A Volume Gauge. A tank containing a liquid has turns of wire wrapped around it, causing it to act like an inductor. The liquid content of the tank can be measured by using its inductance to determine the height of the liquid in the tank. The inductance of the tank changes from a value of L0 corresponding to a relative permeability of 1 when the tank is empty to a value of Lf corresponding to a relative permeability of Km (the relative permeability of the liquid) when the tank is full. The appropriate electronic circuitry can determine the inductance to five significant figures and thus the effective relative permeability of the combined air and liquid within the rectangular cavity of the tank. The four sides of the tank each have width W and height D (Fig. P30.70). The height of the liquid in the tank is d. You can ignore any fringing effects and assume that the relative permeability of the material of which the tank is made can be ignored. (a) Derive an expression for d as a function of L, the inductance corresponding to a certain fluid height, L0 , Lf, and D. (b) What is the inductance (to five significant figures) for a tank$\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{4}$}\right.$full,$\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.$full,$\raisebox{1ex}{$\mathbf{3}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{4}$}\right.$full, and completely full if the tank contains liquid oxygen? Take L₀=0.63000 H. The magnetic susceptibility of liquid oxygen is Xm=1.52* 10-3 .
(c) Repeat part (b) for mercury. The magnetic susceptibility of
mercury is given in Table 28.1. (d) For which material is this
volume gauge more practical?

Step by step solution

01

Important Concepts and Formula

Magnetic flux is given by

$\mathbf{\Phi }\mathbf{=}\mathbf{BA}$

WhereBis the magnetic field and A is the Area

$\mathbf{L}\mathbf{=}\mathbf{N}\frac{\mathbf{\Phi }}{\mathbf{i}}$

Self-Inductance and Flux relation is

Where N is the number of turns $\mathbf{\Phi }$ is the flux and i is the current in the coil

02

Find relation between the width and inductance

We are given at d is the width of the liquid and L is the self-inductance of the coil

The magnetic flux is through the tank both through liquid and air. So total magnetic flux is given by

$\Phi =B_L{A}_L+B_{Air}{A}_{Air}$

Substitute BA by $\frac{\mu Ni}{W}$where is the width and A is the area of the tank and could be substituted by Wxd

Substitute all of this into the equation and get

$\begin{array}{r}{\mathrm{\Phi }}_B=\frac{{\mu }_{0}Ni}{W}\left(\right(D-d\left)W\right)+\frac{K{\mu }_{0}Ni}{W}(W\times d)\\ {\mathrm{\Phi }}_B={\mu }_{0}Ni(D-d+Kd)\end{array}$

Substitute${\Phi }_B$ into$L=\frac{N{\Phi }_B}{i}$ and get

$\begin{array}{r}L={\mu }_0{N}^2\left[\right(D-d)+Kd]\\ =L_0-L_0\frac{d}{D}+L_f\frac{d}{D}\end{array}$

Collecting Like terms we get

$d=\left(\frac{L-L_0}{L_f-L_0}\right)D$

03

For oxygen

Use the given value of$L_0=0.63H$ and$X_m=1.52\times {10}^{-3}$ in the equation

$L=L_0\left(1+X_m\frac{d}{D}\right)$

Input the values of in terms of to obtain the values of inductance

At $d=\frac{1}{4}D,L=0.63024\mathrm{H}$,

At $d=\frac{1}{2}D,L=0.63048H$,

At$d=\frac{3}{4}D,L=0.63072H$

04

For Mercury

Use the given value of$L_o=0.63H$and$X_m=-2.9x{10}^{-5}$ in the equation

$L=L_o(1+X_m\frac{d}{D})$

Input the values of in terms of to obtain the values of inductance

At $d=\frac{1}{4}D,L=0.63000H$,

At $d=\frac{1}{2}D,L=0.62999H$,

At$d=\frac{3}{4}D,L=0.62999H$

05

Comparison

As shown in the results Oxygen has higher inductance which has more spread through the tank so the volume gauge is much better for liquid oxygen

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