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sladkih [1.3K]
2 years ago
10

What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a

feed of 0.02 ipr? What is the cutting time?
Engineering
1 answer:
Studentka2010 [4]2 years ago
8 0

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

so we substitute

MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

Next we find the allowance for approach of the tip of the drill

A = D/2

A = 2/2

= 1 in

now find the time required to drill the hole

Tm = (L + A) / (Fr × Ns)

Lis the the depth of the hole( 3.5 in)

so we substitute our values

Tm = (3.5 + 1) / (0.02 × 229.18  )

Tm = 4.5 / 4.5836

Tm = 0.98 min

Therefore the cutting time is 0.98 min

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bonufazy [111]

Answer:

The total force resisted by the flange bolts is  163.98 N

Explanation:

Solution

The first step is to find  the pipe cross section at the inlet section

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Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²

=50.265 cm² * ( 1 m²/100² cm²)

= 5.0265 * 10^⁻³ m²

Secondly, we find cross section area of  the pipe at the inlet section

A₂ = π /4 D₂²

D₂ =  diameter of the pipe at the inlet section

Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²

= 19.63 cm² * ( 1 m²/100² cm²)

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Now,

we write down the conversation mass relation which is stated as follows:

Q₁ = Q₂

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We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂

So,

V₁ and V₂ are defined as the velocities at the inlet and exit

We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂

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8 0
2 years ago
A 5-mm-thick stainless steel strip (k = 21 W/m•K, rho = 8000 kg/m3, and cp = 570 J/kg•K) is being heat treated as it moves throu
Drupady [299]

Answer:

The temperature of the strip as it exits the furnace is 819.15 °C

Explanation:

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L_c = \frac{V}{A} = \frac{LA}{2A} = \frac{5*10^{-3}}{2} = 0.0025 \ m

The Biot number is given as;

B_i = \frac{h L_c}{k}\\\\B_i = \frac{80*0.0025}{21} \\\\B_i = 0.00952

B_i < 0.1,  thus apply lumped system approximation to determine the constant time for the process;

\tau = \frac{\rho C_p V}{hA_s} = \frac{\rho C_p L_c}{h}\\\\\tau = \frac{8000* 570* 0.0025}{80}\\\\\tau = 142.5 s

The time for the heating process is given as;

t = \frac{d}{V} \\\\t = \frac{3 \ m}{0.01 \ m/s} = 300 s

Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;

T(t) = T_{ \infty} + (T_i -T_{\infty})e^{-t/ \tau}\\\\T(t) = 930 + (20 -930)e^{-300/ 142.5}\\\\T(t) = 930 + (-110.85)\\\\T_{(t)} = 819.15 \ ^0 C

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5 0
3 years ago
What's a disadvantage of highest MERV-rated filters?
Arte-miy333 [17]

Answer:

3) the pressure drop across high MERV filters is significant.

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MERV (Minimum-Efficiency Reporting Value) is used to measure the efficiency of filter to remove particles. A filter of high MERV can filter smaller particles but this causes an increase in reduced air flow that is an increase in pressure drop. High MERV filters capture more particles causing them to get congested faster and thereby increasing pressure drop.

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3 0
3 years ago
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Goshia [24]

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4 0
2 years ago
A seamless pipe 800mm diameter contains a fluid under a pressure of 2N/mm2. If the permissible tensile stress is 100N/mm2, find
Bad White [126]

Answer:

8 mm

Explanation:

Given:

Diameter, D = 800 mm

Pressure, P = 2 N/mm²

Permissible tensile stress, σ = 100 N/mm²

Now,

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\sigma=\frac{\textup{PD}}{\textup{2t}}

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on substituting the respective values, we get

100=\frac{\textup{2\times800}}{\textup{2t}}

or

t = 8 mm

Hence, the minimum thickness of pipe is 8 mm

3 0
2 years ago
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