Compress this in to ratio

The problem asks to calculate the swept volume ($V_s$) and the compression ratio ($CR$) given the bore diameter ($D$), stroke length ($L$), and clearance volume ($V_c$).

Given values: $D = 80 \, \text{mm}$ $L = 90 \, \text{mm}$ $V_c = 0.05 \, \text{m}^3$

First, we need to ensure all units are consistent. Let's convert $D$ and $L$ to meters. $D = 80 \, \text{mm} = 0.08 \, \text{m}$ $L = 90 \, \text{mm} = 0.09 \, \text{m}$

Step 1: Calculate the swept volume ($V_s$). The formula for swept volume is: $V_s = \frac{\pi}{4} D^2 L$ Substitute the values: $V_s = \frac{\pi}{4} (0.08 \, \text{m})^2 (0.09 \, \text{m})$ $V_s = \frac{\pi}{4} (0.0064 \, \text{m}^2) (0.09 \, \text{m})$ $V_s = \frac{\pi}{4} (0.000576 \, \text{m}^3)$ $V_s \approx 0.000452389 \, \text{m}^3$ Rounding to a reasonable number of significant figures: $V_s \approx 0.000452 \, \text{m}^3$

Step 2: Calculate the compression ratio ($CR$). The formula for compression ratio is: $CR = \frac{V_s + V_c}{V_c}$ Substitute the calculated $V_s$ and given $V_c$: $CR = \frac{0.000452389 \, \text{m}^3 + 0.05 \, \text{m}^3}{0.05 \, \text{m}^3}$ $CR = \frac{0.050452389 \, \text{m}^3}{0.05 \, \text{m}^3}$ $CR \approx 1.00904778$ Rounding to two decimal places: $CR \approx 1.01$

The swept volume calculated in the image is $452571.4 \, \text{mm}^3$. Let's verify this. $V_s = \frac{\pi}{4} (80 \, \text{mm})^2 (90 \, \text{mm})$ $V_s = \frac{\pi}{4} (6400 \, \text{mm}^2) (90 \, \text{mm})$ $V_s = \frac{\pi}{4} (576000 \, \text{mm}^3)$ $V_s \approx 452389.34 \, \text{mm}^3$ The image's calculation for $V_s$ is slightly off, but the method is correct. Let's use the more precise value for $V_s$ in $\text{m}^3$.

a) The swept volume is: $\boxed{\text{0.000452} \, \text{m}^3}$

b) The compression ratio is: $\boxed{\text{1.01}}$