The feed rate of single point cutting tool is 3 mm/revolution and the workpiece is rotating at 600 r.p.m. Determine the total machining time to turn the cylindrical surface of length 300 mm of the workpiece.
10 sec
This question asks us to determine the total machining time required to turn a cylindrical surface using a single point cutting tool. We are given the feed rate, workpiece rotation speed, and the length of the surface to be machined.
The total machining time ($T$) for turning a cylindrical surface with a single pass can be calculated using the formula:
$$T = \frac{L}{f \times N}$$
Where:
Let's substitute the given values into the formula:
$$T = \frac{300 \text{ mm}}{3 \text{ mm/revolution} \times 600 \text{ revolutions/minute}}$$
First, calculate the product of feed rate and rotation speed, which gives the feed speed (the distance the tool travels along the workpiece per minute):
Feed speed = $f \times N = 3 \text{ mm/revolution} \times 600 \text{ revolutions/minute} = 1800 \text{ mm/minute}$
Now substitute this back into the machining time formula:
$$T = \frac{300 \text{ mm}}{1800 \text{ mm/minute}}$$
$$T = \frac{300}{1800} \text{ minutes}$$
$$T = \frac{1}{6} \text{ minutes}$$
The options are given in seconds. We need to convert the calculated time from minutes to seconds. There are 60 seconds in 1 minute.
$$T = \frac{1}{6} \text{ minutes} \times 60 \text{ seconds/minute}$$
$$T = 10 \text{ seconds}$$
Therefore, the total machining time required to turn the cylindrical surface is 10 seconds.
Given:
Formula:
$$T = \frac{L}{f \times N}$$
Calculation:
$$T = \frac{300 \text{ mm}}{(3 \text{ mm/rev}) \times (600 \text{ rev/min})} = \frac{300 \text{ mm}}{1800 \text{ mm/min}} = \frac{1}{6} \text{ min}$$
Convert to seconds:
$$T = \frac{1}{6} \text{ min} \times 60 \text{ sec/min} = 10 \text{ seconds}$$
The calculated machining time of 10 seconds matches one of the provided options.
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