Taylor's Tool Life Equation MCQ Quiz in తెలుగు - Objective Question with Answer for Taylor's Tool Life Equation - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

Machining time for turning is given by-

\(T_m=\frac{L_e}{fN}\)

Taylor tool life equation is given by-

VTⁿ = C

\(No\;of\;tool\;regrind=\frac{Machining\;time\;of\;one\;cylinder\;\times\;No\;of\;cylinder}{Tool\;life}\)

Calculation:

Given:

d = 25 mm, L_e = 100 mm, V = 22 m/min, f = 0.046 mm/rev, No. of cylinder = 425

\(V=\frac{\pi DN}{1000}\)

\(V=\frac{\pi\;×\;25\;×\;N}{1000}\)

∴ N = 280.11 rpm

Machining time for turning is given by-

\(T_m=\frac{L_e}{fN}\)

\(T_m=\frac{100}{0.046\;×\;280.11}\Rightarrow 7.76\;min\)

VTn = C

22 × T^0.25 = 55

∴ T = 39.0625 min

\(No.\;of\;tool\;regrind=\frac{Machining\;time\;of\;one\;cylinder\;\times\;No\;of\;cylinder}{Tool\;life}\)

\(No.\;of\;tool\;regrind=\frac{7.76\;\times\;425}{39.0625}\Rightarrow\;84.43\;\approx85\)

Explanation:

Taylor’s Tool Life Equation based on Flank Wear:

VTn = C 

Take log on both side,

log (VTn) = ln C 

⇒ log V + n log T = log C 

So, the variation in cutting speed V with respect to tool life T on a Log-Log scale will be Straight-line.

where, V = cutting speed (m/min)

T = Time (min) (time taken to develop certain flank wear)

n = an exponent that largely depends on tool material

C = constant based on tool and work material and cutting condition.

Values of Exponent ‘n’

Concept:

Taylor's tool life equation:

Taylor has assumed that cutting velocity is the major parameter that is influencing the tool life, hence he established the relationship between cutting velocity and tool life called Taylor's tool life equation.

The Taylor's tool life equation is given as:

 \({{V}} \times {{{T}}^{{n}}} = {{C}}\)

where V = cutting velocity in m/min, T = tool life in min, n = Taylor's exponent 

Additional Information

Concept:

Taylor's tool life equation is given by,

\(N_1{T_1}^n=N_2{T_2}^n\)

Calculation:

Given:

N₁ = 200 rpm, N₂ = 400 rpm, T₁ = 20 mins, T₂ = 5 mins

N₃ = 300 rpm, T₃ = ?

We know the tool life equation constant remains the same.

so, from Taylor's tool life equation, we have

\(N_1{T_1}^n=N_2{T_2}^n\)

\(\Rightarrow 200\times{20}^n=400\times{5}^n \Rightarrow n=0.5\)

Tool life at N₃ = 300 rpm

\(\therefore N_1{T_1}^n=N_3{T_3}^n\)

\(\Rightarrow 200\times{20}^{0.5}=300\times {T_2}^{0.5}\Rightarrow T_2 =8.889\ min\)

Concept:

• Taylor’s Tool Life Equation based on Flank Wear:

VTn = C

Where, V = cutting speed (m/min), T = Time (min) (time taken to develop certain flank wear), n = an exponent that largely depends on tool material, C = constant based on tool and work material and cutting condition.

Calculation:

Given: \(\rm V\sqrt{T} = 300\) ⇒ n = 1/2

• If the cutting speed is reduced by 50%, then the tool life will be enhanced by,

⇒ V' = 0.5V

\(⇒ \rm V'\sqrt{T'} = 300\)

\(⇒ \rm 0.5V\sqrt{T'} = 300\)

\(⇒ \rm V\sqrt{T'} = 600\)

\(⇒T' = \frac{600^2}{V^2}\)     ...(1)

∵ \(\rm V\sqrt{T} = 300\)

\(⇒T = \frac{300^2}{V^2}\)

Using equation 1,

\(⇒\frac{T'}{T} =\frac{600^2}{V^2}\times \frac{V^2}{300^2} = 4\)

\(⇒[\frac{T' -T}{T}] \times 100 =(4 - 1) \times 100% \)

\(⇒[\frac{T' -T}{T}] \times 100 =300\)

So, the tool life will be enhanced by 300%.

Concept:

Taylor's tool life equation is given by - 

VTn = C

where V = Cutting velocity (m/min), T = Tool life (min), n = Taylor's exponent and C = Taylor's constant.

Calculation:

Given:

V₁ = 32 m/min, T1 = 15 min, V2 = 50 m/min, T2 = 12 min, V3 = 72 m/min, T3 = ?

