Question

Water flows into a cylindrical tank (Fig. ) through pipe 1 at the rate of 25 ft/s and leaves through pipes 2 and 3 at 10 ft/s and 12 ft/s, respectively. At 4 is an open air vent. Inside pipe diameters are: D, = 3 in, D,=2in, D, = 2.5 in, D, = 2 in. Calculate (a) dh/dt;

Answer 1

Answer:

Step 1

Water flow into a cylindrical tank through pipe 1 is 25 ft/sec rate. 

And pipe 2 is 10 ft/sec.

And pipe 3 is 12 ft/ sec.

And pipe 4 is open air vent and also given that pipes inside diameter are D= 3 inch. And D= 2 inch and D=2.5 inch And D=2inch.

Calculate ; dh/dt=??

Step 2

Here i provide full solution of this question.

In flow= A₁V₁ = $π \over 4$ $({3 \over 12})^2$ x 25

= $22 \over (7x4)$ x $9 \over 144$ x 25

= $22 \over 28$ x $25 \over 16$

= 1.227 ft³/sec

Q₁= 1.227

and similarly, outflow = $π \over 4$[ $({2 \over 12})^2$ x 10 + $({2.5 \over 12})^2$ x 10]

= $22 \over 7$ x $1 \over 4$ [ $4 \over 144$ x 10 + $2.5x 2.5x 12 \over 144$ ]

= $22 \over 28$ x [$40 \over 144$ + $75 \over 144$ ]

= $22 \over 28$ x $115 \over 144$

= 0.627 ft³/sec

Q₀=0.627

Now let inflow = Q_n= Q₁-Q₀

= 1.227 -0.627

= 0.60 ft³/sec --------------(i)

Cross sectional area of tank = πr² : r=1

then A= π x r = π ft² -----------(ii)

So Q_n= Ax $dh \over dt$

and $dh \over dt$ = $q_n \over A$

From equation (i) and (ii)

$dh \over dt$ = $0.6 \over π$ = $0.6 \over 3.14$ = 0.1910 ft / sec

Hence,

$dh \over dt$ = 0.1910 ft / sec