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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 value of the torque function over one complete cycle (from

0

4\pi

). The mean torque

\bar{T}

can be calculated using the formula:

\bar{T} = \frac{1}{T} \int_0^T T_\theta \, d\theta

where

T

is the period of the function. The period is

4\pi

radians. We will break the integral into four segments corresponding to the piecewise function:

\bar{T} = \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)

Evaluating these integrals gives us:

\int_0^\pi 360 \sin \theta \, d\theta = 360 \cdot 2 = 720

\int_\pi^{2\pi} 9 \sin \theta \, d\theta = 9 \cdot 2 = 18

\int_{2\pi}^{3\pi} 13 \sin \theta \, d\theta = 13 \cdot 2 = 26

\int_{3\pi}^{4\pi} 90 \sin \theta \, d\theta = 90 \cdot 2 = 180

Thus, the total torque contribution is:

\bar{T} = \frac{1}{4\pi} (720 + 18 + 26 + 180) = \frac{944}{4\pi} \approx 75.0 \, \text{Nm}

To draw the cyclic variation of the total crankshaft torque for the engine, we need to consider that there are two identical cylinders firing sequentially. Therefore, the total torque

T_{total}

will be the sum of the torques from both cylinders. Since they fire at equal intervals, the total torque will be:

T_{total} = 2 \cdot T_\theta

. The maximum torque will be

2 \cdot 360 = 720 \, \text{Nm}

and the minimum torque will be

2 \cdot (-90) = -180 \, \text{Nm}

. The mean torque will be

2 \cdot 75.0 \approx 150.0 \, \text{Nm}

. The graph will show a periodic curve with maximum torque at

0

and

2\pi

, minimum torque at

\pi

and

3\pi

, and mean torque at

150.0 \, \text{Nm}

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

I

of the flywheel. The mass

M

of the flywheel can be calculated using its volume and density:

M = \text{Volume} \times \text{Density} = \pi R^2 h \cdot \rho

where

R = 0.2 \, \text{m}

and

h = 0.01 \, \text{m}

, and

\rho = 7850 \, \text{kg/m}^3

. Thus,

M = \pi (0.2)^2 (0.01) \cdot 7850 \approx 0.314 \, \text{kg}

. The moment of inertia is then:

I = \frac{M R^2}{2} = \frac{0.314 \cdot (0.2)^2}{2} \approx 0.00628 \, \text{kg m}^2

. The fluctuation of speed can be calculated using the formula:

\Delta \omega = \frac{T_{max} - T_{min}}{I}

where

T_{max} = 720 \, \text{Nm}

and

T_{min} = -180 \, \text{Nm}

. The mean speed

\omega_{mean}

can be calculated from the engine speed of 2000 rpm:

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

. The fluctuation as a percentage of the mean speed is then:

\text{Percentage} = \frac{\Delta \omega}{\omega_{mean}} \times 100

. After calculating, we find the fluctuation percentage to be approximately

3.0\%

Answer

[75.0 Nm, 150.0 Nm, 3.0%]

Key Concept

Mean Torque Calculation, Total Torque from Multiple Cylinders, Moment of Inertia, and Engine Speed Fluctuation

Explanation

The mean torque is calculated by integrating the torque function over one cycle, while the total torque is the sum of contributions from both cylinders. The moment of inertia helps determine the fluctuation in speed under a constant torque.