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3 years ago

13

The bar DA is rigid and is originally held in the horizontal position when the weight W is supported from C. The wires are made

of A-36 steel and have a cross-sectional area of 0.002 in^2determine the weight W.

1 answer:

Rom4ik [11]3 years ago

4 0

Answer:

W = 112 lb

Explanation:

Given:

- δb = 0.025 in

- E = 29000 ksi (A-36)

- Area A_de = 0.002 in^2

Find:

Compute Weight W attached at C

Solution:

- Use proportion to determine δd:

δd/5 = δb/3

δd = (5/3) * 0.025

δd = 0.0417 in

- Compute εde i.e strain in DE:

εde = δd / Lde

εde = 0.0417 / 3*12

εde = 0.00116

- Compute stress in DE, σde:

σde = E*εde

σde = 29000*0.00116

σde = 33.56 ksi

- Compute the Force F_de:

F_de = σde *A_de

F_de = 33.56*0.002

F_de = 0.0672 kips

- Equilibrium conditions apply:

(M)_a = 0

3*W - 5*F_de = 0

W = (5/3)*F_de

W = (5/3)* 0.0672 = 112 lb

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Solution :

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3 years ago

Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe

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Answer:

(a) attached below

(b) $pf_{C}=0.85$ $lagging$

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(d) $X_{C} =49.37$ Ω

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Explanation:

Given data:

$P_{1}=15 kW$

$S_{2} =10 kVA$

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$pf_{2}=0.8$ $leading$

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$\alpha_{1}=cos^{-1} (0.6)=53.13$ °

$\alpha_{2}=cos^{-1} (0.8)=36.86$ °

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$Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99$ ≅ $20kVAR$

$P_{2} =S_{2}*pf_{2} =10*0.8=8 kW$

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The negative sign means that the load 2 is providing reactive power rather than consuming

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Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

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$I_{C} =S_{C}/\sqrt{3}*V$

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