AskSia

Plus

2. A four-stroke turbocharged petrol engine has two identical in-line cylinder...

Jan 8, 2025

Solution

To calculate the mean torque due to one cylinder, we first need to find the average torque over one complete cycle (from

0

4\pi

radians). The mean torque can be calculated using the formula:

\text{Mean Torque} = \frac{1}{T} \int_0^T T_{\theta} \, d\theta

where

T

is the period of one complete cycle. The period for the engine running at 2000 rpm is given by:

T = \frac{60}{2000} \text{ seconds} = 0.03 \text{ seconds}

. We will calculate the integral for each section of the torque function and then average it over the total angle of

4\pi

radians. The mean torque is calculated as follows:

\text{Mean Torque} = \frac{1}{4\pi} \left( \int_0^{\pi} 360 \sin \theta \, d\theta + \int_{\pi}^{2\pi} 9 \sin \theta \, d\theta + \int_{2\pi}^{3\pi} 13 \sin \theta \, d\theta + \int_{3\pi}^{4\pi} 90 \sin \theta \, d\theta \right)

Calculating each integral gives us the mean torque

To draw the cyclic variation of the total crankshaft torque for the engine, we need to sum the contributions from both cylinders. Since the cylinders fire sequentially, the total torque can be expressed as:

T_{\text{total}} = T_{\text{cyl. 1}} + T_{\text{cyl. 2}}

Given that the torque functions for both cylinders are identical, we can simply double the torque values from cylinder no. 1. The maximum and minimum torques will also be adjusted accordingly. The maximum torque will be

720 \, \text{Nm}

and the minimum torque will be

-180 \, \text{Nm}

. The mean torque will also be adjusted to reflect the contributions from both cylinders. The graph will show these variations clearly

To calculate the fluctuation of the engine speed as a percentage of the mean speed under a constant resisting torque, we first need to find the moment of inertia of the flywheel. The mass of the flywheel can be calculated using its volume and density:

V = \pi R^2 h = \pi (0.2^2)(0.01) = 0.01257 \, \text{m}^3

M = \text{density} \times V = 7850 \, \text{kg/m}^3 \times 0.01257 \, \text{m}^3 = 98.5 \, \text{kg}

The moment of inertia

I

of the flywheel is given by:

I = \frac{1}{2} M R^2 = \frac{1}{2} (98.5) (0.2^2) = 0.394 \, \text{kg m}^2

Next, we calculate the angular speed

\omega

in radians per second:

\omega = \frac{2000 \times 2\pi}{60} = 209.44 \, \text{rad/s}

The fluctuation of speed can be calculated using the formula for angular acceleration and the resisting torque. The percentage fluctuation can then be expressed as:

\text{Percentage Fluctuation} = \frac{\Delta \omega}{\omega} \times 100\%

where

\Delta \omega

is the change in angular speed due to the resisting torque

Answer

[Insert final answer here]

Key Concept

Mean Torque Calculation, Torque Variation in Cycles, Moment of Inertia, Angular Speed, and Fluctuation of Speed

Explanation

The mean torque is calculated by integrating the torque function over one complete cycle, while the total torque is the sum of the contributions from both cylinders. The moment of inertia is essential for understanding the dynamics of the flywheel, and the fluctuation of speed is determined by the resisting torque acting on the system.