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Gelneren [198K]
3 years ago
5

Calculate the Lee for the same wing if we increase the span to 0.245 m. By increasing the span we also increase the glider weigh

t to 0.0523
Engineering
2 answers:
alexira [117]3 years ago
8 0

Answer:

0.21

Explanation:

This would have been a fairly easy one, except for that the first part of the question is missing, and as such, I'd assume a value.

We need to use chord, so, I'm assuming the length of the chord to be 0.045 m

The Area is given by the formula

Area = span * chord

Area = 0.245 * 0.045

Area = 0.011 m²

This area gotten, is what we then divide the glider weight by to get our answer.

Lee = area / weight

Lee = 0.011 / 0.0523

Lee = 0.21

Therefore, using the values of the chord I'd assumed, the Lee of the same wing is 0.21

aleksklad [387]3 years ago
5 0

Answer: 0.2108

Explanation:got it correct

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Answer:

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Explanation:

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

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Z = number of atom in unit cell

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On substituting all the given values , we will get the value of 'a'.

\rho=\frac{2\times 92.91}{6.022\times 10^{23} mol^{-1}\times (0.165 \times 10^{-7} cm)^{3}}

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$h_{2f} = 11.445 \ Btu/lbm$

$h_{2g} = 102.73 \ Btu/lbm$

$u_{2f} = 11.401 \ Btu/lbm$

$u_{2g} = 94.3 \ Btu/lbm$

$T_2=T_{sat}=-2.43^\circ  F$

Change in temperature, $\Delta T = T_2-T_1$

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Now we find the quality,

$h_2=h_f+x_2(h_g-h_f)$

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The final energy,

$u_2=u_f+x_2.u_{fg}$

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   $=38.09297 \ Btu/lbm$

Change in internal energy  

$\Delta u= u_2-u_1$

   = 38.09297-40.5485

  = -2.4556        

5 0
3 years ago
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