7.1 Define the term indicated power of an internal combustion engine - NSC Mechanical Technology Automotive - Question 7 - 2020 - Paper 1

1. Fit a piston with suitable lower crowns.
2. Fit a thicker gasket between the cylinder block and cylinder head.

1. Electric dynamometer.
2. Chassis dynamometer.

To calculate the swept volume (SV) of a single cylinder, we can use the formula:

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

Where:

• D = 70 mm (bore diameter)
• L = 90 mm (stroke length)

Substituting the values:

$SV = \frac{\pi (7.0)^2}{4} \times 9.0 \approx 346.36 \: cm^3$

Given the compression ratio (CR) is 7.5:1, the formula relates swept volume (SV) and clearance volume (CV) as follows:

$CR = \frac{SV + CV}{CV}$

Rearranging gives:

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

Substituting the values:

$CV = \frac{346.36}{7.5 - 1} \approx 53.29 \: cm^3$

To find the new bore diameter (D) when the compression ratio is changed to 9.5:1, we first find the new swept volume (SV) from the original compression ratio:

Using the same formula as before and rearranging:

$SV = new \; CR \times CV - CV$ $= 9.5 \times 53.29 - 53.29 \approx 452.965 \: cm^3$

Now using the swept volume formula:

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

Solving for D:

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

Substituting gives:

$D = \sqrt{\frac{4 \times 452.965}{\pi \times 9.0}} \approx 80.05 \: mm$

Torque (T) can be calculated using the formula:

$T = mass \times gravity \times distance$

Where:

• Mass = 50 kg
• Gravity = 10 m/s²
• Distance = 1 m

Thus:

$T = 50 \times 10 \times 1 = 500 \: N \, m$

Brake power (BP) can be calculated as follows:

$BP = 2 \pi n T$

Using:

• n = 4500/60 (converting to power strokes per second)
• T = 500 N*m

Thus:

$BP = 2 \pi \left( \frac{4500}{60} \right) \times 500 \approx 235.62 \: kW$

Indicated power (IP) can be calculated using the formula:

$IP = P \times L \times A \times n$

Where:

• P = 1450 kPa
• L = 140 mm = 0.14 m
• A is the cross-sectional area of the bore:

$A = \frac{\pi D^2}{4} = \frac{\pi (0.11)^2}{4} \approx 0.0095 \: m^2$

• n = 4500/60 \times 2 for four strokes per cycle.

Substituting gives:

$IP = 1450 \times 0.14 \times 0.0095 \times 37.5 \approx 289.27 \: kW$

Mechanical efficiency can be calculated using the formula:

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

Substituting the values:

$\eta = \frac{235.62}{289.27} \times 100 \approx 81.45\%$