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Alecsey [184]
4 years ago
15

Water flows down a rectangular channel that is 1.2 m wide and 1 m deep. The flow rate is 0.95 m/s. Estimate the Froude number of

the flow.
Engineering
1 answer:
olga_2 [115]4 years ago
8 0

Answer:

The Froude number of the flow in the channel is 0.253

Explanation:

We know that Froude number for a rectangular channel is given by

F_{r}=\frac{v}{\sqrt{gy}}

where,

v = is the velocity of flow

g = is the acceleration due to gravity

y = depth of flow

We know that velocity is calculated from the flow rate as

v=\frac{Q}{Area}\\\\\therefore v=\frac{0.95}{1.2\times 1}=0.792m/s

Applying the given values in the equation we obtain Froude number as followsHence\\\\F_{r}=\frac{0.792}{\sqrt{9.81\times 1}}=0.253

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3.24 Program: Drawing a half arrow (Java) This program outputs a downwards facing arrow composed of a rectangle and a right tria
Lostsunrise [7]

Answer:

Here is the JAVA program:

import java.util.Scanner; // to get input from user

public class DrawHalfArrow{ // start of the class half arrow

public static void main(String[] args) { // starts of main() function body

    Scanner scnr = new Scanner(System.in); //reads input

int arrowBaseHeight = 0; // stores the height of arrow base

int arrowBaseWidth  = 0; // holds width of arrow base

int arrowHeadWidth = 0; // contains the width of arrow head

// prompts the user to enter arrow base height, width and arrow head width

System.out.println("Enter arrow base height: ");

arrowBaseHeight = scnr.nextInt(); // scans and reads the input as int

System.out.println("Enter arrow base width: ");

arrowBaseWidth = scnr.nextInt();

/* while loop to continue asking user for an arrow head width until the value entered is greater than the value of arrow base width */

while (arrowHeadWidth <= arrowBaseWidth) {

    System.out.println("Enter arrow head width: ");

    arrowHeadWidth = scnr.nextInt(); }

//start of the nested loop

//outer loop iterates a number of times equal to the height of the arrow base

 for (int i = 0; i < arrowBaseHeight; i++) {

//inner loop prints the stars asterisks

      for (int j = 0; j <arrowBaseWidth; j++) {

          System.out.print("*");        } //displays stars

          System.out.println();          }

//temporary variable to hold arrowhead width value

int k = arrowHeadWidth;

//outer loop to iterate no of times equal to the height of the arrow head

for (int i = 1; i <= arrowHeadWidth; i++)

{     for(int j = k; j > 0; j--)     {//inner loop to print stars

       System.out.print("*");    } //displays stars

   k = k - 1;

   System.out.println(); } } } // continues to add more asterisks for new line

Explanation:

The program asks to enter the height of the arrow base, width of the arrow base and the width of arrow head. When asking to enter the width of the arrow head, a condition is checked that the arrow head width arrowHeadWidth should be less than or equal to width of arrow base arrowBaseWidth. The while loop keeps iterating until the user enters the arrow head width larger than the value of arrow base width.

The loop is used to output an arrow base of height arrowBaseHeight. So point (1) is satisfied.

The nested loop is being used which as a whole outputs an arrow base of width arrowBaseWidth. The inner loop draws the stars and forms the base width of the arrow, and the outer loop iterates a number of times equal to the height of the arrow. So (2) is satisfied.

A temporary variable k is used to hold the original value of arrowHeadWidth so that it keeps safe when modification is done.

The last nested loop is used to output an arrow head of width arrowHeadWidth. The inner loop forms the arrow head and prints the stars needed to form an arrow head. So (3) is satisfied.

The value of temporary variable k is decreased by 1 so the next time it enters  the nested for loop it will be one asterisk lesser.

The screenshot of output is attached.

