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Dominik [7]
3 years ago
10

What are the advantages of using a network?

Engineering
1 answer:
Rudiy273 years ago
5 0

Answer:

Sharing devices such as printers saves money.

Site (software) licences are likely to be cheaper than buying several standalone licences.

Files can easily be shared between users.

Network users can communicate by email and instant messenger.

Security is good - users cannot see other users' files unlike on stand-alone machines.

Data is easy to backup as all the data is stored on the file server.

I hope I helped

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Three parallel three-phase loads are supplied from a 480V (line-line RMS), 60 Hz three-phase supply. The loads are as follows: L
Travka [436]

Answer:

The total system active power P = P_1 + P_2 + P_3 = 34.91 KW

Explanation:

Load 1: Active power P_1 = 20 HP = 14.91kW;

Reactive power Q1 = P tan(\phi)

                               = 14.91\times tan(cos^{-}0.8) = 11.18 kvar


Load 2: Active power P_2 = 20 kW;

Reactive power Q2 = 0 since the load is purely resistive.

Load 3: Active power P_3 = 0 due to purely capacitiveload

           Reactive power Q_3 = -20 Var

a) since all three loads are connected in parallel therefore

    The total system active power P = P_1 + P_2 + P_3 = 34.91 KW

Total system reactive power Q = Q_1 + Q_2 + Q_3 = 11.18 + 0 -20 = -8.82 kVar

Since Q = 0, the power factor is unity.

Supply current per phase is given by

I = \frac{P}{\sqrt{3}V_{L}}

= \frac{34910}{\sqrt{3}\times 480} = 41.99 A

5 0
3 years ago
A permanent magnet DC motor has an armature resistance of 0.5 Ω and when a voltage of 120 V is applied to the motor it reaches a
BaLLatris [955]

Answer:

12 N-m

Explanation:

The dc motor is operating at 24 V that is its terminal voltage V =24 V

Armature resistance  = 0.2 ohm

No load speed = 240 radian /sec

For motor we know that  as the motor is on no load so  so

Power developed in the motor

Now we know that power = torque× angular speed

So

3 0
4 years ago
Thermal energy storage systems commonly involve a packed bed of solid spheres, through which a hot gas flows if the system is be
hammer [34]

Answer:

A) i) 984.32 sec

ii) 272.497° C

B) It has an advantage

C) attached below

Explanation:

Given data :

P = 2700 Kg/m^3

c = 950 J/kg*k

k = 240 W/m*K

Temp at which gas enters the storage unit  = 300° C

Ti ( initial temp of sphere ) = 25°C

convection heat transfer coefficient ( h ) = 75 W/m^2*k

<u>A) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>

First step determine the Biot Number

characteristic length( Lc ) = ro / 3 = 0.0375 / 3 = 0.0125

Biot number ( Bi ) = hLc / k = (75)*(0.0125) / 40 = 3.906*10^-3

Given that the value of the Biot number is less than 0.01 we will apply the lumped capacitance method

attached below is a detailed solution of the given problem

<u>B) The physical properties are copper</u>

Pcu = 8900kg/m^3)

Cp.cu = 380 J/kg.k

It has an advantage over Aluminum

C<u>) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>

Given that:

P = 2200 Kg/m^3

c = 840 J/kg*k

k = 1.4 W/m*K

3 0
3 years ago
The primary difference between LEED Certification for Buildings and LEED Professional Certifications is that professionals can c
TEA [102]

Answer:false

Explanation:

8 0
3 years ago
A Carnot refrigeration cycle absorbs heat at -12 °C and rejects it at 40 °C. a)-Calculate the coefficient of performance of this
tresset_1 [31]

Answer:

a)COP=5.01

b)W_{in}=2.998 KW

c)COP=6.01

d)Q_R=17.99 KW

Explanation:

Given

T_L= -12°C,T_H=40°C

For refrigeration

  We know that Carnot cycle is an ideal cycle that have all reversible process.

So COP of refrigeration is given as follows

COP=\dfrac{T_L}{T_H-T_L}  ,T in Kelvin.

COP=\dfrac{261}{313-261}

a)COP=5.01

Given that refrigeration effect= 15 KW

We know that  COP=\dfrac{RE}{W_{in}}

RE is the refrigeration effect

So

5.01=\dfrac{15}{W_{in}}

b)W_{in}=2.998 KW

For heat pump

So COP of heat pump is given as follows

COP=\dfrac{T_h}{T_H-T_L}  ,T in Kelvin.

COP=\dfrac{313}{313-261}

c)COP=6.01

In heat pump

Heat rejection at high temperature=heat absorb at  low temperature+work in put

Q_R=Q_A+W_{in}

Given that Q_A=15KW

We know that  COP=\dfrac{Q_R}{W_{in}}

COP=\dfrac{Q_R}{Q_R-Q_A}

6.01=\dfrac{Q_R}{Q_R-15}

d)Q_R=17.99 KW

5 0
4 years ago
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