Engineering Fluid Mechanics. ?

[Ash L]

Homework Statement

Okay so the question goes like this:

In the design problem to follow, the constants in table 1 may be assumed.

Density P = 1000kg/m^3

Viscosity u = 10^(-3) pas

Gravitational Constant g = 9.8m/s^2

There is a tank in this question. Let's call it Tank 1. I can't put a picture of it up but it looks like an inverted cone with the sharp tip sliced off so water can come out of this point. The liquid inside Tank 1 has a height of H1.

The theoretical equations that describe tank 1’s behaviour while draining (with no
inflow) are given below:

A1*dH1/dt = -a1v1(t) Equation (1)

H1(t) =
v[2] (t)/2g Equation (2)
[1]


Q1) Verify the equation for H1(t), equation (2), using the Engineering Bernoulli
Equation.

Q2) Determine an expression for the cross sectional area A1 of the fluid in the tank
as a function of β and H1(t).

Homework Equations

I think the Bernoulli equation is v^2/2g + z + P/g = c I have been sitting down staring at this problem for 5 days and haven't got a clue how to do it. Could someone please help me?

 

[KataKoniK]

Hello! I would be happy to help you with this problem. Let's break it down step by step.

Q1) Verify the equation for H1(t), equation (2), using the Engineering Bernoulli Equation.

First, let's review the Bernoulli equation:

P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2

Where:
P1 and P2 are the pressures at points 1 and 2
ρ is the density of the fluid
v1 and v2 are the velocities at points 1 and 2
g is the gravitational constant
h1 and h2 are the heights at points 1 and 2

Now, let's apply this equation to our problem. Since there is no inflow, we can assume that the velocities at both points are 0. Therefore, the equation becomes:

P1 + ρgh1 = P2 + ρgh2

We can also assume that the pressure at the top of the tank (point 1) is equal to atmospheric pressure, which we can represent as P0. Point 2 is at the opening of the tank, where the water is draining out. At this point, the pressure is also atmospheric, so P2 = P0.

Substituting these values into our equation, we get:

P0 + ρgh1 = P0 + ρgh2

Since P0 is on both sides of the equation, we can cancel it out. We are left with:

ρgh1 = ρgh2

Now, we can solve for h2:

h2 = (h1g)/g

Simplifying this, we get:

h2 = h1/2

This is the same equation as (2) in the problem, so we have verified it using the Bernoulli equation.

Q2) Determine an expression for the cross sectional area A1 of the fluid in the tank as a function of β and H1(t).

To find A1, we will use the equation for the volume of a cone:

V = (1/3)πr^2h

Where:
V is the volume of the cone
r is the radius of the base of the cone
h is the height of the cone

In our problem, the volume of the cone is