In this illustration we'll discuss about
the velocity head of a water jet here we
are given that a bang cube is lowered
into a water stream as shown in Figure
and the velocity of stream related to
the tube is equal to V or we can
consider the water is flowing into bat V
and tube is at rest and we are given
that the closed upper end of the tube is
located height at height H not above the
stream where water is flowing at a speed
V and we are required to find what
height H will the water jet will spurt
and we are given that the closed upper
end of the tube is located a height H
naught now in the solution is quite a
simple case of using Bernoulli's theorem
we can say using Bernoulli's theorem at
stream and at just outside point of
orifice then we can see at just outside
point of orifice the pressure is
atmospheric and the liquid is coming out
at a velocity V 1 and the stream is
having a velocity V and here also the
pressure is p-atmospheric so in this
situation we can use add to the stream
if we use the
ah gee this p-atmospheric plus the
kinetic energy of stream per unit volume
is half Rho v square is equal to at
Point a which is located over here we
can write the kinetic pressure energy is
p-atmospheric plus due to work done by
gravity the gravitational potential
energy is Rho gh naught plus the kinetic
energy is half Rho v1 Square here
p-atmospheric gets cancelled out Rho
also gets cancelled out and the velocity
of water jet just after ejection of
orifice we are getting is v1 which is
root of a v square minus 2 gee H naught
a row I have cancelled out and if we
have got the value of B 1 velocity had
we know under gravity it can be given as
V 1 square by 2 G so the relation we are
getting the height to which water is
getting raised is v square minus 2 gee H
naught divided by 2 G that is the result
of this problem
