An engine develops 100 kW, when rotating at 1800 rpm. Torque required to deliver the power is

Options

A. 531 N-m
B. 570 N-m
C. 520 N-m
D. 551 N-m

Answer

To find the torque required to deliver the power of an engine that develops 100 kW at a speed of 1800 rpm, we can use the relationship between power, torque, and angular velocity. The formula we will use is:

\[ P = \tau \cdot \omega \]

Where:
- \( P \) is the power in watts (W)
- \( \tau \) is the torque in Newton-meters (N·m)
- \( \omega \) is the angular velocity in radians per second (rad/s)

### Step-by-Step Solution:

**Step 1: Convert Power from kW to W**
Given that the engine develops 100 kW, we convert this to watts:
\[ P = 100 \, \text{kW} = 100 \times 10^3 \, \text{W} = 100,000 \, \text{W} \]

**Step 2: Convert RPM to Radians per Second**
The engine speed is given as 1800 rpm. To convert this to radians per second, we use the conversion factor:
\[ \omega = 1800 \, \text{rpm} \times \frac{2\pi \, \text{rad}}{1 \, \text{revolution}} \times \frac{1 \, \text{minute}}{60 \, \text{seconds}} \]

Calculating this gives:
\[ \omega = 1800 \times \frac{2\pi}{60} = 1800 \times \frac{\pi}{30} = 60\pi \, \text{rad/s} \]

**Step 3: Rearrange the Power Formula to Solve for Torque**
We can rearrange the formula \( P = \tau \cdot \omega \) to solve for torque (\( \tau \)):
\[ \tau = \frac{P}{\omega} \]

**Step 4: Substitute the Values into the Torque Formula**
Now we substitute the values we have:
\[ \tau = \frac{100,000 \, \text{W}}{60\pi \, \text{rad/s}} \]

**Step 5: Calculate the Torque**
Calculating the value:
\[ \tau = \frac{100,000}{60\pi} \approx \frac{100,000}{188.4} \approx 530.8 \, \text{N·m} \]

**Step 6: Round to the Nearest Whole Number**
The torque required to deliver the power is approximately:
\[ \tau \approx 531 \, \text{N·m} \]

### Final Answer:
The torque required to deliver the power is approximately **531 N·m**.

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