Question: What is the maximum flow rate delivered to the second tank?

[fizicks101]

Problem Statement:
A water tank (Diameter=10m) is pressurized with a gas at 20MPa. 15m below the surface of the water in the tank, is an exit pipe (Diameter=0.4m, area=A2) to a second pressurized tank of 16MPa. What is the maximum flow rate delivered to the second tank?

I'm not sure what I'm missing here. I used Bernoulli's to solve for V2 (at the outlet pipe) by setting V1 to zero(b/c area of the water surface >> area of exit pipe cross section). I then solved for the mass flow rate (mdot=rho*A2*V2).

V2=sqrt((2/rho)(P1-P2-z*rho*g)),

Any suggestions?

 

[jbsteele]

The maximum flow rate delivered to the second tank can be determined by using Bernoulli's equation and the continuity equation. The Bernoulli's equation states that for an incompressible, inviscid fluid the total energy of a given point in the flow is constant. The total energy consists of pressure energy, kinetic energy and potential energy. We can calculate the velocity of the fluid at the outlet of the pipe (V2) by setting the pressure energy and potential energy at the inlet of the pipe (P1,z1) equal to the pressure energy and potential energy at the outlet of the pipe (P2,z2). V2=sqrt((2/rho)(P1-P2-z1*rho*g))The continuity equation states that the mass flow rate through the pipe is equal to the product of the fluid density (rho), the area of the cross section of the pipe (A2) and the velocity of the fluid (V2). mdot=rho*A2*V2Therefore, the maximum flow rate delivered to the second tank is:mdot=rho*A2*sqrt((2/rho)(P1-P2-z1*rho*g))