8. 15 A manual arc welding cell uses a welder and a fitter. The cell operates 2,000 hriyr. The welder is paid $30/hr and the fitter is paid $25/hr. Both rates include applicable overheads. The cycle time to complete one welded assembly is 15. 4 min. Of this time, the arc-on time is 25%, and the fitter's participation in the cycle is 30% of the cycle time. A robotic arc welding cell is being considered to replace this manual cell. The new cell would have one robot, one fitter, and two workstations, so that while the robot is working at the first sta tion, the fitter is unloading the other station and loading it with new components. The fitter's rate would remain at $25/hr. For the new cell, the production rate would be eight welded assemblies per hour. The arc-on time would increase to almost 52%, and the fitter's participation in the cycle would be about 62%. The installed cost of the robot and worksta tions is $158,000. Power and other utilities to operate the robot and arc welding equipment will be $3. 80/hr, and annual maintenance costs are $3,500. Given a 3-year service life, 15% rate of return, and no salvage value, (a) determine the annual quantity of welded assem blies that would have to be produced to reach the break-even point for the two methods. (b) What is the annual quantity of welded assemblies produced by the two methods work. Ing 2,000 hryr?

Answers

Answer 1

Answer:

The annual quantity of welded assemblies that would have to be produced to reach the break-even point for the two methods is approximately 15,983.

To determine the break-even point between the manual arc welding cell and the robotic cell, we need to calculate the total costs for each method and then equate them.

For the manual arc welding cell:

Labor cost per hour = (welder's hourly rate x arc-on time) + (fitter's hourly rate x fitter's participation in the cycle) = ($30 x 0.25) + ($25 x 0.3) = $11.25

Labor cost per welded assembly = labor cost per hour x cycle time per assembly / 60 = $11.25 x 15.4 / 60 = $2.89

Overhead cost per welded assembly = (labor cost per hour x (1 - arc-on time - fitter's participation in the cycle)) x cycle time per assembly / 60 = ($30 x 0.45) x 15.4 / 60 = $4.68

Total cost per welded assembly = labor cost per welded assembly + overhead cost per welded assembly = $2.89 + $4.68 = $7.57

Total cost per hour = total cost per welded assembly x production rate = $7.57 x 8 = $60.56

Total cost per year = total cost per hour x hours of operation per year = $60.56 x 2,000 = $121,120

For the robotic arc welding cell:

Labor cost per hour = fitter's hourly rate x fitter's participation in the cycle = $25 x 0.62 = $15.50

Labor cost per welded assembly = labor cost per hour x cycle time per assembly / 60 = $15.50 x 15.4 / 60 = $3.97

Overhead cost per welded assembly = power and utility cost per hour + annual maintenance cost / production rate = $3.80 + $3,500 / (8 x 2,000) = $3.80 + $0.22 = $4.02

Total cost per welded assembly = labor cost per welded assembly + overhead cost per welded assembly + (installed cost / (production rate x service life)) = $3.97 + $4.02 + ($158,000 / (8 x 3)) = $3.97 + $4.02 + $6,208.33 = $14.19

Total cost per hour = total cost per welded assembly x production rate = $14.19 x 8 = $113.52

Total cost per year = total cost per hour x hours of operation per year = $113.52 x 2,000 = $227,040

To find the break-even point, we set the total cost of the manual arc welding cell equal to the total cost of the robotic arc welding cell and solve for the annualquantity of welded assemblies:

$121,120 + x($7.57) = $227,040 + x($14.19)

$7.57x - $14.19x = $227,040 - $121,120

$-6.62x = $105,920

x = $105,920 / $6.62

x = 15,982.7

Therefore, the annualquantity of welded assemblies that would have to be produced to reach the break-even point for the two methods is approximately 15,983.

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A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickness is 5 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 10 m2 and 0.15 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window

Answers

Answer:

The percentage loss of the window is $E = 97.3$ %

Explanation:

From the question we are told that

The area of pane of glass is $A = 0.15 m^2$

The thickness is $d = 5mm = (5)/(1000) = 0.005m$

The thickness of the wall is $D = 0.15m$

The area of the wall is $a = 10m^2$

Generally the heat lost as a result of conduction of the window is

$Q_(window) = (j_(glass) * A * (\Delta T) )/(d)$

Where $j_(glass)$ is the thermal conductivity of glass which has a constant value of

$j_(glass) = 0.80 J/(s \cdot m \cdot C^o)$

Substituting values

$Q_(window) = ( 0.80 * 0.15 * (\Delta T) )/(0.005)$

$Q_(window) = 24 \Delta T$

Generally the heat lost as a result of conduction of the wall is

$Q_(wall) = (j_(styrofoam) * A * (\Delta T) )/(d)$

$j_(styrofoam)$ s the thermal conductivity of Styrofoam which has a constant value of $j_(styrofoam) = 0.010J / (s \cdot m \cdot C^o)$

Substituting values

$Q_(wall) = ( 0.010 * 10 * (\Delta T) )/(0.15)$

$Q_(wall) = 0.667 \ \Delta T$

Now the net loss of heat is

$Q_(net) = Q_(window) + Q_(wall)$

Substituting values

$Q_(net) = 24 + 0.667$

$Q_(net) = 24.667$

Now the percentage loss by the window is

$E = (Q_(window) )/(Q_(net)) * 100$

Substituting value

$E = (24)/(24 .667) * 100$

$E = 97.3$ %

Reduce the following lambda-calculus term to the normalform. Show all intermediate steps, with one beta reduction at a time. In the reduction, assume that you are supplied with extra rules thatallow you to reduce the multiplication of two natural numbers into thecorresponding result.(λf.λx.f(f x))(λy.y≠3) 2

Answers

Answer:

Decrease to typical from utilizing lambda-decrease:

The given lambda - math terms is, (λf.λx.f(f(fx)))(λy.y×3)2

The of taking the terms is significant in lambda - math,

For the term, (λy, y×3)2, we can substitute the incentive to the capacity.

