A refreshing gusher

We know that $v_{efflux} = \sqrt{2gh}$

Applying equation of continuity ,

$v_{top} = \frac{r^2}{R^2} \sqrt{2gh} = -\frac{dh}{dt}$

$\Rightarrow \int_{H}^{0} \frac{dh}{\sqrt{h}} = - \frac{r^2 \sqrt{2g}}{R^2} \int_{0}^{t} dt$

$\Rightarrow t = \frac{R^2}{r^2} \sqrt{\frac{2H}{g}} = \fbox{31.9}$

Let the rate of the liquid flowing out of the spigot be $\frac{\mathrm dV}{\mathrm dt}$ = -(rate of volume of the liquid lost as it flows out) . The minus sign is for indicating the decrease in volume . Let $A$ and $a$ be the area of cross-section of the base of the container and the spigot respectively . Let $v$ be the velocity in m/s with which the liquid flows out of the spigot . Thus , the volume of liquid which flows out per second from the spigot = $av$ . Therefore ,

$\frac{\mathrm dV}{\mathrm dt} = -av$ $(Eq. 1)$

Let mass $m$ of the liquid be at height $h$ above the base of the container . It has potential energy = $mgh$ . When this liquid flows out of the spigot , it will have a kinetic energy = $\frac{1}{2}mv^2$ . Hence , using conservation of energy and assuming no loss of energy due to viscosity of liquid or forces of cohesion , we can say that $mgh=\frac{1}{2}mv^2$

$\Rightarrow v=\sqrt{2gh}$ $(Eq.2)$

From $Eq.1$ and $Eq.2$ , we conclude $\frac{\mathrm dV}{\mathrm dt}=-a\sqrt{2gh}$ . Writing the volume of liquid in the container as a function of time , we have $V(t)=Ah(t)$ where $h$ is the height of the liquid in the cylinder as a function of time . Thus , we can write $\frac{\mathrm d\big (V(t)\big )}{dt}=A\frac{\mathrm d\big (h(t)\big )}{dt}$ . Therefore , $A\frac{\mathrm d\big (h(t)\big )}{dt}=-a\sqrt{2gh}\Rightarrow \frac{\mathrm d\big (h(t)\big )}{dt}=-\frac{a}{A}\sqrt{2gh}$ .

Simply put , we have $\frac{\mathrm dh}{\mathrm dt}=-k\sqrt{h}$ where $k=\frac{a}{A}\sqrt{2g}$ .

After variable separation , we have $\frac{\mathrm dh}{\sqrt{h}}=-k$ $\mathrm dt$ . Let the height of the cylinder be $H$ . As the height of the liquid decreases from $H$ to zero , time increases from zero to $t$ . Integrating the above obtained differential equation after applying proper limits , we have

$\int_H^0 \frac{\mathrm dh}{\sqrt{h}}=\int_0^t k \, \mathrm{dt} \Rightarrow -2\sqrt{H}=-kt \Rightarrow t=\frac{2\sqrt{H}}{k}$ .

Substituting for $k$ , we have $t=\frac{A}{a}\sqrt\frac{2H}{g}$ .

Substituting the values , we get $t=\frac{\pi (0.1)^2}{\pi (0.01)^2}\sqrt\frac{2\times 0.5}{9.8}$

$\Rightarrow t=\boxed{31.94}$ seconds .

Bernoulli's Equation says $\frac{1}{2} \rho v_1^2 + \rho g h_1+p_1=\frac{1}{2} \rho v_2^2+\rho g h_2+p_2$ . Because $p_1=p_2=p_{atm}$ in this case and the velocity at the top of the container is $v_1=0$ approximately. Hence, we have $v_2=\sqrt{2g\Delta h}$ . In the case, $\Delta h=h=0.5m$

It follows that the rate of flow at bottom is $Q=v_2 A_0=\sqrt{2gh} A_0$ in which $A_0$ denotes the cross sectional area of the spigot.

Then we have $\frac{dV}{dt}=-\sqrt{2gh} A_0$ , which can be reduced to $\frac{dV}{dt}=-\sqrt{2g} A_0 \sqrt{V/A_1}$ in which $A_1$ denotes the cross sectional area of the vat and $V$ denotes the volume of lemonade.

