Question

Two tanks A and B, of constant cross-sectional area of $10m^{2}and2.5m^2$, respectively, are connected by a 5 cm pipe, 100 m long, with f = 0.03. If the initial difference of water levels is 3 m, how long will it take for $2.5m^3$ of water to flow from A to B? Considering entry and exit losses, it can be grossly assumed that the flow velocity, it can be grossly assumed that the flow velocity, in m/s, through the pipe is $1.75\sqrt{h}$, where h is in m, taking g = 10 m/sec${}^2$, also, may take area of pipe as $2\times {10}^{-3}{m}^2$.

• A. 535 seconds
• B. 516 seconds
• C. 485 seconds
• D. 467 seconds

Answer 1

The correct option is D 467 seconds

Let at any instant the difference in the water levels in the two reservoirs be h and in time dt the water level in the tank A fall by an amount dH. Then, the water level will rise in tank B by an amount $\left(dH\frac{A_1}{A_2}\right)$. Now if dh is the change in difference in head causing the flow



$dh=dH+dH\frac{A_1}{A_2}$

$dH=\frac{dH}{1+\frac{A_1}{A_2}}$

$Now,-A_{1}dH=a.v.dtV=1.75\sqrt{h}$

$-A_1\frac{dh}{1+\frac{A_1}{A_2}}=a\times 1.75\sqrt{h}dt$

$\frac{-A_1{A}_2}{(A_1+A_2)a\left(1.75\right)}{\int }_{h_1}^{h_2}\frac{dh}{\sqrt{h}}={\int }_0^{T}dt$

$T=\frac{2A_1{A}_2}{a(A_1+A_2)\left(1.75\right)}[\sqrt{h_1}-\sqrt{h_2}]$

$h_1\Rightarrow$ The difference of water levels in the reservoirs before flow commences

$h_1=3m$

$h_2\Rightarrow$ The difference of water levels in the reservoirs after flow commences

$LevelfallintankA=\frac{2.5}{10}=0.25$

$LevelriseintankB=\frac{2.5}{2.5}=1m$

$So,h_2=h_1-(1+0.25)$
= 3 - (1.25)

= 1.75 m

Now,

$T=\frac{2\left(10\right)\left(2.5\right)}{(2\times {10}^{-3})(10+2.5)\left(1.75\right)}[\sqrt{3}-\sqrt{1.75}]$

$=1142.86(\sqrt{3}-\sqrt{1.75})=467.66sec$