Question Number 225832 by mr W last updated on 14/Nov/25
Commented by mr W last updated on 14/Nov/25
$${which}\:{tank}\:{will}\:{empty}\:{first}\:{and}\:{why}? \\ $$
Commented by fantastic last updated on 14/Nov/25
$${A}_{\mathrm{1}} {v}_{\mathrm{1}} ={a}_{\mathrm{2}} {v}_{\mathrm{2}} \\ $$$${A}'{v}'={a}_{\mathrm{2}} {v}'' \\ $$$${A}_{\mathrm{1}} {v}_{\mathrm{1}} ={A}'{v}' \\ $$$$\Rightarrow{a}_{\mathrm{2}} {v}''={a}_{\mathrm{2}} {v}_{\mathrm{2}} \\ $$$${v}_{\mathrm{2}} ={v}''\Rightarrow{same}\:{time} \\ $$
Commented by fantastic last updated on 14/Nov/25
![A_1 v_1 =a_2 v_2 A′v′=a_2 v′′ A_1 v_1 =A′v′ ⇒a_2 v′′=a_2 v_2 v_2 =v′′⇒same time [A′=cross section of upper side] [A_1 =cross section of lower side] [a_2 cross section of tap ]](https://www.tinkutara.com/question/Q225836.png)
$${A}_{\mathrm{1}} {v}_{\mathrm{1}} ={a}_{\mathrm{2}} {v}_{\mathrm{2}} \\ $$$${A}'{v}'={a}_{\mathrm{2}} {v}'' \\ $$$${A}_{\mathrm{1}} {v}_{\mathrm{1}} ={A}'{v}' \\ $$$$\Rightarrow{a}_{\mathrm{2}} {v}''={a}_{\mathrm{2}} {v}_{\mathrm{2}} \\ $$$${v}_{\mathrm{2}} ={v}''\Rightarrow{same}\:{time} \\ $$$$\left[{A}'={cross}\:{section}\:{of}\:{upper}\:{side}\right] \\ $$$$\left[{A}_{\mathrm{1}} ={cross}\:{section}\:{of}\:{lower}\:{side}\right] \\ $$$$\left[{a}_{\mathrm{2}} {cross}\:{section}\:{of}\:{tap}\:\right] \\ $$
Answered by mahdipoor last updated on 14/Nov/25
$$\mathrm{P}_{\mathrm{0}} +\frac{\rho\mathrm{u}^{\mathrm{2}} }{\mathrm{2}}+\rho\mathrm{gh}=\mathrm{P}_{\mathrm{0}} +\frac{\rho\mathrm{v}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{0} \\ $$$$\mathrm{and}\:: \\ $$$$\mathrm{A}_{\mathrm{valve}} \mathrm{v}\rho=\mathrm{A}_{\mathrm{surface}} \mathrm{u}\rho \\ $$$$\mathrm{A}_{\mathrm{surface}} \left(\mathrm{h}\right)=\mathrm{a}_{\mathrm{0}} +\mathrm{b}_{\mathrm{0}} \mathrm{h} \\ $$$$\Rightarrow\mathrm{tankX\begin{cases}{\mathrm{b}_{\mathrm{0}} =\frac{\mathrm{A}_{\mathrm{max}} −\mathrm{A}_{\mathrm{min}} }{\mathrm{H}}}\\{\mathrm{a}_{\mathrm{0}} =\mathrm{A}_{\mathrm{min}} }\end{cases}} \\ $$$$\Rightarrow\mathrm{tankY\begin{cases}{\mathrm{b}_{\mathrm{0}} =\frac{\mathrm{A}_{\mathrm{min}} −\mathrm{A}_{\mathrm{max}} }{\mathrm{H}}}\\{\mathrm{a}_{\mathrm{0}} =\mathrm{A}_{\mathrm{max}} }\end{cases}} \\ $$$$\mathrm{u}=\overset{.} {\mathrm{h}} \\ $$$$\mathrm{v}=\frac{\mathrm{a}_{\mathrm{0}} +\mathrm{b}_{\mathrm{0}} \mathrm{h}}{\mathrm{A}_{\mathrm{valve}} }\mathrm{u}=\mathrm{u}\left(\mathrm{a}+\mathrm{bh}\right)=\mathrm{u}.\mathrm{f}\left(\mathrm{h}\right) \\ $$$$\Rightarrow \\ $$$$\sqrt{\mathrm{v}^{\mathrm{2}} −\mathrm{u}^{\mathrm{2}} }=\sqrt{\mathrm{2gh}}=\mid\overset{.} {\mathrm{h}}\mid\sqrt{\mathrm{f}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\mathrm{dt}=−\sqrt{\frac{\mathrm{f}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2gh}}}\mathrm{dh}=−\mathrm{g}\left(\mathrm{h}\right)\mathrm{dh}\Rightarrow \\ $$$$\mathrm{time}=\int_{\mathrm{0}} ^{\:\mathrm{H}} \mathrm{g}\left(\mathrm{h}\right)\mathrm{dh} \\ $$$$… \\ $$
