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Katen [24]
1 year 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]1 year 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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3 0
3 years ago
Explain what a stress-concentration is and how it can be accounted for in the design and analysis of a structural component. Be
Mama L [17]

Answer:

Stress-concentration

        In the machine component the stress become concentrate at the point a particular point and due to this crack formation takes places and after some time some the machine component become fails.High stress concentration means high stress at that point.

It is very important in the design point of view because we have to find out where stress concentration is high and needs to take proper factor of safety to avoid the failure of the machine component.Generally at sharp corner,sudden change and complex shapes stress concentration is high.

3 0
3 years ago
The natural material in a borrow pit has a mass unit weight of 110.0 pcf and a water content of 6%, and the specific gravity of
enot [183]

Answer:

A. 288,030.91 cy

B. The amount of water that must be removed from the natural material is 483541.04254 gallons of water

Explanation:

The natural material in the barrow properties are;

The mass unit weight, γ = 110.0 pcf

The water content, w = 6%

The specific gravity of the soil solids, G_s = 2.63

The desired dry unit weight, \gamma _d = 122 pcf

The water content, w₁ = 5.5 %

The net section volume, V_T = 245,000 cy = 6,615,000 ft³

A.  \gamma _d = W_s/V_T

∴ W_s = V_T × \gamma _d = 6,615,000 ft³ × 122 lb/ft³ = 807030000 lbs

w = (W_w/W_s) ×  100

∴ W_w = (w/100) × W_s = (6/100) × 807030000 lbs = 48421800 lbs

The weight of solids

W = W_s + W_w = 807030000 lbs + 48421800 lbs = 855451800 lbs

V = W/γ = 855451800 lbs/(110.0 lb/ft.³) = 7776834.54545 ft.³ = 288,030.91 cy

V = 288,030.91 cy

The amount of cubic yards of borrow required = 288,030.91 cy

B. The volume of water in the required soil is found as follows;

W_{w1} = (w₁/100) × W_s = (5.5/100) × 807030000 lbs = 44386650 lbs

The amount of water that must be added =  W_{w1} - W_w = 44386650 lbs - 48421800 lbs = -4,035,150 lbs

Therefore, 4,035,150 lbs of water must be removed

The density of water, ρ = 8.345 lbs/gal

Therefore, V = 4,035,150 lbs/(8.345 lbs/gal) = 483541.04254 gal  of water must be removed from the natural material

7 0
3 years ago
A normal shock wave takes place during the flow of air at a Mach number of 1.8. The static pressure and temperature of the air u
Darina [25.2K]

Answer:

The pressure upstream and downstream of a shock wave are related as

\frac{P_{1}}{P_{o}}=\frac{2\gamma M^{2}-(\gamma -1)}{\gamma +1}

where,

\gamma= Specific Heat ratio of air

M = Mach number upstream

We know that \gamma _{air}=1.4

Applying values we get

\frac{P_{1}}{100kPa}=\frac{2\times 1.4\times 1.8^{2}-(1.4 -1)}{1.4 +1}\\\\\frac{P_{1}}{100kPa}=3.61\\\\\therefore P_{1}=361.33kPa(Absloute)

Similarly the temperature downstream is obtained by the relation

\frac{T_{1}}{T_{o}}=\frac{[2\gamma M^{2}-(\gamma -1)][(\gamma -1)M^{2}+2]}{(\gamma +1)^{2}M^{2}}

Applying values we get

\frac{T_{1}}{423}=\frac{[2\times 1.4\times 1.8^{2}-(1.4-1)][(1.4-1)1.8^{2}+2]}{(1.4+1)^{2}\times 1.8^{2}}\\\\\therefore \frac{T_{1}}{423}=1.53\\\\\therefore T_{1}=647.85K=374.85^{o}C

The Mach number downstream is obtained by the relation

M_{d}^{2}=\frac{(\gamma -1)M^{2}+2}{2\gamma M^{2}-(\gamma -1)}\\\\\therefore M_{d}^{2}=\frac{(1.40-1)\times 1.8^{2}+2}{2\times1.4\times 1.8^{2}-(1.4-1)}\\\\\therefore M_{d}^{2}=0.38\\\\M_{d}=0.616

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Jet001 [13]

Answer:

John should detail his Scrum Master.

Explanation:

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3 years ago
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