Mathematical Modelling and Simulation of Aluminium Filling in Conical Pipe and Cylindrical Pipe under High Pressure

Case I: Vertical Cylindrical Pipe

We first consider a case without dissipation. The total force acting on the liquid column, the kinetic energy and potential energy is calculated. The next step is then to analyze the causes of dissipation in the system and then incorporate them in the energy equation.

Total Energy: The potential energy is obtained by considering

the force acting in the liquid column and integrating it by

taking the potential force

$$ E _ {p} = 0 $$

for h = 0 to get the potential

energy as:

$$ E _ {p} = \frac {1}{2} \rho g A h ^ {2} \tag {1} $$

The kinetic energy is given as:

$$ E _ {k} = \frac {1}{2} m v ^ {2} = \frac {1}{2} \rho A (h + H) \dot {h} ^ {2} \tag {2} $$ The total sum of energy in the column is obtained by summing up equations 1 and 2. The differentiated energy together with the 1st principle of thermodynamics gives the balance of energy equation. This gives the energy without dissipation as:

$$ \frac {1}{2} \rho A \left(2 (h + H)\right) \dot {h} \ddot {h} + \dot {h} \dot {h} ^ {2} + 2 g h \dot {h}) = p A \dot {h} $$ (3) To get the total energy equation, energy lost due to dissipation is analyzed and incorporated into equation 3. The losses in the pipe are due to friction on account of the roughness of the wall and also due the singular pressure loss at the entrance or exit of the pipe if liquid rises or if liquid goes down respectively. The energy loss due to friction can be written as:

w W dE p V = ∆  (4)

The pressure loss due to friction as given by Darcy formula is given as
$$\Delta p_w = \frac{\rho f v^2}{2 D_h},$$ which can be written as:
$$\Delta p_w = \frac{\rho f}{2 D_h} (h + H) \dot{h}^2 = \frac{\rho f H}{2 D_h} \left( \frac{h}{H} + 1 \right) \dot{h}^2 = \frac{1}{2} \rho k_w \left( \frac{h}{H} + 1 \right) \dot{h}^2$$
where $k_w = \frac{f h}{D}$ Thus,
$$d E_w = \frac{1}{2} k_w \rho A \left( \frac{h}{H} + 1 \right) \dot{h}^2 |\dot{h}|$$
The singular pressure loss is given as
$$d E_w = \frac{1}{2} k_w \rho \dot{h}^2 |\dot{h}|$$
Equations 5 and 6 are included in Equation 3 to obtain the final equation:
$$\left( (h + H) \ddot{h} + \frac{\dot{h}^2}{2} + g h \right) + \frac{1}{2} \left( k_E + k_w \left( \frac{h}{H} + 1 \right) \dot{h} \right) |\dot{h}| = \frac{p(t)}{\rho}$$

**Case II: Inclined Cylindrical Pipe at an Angle**

Just as is the case with the vertical pipe, the total force acting on the liquid column, the kinetic energy and potential energy is calculated and then the causes of dissipation in the system are analyzed and incorporated in the energy equation.

Total Energy: The total energy is obtained, like in the previous case, by adding the kinetic and potential energy. We also analyze the energy losses due to dissipation and included it to get the governing energy equation. Using the potential energy formula $d E_p = m g d Z$ and considering that $l = \frac{h + H}{\sin(\theta)}$ and $l' = \frac{h}{\sin(\theta)}$, the potential energy in the tube will be given by:
$$E_p = \int m g d Z = \int \rho g A l' d Z = \frac{\rho g A}{\sin(\theta)} \int_0^h z d Z = \frac{1}{2} \rho g A \frac{h^2}{\sin(\theta)}$$

By using the kinetic energy formula and length $l = \frac{h + H}{\sin(\theta)}$ and $l' = \frac{h}{\sin(\theta)}$, for the inclined vertical pipe, the kinetic energy formula is given by:
$$E_k = \frac{1}{2} \rho A \frac{h + H}{\sin(\theta)} \left( \frac{h}{\sin(\theta)} \right)^2$$
Summing up the potential and kinetic energy, the total sum of energy is given by:
$$E = \frac{1}{2} \rho A \left( g \frac{h^2}{\sin(\theta)} + \frac{(h + H) h^2}{\sin^3(\theta)} \right)$$
Differentiating the energy equation, we obtain
$$\dot{E} = \frac{1}{2} \rho A \left( 2 g h \dot{h} + \frac{\dot{h} h^2}{\sin^2(\theta)} + \frac{2 (h + H) h \dot{h}}{\sin^3(\theta)} \right)$$

To get the governing energy equation, we need to analyze and incorporate energy losses due to dissipation into equation 11. To obtain the loss due to friction, $|\dot{V}|$ is replaced with $|\dot{l}|$ into equation 5 and use Darcy's formula.

