91影视

The total capacity can be considered as two parallel capacitors, the first is the one of air and the second is the one of fuel so the total or effective capacity is given by

$C_{eff}=C_a+C_f$ (1)

Air capacitance: We know that the capacitance of parallel plate capacitors, in general, is given by

$C_a=\frac{{蔚}_0{A}_a}{d_a}$ (2)

Where

$$\begin{gathered} A_a=(L-h)w \\ {d}_a=d \end{gathered}$$

Substitution in (2) gives

$C_{伪}=\frac{{蔚}_o\left(L-h\right)蝇}{d}$ (3)

Fuel capacitance: The capacitance of a parallel plate capacitor filled with fuel, in general, is given by

$C_f=\frac{蔚A_f}{d_f}$ (4)

Where:

$$\begin{gathered} A_f=hw \\ {d}_f=d \\ 蔚=K{蔚}_{慰} \end{gathered}$$

Substitution in (4) gives

$C_J=\frac{K{蔚}_{o}hw}{d}$ (5)

Effective capacitance: Also effective capacitance is a capacitance of parallel plates so it is given by

$C_{eff}=\frac{{蔚}_{eff}{A}_{eff}}{d_{eff}}$

Where:

$$\begin{gathered} A_{eff}=Lw \\ {d}_{eff}=d \\ {蔚}_{eff}=K_{eff}{蔚}_{慰} \end{gathered}$$

Substitution in (5) gives

$C_{eff}=\frac{K_{eff}{蔚}_{0}L蝇}{d}$ (6)

Step 4 Substitution in (1) gives

$\frac{K_{eff}{蔚}_{o}Lw}{d}=\frac{K{蔚}_{o}hw}{d}+\frac{{蔚}_o(L鈭�h)w}{d}$ (7)

Simplify (7) to get that.

$K_{eff}L=Kh+(L-h)鈬�K_{eff}L=h(K-1)+L鈬�K_{eff}=h/L(K-1)+1$

The effective constant for K = 1.95

when $\frac{h}{L}=\frac{1}{4}$:

$K_{eff}\frac{1}{4}(1.95-1)+1=1.24$

when$\frac{h}{L}=\frac{1}{2}$:

$K_{eff}\frac{1}{2}(1.95-1)+1=1.24=1.47$

when $\frac{h}{L}=\frac{3}{4}$:

localid="1668315039732" $K_{eff}\frac{3}{4}(1.95-1)+1=1.24=1.74$

The effective constant for K = 33.0.

when $\frac{h}{L}=\frac{1}{4}$:

$K_{eff}\frac{1}{4}(33-1)+1=9$

when $\frac{h}{L}=\frac{1}{2}$:

$K_{eff}\frac{1}{2}(33-1)+1=17$

when $\frac{h}{L}=\frac{3}{4}$:

$K_{eff}\frac{1}{4}(33-1)+1=25$

It's obvious that methanol is more practical than gasoline because we can easily change the capacity better than using gasoline