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ankoles [38]
3 years ago
8

At the end of a power distribution system, a certain feeder supplies three distribution transformer, each one supplying a group

of customers whose connected loads are as under: Transformer Load Demand factor Diversity factor Transformer No. 1 10kW 0.65 1.5 Transformer No. 2 12kW 0.6 3.5 Transformer No. 3 15kW 0.7 1.5 If the diversity factor among the transformer is 1.3; find the maximum load on the feeder.
Engineering
1 answer:
murzikaleks [220]3 years ago
6 0

Answer:

≈ 18.62 kw

Explanation:

Given data :

Diversity factor among transformer = 1.3

No of transformers       load      demand factor   diversity factor

Transformer 1               10 kw      0.65                    1.5

Transformer 2              12 kw      0.6                       3.5

Transformer 3               15 kw      0.7                      1.5

calculate the maximum load on the feeder

demand factor = max demand /  load

Max demand = demand factor * load

max demands:

Transformer 1 = 0.65 * 10 = 6.5

Transformer 2 = 0.6 * 12 = 7.2

Transformer 3 = 0.7 * 15 = 10.5

Diversity factor = ( summation of max demands ) / maximum load on feeder

  1.3 = ( 6.5 +7.2 + 10.5 ) / max load on feeder

hence : max load on feeder = 24.2 / 1.3 = 18.62 kw

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victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0
4 years ago
Which task best fits the role of a planning engineer?
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8 0
3 years ago
10% A steel beam W18x76 spans 32 feet and is subjected to a Moment of 334 kips-ft. Find the load w on the beam. Determine the de
Lilit [14]

Answer:

w = 10.437 kips

deflection at 1/4 span  20.83\E ft

at mid span = 1.23\E ft

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area of cross section = 18*76

length of span = 32 ft

moment = 334 kips-ft

we know that

moment = load *eccentricity

334 = w * 32

w = 10.437 kips

deflection at 1/4 span

\delta = \frac{wa^2b^2}{3EI}

= \frac{10.4375*8^2 *24^2}{3E \frac{BD^3}{12}}

         =\frac{10.437 *8^2*24^2}{3E \frac{18*16^3}{12}}

         = 20.83\E ft

at mid span

\delta = \frac{wl^3}{48EI}

= \frac{10.43 *32^3}{48 *E*\frac{18*16^3}{12}}

\delta = 1.23\E ft

shear stress

\tau = \frac{w}{A} = \frac{10.43 7*10^3}{18*76} =7.3629 psi

6 0
3 years ago
Determine how much concrete you will need for a slab which is 50 feet by 30 feet wide and 1 foot thick
Levart [38]

Answer:

Amount of concrete need to make slab = 1,500 feet³

Explanation:

Given:

Length of slab = 50 feet

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Amount of concrete need to make slab = Volume of cuboid

Volume of cuboid = (l)(b)(h)

Amount of concrete need to make slab = (50)(30)(1)

Amount of concrete need to make slab = 1,500 feet³

6 0
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mass of sludge produced = 32909.09 kg

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6 0
3 years ago
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