Solve the differential equation to obtain $2V^{1/2}=-A_{1}^{-1/2} A_0\sqrt{2g}t+C$ in which $C$ is an unknown constant. In the equation, $A_1=0.1^2\pi, A_0=0.01^2\pi$

Substitute $t=0$ and $V=0.5*0.1^2\pi$ into the equation and then we get $C=2\sqrt{\pi/200}$

Then substitute $V=0$ into the equation and solve for t. Eventually, $t= 31.8s$

You need to find the debit of the water from the spigot first: Firstly, you need to find the velocity:

v= $\sqrt{2*g*h}$

v= $\sqrt{9.8}$

Secondly, you need to find the Debit:

Q=A*v, A=base area of lemon vat= $\pi.r^2= (0.01)^2.\pi$

Q= $(0.01)^2*\sqrt{9.8}*\pi$

Thirdly, you can find the time from the debit and velocity:

t= volume of lemon/debit

Volume of lemon= $\pi*r^2*h= (0.1)^2*0.5*\pi$

$t= (0.1)^2*0.5*\pi/((0.01)^2*\sqrt {98} *\pi)$

$t= 31.94 second$

From Bernoulli's Equation we have:

$\frac{1}{2} \rho v_{1}^2 + \rho g h_{1} + p_{1} = \frac{1}{2} \rho v_{2}^2 + \rho g h_{2} + p_{2}$

where $_{1}$ states the condition at the top of the vat and $_{2}$ states the condition at the bottom of the vat.

Since $p_{1} = p_{2}$ and $v_{1} = 0,$

$\rho g h_{1} = \frac{1}{2} \rho v_{2}^2 + \rho g h_{2}$

$v_{2} = \sqrt{2gΔh}$

We can write the rate of flow at the bottom of the cylinder as below:

$Q = v_{2} A_{spigot}$

$\frac{dV}{dt} = -\sqrt{2gΔh} A_{spigot}$

Because $Δh = \frac{V}{A_{vat}},$

$\frac{dV}{dt} = -\sqrt{2g \frac{V}{A_{vat}}} A_{spigot}$

$\frac{1}{\sqrt{V}} \frac{dV}{dt} = -\sqrt{\frac{2g}{A_{vat}}} A_{spigot}$

$2\sqrt{V} = -\sqrt{\frac{2g}{A_{vat}}} A_{spigot} t + C$

When $t = 0,$ then $V = π \times 0.1^2 \times 0.5.$ Here we obtain $C = \frac{\sqrt{2π}}{10} .$ Hence we can get full equation as below:

$2\sqrt{V} = -\sqrt{\frac{2g}{A_{vat}}} A_{spigot} t + \frac{1}{10} \sqrt{2π}$

When $V = 0,$

$-\sqrt{\frac{2g}{A_{vat}}} A_{spigot} t + \frac{\sqrt{2π}}{10} = 0$

$\sqrt{\frac{2g}{A_{vat}}} A_{spigot} t = \frac{\sqrt{2π}}{10}$

$t = \frac{\sqrt{2π}}{10} \times \sqrt{\frac{A_{vat}}{2g}} \times \frac{1}{A_{spigot}}$

Since $A_{vat} = π \times 0.1^2$ and $A_{spigot} = π \times 0.01^2$

$t = 31.943$

$R,r$ are the radii of he cylinder and the spigot respectively and $\dot{h}, v$ is the change in the height of the water and the speed at the spigot respectively. Due to the conservation of mass,

$\dot{h} \pi R^2 = v \pi r^2$

$\dot{h} \frac{R^2}{r^2} = v$

Also, using the Bernoulli equation,

$P_0+\frac{\rho v^2}{2}+\rho g 0=P_0+\frac{\rho \dot{h}^2}{2}+\rho g h$

Combining,

$\dot{h}^2 \frac{R^4}{r^4}= \dot{h}^2+2g h$

$\Delta t =\sqrt{\frac{R^4-r^4}{r^4 2g}} \int _{h=h_0}^0 \frac{1}{\sqrt{h}} dh$

$\Delta t =-2 \sqrt{\frac{R^4-r^4}{r^4 2g}} \sqrt{h_0}$

Then insert the values.