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3 years ago

The phase sequence of a 3-phase system for which VAN = 120 /90o V and VBN = 120 /210o V is:_______

Engineering

1 answer:

Vadim26 [7]3 years ago

4 0

Answer:

b. abc

Explanation:

The phase rotation and its sequence is order in which voltage is reached to their respective sources. The wave forms of a polyphase flow through AC source to the panel. The three phase system has two possible sequence which can be a-b-c or c-b-a. The correct answer is therefore b.

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Write a function: def solution (S) that, given a string S of letters "L" and "R", denoting the types of shoes in line (left or r

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Answer:

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3 years ago

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An electromagnet is formed when a coil of wire wrapped around an iron core is hooked up to a dry cell battery. The current trave

german

Answer:

true

Explanation:

true An electromagnet is formed when a coil of wire wrapped around an iron core is hooked up to a dry cell battery. The current traveling through the wire sets up a magnetic field around the wire. TRUE or FALSE

3 0

3 years ago

Which of these is a characteristic of a reform movement?

Reika [66]

Answer:

distinguished from more radical social such as revolutionary

7 0

3 years ago

A pump is used to extract water from a reservoir and deliver it to another reservoir whose free surface elevation is 200 ft abov

babunello [35]

Answer:

a) the expected flow rate is 31.4 ft³/s

b) the required brake horsepower is 2808.4 bhp

c) the location of pump inlet to avoid cavitation is -8.4 ft

Explanation:

Given the data in the question;

free surface elevation = 200 ft

total length of pipe required = 1000 ft

diameter = 12 inch

Iron with relative roughness ( k/D ) = 0.0005

H $_{pump$ = 665-0.051Q² [Qinft ]

a) the expected flow rate

given that;

k/D = 0.0005

k/2R = 0.0005

R/k = 1000

now, we determine the friction factor;

1/√f = 2log₁₀( R/k ) + 1.74

we substitute

1/√f = 2log₁₀( 1000 ) + 1.74

1/√f = 6 + 1.74

1/√f = 7.74

√f = 1/7.74

√f = 0.1291989

f = (0.1291989)²

f = 0.01669

Now, Using Bernoulli theorem between two reservoirs;

(p/ρq)₁ + (v²/2g)₁ + z₁ + H $_p$ = (p/ρq)₂ + (v²/2g)₂ + z₂ + h $_L$

0 + 0 + 0 + 665-0.051Q² = 0 + 0 + 200 + flQ²/2gdA²

665-0.051Q² = 200 + flQ²/2gdA²

665-0.051Q² = 200 +[ ( 0.01669 × 1000 × Q² ) / (2 × 32.2 × (π/4)² × 1⁵ )

665 - 0.051Q² = 200 + [ 16.69Q² / 39.725 ]

665 - 200 - 0.051Q² = 0.420138Q²

665 - 200 = 0.420138Q² + 0.051Q²

465 = 0.471138Q²

Q² = 465 / 0.471138

Q² = 986.97196

Q = √986.97196

Q = 31.4 ft³/s

Therefore, the expected flow rate is 31.4 ft³/s

b) the brake horsepower required to drive the pump (assume an efficiency of 78%).

we know that;

P = ρgH $_p$ Q / η

where; H $_p$ = 665 - 0.051(986.97196) = 614.7

we substitute;

P = ( 62.42 × 614.7 × 31.4 ) / ( 0.78 × 550 )

P = 1204804.6236 / 429

P = 2808.4 bhp

Therefore, the required brake horsepower is 2808.4 bhp

c) the location of pump inlet to avoid cavitation (assume the required NPSH=25 ft).

NPSH = ( $P_{atom$ / ρg) - h $_s$ - ( P $_v$ / ρg )

we substitute

25 = ( 2116 / 62.42 ) - h $_s$ - ( 30 / 62.42 )

h $_s$ = 8.4 ft

Therefore, the location of pump inlet to avoid cavitation is -8.4 ft

6 0

3 years ago

In Millikan's oil drop experiment, if the electricfield between the plates was of just the right magnitude, it wouldexactly bala

Vilka [71]

Answer:

The electric field to balance the weight is approximately equals to 3.49x10^5 Newton/Coulumb

Explanation:

In order to be stationary position, magnitude of the total force due to electric field should be equal to the gravitational force that is, $|F_{electrostatic}| = |F_{gravitational}|$

where

$F_{electrostatic}=q.E$

where q is the charge of the droplet and E is the electric field. On the other hand

$F_{gravitational}=m.g=V.d.g=\frac{4}{3}.\pi.r^3.d.g$

where m,V ,d and r are the mass, volume, density and radius of the oil droplet respectively and g is the gravitational acceleration (g=9,80665 m/sn^2). By using the first equation and solving it for the electric field we can write,

$q.E=\frac{4}{3}.\pi.r^3.d.g\\E=\frac{4}{3}.\pi.\frac{r^3.d.g}{q}\\E=\frac{4}{3}.\pi\frac{(1,6x10^{-4})^3.(0,85x10^{-3}).(9,81)}{1,6x10^{-19}}\\E\approx 3,49x10^{5} (N/C)$

(Note: d=0.85 x 10^-3 kg/cm^3 and the unit of electric field is Newton per coulumb (N/C))

8 0

3 years ago