The piping system shown is made of new commercial steel e 0

The piping system shown is made of new commercial steel (e = 0.046mm) and connects two water tanks as shown. If minor losses are ignored and the valves shown are completely open, determine the flow-rate from the first tank.

Solution

Pressure head at inlet = 7 m

Pressure head at outlet (lower pipe) = 0

Pressure head at outlet (upper pipe) = 3 - 1 = 2 m

Friction factor f = e/D = 0.046 / D

Let pressure at junction be P.

dP = fL/D*V^2 / (2g)

For lower pipe from inlet to joint, 7 - P = f*(10)/D*V_in^2 / (2g)

7 - P = 0.046*10/D^2 *V_in^2 / (2g)...............1

For upper pipe, P - 2 = 0.046*(8+1)/D^2*V_out1^2 / (2g)..................2

For lower pipe from joint to outlet, P - 3 = 0.046*8/D^2 * V_out2^2 / (2g)....................3

Subtracting eqn3 from eqn2, 1 = 0.046/(2gD^2)* [9*V_o1^2 - 8*Vo2^2].........4

Adding eqn1 and eqn2, 5 = 0.046/(2gD^2) *[10*Vin^2 + 9*Vo1^2]..................5

Further, by continuity, A*V_in = A*V_o1 + A*V_o2

V_in = V_o1 + V_o2.........Putting it in eqn5,

5 = 0.046/(2gD^2) *[10*(V_o1 + V_o2)^2 + 9*Vo1^2]............6

It cannot be solved further as a crucial variable dia of the pipe is missing.