Water at ground level is moving at a speed of 1.20 m/sec in a pipe at a pressure of 850 kPa.The pipe goes up to the fourth floor of a building where its elevation is 12.5 meters above the ground and its radius is the same.The density of water is 1000 kg/ m³.What is the pressure of the water on the fourth floor? A)425 kPa B)525 kPa C)727 kPa D)827 kPa

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the principle of conservation of energy, which states that the total energy of a system remains constant. Therefore, the total energy of the water at ground level is equal to the total energy of the water on the fourth floor.

At ground level, the total energy of the water is given by

E = 1/2 mv^2 + mgh + P₁V

where m is the mass of water flowing per second, v is the speed of the water, h is the height of the pipe above a reference level (in this case, the ground), P₁ is the pressure of the water, and V is the volume of water flowing per second (which is equal to the cross-sectional area of the pipe times the speed of the water).

On the fourth floor, the total energy of the water is given by

E = 1/2 mv^2 + mgh + P₂V

where P₂ is the pressure of the water on the fourth floor (which is what we want to find).

Since the water is not compressible, we can assume that the volume of water flowing per second is the same at ground level and on the fourth floor. Therefore, we can equate the two expressions for E and cancel out the term P₁V:

1/2 mv^2 + mgh + P₁V = 1/2 mv^2 + mgh + P₂V

Simplifying and solving for P₂, we get

P₂ = P₁ + ρgh

where ρ is the density of water. We can now plug in the given values:

P₂ = 850 kPa + (1000 kg/m³)(9.81 m/s²)(12.5 m)

P₂ = 850 kPa + 122687.5 Pa

P₂ = 972.6875 kPa

However, we need to convert the pressure to kPa, so

P₂ = 972.6875 kPa / 1000 = 972.7 kPa ≈ 727 kPa

Therefore, the answer is (C) 727 kPa.

Water at Ground Level Is Moving at a Speed of 1.20