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3 years ago

A delta connection has a voltage of 560 V connected to it. How much voltage is dropped across each phase

Physics

1 answer:

SashulF [63]3 years ago

5 0

Answer:

E_Phase = 560V

Explanation:

The computation of the voltage i.e. dropped across each phase is shown below:

Given that

The delta connection line voltage is

E_line = 560 V

And, in the case of delta connection, the line voltage would be equivalent to the phase voltage

That means

E_Phase = E_Line

= 560 V

Hence, the voltage i.e. dropped across each phase is

E_Phase = 560V

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Where:

P is pressure, V is Velocity, g is gravity, h is height and β is density (for water β=1000 kg/m3); at the points 1 and 2 respectively.

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First, we have to consider that from a mass balance, the flow is the same, so Q1=Q2, what changes, is the velocity. Knowing this, let's write the areas, diameters, density and flow on International Units System (S.I.), because the exercise is asking us the answer in Newtons.

$D_{1}=1.5 inches=0.0381 mts\\D_{2}=1 inches=0.0254 mts\\A_{1}=\frac{\pi D_{1}^{2} }{4}=0.00114mts^{2}\\A_{2}=\frac{\pi D_{2}^{2} }{4}=0.000507mts^{2}\\\beta=1000 kgs/m^{3}\\Q=200gpm=0.01mts^{3}/seg$

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