Answered by mr W last updated on 14/Nov/25
Commented by mr W last updated on 14/Nov/25
$${assume}\:{the}\:{cross}−{section}\:{of}\:{the} \\ $$$${tap}\:{A}_{\mathrm{0}} \:{is}\:{very}\:{small}\:{compared}\:{with} \\ $$$${the}\:{cross}−{section}\:{of}\:{the}\:{tank}.\:{then} \\ $$$${we}\:{have} \\ $$$${v}=\sqrt{\mathrm{2}{gh}}=\sqrt{\mathrm{2}{g}\left({H}−{x}\right)} \\ $$
Commented by mr W last updated on 15/Nov/25
![Tank X: A_2 <A_1 A_1 =πr_1 ^2 A_2 =πr_2 ^2 let ξ=(x/H), 0≤ξ≤1 r=r_1 +(((r_2 −r_1 )x)/H)=r_1 +(r_2 −r_1 )ξ =r_1 [1+((r_2 /r_1 )−1)ξ]=r_1 (1+λξ) with λ=(r_2 /r_1 )−1=μ−1 and μ=(r_2 /r_1 ). A=πr^2 =πr_1 ^2 (1+λξ)^2 v=(√(2g(H−x)))=(√(2gH(1−ξ))) Q=A_0 v=A_0 (√(2gH(1−ξ))) dV=Adx=Qdt=A_0 (√(2gH(1−ξ))) dt πr_1 ^2 (1+λξ)^2 dx=A_0 (√(2gH(1−ξ))) dt πr_1 ^2 H(1+λξ)^2 dξ=A_0 (√(2gH(1−ξ))) dt ((πr_1 ^2 )/A_0 )(√(H/(2g)))(1+λξ)^2 dξ=(√(1−ξ)) dt let k=(π/A_0 )(√(H/(2g)))=constant kr_1 ^2 (1+λξ)^2 dξ=(√(1−ξ)) dt dt=((kr_1 ^2 (1+λξ)^2 dξ)/( (√(1−ξ)))) ∫_0 ^T dt=∫_0 ^1 ((kr_1 ^2 (1+λξ)^2 dξ)/( (√(1−ξ)))) T=kr_1 ^2 ∫_0 ^1 (((1+λξ)^2 dξ)/( (√(1−ξ)))) =((2kr_1 ^2 )/(15))[(√(1−ξ)){λ^2 (3ξ^2 +4ξ+8)+10λ(ξ+2)+15}]_1 ^0 =((2kr_1 ^2 (8λ^2 +20λ+15))/(15)) =((2kr_1 ^2 [8((r_2 /r_1 )−1)^2 +20((r_2 /r_1 )−1)+15])/(15)) =((2kr_1 ^2 [8((r_2 /r_1 ))^2 +4((r_2 /r_1 ))+3])/(15)) Tank X: T_1 =((2k[8((r_2 /r_1 ))^2 +4((r_2 /r_1 ))+3]r_1 ^2 )/(15)) Tank Y: T_2 =((2k[8((r_1 /r_2 ))^2 +4((r_1 /r_2 ))+3]r_2 ^2 )/(15)) (T_1 /T_2 )=(([8((r_2 /r_1 ))^2 +4((r_2 /r_1 ))+3]r_1 ^2 )/([8((r_1 /r_2 ))^2 +4((r_1 /r_2 ))+3]r_2 ^2 )) (T_1 /T_2 )=((8μ^2 +4μ+3)/([8((1/μ))^2 +4((1/μ))+3]μ^2 )) ⇒(T_1 /T_2 )=((8μ^2 +4μ+3)/(3μ^2 +4μ+8)) since 0≤μ=(r_2 /r_1 )≤1, ⇒(3/8)≤(T_1 /T_2 )≤1 i.e. Tank X will empty first. (T_1 /T_2 )=1 for μ=1 (cylindrical tank) (T_1 /T_2 )=(3/8) for μ=0 (conical tank)](https://www.tinkutara.com/question/Q225848.png)
$${Tank}\:{X}:\:{A}_{\mathrm{2}} <{A}_{\mathrm{1}} \\ $$$${A}_{\mathrm{1}} =\pi{r}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${A}_{\mathrm{2}} =\pi{r}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${let}\:\xi=\frac{{x}}{{H}},\:\mathrm{0}\leqslant\xi\leqslant\mathrm{1} \\ $$$${r}={r}_{\mathrm{1}} +\frac{\left({r}_{\mathrm{2}} −{r}_{\mathrm{1}} \right){x}}{{H}}={r}_{\mathrm{1}} +\left({r}_{\mathrm{2}} −{r}_{\mathrm{1}} \right)\xi \\ $$$$\:\:={r}_{\mathrm{1}} \left[\mathrm{1}+\left(\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }−\mathrm{1}\right)\xi\right]={r}_{\mathrm{1}} \left(\mathrm{1}+\lambda\xi\right) \\ $$$${with}\:\lambda=\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }−\mathrm{1}=\mu−\mathrm{1}\:{and}\:\mu=\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }. \\ $$$${A}=\pi{r}^{\mathrm{2}} =\pi{r}_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{1}+\lambda\xi\right)^{\mathrm{2}} \\ $$$${v}=\sqrt{\mathrm{2}{g}\left({H}−{x}\right)}=\sqrt{\mathrm{2}{gH}\left(\mathrm{1}−\xi\right)} \\ $$$${Q}={A}_{\mathrm{0}} {v}={A}_{\mathrm{0}} \sqrt{\mathrm{2}{gH}\left(\mathrm{1}−\xi\right)} \\ $$$${dV}={Adx}={Qdt}={A}_{\mathrm{0}} \sqrt{\mathrm{2}{gH}\left(\mathrm{1}−\xi\right)}\:{dt} \\ $$$$\pi{r}_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{1}+\lambda\xi\right)^{\mathrm{2}} {dx}={A}_{\mathrm{0}} \sqrt{\mathrm{2}{gH}\left(\mathrm{1}−\xi\right)}\:{dt} \\ $$$$\pi{r}_{\mathrm{1}} ^{\mathrm{2}} {H}\left(\mathrm{1}+\lambda\xi\right)^{\mathrm{2}} {d}\xi={A}_{\mathrm{0}} \sqrt{\mathrm{2}{gH}\left(\mathrm{1}−\xi\right)}\:{dt} \\ $$$$\frac{\pi{r}_{\mathrm{1}} ^{\mathrm{2}} }{{A}_{\mathrm{0}} }\sqrt{\frac{{H}}{\mathrm{2}{g}}}\left(\mathrm{1}+\lambda\xi\right)^{\mathrm{2}} {d}\xi=\sqrt{\mathrm{1}−\xi}\:{dt} \\ $$$${let}\:{k}=\frac{\pi}{{A}_{\mathrm{0}} }\sqrt{\frac{{H}}{\mathrm{2}{g}}}={constant} \\ $$$${kr}_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{1}+\lambda\xi\right)^{\mathrm{2}} {d}\xi=\sqrt{\mathrm{1}−\xi}\:{dt} \\ $$$${dt}=\frac{{kr}_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{1}+\lambda\xi\right)^{\mathrm{2}} {d}\xi}{\:\sqrt{\mathrm{1}−\xi}} \\ $$$$\int_{\mathrm{0}} ^{{T}} {dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{kr}_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{1}+\lambda\xi\right)^{\mathrm{2}} {d}\xi}{\:\sqrt{\mathrm{1}−\xi}} \\ $$$${T}={kr}_{\mathrm{1}} ^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}+\lambda\xi\right)^{\mathrm{2}} {d}\xi}{\:\sqrt{\mathrm{1}−\xi}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{2}{kr}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{15}}\left[\sqrt{\mathrm{1}−\xi}\left\{\lambda^{\mathrm{2}} \left(\mathrm{3}\xi^{\mathrm{2}} +\mathrm{4}\xi+\mathrm{8}\right)+\mathrm{10}\lambda\left(\xi+\mathrm{2}\right)+\mathrm{15}\right\}\right]_{\mathrm{1}} ^{\mathrm{0}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{2}{kr}_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{8}\lambda^{\mathrm{2}} +\mathrm{20}\lambda+\mathrm{15}\right)}{\mathrm{15}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{2}{kr}_{\mathrm{1}} ^{\mathrm{2}} \left[\mathrm{8}\left(\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{20}\left(\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }−\mathrm{1}\right)+\mathrm{15}\right]}{\mathrm{15}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{2}{kr}_{\mathrm{1}} ^{\mathrm{2}} \left[\mathrm{8}\left(\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }\right)^{\mathrm{2}} +\mathrm{4}\left(\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }\right)+\mathrm{3}\right]}{\mathrm{15}} \\ $$$${Tank}\:{X}: \\ $$$$\:{T}_{\mathrm{1}} \:=\frac{\mathrm{2}{k}\left[\mathrm{8}\left(\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }\right)^{\mathrm{2}} +\mathrm{4}\left(\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }\right)+\mathrm{3}\right]{r}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{15}} \\ $$$${Tank}\:{Y}: \\ $$$$\:{T}_{\mathrm{2}} \:=\frac{\mathrm{2}{k}\left[\mathrm{8}\left(\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }\right)^{\mathrm{2}} +\mathrm{4}\left(\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }\right)+\mathrm{3}\right]{r}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{15}} \\ $$$$\frac{{T}_{\mathrm{1}} }{{T}_{\mathrm{2}} }=\frac{\left[\mathrm{8}\left(\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }\right)^{\mathrm{2}} +\mathrm{4}\left(\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }\right)+\mathrm{3}\right]{r}_{\mathrm{1}} ^{\mathrm{2}} }{\left[\mathrm{8}\left(\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }\right)^{\mathrm{2}} +\mathrm{4}\left(\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }\right)+\mathrm{3}\right]{r}_{\mathrm{2}} ^{\mathrm{2}} } \\ $$$$\frac{{T}_{\mathrm{1}} }{{T}_{\mathrm{2}} }=\frac{\mathrm{8}\mu^{\mathrm{2}} +\mathrm{4}\mu+\mathrm{3}}{\left[\mathrm{8}\left(\frac{\mathrm{1}}{\mu}\right)^{\mathrm{2}} +\mathrm{4}\left(\frac{\mathrm{1}}{\mu}\right)+\mathrm{3}\right]\mu^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{T}_{\mathrm{1}} }{{T}_{\mathrm{2}} }=\frac{\mathrm{8}\mu^{\mathrm{2}} +\mathrm{4}\mu+\mathrm{3}}{\mathrm{3}\mu^{\mathrm{2}} +\mathrm{4}\mu+\mathrm{8}} \\ $$$${since}\:\mathrm{0}\leqslant\mu=\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }\leqslant\mathrm{1}, \\ $$$$\Rightarrow\frac{\mathrm{3}}{\mathrm{8}}\leqslant\frac{{T}_{\mathrm{1}} }{{T}_{\mathrm{2}} }\leqslant\mathrm{1} \\ $$$${i}.{e}.\:{Tank}\:{X}\:{will}\:{empty}\:{first}. \\ $$$$\frac{{T}_{\mathrm{1}} }{{T}_{\mathrm{2}} }=\mathrm{1}\:{for}\:\mu=\mathrm{1}\:\left({cylindrical}\:{tank}\right) \\ $$$$\frac{{T}_{\mathrm{1}} }{{T}_{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{8}}\:{for}\:\mu=\mathrm{0}\:\left({conical}\:{tank}\right) \\ $$
Commented by mr W last updated on 15/Nov/25
Commented by mr W last updated on 15/Nov/25