From Taylor tool life - 

VTn = C

For the same tool -

\(V_1T_1^{n}=V_2T_2^{n}\)

\( \frac{V_2}{V_1}=(\frac{T_1}{T_2})^{n}\)

\(\frac{50}{32}=(\frac{15}{12})^{n}\)

\(\frac{25}{16}=(\frac{5}{4})^{n}\)

From the above, n = 2

\(V_3T_3^{n}=V_2T_2^{n}\)

\( \frac{V_2}{V_3}=(\frac{T_3}{T_2})^{n}\)

\( \frac{50}{72}=(\frac{T_3}{12})^{2}\)

\( \frac{25}{36}=(\frac{T_3}{12})^{2}\)

\((\frac{T_3}{12})= \sqrt\frac{25}{36}= \frac56\)

T3 = 10 min

Concept:

Taylor's tool life equation is given by - 

VTn = C

where V = Cutting velocity (m/min), T = Tool life (min), n = Taylor's exponent and C = Taylor's constant.

Calculation:

Given:

n = ?, T₂ = \(\frac{T_1}{8}\), V₂ = 2V₁

From Taylor tool life - 

VTn = C

For the same tool -

\(V_1T_1^{n}=V_2T_2^{n}\)

 V₁T₁ⁿ = 2V₁ × \(\left ( \frac{T_1}{8} \right )^n\)

8ⁿ = 2

n\(\ln \left ( 8 \right )\) = \(\ln \left ( 2 \right )\)

n = \(\frac{1}{3}\)

Explanation:

Taylor’s tool life equation. 

• Tool life of any tool for any work material is governed by machining parameters mainly cutting velocity (V), feed, and depth of cut (t). 
• Cutting velocity affects the maximum and depth of cut minimum. 
• Most widely used tool life equation and expressed in equation form

VTn = C

where V = Cutting velocity, T =Tool life (minutes), n = tool life exponent(generally 0.5-0.8), C = tool life constrant (Empirical)  

Additional Information

Tool Life
Tool life generally indicates, the amount of satisfactory performance or service rendered by a fresh tool or a cutting point till it is declared failed.

• In R and D the tool life is the actual machining time by which a fresh cutting tool satisfactorily works before it needs replacement or reconditioning.
• The modern tools hardly fail prematurely or abruptly by mechanical breakage or rapid plastic deformation.
• Those fail mostly by wearing a process that systematically grows slowly with machining time.
• In that case, tool life means the span of actual machining time by which a fresh tool can work before attaining the specified limit of tool wear. 
• In industries, tool life is the length of time of satisfactory service or amount of acceptable output provided by a fresh tool prior to it is required to replace or reconditioned.
• Assessment of tool life for R and D purposes, tool life is always assessed or expressed by the span of machining time in minutes, whereas, in industries besides machining time in minutes some other means are also used to assess tool life, depending upon the situation, such as

1. no. of pieces of work machined
2. the total volume of material removed
3. the total length of the cut.

Concept:

\({\rm{No}}.{\rm{of\;pieces\;in\;one\;tool\;life}} =\rm \frac{{Tool\;life}}{{Machining\;time\;required\;for\;one\;piece}}\)

Calculation:

Given:

C = 140, n = 0.18, length = 120 mm, diameter (D) = 80 mm, velocity (V) = 100 m/min

Feed (f) = 0.2 mm/rev

VTⁿ = C

100 × (T)^0.18 = 140

T = 6.483 min

Now,

\({\rm{Machining\;time\;per\;piece}} = \frac{L}{{fN}}\)

\({\rm{Machining\;time\;required\;per\;piece}} = \frac{{\pi DL}}{{fV}} = \frac{{\pi \times 80 \times 120}}{{0.2 \times 100 \times 1000}}\)

∴ Machining time required per piece = 1.5079 min

Now,

\({\rm{No}}.{\rm{\;of\;pieces}} = \frac{{6.483}}{{1.5079}}\)

∴ No. of pieces = 4.3 ≈ 5 (next higher integer)

Explanation:

Given:

n = 0.5, C = 550

Calculation:

Let initial speed of tool = V₁

And, initial tool life = T₁

Now, cutting speed is reduced by 35%,

Thus, V₂ = (0.65) V₁

Taylor’s equation for tool wear,

VTⁿ = C

VT^0.5 = 550

V₁T₁^0.5 = 550     …1)

Also, V₂T₂^0.5 = 550

(0.65 V₁) T₂^0.5 = 550

V₁T₂^0.5 = 846.15     …2)

Dividing 2) by 1)

\(\frac{{{V_1}T_2^{0.5}}}{{{V_1}T_1^{0.5}}} = \frac{{846.15}}{{550}}\)

\(\left( {\frac{{{T_2}}}{{{T_1}}}} \right) = 2.366\)

Now,

\(\frac{{{T_2} - {T_1}}}{{{T_1}}} = 1.366\)

Percentage increase in tool life is

\(\left( {\frac{{{T_2} - {T_1}}}{{{T_1}}}} \right) \times 100 = 1.366 \times 100\)

∴ Percentage increase in tool life = 136.6%