Bernoulli energy equation

[logistic_guy]

here is the question

Water at $\displaystyle 120$ kPa (gage) is flowing in a horizontal pipe at a velocity of $\displaystyle 1.15$ m/s. The pipe makes a $\displaystyle 90^{\circ}$ angle at the exit and the water exits the pipe vertically into the air. The maximum height the water jet can rise is

(a) $\displaystyle 6.9$ m
(b) $\displaystyle 7.8$ m
(c) $\displaystyle 9.4$ m
(d) $\displaystyle 11.5$ m
(e) $\displaystyle 12.3$ m


my attemb
this question can be solved by Bernoulli equation
$\displaystyle \frac{v^2}{2} + gz + \frac{p}{\rho}$
the density $\displaystyle \rho$ isn't given
is it correct to assume density of water $\displaystyle 1000$ kg/m$\displaystyle ^3$?

 

[khansaheb]

here is the question

Water at $\displaystyle 120$ kPa (gage) is flowing in a horizontal pipe at a velocity of $\displaystyle 1.15$ m/s. The pipe makes a $\displaystyle 90^{\circ}$ angle at the exit and the water exits the pipe vertically into the air. The maximum height the water jet can rise is

(a) $\displaystyle 6.9$ m
(b) $\displaystyle 7.8$ m
(c) $\displaystyle 9.4$ m
(d) $\displaystyle 11.5$ m
(e) $\displaystyle 12.3$ m


my attemb
this question can be solved by Bernoulli equation
$\displaystyle \frac{v^2}{2} + gz + \frac{p}{\rho}$
the density $\displaystyle \rho$ isn't given
is it correct to assume density of water $\displaystyle 1000$ kg/m$\displaystyle ^3$?

Yes...

 

[logistic_guy]

Yes...

thank

$\displaystyle \frac{v_1^2}{2} + gz_1 + \frac{p_1}{\rho} = \frac{v_2^2}{2} + gz_2 + \frac{p_2}{\rho}$

i'm very good in physics so i understand if we throw object upward its velocity will be zero at the highest point
so $\displaystyle v_2 = 0$

$\displaystyle \frac{v_1^2}{2} + gz_1 + \frac{p_1}{\rho} = gz_2 + \frac{p_2}{\rho}$

i'll chose the initial height of water $\displaystyle z_1$ to be ground so $\displaystyle z_1 = 0$

$\displaystyle \frac{v_1^2}{2} + \frac{p_1}{\rho} = gz_2 + \frac{p_2}{\rho}$

is my analize correct?

 

[khansaheb]

thank

$\displaystyle \frac{v_1^2}{2} + gz_1 + \frac{p_1}{\rho} = \frac{v_2^2}{2} + gz_2 + \frac{p_2}{\rho}$

i'm very good in physics so i understand if we throw object upward its velocity will be zero at the highest point
so $\displaystyle v_2 = 0$

$\displaystyle \frac{v_1^2}{2} + gz_1 + \frac{p_1}{\rho} = gz_2 + \frac{p_2}{\rho}$

i'll chose the initial height of water $\displaystyle z_1$ to be ground so $\displaystyle z_1 = 0$

$\displaystyle \frac{v_1^2}{2} + \frac{p_1}{\rho} = gz_2 + \frac{p_2}{\rho}$

is my analize correct?

Yes ..., and continue...

 

[logistic_guy]

Yes ..., and continue...

thank

$\displaystyle z_2 = \frac{1}{g}(\frac{v_1^2}{2} + \frac{p_1}{\rho} - \frac{p_2}{\rho}) = \frac{1}{g}(\frac{v_1^2}{2} + \frac{p_1 - p_2}{\rho})$

$\displaystyle = \frac{1}{9.8}(\frac{1.15^2}{2} + \frac{120000 - 101300}{1000}) = 1.98$ m

not one of the options

what i do wrong?