Question

The water pressure in the pipes going into an office building is 4 atm at street level. In other words, the waterpressure at street level is 3 atm above. The water enters the 5.0 cm diameter street level pipe at a speedof 1 m/s. The pipe tapers down to a diameter of 3.0 cm at the top floor, 15 m above street level where afaucet is left on. What is the flow velocity and pressure in the pipe at the top floor. Assume no branch pipes and ignore viscosity.

Answer 1

In the pipe at the top floor:

The flow velocity is
`v_2= 2.78 (m)/(s)\\`

The pressure is
`P_2 = 2.52 \ atm`

Step-by-step explanation:

Following the Bernoulli's equation, the total pressure both, at street level and at the top floor must be the same. Therefore, we write the equation to solve the problem as follows:

`P_(1)+ \rho gh_1+(1)/(2) \rho v_1^2=P_(2)+ \rho gh_2+(1)/(2) \rho v_2^2\\`

First, to find the flow velocity at the top floor, we know that volume flow must be the same on both sides of the pipe, therefore we can write as follows:

`Q_1=Q_2\\A_1v_1=A_2v_2\\v_2 = (A_1v_1)/(A_2)\\`

And for the relationship between the areas we have:

`(A_1)/(A_2)=(\pi r_1^2)/(\pi r_2^2)=((r_1)/(r_2))^2`

Then, we replace the previous expressions in the main equation as follows:

`v_2=((r_1)/(r_2))^2v_1=((5)/(3))^2 * 1(m)/(s)=2.78 (m)/(s)\\P_2=P_1+(1)/(2)\rho v_1^2-\rho gh_2- (1)/(2) \rho v_2^2\\P_2=4 atm* 101235 (Pa)/(atm)+997(kg)/(m^3)((1)/(2)( 1(m)/(s))^2-9.8 (m)/(s^2) 15m- (1)/(2)( 2.78(m)/(s))^2)\\P_2= (255386.89 Pa)/(101325(Pa)/(atm))= 2.52 atm\\P_2 = 2.52 atm`

Answer 2

v₂ = 2.78 m / s , P₂ = 2.55 10⁵ Pa

Step-by-step explanation:

For this exercise in fluid mechanics, we use Bernoulli's equation, which is an expression of energy conservation.

P + ½ ρ v² +ρg y = cte.

And the continuity equation

Q = A v

Let's define two points the street level, point 1 and the level of the third floor point 2, let's see what they give us

At point 1 we have the pressure (P₁ = 4 atm), the diameter of the pipe (d₁ = 5.0 cm) the speed with which it arrives (v₁ = 1.0 m / s)

At point 2 they give the pressure (P₂ = 1 atm), pipe diameter (d₂ = 3.0 cm) and height (y₂ = 15m)

Ask for speed, let's use the continuity equation at the two points

A₁ v₁ = A₂ v₂

The area and the pipe are

A = π r² = π (d/2)²

π / 4 d₁² v₁ = π / 4 d₂² v²

v₂ = v₁ d₁² / d₂²

v₂ = 1 5.0² / 3.0²

v₂ = 2.78 m / s

Let's reduce the magnitudes to the SI System

P₁ = 4 atm (1,013 10⁵ Pa / 1 atm) = 4,052 10⁵ Pa

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

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

calculate

P₂ = 4,052 10⁵+ ½ 1000 (1² - 2.78²) + 1000 9.8 (0 - 15)

P₂ = 4.052 10⁵– 3364.2 - 147000

P₂ = 254835.8 Pa

P₂ = 2.55 10⁵ Pa