Question

Assuming that Bernoulli's equation applies, compute the volume of water ΔV that flows across the exit of the pipe in 1.00 s . In other words, find the discharge rate ΔV/Δt.

Answer 1

Answer:

12.65m³/ sec

Explanation:

The process flow shown in the attachment along with this explanation, with point 2 taken in the liquid at the entrance to the nozzle and point 3 at the exit of the nozzle.

Since A₁ is very large compared to A₂, v₁ ≅ 0. The pressure p₂ is greater tha₁n 1atm (101.3kN/m3) by the head of fluid of Hm. The pressure p₃ which is at point 3 is at 1atm. Using point 2 as a datum, z₂=0 and z₃=0. From bernoulli’s equation

z₂g+ v₂²/2 + p₂/ρ = z₃g+ v₃²/2 + p₃/ρ

we can rearrange the above Bernoulli’s equation as

z₂g+ v₂²/2 + p₂-p₃/ρ = z₃g+ v₃²/2

0+0+p₂-p₃/ρ = 0 + v₃²/2

Solving for v₂

 v₃ ⇒$\sqrt{2(p_{2} -p_{3}) /density}$ ..........Equation 1

since p₂-p₁ = Hρg and p₁ = p₃ ( both at 1atm)

H =  p₂-p₃/ρg

p₂-p₃ = Hρg...............................Equation 2

put Equation 2 into Equation 1

v₃ = $\sqrt{2gH}$  .....................Equation 3

v₃ = $\sqrt{2*9.81*8} = 12.65m/s$

The volumetric flowrate in 1.00sec is given as

                         flow rate = v₃A₃= 12.65×$1.6 *10^{-2}$

                                         =0.202m³/sec