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2 years ago

10

Determine the design angle ϕ (0∘≤ϕ≤90 ∘) between struts AB and AC so that the 400 lb horizontal force has a component of 600 lb

which acts up to the left, in the same direction as from B towards A. Take θ = 30 ∘.

1 answer:

Svetllana [295]2 years ago

4 0

Answer:

design angle ∅ = 4.9968 ≈ 5⁰

Explanation:

First calculate the force Fac :

Fac = $\sqrt{400^2 + 650^2 - 2(400)(650)cos30}$

= $\sqrt{160000 + 422500 - 80210}$

= 708.72 Ib

using the sine law to determine the design angle

$\frac{sin}{400} = \frac{sin 30}{Fac}$

hence ∅ = $sin^{-1} (\frac{sin 30 *400}{708.72} )$

= $sin^{-1} 0.0871$ = 4.9968 ≈ 5⁰

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Answer:

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(b) Change in specific entropy=0.0044 $\frac{KJ}{Kg-K}$

Explanation:

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$h_2=167.5+0.83(2537.4-167.5)\frac{KJ}{Kg}$

$h_2=2134.57\frac{KJ}{Kg}$

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$s_2=6.6944\frac{KJ}{Kg-K}$

(a)

Work out put = $h_1-h_2$

=2827.4-2134.57 $\frac{KJ}{Kg}$

Work out put =692.83 $\frac{KJ}{Kg}$

(b) Change in specific entropy

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Change in specific entropy =0.0044 $\frac{KJ}{Kg-K}$

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Answer:

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