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3) Back to the horizontal Venturi tube; it has a funnel
in the middle (between the wide pipe on the left and the
narrow pipe on the right, for example). The constant speed
of the liquid in the left pipe is v1 while the constant
speed in the right pipe is v2. The right pipe is open and
the escaping liquid goes into the atmosphere. It means that
P2=1.013*10^5 Pa. But the pressure in the left pipe, P1,
must be larger. In other words, a constant force F1 must be
applied by the pushing piston on the left side, even when
there is no change in elevation. Is the "P*dV work" done by
the piston equal to the change of kinetic energy of the
dV parcel of water, as it travels through the funnel part
of the setup? No it is not. Here is an illustration.
Left pipe, A1=1 m^2, v1=0.1m/s, rho=1000 kg/m^3
Right pipe, A2=0.1 m^2.
Therefore: v2=v1*10=1 m/s. (non-compressibility)
Suppose the parcel of liquid has dV=1000 cm^3; this
corresponds to dx1=1 mm (in the left pipe) and dx2= 10 mm
(in the right pipe). The mass of the parcel is 1 kg and
the kinetic energy gained by it is KE2-KE1=0.505 J.
The parcel escaping from the funnel, in one second, has 0.595 J
more KE than the parcel entering the funnel in one second.
How much work is done during that time of one second?
The pressure after the funnel P2=101300 Pa. Using the
Bernoulli formula I find P1=P2+4851Pa=106151 Pa.
The left pushing force (atmosphere plus my hand) F1=P1*A1=
106151 N.
The work of that force is F1*dx1=+106.151 J
The right pushing force (atmosphere only) is F2=P2*A2
10130 N. The work of that force is F2*dx2=-101.3 J.
The net work done is 4.85 J.
We have a paradoxical situation: the net work done on the
ideal liquid is 4.85 J, kinetic energy is increased by
0.505 J while potential energy remains constant. The
4.346 J of work is unaccounted for. Where am I wrong?