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Katen [24]
10 months ago
6

Where loads are likely to be on continuously, the calculated load for branch circuits and feeders must be figured at (100%) (125

%) (150%) of the continuous load, (plus) (minus) the non-continuous load. This requirement generally (does) (does not) apply to residential wiring.
Engineering
1 answer:
Anna [14]10 months ago
4 0

Where loads are likely to be on continuously, the calculated load for branch circuits and feeders must be figured at 125%.

Section 210.19(A)(1) permits the bigger of the two values listed below to be utilized as the connectors 's ultimate size for sizing an ungrounded branch circuit conductor:

Without any extra adjustments or corrections, either 125% of the continuous load, OR

When adjustment and corrective factors are applied, the load is 100% (not 125% as stated previously).

This will be the same in the 2020 NEC. The introduction of new exception 2 is what has changed. To comprehend this new exception, one must study it very carefully. A part of a branch circuit connected to pressure connectors (such as power distribution blocks) that complies with 110.14(C)(2) may now be sized using the continuous load plus the noncontiguous load instead of 125% of the continuous load thanks to the new exception.

To know more about connectors click here:

brainly.com/question/16987039

#SPJ4

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A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte
Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

a(t)=5(1-\frac{t}{15})

a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})

Integrating both sides we get

\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m

Thus the remaining 125 meters will be covered with a constant speed of

v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s

in time equalling t_{2}=\frac{125}{37.5}=3.33seconds

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0
2 years ago
A 3-phase , 1MVA, 13.8kV/4160V, 60 Hz, transformer with Y-Delta winding connection is supplying a3-phase, 0.75 p.u. load on the
Tanya [424]

Answer:

a) 23.89 < -25.84 Ω

b) 31.38 < 25.84 A

c) 0.9323 leading

Explanation:

A) Calculate the load Impedance

current on load side = 0.75 p.u

power factor angle = 25.84

I_{load} = 0.75 < 25.84°

attached below is the remaining part of the solution

<u>B) Find the input current on the primary side in real units </u>

load current in primary = 31.38 < 25.84 A

<u>C) find the input power factor </u>

power factor = 0.9323 leading

<em></em>

<em>attached below is the detailed solution </em>

8 0
2 years ago
Compute the theoretical density of ZnS given that the Zn-S distance and bond angle are 0.234 nm and 109.5o, respectively. The at
andriy [413]

Answer: the theoretical density is 4.1109 g/cm³

Explanation:  

first the image of one set of ZnS bonding in the crystal structure, we calculate the value of angle θ

θ + ∅ + 90° = 180°

θ = 90° - ∅

θ = 90° - ( 109.5° / 2 )

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next we calculate the value of x from the geometry

given that;  distance angle d = 0.234

x = dsinθ

= 0.234 × sin35.25°)

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next we calculate the length of the unit cell

a = 4x

a = 4(0.135)

a = 0.54 nm = 0.54 × 10⁻⁷ cm

next we calculate number of formula units

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= 1 + 3

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next we calculate the theoretical density using  this equation

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= [n'∑(Ac + AA)] / [(a)³NA]

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∑AA is the sum of weights of all anions in the formula unit( 32.06 g/mol)

Na is the Avogadro’s number( 6.023 × 10²³ units/mole)

so we substitute

P = [4( 65.41 + 32.06)] / [ ( 0.54 × 10⁻⁷ )³ × (6.023 × 10²³)]

= 389.88 / 94.84

= 4.1109 g/cm³

therefore the theoretical density is 4.1109 g/cm³

5 0
2 years ago
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