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nalin [4]
3 years ago
7

A three-phase transformer bank consists of 3 single-phase transformers to handle 400 kVA witha 34.5kV/13.8kV voltage ratio. Find

the rating of each individual single-phase transformer in the bank (high voltage, low voltage, turns ratio, and apparent power) for the following transformer connections:
Engineering
1 answer:
Contact [7]3 years ago
6 0

Answer:

(a) high voltage = 19.9 kV, low voltage = 7.97 kV, turns ratio = 2.50:1, apparent power = 133 kVA

(b) high voltage = 19.9 kV, low voltage = 13.8 kV, turns ratio = 1.44:1, apparent power = 133 kVA

(c) high voltage = 34.5 kV, low voltage = 7.97 kV, turns ratio = 4.33:1, apparent power = 133 kVA

(d) high voltage = 34.5 kV, low voltage = 13.8 kV, turns ratio = 2.50:1, apparent power = 133 kVA

Explanation:

The turn ratio can be estimated by taking the ratio of the values of the high voltage and the low voltage. For example, the turn ratio for (a) is 19.9 kV/7.97 kV = 2.50:1. Similarly, the apparent power for each transformer is calculated as 400 kVA/3 = 133 kVA. Furthermore, the high and low voltages for the Δ are the same for the three-phase transformer i.e. 34.5 kv/13.8 kV. The high and high voltages for Y connection is calculated as 34.5 kV/1.73 = 19.9 kV and 13.8 kV/1.73 = 7.97 kV.

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When you come to an intersection, follow the _________ before you proceed.
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(20 points) A 1 mm diameter tube is connected to the bottom of a container filled with water to a height of 2 cm from the bottom
zzz [600]

Solution :

Given :

h = 2 cm

Diameter of the tube , d = 1 mm

Diameter of the hose, D = 6 mm

Between 1 and 2, by applying Bernoulli's principle, we get

As point 1 is just below the free surface of liquid, so

$P_1=P_{atm} \text{ and} \ V_1=0$

$\frac{P_{atm}}{\rho g}+\frac{v_1^2}{2g} +h = \frac{P_2}{\rho g}$

$\frac{101.325}{1000 \times 9.81}+0.02 =\frac{P_2}{\rho g}$

$P_2 = 111.35 \ kPa$

Therefore, 111.325 kPa is the gas supply pressure required to keep the water from leaking back into the tube.

Velocity at point 2,

$V_2=\sqrt{\left(\frac{111.135}{\rho g}+0.02}\right)\times 2g$

   = 1.617 m/s

Flow of water,  $Q_2 = A_{tube} \times V_2$

                               $=\frac{\pi}{4} \times (10^{-3})^2 \times 1.617 $

                               $1.2695 \times 10^{-6} \ m^3/s$

Minimum air flow rate,

$Q_2 = Q_3 = A_{hose} \times V_3$

$V_3 = \frac{Q_2}{\frac{\pi}{4}D^2}$

$V_3 = \frac{1.2695 \times10^{-6}}{\pi\times 0.25 \times 36 \times 10^{-6}}$

    = 0.0449 m/s

b). Reynolds number in hose,

$Re = \frac{\rho V_3 D}{\mu} = \frac{V_3 D}{\nu}$

υ for water at 25 degree Celsius is $8.9 \times 10^{-7} \ m^2/s$

υ for air at 25 degree Celsius is $1.562 \times 10^{-5} \ m^2/s$

$Re_{hose}=\frac{0.0449 \times 6 \times 10^{-3}}{1.562 \times 10^{-5}}$

           = 17.25

Therefore the flow is laminar.

Reynolds number in the pipe

$Re = \frac{V_2 d}{\nu} = \frac{1.617 \times 10^{-3}}{8.9 \times 10^{-7}}$

                = 1816.85, which is less than 2000.

So the flow is laminar inside the tube.

3 0
3 years ago
In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by
Stells [14]

Answer:

\dfrac{\bar{h}}{h}=\dfrac{5}{4}

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Given that

Nu_x=0.035Re_x^{0.8} Pr^{1/3}

We know that

Rex=ρvx/μ

So

Nu_x=0.035Re_x^{0.8} Pr^{1/3}

Nu_x=0.035\times\left(\dfrac{\rho vx}{\mu}\right)^{0.8}Pr^{1/3}

All other quantities are constant only x is a variable in the above equation .so lets take all other quantities as a constant C

Nu_x=C.x^{0.8}=C.x^{4/5}

We also know that

Nux=hx/K

C.x^{4/5}=\dfrac{hx}{k}

m is the constant

h=mx^{-1/5}

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The average value of h given as

\bar{h}=\dfrac{\int_{0}^{L}hdx}{L}

\bar{h}=\dfrac{5m}{4}\times\dfrac{L^{4/5}}{L}

\bar{h}=\dfrac{5m}{4}L^{-1/5}             ---------1

The value of local heat transfer coefficient at x=L

h=mx^{-1/5}

h=mL^{-1/5}            -----------2

From 1 and 2 we can say that

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8 0
4 years ago
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rusak2 [61]

Answer:

f = 0.04042

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e = 10 m

change in  P = 235 N/m²

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R = 188.9 Nm/kgk

we solve this using this formula;

P = ρcos*R*T

we put in the values into this equation

600x10³ = ρcos * 188.9 * 273

600000 = ρcos51569.7

ρcos = 600000/51569.7

=11.63

from here we find the head loss due to friction

Δp/pg = feμ²/2D

235/11.63 = f*10*4/2*40x10⁻³

20.21 = 40f/0.08

20.21*0.08 = 40f

1.6168 = 40f

divide through by 40

f = 0.04042

5 0
3 years ago
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