Bernoulli energy equation

logistic_guy

Senior Member
Joined
Apr 17, 2024
Messages
1,370
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\(\displaystyle ^3\)?
 
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...
 
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?😣
 
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...
 
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?☹️
 
Top