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3 years ago

7

Calculate the pressure drop in a duct (measured by a differential oil manometer) if the differential height between the two flui

d columns is 3.2 cm and the density of oil is 860 kg/m3 .

1 answer:

Burka [1]3 years ago

6 0

Answer:

The pressure drop is 269.7N/m^2

Explanation:

∆P = ∆h × rho × g

∆h = 3.2cm = 3.2/100 = 0.032m, rho = 860kg/m^3, g = 9.8m/s^2

∆P = 0.032×860×9.8 = 269.7N/m^2

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A Michelson interferometer operating at a 500 nm wavelength has a 3.73-cm-long glass cell in one arm. To begin, the air is pumpe

LekaFEV [45]

Answer:

The number of bright-dark fringe is 42

Solution:

As per the question:

Wavelength of light, $\lambda = 500\ nm = 500\times 10^{- 9}\ m$

Length of the glass cell, x = 3.73 cm = 0.0373 m

Refractive index, $\mu = 1.00028$

Now,

To calculate the bright-dark fringe shifts, we use the formula given below:

$d_{m} = \frac{2x}{\lambda }\times (\mu - 1)$

Now, substituting the appropriate values in the above formula:

$d_{m} = \frac{2\times 0.0373}{500\times 10^{- 9}}\times (1.00028 - 1)$

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3 years ago

A cross beam in a highway bridge experiences a stress of 14 ksi due to the dead weight of the bridge structure. When a fully loa

zlopas [31]

Answer:

a) 2.452

b) 1.256

Explanation:

Stress due to dead weight. = 14 Ksi

Stress due to fully loaded tractor-trailer = 45Ksi

ultimate tensile strength of beam = 76 Ksi

yield strength = 50 Ksi

endurance limit = 38 Ksi

Determine the safety factor for an infinite fatigue life

a) If mean stress on fatigue strength is ignored

β = ( 45 - 14 ) / 2

= 15.5 Ksi

hence FOS ( factor of safety ) = endurance limit / β

= 38 / 15.5 = 2.452

b) When mean stress on fatigue strength is considered

β2 = 45 + 14 / 2

= 29.5 Ksi

Ratio = β / β2 = 15.5 / 29.5 = 0.5254

Next step: applying Goodman method

Sa = [ ( 0.5254 * 38 *76 ) / ( 0.5254*76 + 38 ) ]

= 19.47 Ksi

hence the FOS ( factor of safety ) = Sa / β

= 19.47 / 15.5 = 1.256

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3 years ago

6.03 Discussion: Then & Now - Safety

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3 years ago

The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp

Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

$Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1$

The Lower Surface Cp is given as

$Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1$

The difference of the Cp over the airfoil is given as

$\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32$

Now the Lift Coefficient is given as

$C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192$

Now the coefficient of moment about the leading edge is given as

$C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306$

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

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3 years ago