A single-phase inductive load draws a 10 MW at 0.6 power factor lagging. Calculate the reactive power of a capacitor to be conne
cted in parallel with the load to raise the power factor to 0.89. Type your answer on the box, and make sure your answer have two decimal digits and the correct sign.
1 answer:
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Answer:
Explanation:
Lets take the numerator of the fraction to be = x
So the denominator of the fraction is 4 more than the numerator = x+4
The fraction is ;
Now add 4 to the numerator and add 7 to the denominator as;
This new fraction is equal to 1 half =1/2
write the equation as;
perform cross-product
2(x+4 )=1( x+11 )
2x+8 = x + 11
2x-x = 11-8
x=3
The original fraction is;
Answer:
P=361.91 KN
Explanation:
given data:
brackets and head of the screw are made of material with T_fail=120 Mpa
safety factor is F.S=2.5
maximum value of force P=??
<em>solution:</em>
to find the shear stress
T_allow=T_fail/F.S
=120 Mpa/2.5
=48 Mpa
we know that,
V=P
<u>Area for shear head:</u>
A(head)=π×d×t
=π×0.04×0.075
=0.003×πm^2
<u>Area for plate:</u>
A(plate)=π×d×t
=π×0.08×0.03
=0.0024×πm^2
now we have to find shear stress for both head and plate
<u>For head:</u>
T_allow=V/A(head)
48 Mpa=P/0.003×π ..(V=P)
P =48 Mpa×0.003×π
=452.16 KN
<u>For plate:</u>
T_allow=V/A(plate)
48 Mpa=P/0.0024×π ..(V=P)
P =48 Mpa×0.0024×π
=361.91 KN
the boundary load is obtained as the minimum value of force P for all three cases. so the solution is
P=361.91 KN
note:
find the attached pic
