Question

A single cylinder, four stroke cycle oil engine is fitted with a rope brake. The diameter of the brake wheel is 600 mm and the rope diameter is 26 mm. The dead load on the brake is 200 N and the spring balance reads 30 N. If the engine runs at 450 r.p.m., what will be the brake power of the engine?

(a) 2 kW

(b) 2.5 kW

(c) 3 kW

(d) 3.5 kW

The question was posed to me in exam.

The origin of the question is Engine Power in division Measurements and Testing/Performance Parameters and Characteristics of IC Engine

Answer 1

To calculate the brake power (BP) of the engine, we can use the formula for brake power based on the rope brake method:

\[

BP = \frac{2 \pi N T}{60}

\]

Where:

- \(N\) = engine speed in revolutions per minute (rpm)

- \(T\) = brake torque (N·m)

Step 1: Calculate the brake torque (\(T\)).

The torque is calculated using the rope brake tension difference. The total tension on the rope is the sum of the dead load and the spring balance reading.

\[

T = (F_1 - F_2) \times R

\]

Where:

- \(F_1\) = dead load = 200 N

- \(F_2\) = spring balance reading = 30 N

- \(R\) = radius of the brake wheel = half of the diameter of the brake wheel = \(600 \, \text{mm} / 2 = 300 \, \text{mm} = 0.3 \, \text{m}\)

Thus,

\[

T = (200 - 30) \times 0.3 = 170 \times 0.3 = 51 \, \text{N·m}

\]

Step 2: Use the formula to calculate brake power (BP).

Now we can substitute the known values into the formula:

- \(N = 450 \, \text{rpm}\)

- \(T = 51 \, \text{N·m}\)

\[

BP = \frac{2 \pi \times 450 \times 51}{60} = \frac{2 \times 3.1416 \times 450 \times 51}{60} \approx 2.4 \, \text{kW}

\]

So, the closest answer is:

(b) 2.5 kW