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2 years ago

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A gun drilling operation is used to drill a 9/64-in diameter hole to a certain depth. It takes 4.5 minutes to perform the drilli

ng operation using high-pressure fluid delivery of coolant to the drill point. The current spindle speed = 4000 rev/min, and feed = 0.0017 in/rev. In order to improve the surface finish in the hole, it has been decided to increase the speed by 20% and decrease the feed by 25%.
How long will it take to perform the operation at the new cutting conditions?

1 answer:

hichkok12 [17]2 years ago

5 0

Answer:

fr = Nf = 4000 rev/min (0.0017 in/rev) = 6.8 in/min Hole depth

d = 4.5 min (6.8 in/min) = 30.6 in New speed v = 4000(1 + 0.20) = 4800 rev/min

New feed f = 0.0017(1- 0.25) = 0.001275 in/min

New feed rate fr = 4800(0.001275) = 6.12 in/min

New drilling time Tm = 30.6/6.12 in/min = 5.0 min

Explanation:

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At a certain college, 30% of the students major in engineering, 20% play club sports, and 10% both major in engineering and play

hodyreva [135]

At a certain college, 30% of the students major in engineering, 20% play club sports, and 10% both major in engineering and play club sports. A student is selected at random.

a) What is the probability that the student is majoring in engineering?

b) What is the probability that the student plays club sports?

c) Given that the student is majoring in engineering, what is the probability that the

student plays club sports?

d) Given that the student plays club sports, what is the probability that the student is

majoring in engineering?

e) Given that the student is majoring in engineering, what is the probability that the

student does not play club sports?

f) Given that the student plays club sports, what is the probability that the student is not

majoring in engineering?

Answer and Explanation

The venn diagram for the question is in the attachment.

Percentage majoring in engineering = 30% = 0.3

Percentage that plays club sport = 20% = 0.2

Percentage that major in engineering and play sport = 10% = 0.1

Percentage majoring in engineering & do not play club sport = 30 - 10 = 20% = 0.2

Percentage that plays club sport & do not major in engineering = 20 - 10 = 10% = 0.1

Total percentage = 100%

a) probability that student is majoring in Engineering P(E) = 30/100 = 0.3

b) probability that student plays club sport = 20/100 P(S) = 0.2

c) probability that the

student plays club sport given that the student is majoring in engineering, P(S|E) = (P(E and S))/(P(E))

P(E and S) = 10/100 = 0.1, P(E) = 0.3

P(S|E) = 0.1/0.3 = 0.3333

d) probability that the

student is majoring in engineering given that the student plays club sport, P(E|S) = (P(S and E))/(P(S))

P(S and E) = 10/100 = 0.1, P(S) = 0.2

P(E|S) = 0.1/0.2 = 0.5

e) probability that the

student does not play club sport given that the student is majoring in engineering, P(S'|E) = (P(E and S'))/(P(E))

P(E and S') = 20/100 = 0.2, P(E) = 0.3

P(S'|E) = 0.2/0.3 = 0.667

f) probability that the

student is not majoring in engineering given that the student plays club sport, P(E'|S) = (P(S and E'))/(P(S))

P(S and E') = 10/100 = 0.1, P(S) = 0.2

P(E|S) = 0.1/0.2 = 0.5

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2 years ago

The purification of hydrogen gas is possible by diffusion through a thin palladium sheet. Calculate the number of kilograms of h

diamong [38]

Answer:

$M=0.0411 kg/h or 4.1*10^{-2} kg/h$

Explanation:

We have to combine the following formula to find the mass yield:

$M=JAt$

$M=-DAt(ΔC/Δx)$

The diffusion coefficient : $D=6.0*10^{-8} m/s^{2}$

The area : $A=0.25 m^{2}$

Time : $t=3600 s/h$

ΔC: $(0.64-3.0)kg/m^{3}$

Δx: $3.1*10^{-3}m$

Now substitute the values

$M=-DAt(ΔC/Δx)$

$M=-(6.0*10^{-8} m/s^{2})(0.25 m^{2})(3600 s/h)[(0.64-3.0kg/m^{3})(3.1*10^{-3}m)]$

$M=0.0411 kg/h or 4.1*10^{-2} kg/h$

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2 years ago

Evan notices a small fire in his workplace. Since the fire is small and the atmosphere is not smoky he decides to fight the fire

Norma-Jean [14]

Answer:

not calling the firemean

Explanation:

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2 years ago

A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calcula

makkiz [27]

Answer:

F_y = 151319.01N = 15.132 KN

Explanation:

From the linear momentum equation theory, since flow is steady, the y components would be;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

We are given;

Length; L = 5ft = 1.52.

Initial diameter;d1 = 12in = 0.3m

Exit diameter; d2 = 24 in = 0.6m

Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s

Initial pressure;p1 = 30 psi = 206843 pa

Thus,

initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²

Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²

Now, we know that volume flow rate of water is given by; Q = A•V

Thus,

At exit, Q2 = A2•V2

So, 0.28 = 0.28•V2

So,V2 = 1 m/s

When flow is incompressible, we often say that ;

Initial mass flow rate = exit mass flow rate.

Thus,

ρ1 = ρ2 = 1000 kg/m³

Density of water is 1000 kg/m³

And A1•V1 = A2•V2

So, V1 = A2•V2/A1

So, V1 = 0.28 x 1/0.07

V1 = 4 m/s

So, from initial equation of y components;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

Where F_y is vertical force of enlargement pressure and P2 = 0

Thus, making F_y the subject;

F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2

Plugging in the relevant values to get;

F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)

F_y = 151319.01N = 15.132 KN

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2 years ago

An adiabatic gas turbine expands air at 1300 kPa and 500◦C to 100 kPa and 127◦C. Air enters the turbine through a 0.2-m2 opening

Viktor [21]

Given:

Pressure, $P_{1}$ = 1300 kPa

Temperature, $T_{1}$ = $500^{\circ}$

$P_{2}$ = 100 kPa

$T_{2} = 127^{\circ}$

velocity, v = 40 m/s

A = 1 $m^{2}$

Solution:

For air propertiess at

$P_{1}$ = 1300 kPa

$T_{1}$ = $500^{\circ}$

$h_{1}$ = 793kJ/K

$v_{1}$ = $0.172\frac{m^{3}}{kg}$

and also at

$P_{2}$ = 100 kPa

$T_{2} = 127^{\circ}$

$h_{2}$ = 401 KJ/K

$v_{2}$ = $1.15\frac{m^{3}}{kg}$

a) Mass flow rate is given by:

$m' = \frac{Av}{v_{1}}$

Now,

$m = \frac{0.2\times 40}{0.172}$ = 46.51 kg/s

b) for the power produced by turbine, $P = m'(h_{1} - h_{2})$

$P = 46.51\times(793 - 401)$ = 18.231 MW

5 0

2 years ago