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iris [78.8K]
3 years ago
15

What is the role of the architects in modern development​

Engineering
1 answer:
pshichka [43]3 years ago
7 0

Answer:

to plan, design, and oversee the construction of a building

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Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe
NemiM [27]

Answer:

(a) attached below

(b) pf_{C}=0.85 lagging

(c) I_{C} =32.37 A

(d) X_{C} =49.37 Ω

(e) I_{cap} =9.72 A and I_{line} =27.66 A

Explanation:

Given data:

P_{1}=15 kW

S_{2} =10 kVA

pf_{1} =0.6 lagging

pf_{2}=0.8 leading

V=480 Volts

(a) Draw the power triangle for each load and for the combined load.

\alpha_{1}=cos^{-1} (0.6)=53.13°

\alpha_{2}=cos^{-1} (0.8)=36.86°

S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA

Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99 ≅ 20kVAR

P_{2} =S_{2}*pf_{2} =10*0.8=8 kW

Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99 ≅ -6 kVAR

The negative sign means that the load 2 is providing reactive power rather than consuming  

Then the combined load will be

P_{c} =P_{1} +P_{2} =15+8=23 kW

Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR

(b) Determine the power factor of the combined load and state whether lagging or leading.

S_{c} =P_{c} +jQ_{c} =23+14j

or in the polar form

S_{c} =26.92°

pf_{C}=cos(31.32) =0.85 lagging

The relationship between Apparent power S and Current I is

S=VI^{*}

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

I_{C} =S_{C}/\sqrt{3}*V

I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

Q_{C} =3*V^2/X_{C}

X_{C} =3*V^2/Q_{C}

X_{C} =3*(480)^2/14*10^3 Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is  

I_{cap} =V/X_{C} =480/49.37=9.72 A

Line current flowing from the source is

I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A

8 0
3 years ago
Whats viruses c liver?
PtichkaEL [24]

Answer:

Hepatitis C is a viral infection that causes liver inflammation, sometimes leading to serious liver damage. The hepatitis C virus (HCV) spreads through contaminated blood.

5 0
2 years ago
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Explain the difference between thermoplastics and thermosets giving structure property correlation.
Misha Larkins [42]

Answer:

Explanation:

Thermosetting polymers are infusible and insoluble polymers. The reason for such behavior is that the chains of these materials form a three-dimensional spatial network, intertwining with strong equivalent bonds. The structure thus formed is a conglomerate of interwoven chains giving the appearance and functioning as a macromolecule, which as the temperature rises, simply the chains are more compacted, making the polymer more resistant to the point where it degrades.

Macromolecules are molecules that have a high molecular mass, formed by a large number of atoms. Generally they can be described as the repetition of one or a few minimum units or monomers, forming the polymers. In contrast, a thermoplastic is a material that at relatively high temperatures, becomes deformable or flexible, melts when heated and hardens in a glass transition state when it cools sufficiently. Most thermoplastics are high molecular weight polymers, which have associated chains through weak Van der Waals forces (polyethylene); strong dipole-dipole and hydrogen bond interactions, or even stacked aromatic rings (polystyrene). Thermoplastic polymers differ from thermosetting polymers or thermofixes in that after heating and molding they can overheat and form other objects.

Thermosetting plastics have some advantageous properties over thermoplastics. For example, better resistance to impact, solvents, gas permeation and extreme temperatures. Among the disadvantages are, generally, the difficulty of processing, the need for curing, the brittle nature of the material (fragile) and the lack of reinforcement when subjected to tension. But even so in many ways it surpasses the thermoplastic.

The physical properties of thermoplastics gradually change if they are melted and molded several times (thermal history), these properties are generally diminished by weakening the bonds. The most commonly used are polyethylene (PE), polypropylene (PP), polybutylene (PB), polystyrene (PS), polymethylmethacrylate (PMMA), polyvinylchloride (PVC), ethylene polyterephthalate (PET), Teflon (or polytetrafluoroethylene, PTFE) and nylon (a type of polyamide).

They differ from thermosets or thermofixes (bakelite, vulcanized rubber) in that the latter do not melt when raised at high temperatures, but burn, making it impossible to reshape them.

Many of the known thermoplastics can be the result of the sum of several polymers, such as vinyl, which is a mixture of polyethylene and polypropylene.

When they are cooled, starting from the liquid state and depending on the temperatures to which they are exposed during the solidification process (increase or decrease), solid crystalline or non-crystalline structures may be formed.

This type of polymer is characterized by its structure. It is formed by hydrocarbon chains, like most polymers, and specifically we find linear or branched chains

4 0
3 years ago
While discussing VIN numbers, Technician A says that the first digit of the VIN identifies the country where the vehicle was man
ruslelena [56]
Usually the first digit of the vin id’s the country it was built. So technician A would be correct. That’s usually how it is. Hope this helps. Please let me know if this is incorrect
4 0
3 years ago
Which of the following would not be considered hot work? A chipping B soldering C
tankabanditka [31]
I believe the answer is D: brazing
Hope this helps you have a good night
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