Question

A four cylinder 4-stroke gas engine has cylinder diameter of 20 cm, stroke-bore ratio is 1.5; clearance volume \(2500 \text{ cm}^3\), engine speed 300 RPM, mean effective pressure \(5 \text{ kg}\cdot\text{cm}^{-2}\) and mechanical efficiency 70 per cent. Calculate IHP of the engine in kW.

• A. 23.45
• B. 31.43
• C. 46.90
• D. 62.86

Hint:

Always distinguish between 2-stroke ($N' = N$) and 4-stroke ($N' = N/2$) engines when calculating power strokes.
Keep units consistent: pressure in $\text{kg/cm}^2$, area in $\text{cm}^2$, and stroke length in meters.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Indicated Horsepower (IHP) represents the total power developed inside the engine cylinders by the combustion of fuel.
It is calculated from the mean effective pressure, cylinder dimensions, and the frequency of power strokes.
Key Formula or Approach:
The standard metric formula for Indicated Horsepower (IHP) is:
\[ IHP = \frac{P_m \cdot L \cdot A \cdot N' \cdot n}{4500} \] where:
$P_m$ = Mean effective pressure in $\text{kg/cm}^2$
$L$ = Stroke length in meters
$A$ = Cross-sectional area of cylinder in $\text{cm}^2$
$N'$ = Number of power strokes per minute ($N/2$ for a 4-stroke engine)
$n$ = Number of cylinders
$4500$ = Conversion constant ($1\text{ metric hp} = 4500\text{ kg-m/min}$)

Step 2: Detailed Explanation:
Let us determine each parameter from the given values:
Cylinder diameter, $D = 20\text{ cm}$
Stroke-bore ratio, $\frac{L_c}{D} = 1.5 \implies L_c = 1.5 \times 20 = 30\text{ cm} = 0.30\text{ m}$
Cylinder area, $A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (20)^2 \approx 314.16\text{ cm}^2$
Rotational speed, $N = 300\text{ RPM}$
Since it is a 4-stroke engine, power strokes per cylinder per minute:
\[ N' = \frac{N}{2} = \frac{300}{2} = 150 \text{ strokes/min} \] Number of cylinders, $n = 4$
Mean effective pressure, $P_m = 5\text{ kg/cm}^2$
Calculate IHP in metric horsepower:
\[ IHP = \frac{5 \times 0.30 \times 314.16 \times 150 \times 4}{4500} \] \[ IHP = \frac{282744}{4500} \approx 62.83 \text{ hp (metric)} \] Converting metric horsepower to kilowatts (using the standard conversion $1\text{ hp (metric)} \approx 0.7355\text{ kW}$ or standard engine conversion $1\text{ hp (SAE)} \approx 0.746\text{ kW}$):
\[ IP_{\text{kW}} = 62.83 \times 0.746 \approx 46.90 \text{ kW} \]

Step 3: Final Answer:
The indicated power (IHP) of the engine is $46.90\text{ kW}$.