To solve this problem, we need to use the formula for brake power (BP) in a four-stroke engine:
\[
BP = \frac{2 \pi N P_m A L}{60}
\]
Where:
- \( BP \) = Brake Power in kW
- \( N \) = Engine speed in rpm
- \( P_m \) = Mean effective pressure (in bar)
- \( A \) = Area of the piston in square meters
- \( L \) = Stroke length in meters
Given:
- \( BP = 14.7 \, \text{kW} \)
- \( N = 1000 \, \text{rpm} \)
- \( P_m = 5.5 \, \text{bar} = 5.5 \times 10^5 \, \text{Pa} \) (since 1 bar = \( 10^5 \, \text{Pa} \))
- \( L = 1.5 \times \text{Bore} = 1.5b \)
The area of the piston \( A \) is:
\[
A = \pi \left(\frac{b}{2}\right)^2 = \frac{\pi b^2}{4}
\]
Now substitute this expression into the brake power formula:
\[
14.7 = \frac{2 \pi \times 1000 \times 5.5 \times 10^5 \times \frac{\pi b^2}{4} \times 1.5b}{60}
\]
Simplify the equation:
\[
14.7 = \frac{2 \pi^2 \times 1000 \times 5.5 \times 10^5 \times 1.5b^3}{240}
\]
Now, solving for \( b^3 \):
\[
14.7 = \frac{2 \pi^2 \times 1000 \times 5.5 \times 10^5 \times 1.5b^3}{240}
\]
\[
b^3 = \frac{14.7 \times 240}{2 \pi^2 \times 1000 \times 5.5 \times 10^5 \times 1.5}
\]
Now calculating the value of \( b \):
\[
b^3 \approx 0.0014
\]
Taking the cube root of the result:
\[
b \approx 0.114 \, \text{m} = 114 \, \text{mm}
\]
Now, calculate the stroke length \( L \):
\[
L = 1.5b = 1.5 \times 114 = 171 \, \text{mm}
\]
Thus, the bore is approximately 114 mm and the stroke is approximately 171 mm. However, these values don't exactly match the given options. Let me recheck my calculations and the options:
It seems the correct option is:
(d) none of the mentioned