Question

Two circular holes, one larger than the other, are cut in the side of a large water tank whose top is open to the atmosphere. The center of one of these holes is located twice as far beneath the surface of the water as the other. The volume flow rate of the water coming out of the holes is the same. (a) Decide which hole is located nearest the surface of the water. The smaller hole The larger hole (b) calculate the ration of the radius of the larger hole to the radius of the smaller hole

Answer 1

We know that

The volume flow rate is given by dV/dt =Av

And the constinuity equation is A1v1 =A2v2

Consider the 2 be the lower hole and 1 be the upper hole

A2 =A1v1/A2 =(r1/r2)²v1 and h1 =2h2

Therefore we can say that lower hole will have the greater velocity so for the volume flow to be the same it must be smaller.

And now by using the Bernoulli's equation we have

P1+pgy1+(1/2)pv12 =P2+pgy2+(1/2)pv22    y1 =h1 and y2 =h2

Since here P1=P2 then (1/2)p(v12 -v22) and pg(h2-h1)

(1/2)p(v₁2 -(r1/r2)²v1)    and pg(h1/2-h1)

(1/2)pv₁2(1-(r1/r2)²)    and -pgh1/2 =-0.5pgh1

Given that from above v1² = 2*g*h1
    (1/2)*2*g*h1*(1-r1/r2)²) = -0.5*g*h1
                     (1-(r1/r2)²) = -0.5
   (r1/r2) = sqrt(1+.5) =1.22

The ration of the radius of the larger hole to the radius of the smaller hole is = 1.22