8 0
3 years ago
Using Pascal’s Law and a hydraulic jack, you want to lift a 4,000 lbm rock. The large cylinder has a diameter of 6 inches.
jolli1 [7]

Answer:

a diameter of D₂ = 0.183 inches would be required

Explanation:

appyling pascal's law

P applied to the hydraulic jack = P required to lift the rock

F₁*A₁ = F₂*A₂

since A₁= π*D₁²/4 ,  A₂= π*D₂²/4

F₁*π*D₁²/4 = F₂* π*D₂²/4

F₁*D₁²=F₂*D₂²

D₂ = D₁ *√(F₁/F₂)

replacing values

D₂ = D₁ *√(F₁/F₂) =  6 in * √(120 lbf/(4000 lbm * 32.174 (lbf/lbm)) = 0.183 inches

6 0
3 years ago
A driver traveling in her 16-foot SUV at the speed limit of 30 mph was arrested for running a red light at 15th and Main, an int
Lynna [10]

Answer:

(a) Yes

(b) 102.8 ft

Explanation:

(a)First let convert mile per hour to feet per second

30 mph = 30 * 5280 / 3600 = 44 ft/s

The time it takes for this driver to decelerate comfortably to 0 speed is

t = v / a = 44 / 10 = 4.4 (s)

given that it also takes 1.5 seconds for the driver reaction, the total time she would need is 5.9 seconds. Therefore, if the yellow light was on for 4 seconds, that's not enough time and the dilemma zone would exist.

(b) At this rate the distance covered by the driver is

s = v_0t + \frac{at^2}{2}

s =44*1.5 + 44(4.4) - \frac{10*4.4^2}{2} = 162.8 (ft)

Since the intersection is only 60 feet wide, the dilemma zone must be

162.8 - 60 = 102.8 ft

4 0
3 years ago
) Assuming that the net environmental impact and all other considerations are equal to produce, package, and distribute both pro
DanielleElmas [232]

Answer:

biodegradable product

Explanation:

The biodegradable product will be the best product to buy on the market. This is due to the less harmful impact the product has on the environment. The plastics will degrade easily unlike the non-biodegradable products that are fossil fuel derivatives.

7 0
3 years ago
An overhead 25m long, uninsulated industrial steam pipe of 100mm diameter is routed through a building whose walls and air are a
DedPeter [7]

Answer:

a) he rate of heat loss from the steam line is 18.413588 kW

b) the annual cost of heat loss from line is $12904.25

Explanation:

a)

first we find the area

A = πdL

d is the diameter (0.1m) and L is the length (25m)

so

A = π ×  0.1 × 25

A = 7.85 m²

Now rate of heat loss through convection

qconv = hA(Ts -Ta)

h is the convective heat transfer coefficient (10 W/m²K), Ts is surface temperature (150°), Ta is temperature of air (25°)

so we substitute

qconv = 10 W/m²K × 7.85 m² × ( 150° - 25°)

qconv = 9817.477 J/s

Now heat lost through radiation

qrad = ∈Aα ( Ts⁴ - Ta⁴)

∈ is the emissivity (0.8), α is the boltzmann constant ( 5.67×10⁻⁸m⁻²K⁻⁴ ),

first we shall covert our temperatures from Celsius to kelvin scale

Ts is surface temperature (150 + 273K ), Ta is temperature of air (25 + 273K)

so we substitute

qrad = 0.8 × 7.854 × 5.67×10⁻⁸ × ( (423)⁴ - (298)⁴ )

qrad = 3.5625×10⁻⁷ × 2.413×10¹⁰

qrad = 8596.112 J/s

Now to get the total rate of heat loss through convection and radiation, we say

q = qconv + qrad

q = 9817.477 + 8596.112

q = 18413.588 J/s ≈ 18.413588 kW

Therefore the rate of heat loss from the steam line is 18.413588 kW

b)

annual cost of heat lost rate

A = C × q/n × ( 3600 × 24 × 365 )

C is the cost of heat per MJ( $0.02/10⁶) n is broiler efficiency ( 0.9)

so we substitute

A = 0.02/10⁶  × 18413.588/0.9 × ( 3600 × 24 × 365 )

A = $12904.25

Therefore the annual cost of heat loss from line is $12904.25

4 0
3 years ago
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