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3 years ago

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In a surface grinding operation, the wheel diameter = 8.0 in, wheel width = 1.0 in, wheel speed = 6000 ft/min, work speed = 40 f

t/min, infeed = 0.002 in, and crossfeed = 0.20 in. The number of active grits per square inch of wheel surface = 400. Determine
(a) the average length per chip,
(b) the metal removal rate
(c) the number of chips formed per unit and the average volume per chip?
(d) If the tangential cutting force on the workpiece is 50 lb, what is the specific energy calculated for this job?

1 answer:

Katen [24]3 years ago

7 0

Answer: the answer will be d because it is the right one to be

Explanation:

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1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

a) The total number of users that can be accomodated in the system is:

$n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )$

$n = 1000\,users$

b) The length of the side of each cell is:

$l = \sqrt{1\,km^{2}}$

$l = 1\,km$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

c) The total number of users that can be accomodated in the system is:

$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

The length of each side of the cell is:

$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

8 0

3 years ago

A cylindrical specimen of steel has an original diameter of 12.8 mm. It is tested in tension its engineering fracture strength i

Mama L [17]

Answer:

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

Given that;

Original diameter $d_{o}$ = 12.8 mm

Final diameter $d_{f}$ = 10.7

Engineering stress $\alpha _{E}$ = 460 Mpa

a) determine The ductility in terms of percent reduction in area;

Ai = π/4( $d_{o}$ )² ; Ag = π/4( $d_{f}$ )²

% = π/4 [ ( ( $d_{f}$ )² - ( $d_{o}$ )²) / ( π/4 ( $d_{o}$ )²) ]

= ( ( $d_{f}$ )² - ( $d_{o}$ )²) / ( $d_{o}$ )² × 100

we substitute

= [( (10.7)² - (12.8)²) / (12.8)² ] × 100

= [(114.49 - 163.84) / 163.84 ] × 100

= - 0.3012 × 100

= -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress $\alpha _{T}$ = $\alpha _{E}$ ( 1 + $E_{E}$ )

$E_{E}$ is engineering strain

$E_{E}$ = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49

$E_{E}$ = 0.431

so we substitute the value of $E_{E}$ into our initial equation;

True stress $\alpha _{T}$ = 460 ( 1 + 0.431)

True stress $\alpha _{T}$ = 460 (1.431)

True stress $\alpha _{T}$ = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

6 0

2 years ago

The ice on the rear window of an automobile is defrosted by attaching a thin, transparent, film type heating element to its inne

pshichka [43]

Answer:

A)Q = 1208.33 W/m²

B)K = 0.138 W/m.K

Explanation:

We are given;

inside air temperature;T_∞,i =25 °C = 25 + 273 = 298K

outside air temperature;T_∞,o = -10°C = - 10 + 273 = 263K

Inner surface temperature;T_s,i = 15 °C = 15 + 273 = 288K

Thickness, L = 4mm = 0.004m

convection heat transfer coefficient ; hi = 25 W/(m².K)

A) From an energy balance at the inner surface and the thermal circuit, the electric power required per unit window area is given as;

Q = [(T_s,i - T_∞,o)/((L/k) + (1/hi))] - [(T_∞,o - T_s,i)/(1/hi)]

Plugging in the relevant values with k for glass as 1.4 W/m.k, we have;

Q = [(288 - 263)/((0.004/1.4) + (1/25))] - [(263 - 288)/(1/25)]

Q = 583.33 + 625

Q = 1208.33 W/m²

B) The formula for thermal conductivity is;

K = (QL)/(AΔT)

Where;

K is the thermal conductivity in W/m.K

Q is the amount of heat transferred through the material

L is the distance between the two isothermal planes

A is the area of the surface in square meters

ΔT is the difference in temperature in Kelvin

ΔT = 298K - 263K = 35K

Now, since we have value of heat per unit area to be Q = 1208.33 W/m², let's rearrange the equation to reflect that; Thus ;

k = (Q/A) x (L/ΔT)

K = 1208.33 x (0.004/35)

K = 0.138 W/m.K

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3 years ago

"A communication between two devices is over the maximum limit of an ethernet frame size. The Transmission Control Protocol (TCP

Sliva [168]

COmmunication Devices

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TCP/IP, or the Transmission Control Protocol/Internet Protocol, is a suite of communication protocols used to interconnect network devices on the internet. TCP/IP can also be used as a communications protocol in a private network (an intranet or an extranet).

The sequence number in a header is used to keep track of which segment out of many this particular segment might be. The next field, the acknowledgment number, is a lot like the sequence number. The acknowledgment number is the number of the next expected segment.

The 32 bit sequence number field defines the number assigned to the first byte of data contained in this segment. TCP is a stream transport protocol. To ensure connectivity, each byte to be transmitted is numbered.

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Bending is defined as? A. the application of a load tending to distort a member in one direction. B. the application of opposing

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2 years ago