Question

Suppose that two tanks, $\mathbf{1}$and $\mathbf{2}$, each with a large opening at the top, contain different liquids. A small hole is made in the side of each tank at the same depth h below the liquid surface, but the hole in tank $\mathbf{1}$has half the cross-sectional area of the hole in tank localid="1661534674200" $\mathbf{2}$.

(a) What is the ratio localid="1661534677105" ${\mathbf{\rho }}_{\mathbf{1}}\mathbf{/}{\mathbf{\rho }}_{\mathbf{2}}$of the densities of the liquids if the mass flow rate is the same for the two holes?

(b) What is the ratio localid="1661534679939" ${\mathbf{\text{R}}}_{\mathbf{\text{V1}}}{\mathbf{\text{/R}}}_{\mathbf{\text{V2}}}$from the two tanks?

(c) At one instant, the liquid in tank localid="1661534683116" $\mathbf{1}$is localid="1661534686236" $\mathbf{12}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$above the hole. If the tanks are to have equal volume flow rates, what height above the hole must the liquid in tank localid="1661534690073" $\mathbf{2}$be just then?

Short answer

Answer

1. The ratio of the densities of the liquids through two holes is $\mathbf{2}$.

2. The ratio of volume flow rates from the two tanks is$\mathbf{1}\mathbf{/}\mathbf{2}$

3. The height of liquid above the hole of tank 2, ${\mathit{h}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathit{c}\mathit{m}$

Step-by-step solution

Given data

$\mathrm{A}1=\frac{1}{2}{\mathrm{A}}_2$

Determining the concept

Use Bernoulli’s equation to find the ratio of densities and the equation of continuity to find the ratio of volume flow rate. To calculate the height of the water level, use both equations together. According to Bernoulli’s equation, the speed of a moving fluid increases, and the pressure within the fluid decreases.

Formulae are as follows:

• ${\mathrm{R}}_{\mathrm{v}}=\mathrm{Av}$

• ${\mathrm{\rho }}_1{\mathrm{v}}_1{\mathrm{A}}_1={\mathrm{\rho }}_2{\mathrm{v}}_2{\mathrm{A}}_2$

Where,Ais the area,vis velocity,Ris the rate, and$P$is density.

(a) Determining the ratio ρ1ρ2  of the densities of the liquids through two holes

It is given that the mass flow rate of the two holes is constant.

So,

$\begin{array}{c}{\mathrm{\rho }}_1{\mathrm{v}}_1{\mathrm{A}}_1={\mathrm{\rho }}_2{\mathrm{v}}_2{\mathrm{A}}_2\\ \frac{{\mathrm{\rho }}_1}{{\mathrm{\rho }}_2}=\frac{{\mathrm{v}}_2{\mathrm{A}}_2}{{\mathrm{v}}_1{\mathrm{A}}_1}\end{array}$

It is known that the hole is made at the same depth for each tank.

So,

${\mathrm{v}}_1={\mathrm{v}}_2$

Because$\mathrm{v}=\sqrt{2\mathrm{gh}}$,

$\begin{array}{l}\frac{{\mathrm{\rho }}_1}{{\mathrm{\rho }}_2}=\frac{{\mathrm{A}}_2}{{\mathrm{A}}_1}\\ \frac{{\mathrm{\rho }}_1}{{\mathrm{\rho }}_2}=2\end{array}$

Hence,the ratio of the densities of the liquids through two holes is$\mathbf{2}$.

(b) Determining the ratio Rv1Rv2  of volume flow rates from the two tanks

The equation of continuity is written as,

${\mathrm{R}}_{\mathrm{v}}=\mathrm{Av}$

So, the ratio of mass flow rate will become,

$\begin{array}{c}\frac{{\mathbf{R}}_{\mathbf{v}\mathbf{1}}}{{\mathbf{R}}_{\mathbf{v}\mathbf{2}}}\mathbf{=}\mathbf{}\frac{{\mathbf{A}}_{\mathbf{1}}{\mathbf{v}}_{\mathbf{1}}}{{\mathbf{A}}_{\mathbf{2}}{\mathbf{v}}_{\mathbf{2}}}\\ \mathbf{=}\frac{{\mathbf{A}}_{\mathbf{1}}}{{\mathbf{A}}_{\mathbf{2}}}\mathbf{}\\ \frac{{\mathbf{R}}_{\mathbf{v}\mathbf{1}}}{{\mathbf{R}}_{\mathbf{v}\mathbf{2}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{}\end{array}$

Hence, the ratio of volume flow rates from the two tanks is $\mathbf{1}\mathbf{/}\mathbf{2}$.

(c) Determining the height of liquid above the hole of tank 2

Now, for this part,

$\frac{{\mathrm{R}}_{\mathrm{v}1}}{{\mathrm{R}}_{\mathrm{v}2}}=1$

So,

$1=\frac{{\mathrm{A}}_1{\mathrm{v}}_1}{{\mathrm{A}}_2{\mathrm{v}}_2}$

Rearranging this equation as,

$\begin{array}{c}\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\frac{{\mathrm{A}}_2}{{\mathrm{A}}_1}\\ \frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=2\end{array}$

But, it is known that,

$\mathrm{v}=\sqrt{2\mathrm{gh}}$

So,

$\begin{array}{c}2=\frac{\sqrt{2{\mathrm{gh}}_1}}{\sqrt{2{\mathrm{gh}}_2}}\\ 2=\frac{\sqrt{{\mathrm{h}}_1}}{\sqrt{{\mathrm{h}}_2}}\\ {\mathrm{h}}_2=\frac{{\mathrm{h}}_1}{4}\\ {\mathrm{h}}_2=\frac{12\text{\hspace{0.17em}cm}}{4}\\ {\mathrm{h}}_2=3\text{\hspace{0.17em}cm}\end{array}$

Hence, the height of liquid above the hole of tank 2,${\mathrm{h}}_2=3\text{\hspace{0.17em}cm}$.