During rough turning of 200 mm diameter steel bar at 300 rpm using HSS tool, the tool failure occurred after removing 1000 cc of material. When the speed was reduced to 250 rpm, the tool failure occurred after removing 5250 cc of material. Assuming that Taylor's equation applies, the expected tool life in 'cc' at a cutting speed of 275 rpm would be

3125

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2633.9

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2207.4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1642.2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 2207.4
$V_{1} T_{1}^{n} = V_{2} T_{2}^{n}$

$\frac{π D N_{1}}{1000} T_{1}^{n} = \frac{π D N_{2}}{1000} T_{2}^{n}$

$N_{1} T_{1}^{n} = N_{2} T_{2}^{n}$

$300 (1000)^{n} = 250 (5250)^{n}$

$\frac{300}{250} = {(\frac{5250}{1000})}^{n}$

$\Rightarrow x = 0.1099$

$N_{1} T_{1}^{n} = N_{3} T_{3}^{n}$

$300 (1000)^{0.1099} = 275 (x)^{0.1099}$

$\Rightarrow x = 2207.21 cc$

Suggest Corrections

Similar questions

Q. During the turning of a 20 mm diameter steel bar at a spindle speed of 400 rpm, a tool life of 20 minute is obtained. When the same bar is turned at 200 rpm, the tool life becomes 60 minute. Assume that Taylor's tool life equation is valid. When the bar is turned at 300 rpm, the tool life (in minute) is approximately

Linked Data:
Blind hole $10 m m$ diameter, $50 m m$ deep are being drilled in steel block, drilling spindle speed is $600 r p m$ , feed $0.2 m m / r e v,$ point angle of drill is $120$

During the above operation, the drill wears out after producing $200$ holes.Taylor's tool life equation is of the form $V T^{0.3} = C,$ where $V =$ cutting speed in $m / m i n$ and $T =$ tool life in min. Taylor's constnat $C$ will be