How Does Adding a Nozzle Affect Water Efflux Velocity and Energy Conservation?

[surrelative]

nozzle complex!

i have a tank having water filled till an elevation 'h'. Now at the bottom of this is a small opening of circular cross-section with a diameter say 'd1'. This arrangement is kept in free atmosphere so a uniform pressure Patm=1.013 bar applies everywhere. This is case 1 and in order to calculate velocity of efflux we can simply apply bernaulli's as given by: (let density of water be: q and neglecting any viscous effects)

Patm/(q*g) + h = Patm/(q*g) + v1^2/(2*g)
which gives v1 = (2gh)^1/2
now this happens due to complete conversion of the potential energy stored in water into kinetic energy out of the effluxing pipe i.e (2gh)^1/2 is the max velocity the fluid can achieve having a head 'h'

Now for case 2 if we connect a nozzle in front of this pipe with a converging circular cross-section of diameter 'd2' (d2<d1), then according to bernaulli's the velocity of water through the pipe should increase, but we already found that v1 is the max velocity that can be achieved by complete conversion of potential energy of water stored in the tank. I don't think vacuum will develop at the at the discharge of the nozzle, it being fully exposed to atmosphere.

So what will be the velocity of efflux for case 2 i.e through the nozzle discharge? and if it is supposed to be greater than v1, then how we conserve energy at the water surface in the tank and at the nozzle discharge in case2?

i have a strong feeling I'm missing something very basic. please help!

 

[spanky489]

mass flux stays the same, so you can calculate the 2nd speed with

\dot{V} = A_{1} \cdot v_{1}= A_{2} \cdot v_{2}

 

[timthereaper]

In the right part of Bernoulli's equation, wouldn't the pressure decrease because the water is moving faster?

 

[surrelative]

yeah, the mass efflux remains constant.

i.e dm/dt=rho*A1*v1=rho*A2*v2
let us consider a interval t for clculating the efflux so that we can write,
m=rho*A1*v1*t=rho*A2*v2*t

my question is:
the K.E in the first case is, K.E1 = m*v1^2/2 (which is = m*g*h by conserving energy)
now if the fluid starts to move faster on adding the nozzle, with a velocity v2, the K.E becomes, K.E2=m*v2^2/2 (m being constant following the continuity equation)

So as v2>v1
according to the results we got K.E2 > K.E1
how is this possible as K.E1 =potential energy of the liquid stored in tank?

@timthereaper: the bernaulli's as stated above is for the case when there is only a tap with a constant circular cross-section and not for a nozzle. The pressure on the RHS in that equation is only the atm pr as there isn't any residual static pressure.