To determine the cut-off ratio for a diesel engine, it's essential to understand the fundamental relationships between the various volumes involved in its operational cycle. These include the clearance volume, total cylinder volume, stroke volume, and the volume at which fuel injection ceases.

Diesel Engine Fundamentals

A diesel engine operates on a cycle characterized by constant pressure heat addition. Understanding the key volumetric ratios is crucial for analyzing its performance:

• Compression Ratio (\(r\)): This critical parameter is defined as the ratio of the total cylinder volume (\(V_1\)) at the beginning of the compression stroke to the clearance volume (\(V_2\)) at the end of the compression stroke. It quantifies the extent to which the air is compressed before combustion begins.
  $$\text{Compression Ratio}, r = \frac{V_1}{V_2}$$
• Cut-off Ratio (\(\rho\)): Specific to diesel engines, the cut-off ratio is the ratio of the volume after the constant pressure heat addition process (\(V_3\)) to the clearance volume (\(V_2\)). It indicates the point in the stroke where the fuel supply is cut off.
  $$\text{Cut-off Ratio}, \rho = \frac{V_3}{V_2}$$
• Stroke Volume (\(V_s\)): This represents the volume swept by the piston as it moves from one end of its travel to the other. It is the difference between the total cylinder volume and the clearance volume.
  $$V_s = V_1 - V_2$$

Volume Relationships for Cut-off Ratio

We are provided with the engine's compression ratio (\(r\)) and the information that the cut-off occurs at a specific percentage of the stroke. Let's derive the formula for the cut-off ratio based on these given parameters:

• From the definition of the compression ratio, we can express the total cylinder volume in terms of the clearance volume:
  $$V_1 = r \cdot V_2$$
• The stroke volume (\(V_s\)) can then be expressed in terms of the clearance volume (\(V_2\)) and the compression ratio (\(r\)):
  $$V_s = V_1 - V_2$$ Substituting \(V_1 = r \cdot V_2\): $$V_s = (r \cdot V_2) - V_2 = (r-1)V_2$$
• The problem states that the fuel cut-off occurs at 5% of the stroke. This means that the volume increase during the constant pressure heat addition phase, starting from \(V_2\), is 5% of the stroke volume:
  $$V_3 - V_2 = 0.05 \cdot V_s$$
• Now, substitute the expression for \(V_s\) from the previous step into this equation:
  $$V_3 - V_2 = 0.05 \cdot (r-1)V_2$$
• To find the cut-off ratio, \(\rho = V_3 / V_2\), we first rearrange the equation to solve for \(V_3\):
  $$V_3 = V_2 + 0.05 \cdot (r-1)V_2$$ Factor out \(V_2\) from the right side: $$V_3 = V_2 [1 + 0.05 (r-1)]$$
  Finally, divide both sides by \(V_2\) to obtain the formula for the cut-off ratio:
  $$\rho = \frac{V_3}{V_2} = 1 + 0.05 (r-1)$$

Cut-off Ratio Calculation

We are given the following values for the diesel engine:

• Compression Ratio, \(r = 18\)
• Cut-off takes place at 5% of the stroke, which is equivalent to a fraction of 0.05.

Using the derived formula for the cut-off ratio:

Substitute the given value of the compression ratio, \(r = 18\):

Therefore, the cut-off ratio for this diesel engine is 1.85.