Question

Water is to be withdrawn from an 8-m-high water reservoir by drilling a 2.2-cm-diameter hole at the bottom surface. Disregarding the effect of the kinetic energy correction factor, determine the flow rate of water through the hole if (a) the entrance of the hole is well-rounded and (b) the entrance is sharp-edged.

Answer 1

Answer:

(a) well-rounded entrance = 0.00469 m³/s

(b) sharp-edged entrance = 0.00389 m³/s

Explanation:

Assumptions are

1. The flow is steady and incompressible

2. The reservoir is open to atmospheric pressure

3. The effect of kinetic energy correction factor is disregarded

Note,

Loss coefficient KL for well-rounded entrance is 0.03 and for

sharp-edged entrance is 0.5

Height of Reservoir Z1 is 8 m

Reference height Z2 is zero

P1 = P2 = P atm

Diameter of the hole = 2.2 cm = 0.022 m

Answer 2

Answer / Explanation:

To solve this question, we need to first understand that water is to be withdrawn from a water reservoir by drilling a hole at the bottom surface. the flow rate of water through the hole is to be determined for the well rounded head and and sharp edged entrance cases.

Moving forward in the solution, we assume that:

(a) The flow is steady and in-compressible

(b) The reservoir is opened to the atmosphere so that the pressure becomes atmospheric pressure at the free surface

(c) The effect of the kinetic energy correction factor is disregarded and so therefore, α = 1.

To properly analyse the assumptions, we could say that:

The loss coefficient Kl = o.5 for the sharp edged entrance and and Kl = 0.03 for the well rounded entrance. We then take the first point at the free surface of the reservoir and the second point at the exit hole . We also take the reference level at the exit of the hole. That is ( Z₂ = 0). Noting that the fluid at both point is opened to the atmosphere. Therefore, ( P₁ = P₂ = Patm) and that the fluid velocity at the free surface is zero (0). That is ( V₁ = 0). Where the energy equation for the control volume between these two points is simplified by the equation below:

P₁ / Pg +  α₁ . V₁²/2g + Z₁ + Hpump. u = P₂/Pg +  α₂ . V₂²/2g + Z₂ + Hturbine. e

= Z₁ = α₂ . V₂²/2g + Hl

Where the head loss is expresses as Hl = Kl. V²/2g .

Now substituting and solving further for V₂, we have:

Z₁ = α₂ .V₂²/2g +  Kl. V²/2g  = 2gz₁ = V₂² ( α₂ + Kl )

V₂ = √  2gz₁ / α₂ + Kl  = √  2gz₁ / 1 + Kl

Now, noting that since α = 1, we should also not that in the special Kl = 0, it reduces to the Torricelli Equation where, V₂ = √ 2gz₁ as expected.

Then the volume flow rate becomes:

V = AcV₂ = π D²hole / 4 √  2gz₁ / 1 + Kl

Now, if we go ahead to substitute the numerical values, the flow rate for both cases will then be:

For well rounded entrance:

V =  π D²hole / 4 √ 2gz₁ / 1 + Kl = π ( 0.022m)² / 4 √ 2 ( 9.81m/s²) (3m) / 1+0.03

= 1.34 x 10 ⁻³ m³/s

While,

For sharp edged entrance:

V = π D²hole / 4 √ 2gz₁ / 1 + Kl =  π ( 0.022m)² / 4  √ 2 ( 9.81m/s²) (3m) / 1+0.5

= 1.11 X 10 ⁻³ m³/s

So therefore, we can conclude that the flow rate in the case of friction-less flow (Kl = 0) is 1.36 X 10 ⁻³ m³/s.

We should also note that the friction less losses also cause the flow rate to decrease by 1.5% for well rounded entrance and 18.5 for the sharp edged entrance.