Mechanical Engineering - Production Engineering - Discussion

Discussion Forum : Production Engineering - Section 4 (Q.No. 5)
5.
In a single point turning operation with a cemented carbide and steel combination having a Taylor exponent of 0.25, if the cutting speed is halved, then tool life will become
half
two times
eight times
sixteen times
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
8 comments Page 1 of 1.

SAMEER ansari said:   5 years ago
Thanks all for explaining this.

VIPIN said:   5 years ago
VT^n = constant.
n = .25,

initial velocity = V,
final velocity = V/2,
intial time = T1,
FINAL TIME = T2,
VT1^(.25) = (V/2)T2^(.25).
T2 = 16T1.

Karthik v k said:   6 years ago
Thanks.

Abdul said:   8 years ago
VT^0.25 = 0.5Vt^.25.

t/T = 2^4 = 16.

t = 16T.

Rahul Soneriya said:   8 years ago
We know that:

V1T1^n = V2T2^n........1.

Given that n = 0.25 and V2 = V1/2.

Put the above values in equation 1.

V1T1^0.25 = V1/2*T2^0.25.

T2/T1 = (2)^4 = 16.

T2 = 16T1 answer.

DANU said:   9 years ago
V1(T1)^n = V2(T2)^n.
V1/V2 = (T2/T1)^n.

NOW V2 = V1/2.

2V1/V1 = (T2/T1)^0.25.
T2/T1 = 2^(1/0.25).

= 2^4.
T2 = 16 T1.

SYED KHAJA said:   9 years ago
C = (VT)^N.
C = (VT)0.25.
Now V = V/2.

C = (V/2*T)^0.25.
C = (VT)^0.25/16.
16C = (VT)^.25.

Manu said:   1 decade ago
VT^n = C 2V1/V2 = (T2/T1)^0.25.

T2/T1 = 16.

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