Question 7.3: Hydraulic Capacitance of a Conical Tank Derive the capacitan...

a. From Figure 7.5b, the radius r of the cross-section A at an arbitrary height is

r=h \tan \alpha=h \frac{R}{H}.

Thus, the volume of the liquid is

V(h)=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \frac{\pi R^{2}}{H^{2}} h^{3}

and the stored mass is

m=\rho V(h)=\frac{1}{3} \frac{\rho \pi R^{2}}{H^{2}} h^{3}.

Note that the pressure caused by the height of the liquid is

p=p_{\mathrm{a}}+\rho g h,

which gives

\frac{\mathrm{d} p}{\mathrm{~d} h}=\rho g.

Thus, the capacitance of the conical tank is

C=\frac{\mathrm{d} m}{\mathrm{~d} p}=\frac{\mathrm{d} m}{\mathrm{~d} h} \frac{\mathrm{d} h}{\mathrm{~d} p}=\left(\frac{\rho \pi R^{2}}{H^{2}} h^{2}\right)\left(\frac{1}{\rho g}\right)=\frac{\pi R^{2} h^{2}}{H^{2} g} .

b. The hydraulic capacitance can also be derived directly using C=A(h) / g, which yields

C=\frac{\pi r^{2}}{g}=\left(\frac{\pi}{g}\right)\left(\frac{R^{2} h^{2}}{H^{2}}\right)=\frac{\pi R^{2} h^{2}}{H^{2} g}.