Thus the energy loss due to friction is given as:
$$E_w = \frac{\rho \lambda A (h + H) h^2 |\dot{l}|}{2 D_h \sin^4(\theta)} = \frac{\rho \lambda \pi r^2 H \left( 1 + \frac{h}{H} \right) h^2 |\dot{l}|}{4 r \sin^4(\theta)} = \rho A k_w \left( 1 + \frac{h}{H} \right) h^2 |\dot{l}| 2 \sin^4(\theta)$$

With $k_w = \frac{\lambda H}{2 r}$

Similarly, the singular pressure loss at the entrance of the pipe is obtained by replacing $|\dot{V}|$ with $|\dot{l}|$ in equation 7 to obtain:
$$\Delta p_E = \frac{1}{2} k_E v^2 = \frac{1}{2} \rho k_E l^2 = \frac{1}{2} \rho k_E \frac{h^2}{\sin^2(\theta)}$$

The final equation energy equation with dissipation is obtained by incorporation equations 12 and 13 into equation 11 and simplifying to get:
$$\left( g h + \frac{\dot{h}^2}{2 \sin^2(\theta)} + \frac{(h + H) h}{2 \sin^2(\theta)} + k_E \frac{\dot{h} |\dot{l}|}{2 \sin^2(\theta)} \right) + \left( k_w \left( 1 + \frac{h}{H} \right) \frac{\dot{h} |\dot{l}|}{2 \sin^2(\theta)} + k_E \frac{\dot{h} |\dot{l}|}{2 \sin^2(\theta)} \right) = \frac{p(t)}{\rho}$$

Case III: Vertical Conical Pipe

The same procedure is done for the vertical conical pipe as was in the previous cases. To get the potential energy, we use the potential energy formula and the radius, $r = R_0 + \frac{z}{\tan \alpha}$

to obtain:

$$ E _ {p} = \rho g \pi \left(\frac {R _ {o} ^ {2} h ^ {2}}{2} + \frac {2 R _ {o} h ^ {3}}{3 \tan \alpha} + \frac {h ^ {4}}{4 \tan^ {2} \alpha}\right) $$

2 2 3 4

2 2 2 3tan 4tan (15)

The Kinetic energy is obtained by getting the kinetic of a

small region z, then integrating over the whole region from

-H to h. We get a function for the velocity of the small region

v(z) using the flow rate Q equation

$$ Q = \frac {d v}{d t}. $$ The volume is given as:

( ) ( ) 2 2 2 3 3 2 0 tan tan 3tan

$$ = \pi \int_ {- H} ^ {h} \left(R _ {o} + \frac {z}{\tan \alpha}\right) ^ {2} d z = \pi \left[ R _ {0} ^ {2} (h + H) + \frac {R _ {o} \left(h ^ {2} - H ^ {2}\right)}{\tan \alpha} + \frac {h ^ {3} + H ^ {3}}{3 \tan \alpha} \right] $$ $$ V = \pi \int_ {- H} ^ {h} \left(R _ {o} + \frac {z}{\tan \alpha}\right) ^ {2} d z = \pi \left[ R _ {0} ^ {2} (h + H) + \frac {R _ {o} \left(h ^ {2} - H ^ {2}\right)}{\tan \alpha} + \frac {h ^ {3} + H ^ {3}}{3 \tan \alpha} \right] $$ h o (16) To obtain the velocity equation, equation 14 is first differentiated to get the flow rate Q and the result divided by the cross-sectional area and then

2 $$ d V = A (z) d z = \pi \left(R _ {o} + \frac {z}{\tan \alpha}\right) ^ {2} d z $$ is replaced into the result, ( ) integrated and simplified to obtain the kinetic energy in the whole pipe as:

$$ = \frac {1}{2} \rho \pi \left(R _ {0} + \frac {h}{\tan \alpha}\right) ^ {3} \left(\frac {h + H}{R _ {o} + \frac {h}{\tan \alpha}}\right) h $$

3 2 0 1 2 tan tan h h H E R h h R ρπ α α  (17)

K

o The total energy is obtained by summing Equations 14 and 16 and differentiating the result to obtain and using the 1st principle of thermodynamics dE pdv = . The total energy equation is thus given as:

$$ \dot {h} + \frac {3}{2} \left(\frac {h + H}{R _ {o} - \frac {H}{\tan \alpha}}\right) \left(\frac {\dot {h} ^ {2}}{\tan \alpha}\right) + \frac {1}{2} \left(R _ {0} + \frac {h}{\tan \alpha}\right) \left(\frac {\dot {h} ^ {2} + 2 (h + H) \dot {h}}{R _ {0} - \frac {H}{\tan \alpha}}\right) = \frac {1}{2} $$ ¨ 2 2   ( )

2 3 1 2 tan 2 tan tan tan o h h H h h H h h p gh R H H R R α α ρ α α

0

0 (18) The possible energy losses due to dissipation are analyzed and incorporated into Equation 17.

