Flow between two tanks 20 Case I A fluid flows between two t

Flow between two tanks. (20) Case I: A fluid flows between two tanks A and B because P_A > P_B. The tanks are at the same elevation and there is no pump in the line. The connecting line has a cross-sectional area S_1 and the mass rate of flow is w for a pressure drop of P_A - P_B. Case II: It is desired to replace the connecting line by two lines, each with cross section S_11 = Yz (Si). What pressure difference (Pa - Pb)ii is needed to give the same total mass flow rate as in Case 1 ? Assume turbulent flow and use the given Blass formula for the friction factor. Neglect entrance and exit losses. Belasis Formula for friction factor, f = 0.0791/(Re)1/4

Solution

Friction loss DP = rho *g * f(L/D) *( V²/2g), where f is the turbukent friction factor = Fn (Re)

In original case, w = mass flow rate for pressure drop Pa-Pb - friction loss=Pa-Pb - rho*g *f(L/D) ( V²/2g)

In new scenario, D = D/1.414

frictional factor term ncrease by 1.414, but mass flow rate for each pipe is 1/2 of original, so Vdecreases by 1/2, hence in friction drop terms 1/4 factor enters, total factor 1.414/4 for each pipe. Pressure drops by 0.3535 for each pipe. in order to get the same velocity as before in the pipe, need to scale up the pressure diff by 4/1.414

( assume friction factor constant in turbulent flow).

If friction factor changes by Re, then need to compute with a program.