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3 years ago

Find the percent change in cutting speed required to give an 80% reduction in tool life when the value of n is 0.12.

Engineering

1 answer:

vaieri [72.5K]3 years ago

6 0

Answer:21.3%

Explanation:

Given

80 % reduction in tool life

According to Taylor's tool life

$VT^n$ =c

where V is cutting velocity

T=tool life of tool

80 % tool life reduction i.e. New tool Life is 0.2T

Thus

$VT^{0.12}=V'\left ( 0.2T\right )^{0.12}$

$V'=\frac{V}{0.2^{0.12}}$

$V'=\frac{V}{0.824}$ =1.213V

Thus a change of 21.3 %(increment) is required to reduce tool life by 80%

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One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Deter

lukranit [14]

The pressure of water is 7.3851 kPa

<u>Explanation:</u>

Given data,

V = 150× $10^{-3}$ $m^{3}$

m = 1 Kg

$P_{1}$ = 2 MPa

$T_{2}$ = 40°C

The waters specific volume is calculated:

$v_{1}$ = V/m

Here, the waters specific volume at initial condition is $v_{1}$ , the containers volume is V, waters mass is m.

$v_{1}$ = 150× $10^{-3}$ $m^{3}$ /1

$v_{1}$ = 0.15 $m^{3}$ / Kg

The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 $m^{3}$ / Kg and 0.13 $m^{3}$ / Kg.

$T_{1}$ = 350+(400-350) $\frac{0.15-0.13}{0.1522-0.1386}$

$T_{1}$ = 395.17°C

Hence, the initial temperature is 395.17°C.

The volume is constant in the rigid container.

$v_{2}$ = $v_{1}$ = 0.15 $m^{3}$ / Kg

In saturated water labels for $T_{2}$ = 40°C.

$v_{f}$ = 0.001008 $m^{3}$ / Kg

$v_{g}$ = 19.515 $m^{3}$ / Kg

The final state is two phase region $v_{f}$ < $v_{2}$ < $v_{g}$ .

In saturated water labels for $T_{2}$ = 40°C.

$P_{2}$ = $P_{Sat}$ = 7.3851 kPa

$P_{2}$ = 7.3851 kPa

7 0

3 years ago

What is engine knock? What cause the engine knock problem?

antiseptic1488 [7]

Answer:

When the uneven burning of the fuel takes place due to the incorrect air/fuel mixture inside the engine cylinder, a knocking sound is observed. This is called as the engine knocking.

Explanation:

When the uneven burning of the fuel takes place due to the incorrect air/fuel mixture inside the engine cylinder, a knocking sound is observed. This is called as the engine knocking.

The engine knock problem can be caused due to the following reason

a) When the octane rating of the fuel used is low.

b) The deposition of the carbon around the cylinder walls takes place.

c) The spark plug used in the vehicle is not correct.

3 0

3 years ago

Write a program to input 6 numbers. After each number is input, print the biggest of the numbers entered so far.

likoan [24]

Answer:

P<u>rogram:</u>

# Enter Numbers #

number1 = int(input("Enter number: " ))

print("Largest: " + string(number1))

#for num 2 #

number2 = int(input("Enter a number: "))

if number2 > number1:

print("Largest: " + string(number2))

else:

print("Largest: " + string(num1))

#for num 3 #

number3 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3)))

#for num 4 #

number4 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4)))

#for num 5 #

number5 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4, number5)))

#for num 6 #

number6 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4, number5, number6)))

# END #

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3 years ago

The properties of the air in the inlet section with A1 = 0.25ab m2 in a converging-diverging channel are given as U1 = 25a,b m/s

NeX [460]

Answer:

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Explanation:

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Kitty [74]

Answer:

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Explanation:

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3 years ago