The energy loss due to friction is obtained, 2 3 2 0 2 2 1 1 2 tan tan o w w R h h h dE k h R H ρ π α α     = + + +          and the singular pressure loss, 2 2 1 ( ) 2 tan E E o h dE k R h h ρπ α = +  is incorporated into equation 17 and simplified to get the final energy equation:

$$ \begin{array}{l} i + \frac {3}{2} \left(\frac {h + H}{R _ {o} - \frac {H}{\tan \alpha}}\right) \left(\frac {\dot {h} ^ {2}}{\tan \alpha}\right) + \frac {1}{2} \left(R _ {0} + \frac {h}{\tan \alpha}\right) \left(\frac {\dot {h} ^ {2} + 2 (h + H) \dot {h}}{R _ {0} - \frac {H}{\tan \alpha}}\right) \\ + \frac {1}{2} k _ {w} \left(1 + \frac {h}{H}\right) \dot {h} | \dot {h} | + \frac {1}{2} k _ {E} \dot {h} | \dot {h} | = \frac {p}{\rho} \\ \end{array} $$ ¨ 2 2   ( )

2 3 1 2 tan 2 tan tan tan 1 1 1 2 2

h h H h h H h h gh R H H R R

0 α α α α

0 o h p k h h k h h H

  w E

ρ (19)

Case of Cylindrical Pipe

We begin the discussion of the numerical results for the conical pipe inclined at an angle. The modeling was done in Ansys Fluent in which case we used a pressure-based, steady solver. The model used was k-epsilon, standard and enhanced wall treatment. Using Bernoulli’s equation, the kinetic energy was calculated using computed pressure and velocity values from Ansys Fluent. The Bernoulli equation gives $$ p _ {1} + \frac {1}{2} \rho v _ {1} ^ {2} = p _ {2} + \frac {1}{2} \rho v _ {2} ^ {2} $$ and since 2 2 1 2 v v < it implies that 2 1v can be neglected. This leads to ( ) 1 2 2 2 2 P P v ρ − = + For a real fluid, we have $$ p _ {1} + \frac {1}{2} \rho v _ {1} ^ {2} = p _ {2} + \frac {1}{2} \rho u _ {2} ^ {2} + \frac {1}{2} k _ {E} \rho v _ {2} ^ {2} \tag {20} $$ Neglecting 2 1v in Eqn 20, calculating KE, we obtain

1 2 E p P u k v

2 1 2 2 ρ $$ = \frac {p _ {1} - P _ {2} - 1}{o v ^ {2}} $$

2 2 ρ From the simulation in Fluent, for the cylindrical pipe $$ v _ {2} ^ {2} = + \frac {2 (7 3 7 7 . 2 4 3 - 8 1 5 . 6 4 1 3)}{2 4 6 0} = 5. 3 3 2 8 $$ $$ k _ {E} = 1 - \frac {0 . 7 0 9 3 2 9 2 ^ {2}}{5 . 3 3 2 8 3} $$ The contours of the total pressure and the Y-velocity are plotted (Figure 3 & 4).

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Case of Conical Pipe

The same procedure was done for the case of the conical, the was calculated.

For the simulation in Fluent, for the conical pipe $$ v _ {2} ^ {2} = \frac {2 (7 2 5 2 . 4 1 0 8 - 2 1 9 0 . 4 4 9 5)}{2 4 6 0} = 4. 1 1 5 4 $$ $$ k _ {E} = 1 - \frac {1 . 3 3 9 8 ^ {2}}{4 . 1 1 5 4} = 0. 3 $$ The contours of total pressure and Y-velocity for the conical pipe were plotted (Figures 5 & 6).

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Piecewise Function

A piecewise function was used to define the pressure. The pressure was given as a linear function as

0, 1 ( ) 817.77 817.77, 1 10 7360, 10

t p t t t t $$ () = \left\{ \begin{array}{l l} 0, & t < 1 \\ 8 1 7. 7 7 t - 8 1 7. 7 7, & 1 < t < 1 \\ 7 3 6 0, & t > 1 0 \end{array} \right. $$ Initially we had used a jump in pressure given as

0, 1 ( ) 7360 1

t p t t $$ ) = \left\{ \begin{array}{l l} 0, & t < 1 \\ 7 3 6 0 & t < 1 \end{array} \right. $$ To compare the effects of the different pressure function on the height function equation, the graph was plotted (Figure 7).

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