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klemol [59]
4 years ago
9

A trapezoidal ditch is designed with a bottom width of 3 space f t and side slopes of m equals 1 on both sides. The channel is m

ade of concrete (n equals 0.013) and has a longitudinal slope of 1 percent sign. If the flow rate on a particular day is 125 space f t cubed divided by s, determine the depth of flow in the channel. Assume steady uniform flow.
Engineering
1 answer:
meriva4 years ago
7 0

Answer:

d = 3.25 ft

Explanation:

n = 0.013

side slope, m = 1

Flow rate, Q = 125 ft³/s

Width, w = 3 ft

Longitudinal slope, S = 1% = 0.01

Let the depth = d

Area, A = wd = 3d

The equation for the flow rate can be given as:

Q = \frac{1.49}{n} AR_{h} ^{2/3} S^{1/2}...........................(1)

R_{h} = \frac{Area}{Perimeter} \\R_{h} = \frac{3d}{3 + 2d}\\

Substituting all appropriate values into equation (1)

125 = \frac{1.49}{0.013} *3d*(\frac{3d}{3+2d})^{2/3} * 0.01^{1/2}

125 = \frac{71.52d^{5/3} }{(3+2d)^{2/3} } \\1.75(3+2d)^{2/3} = d^{5/3}\\(3+2d)^{2/3} = 0.57d^{5/3}\\3+2d =  (0.57d^{5/3})^{3/2} \\3+2d = 0.43d^{5/2}\\(3+2d)^{2} = 0.187d^{5} \\9 + 6d + 4d^{2} =  0.187d^{5}\\ 0.187d^{5}- 4d^{2}-6d -9 = 0\\d^{5}-21.39d^{2}-32.09d -48.13 = 0\\

On solving the equation above

d = 3.25 ft

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Answer:

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3 years ago
Who developed the process of blueprinting?
VikaD [51]
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A helical compression spring is made with oil-tempered wire with wire diameter of 0.2 in, mean coil diameter of 2 in, a total of
Naya [18.7K]

Answer:

a. Solid length Ls = 2.6 in

b. Force necessary for deflection Fs = 67.2Ibf

Factor of safety FOS = 2.04

Explanation:

Given details

Oil-tempered wire,

d = 0.2 in,

D = 2 in,

n = 12 coils,

Lo = 5 in

(a) Find the solid length

Ls = d (n + 1)

= 0.2(12 + 1) = 2.6 in Ans

(b) Find the force necessary to deflect the spring to its solid length.

N = n - 2 = 12 - 2 = 10 coils

Take G = 11.2 Mpsi

K = (d^4*G)/(8D^3N)

K = (0.2^4*11.2)/(8*2^3*10) = 28Ibf/in

Fs = k*Ys = k (Lo - Ls )

= 28(5 - 2.6) = 67.2 lbf Ans.

c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.

For C = D/d = 2/0.2 = 10

Kb = (4C + 2)/(4C - 3)

= (4*10 + 2)/(4*10 - 3) = 1.135

Tau ts = Kb {(8FD)/(Πd^3)}

= 1.135 {(8*67.2*2)/(Π*2^3)}

= 48.56 * 10^6 psi

Let m = 0.187,

A = 147 kpsi.inm^3

Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi

Ssy = 0.50 Sut

= 0.50(198.6) = 99.3 kpsi

FOS = Ssy/ts

= 99.3/48.56 = 2.04 Ans.

7 0
3 years ago
(Gas Mileage) Drivers are concerned with the mileage their automobiles get. One driver has kept track of several trips by record
Effectus [21]

Answer:

import java.util.*;

public class Main {

   

   public static void main(String[] args) {

     

       double milesPerGallon = 0;

       int totalMiles = 0;

       int totalGallons = 0;

       double totalMPG = 0;

       

       Scanner input = new Scanner(System.in);

 

       while(true){

           System.out.print("Enter the miles driven: ");

           int miles = input.nextInt();

           if(miles <= 0)

               break;

           else{

               System.out.print("Enter the gallons used: ");

               int gallons = input.nextInt();

               totalMiles += miles;

               totalGallons += gallons;

               milesPerGallon = (double) miles/gallons;

               totalMPG = (double) totalMiles / totalGallons;

               System.out.printf("Miles per gallon for this trip is: %.1f\n", milesPerGallon);

               System.out.printf("Total miles per gallon is: %.1f\n", totalMPG);

           }

       }

   }  

}

Explanation:

Initialize the variables

Create a while loop that iterates until the specified condition is met inside the loop

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6 0
3 years ago
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