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goldfiish [28.3K]
3 years ago
8

Which of the following statements do not correctly describe pull manufacturing? (1). Material flow is determined by the need of

the downstream workstation (2). Control is transferred from the beginning of the line to the end (3). A single piece flow is ideal for a push system and not for a pull system (4). Tries to minimize the number of items pulled at the same time Select one: a. (3) b. (1) and (3) c. (2) and (4) d. (4)
Engineering
1 answer:
weeeeeb [17]3 years ago
5 0

Answer: option B.(1) and (3)

(1). Material flow is determined by the need of the downstream workstation

(3). A single piece flow is ideal for a push system and not for a pull system

Explanation:

A pull system is a Lean manufacturing principle created to reduce waste in the production process. A pull system is to build products based on actual demand and not on forecasts.

One of the most important principles in the Lean Philosophy is the Pull Principle. The Pull principle is based on the assumption that one should only produce what is asked for by the customer. The Pull principle can be applied in more than just production, it can be applied in the office and even in life.

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Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi
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Answer:

\Delta V = 209.151\,L, \Delta C = 217.517\,USD

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Where C_{D} is the drag coefficient and A is the frontal area, respectively. The work loss due to drag forces is:

W = F_{D}\cdot \Delta s

The reduction on amount of fuel is associated with the reduction in work loss:

\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s

Where F_{D,1} and F_{D,2} are the original and the reduced frontal areas, respectively.

\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s

The change is work loss in a year is:

\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)

\Delta W = 2.043\times 10^{9}\,J

\Delta W = 2.043\times 10^{6}\,kJ

The change in chemical energy from gasoline is:

\Delta E = \frac{\Delta W}{\eta}

\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}

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The changes in gasoline consumption is:

\Delta m = \frac{\Delta E}{L_{c}}

\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }

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\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )

\Delta C = 217.517\,USD

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