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alexdok [17]
3 years ago
5

why is it important for the client and the engineering team to maintain open communications even after the delivery of the produ

ct
Engineering
1 answer:
goblinko [34]3 years ago
3 0

Answer:

It is important for the client and the engineering team to maintain open communications even after the deliver of the product because

1) One of doctrine of Six Sigma which is a tool used to improve processes requires the input of effort meant to attain a stable process that can be easily predicted and geared towards the reduction of variation in the process is a pre requisite for the business to succeed

2) Communication can also help in product monitoring and control as well as ensuring that the product performance requirements are met

3) Communication between the client and the engineering team can serve as a means of reporting failures and collecting data for analysis

Explanation:

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Air-conditioners consume a significant amount of electrical energy in buildings. Split air conditioner is a unitary system where
nydimaria [60]

Answer:

Evaporator,Compressor,Condensor ,Expanding valve

Explanation:

Split air conditioning :

  Split air conditioning means that, condensor unit or some time called outdoor unit is split from evaporator.It means that evaporator and condensor are placed at some distance.

The four component of split air conditioning system are as follows

1.Evaporator

 It absorb heat from room and produces the cooling effect.

2.Compressor

 It compresses the refrigerant which exits from evaporator.

3.Condensor

 It rejects the heat and cool the evaporator.

4.Expanding valve

  It allows to refrigerant to cool up to evaporator pressure.

6 0
3 years ago
1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil
kakasveta [241]

Answer:

a) n = 1000\,users, b)\Delta t_{min} = \frac{1}{30}\,h, \Delta t_{max} = \frac{\sqrt{2} }{30}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h, c) n = 10000000\,users, \Delta t_{min} = \frac{1}{3000}\,h, \Delta t_{max} = \frac{\sqrt{2} }{3000}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

Explanation:

a) The total number of users that can be accomodated in the system is:

n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )

n = 1000\,users

b) The length of the side of each cell is:

l = \sqrt{1\,km^{2}}

l = 1\,km

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{30}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{30}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h

c) The total number of users that can be accomodated in the system is:

n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )

n = 10000000\,users

The length of each side of the cell is:

l = \sqrt{100\,m^{2}}

l = 10\,m

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{3000}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

8 0
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The Fisher effect says that _______ . Group of answer choices the nominal interest rate adjusts one for one with the inflation r
Greeley [361]

Answer:

what wrong subject

Explanation:

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Ruler game, HELPPPPP
viktelen [127]
D! :D
Hope I helped!!
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2 years ago
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Heres a question that needs to be answered fast im running out of time
earnstyle [38]

Answer:

9

Explanation:

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3 years ago
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