# Converting a normal water valve into a gravity feed valve

It is time to calculate the force of the coil when the coil is energized (Fcoil).
This calculation is not intuitive so we need to do some simplifications.

We assume that the magnetic field inside the energized solenoid is uniform.
Since it is uniform we can apply the Amperes law in order to calculate the magnetic field B inside the solenoid.

We choose a rectangular surface S with l : length and h : height.
Take a look at the dia diagram made in the first page.

We apply the Amperes law into surface S
$\oint\limits_S {\vec Bd\vec S} =\mu _o {\rm I}_{enclosed}=\mu _o \left({\frac{N}{\lambda }l}\right)I$ (1)
$\oint\limits_S {\vec Bd\vec S} = \int {\vec Bd\vec l} + \int {\vec Bd\vec h} + \int {\vec Bd\vec l} +\int {\vec Bd\vec h}=Bl+0+0+0= Bl$ (2)
Using (1)+(2) we can find the magnetic field inside the solenoid : $B =\mu _o \frac{N}{\lambda }I$ (3)

From the magnetic flux definition we have :
$\Phi _{one\_turn} =\iint {\vec {\rm B}d\vec A} = BA\mathop =\limits^{(3)} \mu _o \frac{N}{\lambda }I(t)(\pi R^2 )$ (4)

Using the (4) we get : $\Phi _{total} = N\Phi _{one\_turn} = \mu _o \frac{{N^2 }}{\lambda }(\pi R^2 )I(t)$ (5)

From the faraday law we know that :
$EMF = V_L = - \frac{{d\Phi _{total} }}{{dt}}\mathop \Rightarrow \limits^{(5)} V_L = \frac{{\mu _o N^2 }}{\lambda }(\pi R^2 )\frac{{dI(t)}}{{dt}}$  (6)

We know that :
$V_L = L\frac{{dI(t)}}{{dt}}$  (7)

So using (6) and (7) we get : $L =\frac{{\mu _o N^2 }}{\lambda }(\pi R^2 )$ (8)

The energy of the magnetic field is :

$W = \int {dw = \int {Pdt = \int {V_L } } } Idt\mathop =\limits^{(7)} \int {L\frac{{dI(t)}}{{dt}}I(t)dt = \frac{1}{2}} LI^2$  (9)

Using (8) and (9) we get :
$W_{empty} = \frac{1}{2}\frac{{\mu _o N}}{\lambda }^2 (\pi R^2 )I^2$  (10)
Notice that this is the energy of the magnetic field when the core is outside the coil (0V case)

When the voltage is applied then the core is fully inside the coil. in that case the energy becomes :

$W_{filled} = \frac{1}{2}\frac{{\mu _{new} N}}{\lambda }^2 (\pi R^2 )I^2$  (11)
$\Delta W = W_{filled} - W_{empty} = \frac{1}{2}\frac{{N^2 }}{\lambda }(\pi R^2 )I^2 (\mu _{new} - \mu _o )$  (12)

Finally we can calculate the force applied at the core from the magnetic field using the (12) :

$F_{coil} = \frac{{\Delta W}}{\lambda } = \frac{1}{2}\frac{{N^2 }}{{\lambda ^2 }}(\pi R^2 )I^2 (\mu _{new} - \mu _o )$ (13)

Let's now look at the rest of the forces.

On the 0V state we have :

Fspring + FLiquid1 = Fw + Fstraw

Where :
Fspring is the force of the spring. Using the Hookes law we have Fspring = -kξ

Fw is the weight. Fw=mg

FLiquid1 can be calculated by Pascals law.
P2-P1=pgH  => (using P=F/A) => FLiquid1 = [(FL2/A2) - pgh]A1

Notice that : FL2= FLiquid2 - Fsurface_y

Fstraw is the reaction force of the straw

At 12V state we have :

Fspring + FLiquid1 = Fw + Fcoil

The condition in order to work our system is that the
Fcoil > Fspring + FLiquid1 - Fw

That's all for now !

Stay tuned !