Steady state flow: series pipeline

[lleepptt]

Homework Statement

Water is discharged from a reservoir through a pipe 700m long. For the first 100m of its length the pipe is 100mm in diameter and then suddenly enlarged to 150mm diameter for the remaining 600m. The pipe terminates in a nozzle which discharges a jet 25mm in diameter at a point 15m below the reservoir surface. Head loss in the nozzle is h(ln)= (0.063*vj^2)/(2*g) where vj is the jet velocity. Assuming lamda=0.02 and a sharp-edged entry to the pipe (K = 0.5), determine the discharge. [ans. 7.67 l/s]

Homework Equations

bernoulli equation: p1/2g + u1^2/2g + z1 = p2/2g + u2^2/2g + z2 + hln + hp
where hln=K*u^2/2g
and hp=lamda*L*u^2/2*g*D

The Attempt at a Solution

I have been working at this for about five hours now; I honestly have no clue what I am supposed to do now. I think I am supposed to use the bernoulli equation to find "vj" by having the left hand side for the pressure/velocity/z for the reservoir and the right hand side for the nozzle. But when negate the pressure (as they are both open to the atmosphere and thus cancel out?), negate the velocity of the reservoir as I assume it's close to zero, and negate z for the nozzle as I pass the datum line that z is measured from along the centre of the pipe, I never get the right answer for vj when I rearrange the equation. I worked out using the answer for Q (discharge) that is given that vj=15.625m/s as Q=vA. But from here I don’t know where to go, I can't the right answer no matter what I do. Please help.

 

[LawrenceC]

You must determine the loss, hp, at each segment and relate it to the nozzle exit velocity. Since the flow is incompressible, you know the velocity in the two sections in terms of the nozzle exit velocity. Therefore apply the hp terms as follows. Answer is correct.

hp1 = (lambda*L1*/(2*g*D1)) * (Vnozzle*A1/Anozzle)*(Vnozzle*A1/Anozzle)
hp2 = (lambda*L2*/(2*g*D2)) * (Vnozzle*A2/Anozzle)*(Vnozzle*A2/Anozzle)