Can someone please explain the relation: [v/(ro-r)]=[vmax/ro] where ro is .5 in?

Question

Can someone please explain the relation: [v/(ro-r)]=[vmax/ro] where ro is .5 in? It's for the second part and I don't understand why ro isn't 6in (radius of pipe 4)

Water flows steadily into a tank through pipes 1 and 2 and discharges at a steady rate out of the tank through pipes 3 and 4. The mean velocity of inflow and outflow in pipes 1, 2, and 3 is 50 ft/s, and the hypothetical outflow velocity in pipe 4 varies linearly from zero at the wall to a maximum at the center of the pipe. What is the discharge from pipe 4, and what is the maximum velocity in pipe 4?

Explanation / Answer

Q1 + Q2 = Q3 + Q4

A1*V1 + A2*V2 = A3*V3 + A4*V4

pi/4*d1^2*V1 + pi/4*d2^2*V2 = pi/4*d3^2*V3 + pi/4*d4^2*V4

d1^2*V1 + d2^2*V2 = d3^2*V3 + d4^2*V4

1^2*50 + 2^2*50 = 1.5^2*50 + 12^2*V4

V4 = 0.95486 ft/s

This is average veloicty.

Discharge = pi/4*d4^2*V4

= 3.14/4*(12/12)^2 *0.95486

= 0.75 ft^3/s

Since v varies linearly with r we have the general equation of line, v = a*r + b where a and b are constants.

At r = 0 we have v = v_max . Thus, v_max = a*0 + b or b = v_max

Also, at r = r0, we have v = 0. Thus, 0 = a*r0 + b = a*r0 + v_max or a = -v_max / r0

Therefore, v = (-v_max / r0)*r + v_max

Take a ring of width dr at radius r.

Area of ring = 2*pi*r*dr

Flow through the ring dq = (2*pi*r*dr)*((-v_max / r0)*r + v_max)

Now integrating it from r = 0 to r = r0 we get

Q = 2*pi*[(-v_max / r0)*r0^3 /3 + v_max*r0^2 /2]

2*pi*[(-v_max / r0)*r0^3 /3 + v_max*r0^2 /2] = 0.75

Putting r0 = 12/2 = 6 in = 6/12 ft = 0.5 ft we get

2*3.14*[(-v_max / 0.5)*0.5^3 /3 + v_max*0.5^2 /2] = 0.75

v_max = 2.864 ft/s

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