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3 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]3 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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At T = 100 °C the second and third virial coefficients are

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C = 25200 cm^6 mo1^-2

Now according isothermal work of one mole methyl gas is

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a= $v_2\\$

b= $v_1$

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$\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\ \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\$

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$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a= $v_2\\$

b= $v_1$

Now calculate V1 and V2 at given condition

$\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}$

Substitute given values $P_1\\$ = 1 x 10^5 , T = 373.15 and given values of coefficients we get

$10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2$

Solve for V1 by iterative or alternative cubic equation solver we get

$v_1=30780 cm^3/mol$

Similarly solve for state 2 at P2 = 50 bar we get

$v_1=241.33 cm^3/mol$

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$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a=241.33

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After performing integration we get work done on the system is

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get

dV=RT(-1/p^2+0+C')dP

Hence work done on the system is

$W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP$

a= $v_2\\$

b= $v_1$

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series

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Question

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