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3 years ago

A plastic molding machine produces a product whose annual demand is in the millions.

Engineering

1 answer:

BabaBlast [244]3 years ago

8 0

Answer:

a. 78.4 pieces/hr

b. $0.1806/min

c. $1.34/piece

Explanation:

(a) With a cycle time Tc = 45 sec = 0.75 min.

Production rate, Rp = 60/0.75 = 80 pieces/hr.

Let us factor in the 98% proportion uptime, so Production rate, Rp = 0.98*(80) = 78.4 pieces/hr

Quantity of product annualy = 6000 *(78.4) = 470,400 pieces/yr

(b) Equipment cost rate, Ceq = 500,000(1.30)/(60 x 10 x 6000) = $0.1806/min.

Cost rate of labour ,CL = 18.00(0.20) = $3.60/hr = $0.06/min

Conclusively, cost per piece, Cpc = 1.20(0.88) + (0.06 + 0.1806)(0.75) + 0.10 = $1.34/piece

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Len [333]

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3 years ago

Air is compressed isothermally from 13 psia and 55°F to 80 psia in a reversible steady-flow device. Calculate the work required,

solong [7]

Answer:64.10 Btu/lbm

Explanation:

Work done in an isothermally compressed steady flow device is expressed as

Work done = P₁V₁ In { P₁/ P₂}

Work done=RT In { P₁/ P₂}

where P₁=13 psia

P₂= 80 psia

Temperature =°F Temperature is convert to °R

T(°R) = T(°F) + 459.67

T(°R) = 55°F+ 459.67

=514.67T(°R)

According to the properties of molar gas, gas constant and critical properties table, R which s the gas constant of air is given as 0.06855 Btu/lbm

Work = RT In { P₁/ P₂}

0.06855 x 514.67 In { 13/ 80}

=0.06855 x 514.67 In {0.1625}

= 0.06855 x 514.67 x -1.817

=- 64.10Btu/lbm

The required work therefore for this isothermal compression is 64.10 Btu/lbm

8 0

2 years ago

Vending machine controller (adapted from Katz, "Contemporary Logic Design") Design and implement a finite state machine that con

babunello [35]

Answer:

Check the explanation

Explanation:

A vending machine controller is that type of machine that comes with a single serial port on the same chip as the programmable processor. The controller comprises of a port arbitrator that selectively attaches or links one of a number of serially communicating devices to this single serial port.

Kindly check the attached image to get the step by step explanation to the above question.

6 0

3 years ago

The 10 foot wide circle quarter gate AB is articulated at A. Determine the contact force between the gate and the smooth surface

slamgirl [31]

Answer:

$F = 641,771.52 \dfrac{lb-ft}{s^2}$

Explanation:

Given that

R=8 ft

Width= 10 ft

We know that hydro statics force given as

F=ρ g A X

ρ is the density of fluid

A projected area on vertical plane

X is distance of center mass of projected plane from free surface of water.

Here

X=8/2 ⇒X=4 ft

A=8 x 10=80 $ft^2$

So now putting the values

F=ρ g A X

F=62.4(32.14)(80)(4)

$F = 641,771.52 \dfrac{lb-ft}{s^2}$

4 0

3 years ago

Close to 16 billion pounds of ethylene glycol (EG) were produced in 2013. It previously ranked as the twenty-sixth most produced

bekas [8.4K]

Answer:

a) 0.684

b) 0.90

Explanation:

Catalyst

EO + W → EG

a) calculate the conversion exiting the first reactor

CAo = 16.1 / 2 mol/dm^3

Given that there are two stream one contains 16.1 mol/dm^3 while the other contains 0.9 wt% catalyst

Vo = 7.24 dm^3/s

Vm = 800 gal = 3028 dm^3

hence Im = Vin/ Vo = (3028 dm^3) / (7.24dm^3/s) = 418.232 secs = 6.97 mins

next determine the value of conversion exiting the reactor ( Xai ) using the relation below

KIm = $\frac{Xai}{1-Xai}$ ------ ( 1 )

make Xai subject of the relation

Xai = KIm / 1 + KIm --- ( 2 )

where : K = 0.311 , Im = 6.97 ( input values into equation 2 )

Xai = 0.684

B) calculate the conversion exiting the second reactor

CA1 = CA0 ( 1 - Xai )

therefore CA1 = 2.5438 mol/dm^3

Vo = 7.24 dm^3/s

To determine the value of the conversion exiting the second reactor ( Xa2 ) we will use the relation below

XA2 = ( Xai + Im K ) / ( Im K + 1 ) ----- ( 3 )

where : Xai = 0.684 , Im = 6.97, and K = 0.311 ( input values into equation 3 )

XA2 = 0.90

4 0

3 years ago