What are two advantages of forging when compared to machining a part from a billet?

Answers

Answer 1

Answer:

Answer:

Less material waste and time.

Explanation:

Two advantages of forging vs machining would be that with forging there is much less waste of material. With machining you remove a large amount of material turning into not so valuable chips.

There is also a time factor, as machining can be very time intensive. This depends on the speed of the machining, newer machines tend to be very fast, and forging requires a lengthy heating, but for large parts the machining can be excessively long.

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The compressibility factor provides a quick way to assess when the ideal gas law is valid. Use a solver to find the minimum temperature where the fluid has a vapor phase compressibility factor greater than 0.95 at 3 MPa. Report the value in oC, without units.

Answers

Answer:

The answer is

Explanation:

The compressibility factor

___________ are used in an automotive shop and can be harmful to the environment if not disposed of properly.Oil filters
Fluorescent lamps
Mercury-containing lamps
All of the above

Answers

Answer: D all above

Explanation:

Jus done it

A conical funnel of half-angle θ = 30 drains through a small hole of diameter d = 6:25 mm at the vertex. The speed of the liquid leaving the funnel is V= √ 2gy where y is the height of the liquid free surface above the hole. The funnel initially is filled to height y0 = 300 mm. Obtain an expression for the time, t, for the funnel to completely drain, and evaluate. Find the time to drain from 300 mm to 150 mm (a change in depth of 150 mm), and from 150 mm to completely empty (also a change in depth of 150 mm). Can you explain the discrepancy in these times?

Answers

Answer:

$3175.424*y^3 + 0.45153*y + 55.552*y^2 = t + 12.0347055$

t = 12.03

t = 81.473

velocity of the fluid decreases with level of fluid. Hence, in later stages the time taken is greater than in earlier stages

Explanation:

Given:

- The half angle θ = 30°

- The diameter of the small hole d = 6.25 mm

- The flow rate out of the funnel Q = A*√ 2gy

- The volume of frustum of cone is given by:

$V = (\pi )/(12)*y*(D^2 + d^2 + d*D})$

Where,

D: Is the larger diameter of the frustum

d: Is the smaller diameter of the frustum

y: The height of the liquid free surface from small diameter d base.

Find:

- Obtain an expression for the time, t, for the funnel to completely drain, and evaluate.

- Find the time to drain from 300 mm to 150 mm (a change in depth of 150 mm) and from 150 mm to completely empty (also a change in depth of 150 mm)

- Can you explain the discrepancy in these times?

Solution:

- We will use rate of change analysis by considering the rate of change of volume, and then apply the chain rule.

- The Volume of the frustum is a function of d , D , y. V = f ( d , D , y ). In our case the diameter of base d remains constant. Then we are left with:

V = f ( D , y )

- Determine a relationship between y and d. We will use half angle θ to determine the relationship between D and y by applying trigonometric ratios:

tan ( θ ) = D / 2*y

D = 2*y*tan ( θ )

- We developed a relationship between D and y in terms of half angle which remains constant. So the Volume is now only a function of one variable y:

V = f ( y )

- The volume of frustum of the cone can be written as:

$V = (\pi )/(12)*y*(D^2 + d^2 + d*D})$

Substituting the relationship for D in terms of y we have:

$V = (\pi )/(12)*y*(4*y^2*tan(Q) ^2 + d^2 + 2*d*y*tan(Q)})$

- Now by rate of change of Volume analysis we have:

dV / dt = [dV / dy] * [dy / dt]

- Computing dV / dy, where V = f(y) only:

$V = (\pi )/(12)*(4*y^3*tan(Q) ^2 + y*d^2 + 2*d*y^2*tan(Q)})\n\n(dV)/(dy) = (\pi )/(12)*(12*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q)})\n\n$

- Where, dV/dt = Volume flow rate:

$(dV)/(dt) = - Q\n(dV)/(dt) = - A*V\n(dV)/(dt) = - (\pi*d^2 )/(4) *√(2*g*y)$

- Then from Chain rule we have:

[dy / dt] = [dV / dt] / [dV / dy]

$(dy)/(dt) = (- (\pi*d^2 )/(4) *√(2*g*y))/((\pi )/(12)*(12*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q))\n\n} \n(dy)/(dt) = (-d^2 *√(2*g*y))/((4*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q))\n\n}$

- Separate variables:

$\frac {(4*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q)} {√(2*g*y)} .dy = {-d^2} .dt\n\n\frac { 1 } { √(2*g) }*(4*y^1^.^5*tan(Q) ^2 + y^-^0^.^5d^2 + 4*d*y^0^.^5*tan(Q))= {-d^2} .dt\n$

- Integrate both sides:

$\frac { 1 } { √(2*g) }*(4*y^1^.^5*tan(Q) ^2 + y^-^0^.^5d^2 + 4*d*y^0^.^5*tan(Q))= {-d^2} .dt\n\n\frac { 1 } { √(2*g) }*((8)/(5)*y^2^.^5*tan(Q) ^2 + 2*d^2*y^0^.^5 + (8)/(3)*d*y^1^.^5*tan(Q)) = -d^2*t + C\n\n\frac { 1 } { √(2*y*g) }*((8)/(5)*y^3*tan(Q) ^2 + 2*d^2*y + (8)/(3)*d*y^2*tan(Q)) = -d^2*t + C\n\n0.1204*y^3 + 0.000017638*y + 0.00217*y^2 = -0.0000390625*t + C\n\n3175.424*y^3 + 0.45153*y + 55.552*y^2 = -t + C$

- Evaluate @ t = 0 , y = 0.3 m

$3175.424*(0.15)^3 + 0.45153*(0.15) + 55.552*(0.15)^2 = 0 + C\n\nC = 12.0347055\n\n3175.424*y^3 + 0.45153*y + 55.552*y^2 = t + 12.0347055$

- Time taken from y = 300 to 150 mm:

$3175.424*0.15^3 + 0.45153*0.15 + 55.552*0.15^2 = -t + 90.871587\n\nt = 0 - -12.0347055\n\nt = 12.0347055 s\n$

- Time taken from y = 150 to 0 mm:

t = t_(0-300) - ( t_(0-150) = 90.871587 - 12.0347055 = 81.4721 s

- The discrepancy is time can be explained by the velocity of fluid coming out of bottom is function of y. The velocity of the fluid decreases as the level of fluid decreases hence the time taken from y =150 mm to 0 mm is larger than y = 300 mm to y = 150 mm.

Calculate the "exact" alkalinity of the water in Problem 3-2 if the pH is 9.43.Calculate the "approximate" alkalinity (in mg/L as CaCO3 ) of a water containing 120 mg/L of bicarbonate ion and 15 mg/L of carbonate ion.

Answers

Answer:

A) approximate alkalinity = 123.361 mg/l

B) exact alkalinity = 124.708 mg/l

Explanation:

Given data :

A) determine approximate alkalinity first

Bicarbonate ion = 120 mg/l

carbonate ion = 15 mg/l

Approximate alkalinity = ( carbonate ion ) * 50/30 + ( bicarbonate ion ) * 50/61

= 15 * (50/30) + 120*( 50/61 ) = 123.361 mg/l as CaCO3

B) calculate the exact alkalinity of the water if the pH = 9.43

pH + pOH = 14

9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57

[OH^- ] = 10^-4.57 = 2.692*10^-5 moles/l

[ OH^- ] = 2.692*10^-5 * 179/mole * 10^3 mg/g = 0.458 mg/l

[ H^+ ] = 10^-9.43 * 1 * 10^3 = 3.7154 * 10^-7 mg/l

therefore the exact alkalinity can be calculated as

= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )

= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )

= 124.708 mg/l

) A flow is divided into two branches, with the pipe diameter and length the same for each branch. A 1/4-open gate valve is installed in line A, and a 1/3-closed ball valve is installed in line B. The head loss due to friction in each branch is negligible compared with the head loss across the valves. Find the ratio of the velocity in line A to that in line B (include elbow losses for threaded pipe fittings).

Answers

Answer:

Va / Vb = 0.5934

Explanation:

First step is to determine total head losses at each pipe

at Pipe A

For 1/4 open gate valve head loss = 17 *Va^2 / 2g

elbow loss = 0.75 Va^2 / 2g

at Pipe B

For 1/3 closed ball valve head loss = 5.5 *Vb^2 / 2g

elbow loss = 0.75 * Vb^2 / 2g

Given that both pipes are parallel

17 *Va^2/2g + 0.75*Va^2 / 2g = 5.5 *Vb^2 / 2g + 0.75 * Vb^2 / 2g

∴ Va / Vb = 0.5934

Calculate the amount of power (in Watts) required to move an object weighing 762 N from point A to point B within 29 seconds. Distance between point A and point B is 5.0 meters and they are at the same level. (Ignore the frictional losses)