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kaheart [24]
3 years ago
11

The question is in the following image.

Engineering
1 answer:
Leona [35]3 years ago
3 0

Answer: Statement "E" is true

Explanation:

A - Statement is not true because, though static methods can not directly access any instance variables or methods. But they can access them by using their object reference.

B - Statement is not true because a non static method can also access a static variable.

C - Statement is not true because according to the rules for instance variables in java, it can be marked as final.

D - Statement is not true because though it can be declared as final by initialization of variable mandatory and initialization of class loading, there is no rule that says it must be declared as final.

E - This statement is true because in the constructor, static variables are not associated with each object of the class since it is shared by each object. Therefore, if you initialize static variable into constructor, it means that you are trying to associate with a particular member of class and it's not possible to do so!

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Water flows in a pipeline. At a point in the line where the diameter is 7 in., the velocity is 12 fps and the pressure is 50 psi
PolarNik [594]

Answer:

a)   P₂ = 3219.11 lbf / ft² , b)    P₂ = 721.91 lbf / ft² , c)  P₂ = 5707.31 lbf / ft²

Explanation:

For this exercise we can use the fluid mechanics equations, let's start with the continuity equation, index 1 is for the starting point and index 2 for the end point of the reduction

     A₁ v₁ = A₂ v₂

     v₂ = v₁ A₁ / A₂

The area of ​​a circle is

    A = π r² = π/4  d²

     v₂ = v₁ (d₁ / d₂)²

Let's calculate

    v₂ = 12 (7/3)²

    v₂ = 65 feet / s

Now let's use Bernoulli's equation

     P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

     P₁ - P₂ = ρ g (y₂ –y₁) + ½ ρ (v₂² - v₁²)

Case 1. The pipe is horizontal, so

      y₁ = y₂

      P₁ - P₂ = ½ ρ  (v₂² –v₁²)

      P₂ = P₁ - ½ ρ (v₂² –v₁²)

     ρ = 62.43 lbf / ft³

     P₁ = 50 psi (144 lbf/ ft² / psi) = 7200 lbf / ft²

    P₂ = 7200 - ½ 62.43 / 32 (65² -12²)

    P₂ = 7200 - 3980.89

    P₂ = 3219.11 lbf / ft²

Case 2 vertical pipe with water flow up

        y₂ –y₁ = 40 ft

        P₁ - P₂ = ρ g (y₂ –y₁) + ½ rho (v₂² - v₁²)

        7200 - P₂ = 62.43 (40) + ½ 62.43 / 32 (65 2 - 12 2) =

        P₂ = 7200 - 2497.2 - 3980.89

         P₂ = 721.91 lbf / ft²

Case 3. Vertical water pipe flows down

         y₂ –y₁ = -40

         P₂ = 7200 + 2497.2 - 3980.89

         P₂ = 5707.31 lbf / ft²

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