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3 years ago

What is the difference between a single-model production line and a mixed-model production line?

1 answer:

6 0

The unique model production line is responsible for producing identical pieces. For this purpose the balancing of the assembly line is only responsible for assembling a model throughout the line.

This is a considerable difference compared to the mixed model assembly line where many models are assembled during the same production line, that is, it produces parts or products that have slight changes accommodated in them, with slight variations in their model or products of soft variety

The choice of the type of production depends on the type of company and its own demand, always prioritizing the efficiency in the operation. Generally, the mixed model tends to be chosen when demand is very large and customer demand is required to be met. In others it is considered a plant model in which half of the line is mixed and the other one is the only model in order to keep the efficiency balanced.

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A helical compression spring is made with oil-tempered wire with wire diameter of 0.2 in, mean coil diameter of 2 in, a total of

Naya [18.7K]

Answer:

a. Solid length Ls = 2.6 in

b. Force necessary for deflection Fs = 67.2Ibf

Factor of safety FOS = 2.04

Explanation:

Given details

Oil-tempered wire,

d = 0.2 in,

D = 2 in,

n = 12 coils,

Lo = 5 in

(a) Find the solid length

Ls = d (n + 1)

= 0.2(12 + 1) = 2.6 in Ans

(b) Find the force necessary to deflect the spring to its solid length.

N = n - 2 = 12 - 2 = 10 coils

Take G = 11.2 Mpsi

K = (d^4*G)/(8D^3N)

K = (0.2^4*11.2)/(8*2^3*10) = 28Ibf/in

Fs = k*Ys = k (Lo - Ls )

= 28(5 - 2.6) = 67.2 lbf Ans.

c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.

For C = D/d = 2/0.2 = 10

Kb = (4C + 2)/(4C - 3)

= (4*10 + 2)/(4*10 - 3) = 1.135

Tau ts = Kb {(8FD)/(Πd^3)}

= 1.135 {(8*67.2*2)/(Π*2^3)}

= 48.56 * 10^6 psi

Let m = 0.187,

A = 147 kpsi.inm^3

Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi

Ssy = 0.50 Sut

= 0.50(198.6) = 99.3 kpsi

FOS = Ssy/ts

= 99.3/48.56 = 2.04 Ans.

7 0

3 years ago

A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

$E = G \times H$

Where G represents complex rated power and H is the inertia constant of turbo-generator.

$E = 100 \times 8 \\\\E = 800 \: MJ$

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}$

Where M is given by

$M = \frac{E}{180 \times f}$

$M = \frac{800}{180 \times 50}$

$M = 0.0889 \: MJ \cdot s/ elec \: \: deg$

So, the rotor acceleration is

$P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}$

$30 = 0.0889 \frac{d^2 \delta}{dt^2}$

$\frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}$

$\frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2$

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$\Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2$

Where t is given by

$1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2 \\\\t = 0.2 \: sec$

So,

$\Delta \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2$

$$ \Delta \delta = 6.75 \: elec \: deg$

The change in torque in rpm/s is given by

$\Delta \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ }$

$\Delta \delta =28.12 \: \: rpm/s$

The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by

$Rotor \: speed = \frac{120 \cdot f}{P} + (\Delta \delta)\cdot t$

Where P is the number of poles of the turbo-generator.

$Rotor \: speed = \frac{120 \cdot 50}{4} + (28.12)\cdot 0.2$

$Rotor \: speed = 1500 + 5.62$

$$ Rotor \: speed = 1505.62 \:\: rpm$

4 0

3 years ago

Which of the following are the same as 1545.5347

Reptile [31]

Explanation:

D the answer is D and anyone can answer this Question

5 0

2 years ago

Need help fast 50 points Project: Creating a Morphological Matrix

Leokris [45]

Answer:

The fundamental difference between effective and less effective matrix organizations is whether the tension between different perspectives is creative or destructive. While various processes, systems and tools can help, what matters most is what top leadership says and does and how that flows through the organization in shared targets, clear accountabilities, live team interactions and team-building transparency and behaviors.

Getting matrix management right is linked inextricably to an organization’s culture - the only sustainable competitive advantage. Key components of a culture can be grouped into behaviors, relationships, attitudes, values and the environment.

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My colleague Joe Durrett has worked all sides of matrix organizations in marketing at Procter & Gamble, sales and general management at Kraft General Foods and CEO of Information Resources, Broderbund Software and Advo. He has seen matrices at their best and at their worst and offered his perspective for this article along with his partners John Lawler and Linda Hlavac. The 12 ways to make matrix organizations more effective were built on their ideas.

Explanation:

7 0

2 years ago

A complete circuit includes all of the following except a/an

lbvjy [14]

The correct answer is ( C )

8 0

3 years ago