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miss Akunina [59]
3 years ago
14

A horizontal curve was designed for a four-lane highway for adequate SSD. Lane widths are 12 ft, and the superelevation is 0.06

and was set assuming maximum fs. If the necessary sight distance required 52 ft of lateral clearance from the roadway centerline, what design speed was used for the curve

Engineering
2 answers:
Law Incorporation [45]3 years ago
7 0

Answer:

80 mi/h

Explanation:

To get the adequate design speed the assumed design speed ( Fs ) for the Horizontal curve has to produce the<em> </em><u><em>necessary Middle ordinate distance that is equal to the provided Middle ordinate</em></u>

The provided middle ordinate is gotten form the formula

Ms = Available clearance - lane width - lane width / 2  ( equation 1 )

available clearance = 52 ft

lane width = 12 ft

hence equation 1 becomes

Ms = 52 - 12 - 12/2

      = 40 - 6 = 34 ft

To check if the assumed design speed will produce the necessary middle ordinate distance that is equal to the provided middle ordinate the formula below is used

Ms₍necessary₎  = Rv ( 1 - cos\frac{90SSD}{\pi Rv  } ) (equation 2 )

Assuming a design speed ( Fs ) of 80 mi/h and referring to the engineering table and also considering the super elevation of 0.06 ( 6% ) the values of

Rv = 3050 ft

SSD = 910 ft

substituting these values into (equation 2) Ms ( necessary ) becomes

= 3050 ( 1 - cos \frac{90*910}{\pi * 3050 } )

= 3050 ( 1 - cos \frac{81900}{\pi * 3050 } )

= 33.876 ft

33.876 ft is approximately equal to the provided middle ordinate of 34 ft hence the design speed used would be 80mi/h

vampirchik [111]3 years ago
3 0

Answer and Explanation:

The answer is attached below

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23.3808 kW

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Q = Volumetric flow rate = 0.2 m³/s

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Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

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Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W

∴ Power input to the pump 20.7088 kW

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3 years ago
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