:

$T = \frac{(46.08 \times 10^3)}{2\pi \times 41.67} \approx 176 \, Nm$

Mechanical efficiency (η) can be calculated using:

$\eta = \frac{BP}{IP} \times 100$

Substituting the known values:

$\eta = \frac{46.08}{52.39} \times 100 \approx 87.96 \%$