In this illustration, we'll analyze a u-tube,
as a water siphon. we are given that the bend
of an inverted u-tube is kept 1 meter above,
a water surface with 1 arm dipped in water.
and the other end b of u-tube of, its second
arm is 7 meter below the water surface as
we can see. and the water issues, from end
b of siphon, as a free jet. at atmospheric
pressure. we are required to determine the
speed of jet at end b and the pressure of
water in the, bend of u-tube, and the atmospheric
pressure and density of water is given to
us. so here in solution. we use. bernoulli's
theorem. at, free surface. of water. and at,
end b. then here we can see on free surface
of water we consider water to be at rest so
pressure is only p atmospheric. and b, pressure
is p atmospheric. and, the potential energy
per unit volume of the fluid with respect
to this water level is minus, ro g h. plus
the kinetic energy of fluid per unit volume
is half ro v square. here p atmospheric gets
cancelled out and this gives us the value
of. velocity of water jet that is equal to.
root 2 gee h, and here h we can substitute
as 7 meter so this gives us root of. 2 multiplied
by 10 multiplied by 7. that is root of 140.
which is 11 point, 8 3 meter. per second.
that is the result of this problem for the,
velocity of jet coming out. now to find the
pressure in the bend again we can use. now,
we use, bernoulli's theorem, at water surface,
and bend, then here we get, at water surface
again the pressure is p atmospheric. and as
the cross sectional area of tube is same the
flow velocity of water will be same as that
of flow velocity at b. so here pressure if
we consider at p, plus, here, the gravitational
potential energy is, ro g h 1 and h 1 we consider
as 1 meter plus, kinetic energy is half ro
v square, so, this gives us the value of pressure
in the bend that is p atmosphere, minus, both
of these terms, when we subtract half ro v
square we can write as ro g h. so this will
be ro, g, h plus, h 1. if we substitute the
values this is, 10 to power 5. minus, ro is
thousand g is 10. so this will be 7 plus 1,
8 into 10 to power, 4. so this will numerically
be given as 2 into 10 to power 4 pascal, that
is a result of this, problem.