Therefore apply beta-decrease on “(λy, y×3)2,“ will return 2 × 3 = 6

Presently the tem becomes, (λf λx f(f(fx)))6

The main term, (λf λx f(f(fx))) takes a capacity and a contention and substitute the contention in the capacity.

Here it is given that it is conceivable to substitute, the subsequent increase in the outcome.

In this way by applying next level beta - decrease, the term becomes f(f(f(6))), which is in ordinary structure.

Considering only (110), (1 1 0), (101), and (10 1 ) as the possible slip planes, calculate the stress at which a BCC single crystal will yield if the critical resolved shear stress is 50 MPa and the load is applied in the [100] direction.

Answers

Solution :

i. Slip plane (1 1 0)

Slip direction -- [1 1 1]

Applied stress direction = ( 1 0 0 ]

τ = 50 MPa ( Here slip direction must be perpendicular to slip plane)

τ = σ cos Φ cos λ

$\cos \phi = ((1,0,0) \cdot (1,1,0))/(1 * \sqrt2)$

$=(1)/(\sqrt2 )$

$\cos \lambda = ((1,0,0) \cdot (1,-1,1))/(1 * \sqrt3)$

$=(1)/(\sqrt3 )$

τ = σ cos Φ cos λ

∴ $50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3)$

σ = 122.47 MPa

ii. Slip plane --- (1 1 0)

Slip direction -- [1 1 1]

$\cos \phi = ((1, 0, 0) \cdot (1, -1, 0))/(1 * \sqrt2) =(1)/(\sqrt2)$

$\cos \lambda = ((1, 0, 0) \cdot (1, 1, -1))/(1 * \sqrt3) =(1)/(\sqrt3)$

τ = σ cos Φ cos λ

∴ $50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3)$

σ = 122.47 MPa

iii. Slip plane --- (1 0 1)

Slip direction --- [1 1 1]

$\cos \phi = ((1, 0, 0) \cdot (1, 0, 1))/(1 * \sqrt2) =(1)/(\sqrt2)$

$\cos \lambda = ((1, 0, 0) \cdot (1, 1, -1))/(1 * \sqrt3) =(1)/(\sqrt3)$

τ = σ cos Φ cos λ

∴ $50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3)$

σ = 122.47 MPa

iv. Slip plane -- (1 0 1)

Slip direction ---- [1 1 1]

$\cos \phi = ((1, 0, 0) \cdot (1, 0, -1))/(1 * \sqrt2)=(1)/(\sqrt2)$

$\cos \lambda = ((1, 0, 0) \cdot (1, -1, 1))/(1 * \sqrt3) =(1)/(\sqrt3)$

τ = σ cos Φ cos λ

∴ $50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3)$

σ = 122.47 MPa

∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0

A horizontal curve on a single-lane highway has its PC at station 1+346.200 and its PI at station 1+568.70. The curve has a superelevation of 6.0% and is designed for 120 km/h. The limiting value for coefficient of side friction at 120 km/h is 0.09. What is the station of the PT? Remember that 1 metric station = 1000 m.

Answers

Answer:

The solution is given in the attachments.

For a p-n-p BJT with NE 7 NB 7 NC, show the dominant current components, with proper arrows, for directions in the normal active mode. If IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA, calculate the base transport factor, emitter injection efficiency, common-base current gain, common-emitter current gain, and ICBO. If the minority stored base charge is 4.9 * 10-11 C, calculate the base transit time and lifetime.

Answers

Answer:

=> base transport factor = 0.98.

=> emitter injection efficiency = 0.99.

=> common-base current gain = 0.97.

=> common-emitter current gain = 32.34.

=> ICBO = 1 × 10^-6 A.

=> base transit time = 0.325.

=> lifetime = 1.875.

Explanation:

(Kindly check the attachment for the diagram showing the dominant current components, with proper arrows, for directions in the normal active mode).

The following parameters or data are given for a p-n-p BJT with NE 7 NB 7 NC and they are: IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA.

(1). The base transport factor = ICp/IEp=9.8/10 = 0.98.

(2). emitter injection efficiency =IEp/ IEp + ICn = 10/10 + 0.1 = 0.99.

(3).common-base current gain = 0.98 × 0.99 = 0.9702.

(4).common-emitter current gain =0.97 / 1- 0.97 = 32.34.

(5). Icbo = Ico = 1 × 10^-6 A.

(6). base transit time = 1248 × 10^-2 × (1.38× 10^-23/1.603 × 10^-19). = 0.325.

(7).lifetime;

= > 2 = √0.325 + √ lifetime.

= Lifetime = 2.875.

Kim is working on the cost estimate and feasible design options for a building. Which stage of a construction plan is Kim working on now? A. design development B. schematic design C. mechanical D. structural

Answers

Kim is working on the cost estimate. The stage of a construction plan is Kim working on now is schematic design. The correct option is B.

What is a schematic design?

A schematic design is an outline of a house or a building or another construction thing. The schematic design makes the outline map of the exterior or interior of the building. It is the foremost phase of designing something.

The design expert discusses the project three-dimensionally at this point in the process. To define the character of the finished project and an ideal fulfillment of the project program, a variety of potential design concepts are investigated.

The schematic design consists of a rough sketch with markings and measurements.

Therefore, the correct option is B. schematic design.

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Answer:

B. schematic design

Explanation:

This correct for Plato/edmentum

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