Wyliczyć v2 z tego równania p- gęstość wody 997kg/m3 v1-0 p1- 101kPa g- przyspieszenie ziemskie 9,81.... Question from @mrocznyko

June 2022 1 12 Report

Wyliczyć v2 z tego równania
p- gęstość wody 997kg/m3
v1-0
p1- 101kPa
g- przyspieszenie ziemskie 9,81m/s2
z1-3
v2-?
p2-109kPa
z2-0

basetla

$dane:\\\rho = 997\frac{kg}{m^{3}}\\p_1 = 101 \ kPa = 101 \ 000 \ Pa\\p_2 = 109 \ kPa = 109 \ 000 \ Pa\\z_1 = 3\\z_2 = 0\\\nu_1 = 0\\g = 9,81\frac{m}{s^{2}}\\szukane:\\\nu_2 = ?\\\\Rozwiazanie\\\\\frac{\rho\cdot\nu_1^{2}}{2} + p_1+z_1\cdot\rho\cdot g = \frac{\rho\cdot \nu_2^{2}}{2}+p_2+z_2\cdot\rho\cdot g\\\\dla \ \ \nu_1 = 0 \ i \ \ z_2 = 0, \ rownanie \ ma \ postac:\\\\0+p_1+z_1\cdot \rho\cdot g = \frac{\rho\cdot\nu_2^{2}}{2}+p_2+0$

$\frac{\rho\cdot\nu_2^{2}}{2}+p_2 = p_1 + z_1\cdot\rho\cdot g \ \ /\cdot\frac{2}{\rho}\\\\\nu_2^{2} +\frac{2p_2}{\rho} = \frac{2p_1}{\rho}+2z_1\cdot g\\\\\nu_2^{2} = \frac{2p_1}{\rho} - \frac{2p_2}{\rho} + 2\cdot z_1\cdot g\\\\\nu_2^{2} = \frac{2}{\rho}(p_1-p_2)+2z_1g \ \ |()\sqrt{}\\\\\nu_2 = \sqrt{\frac{2}{\rho}(p_1-p_2)+2z_1g}\\\\\nu_2 = \sqrt{\frac{2}{997}(101000-109000)+2\cdot3\cdot9,81}\\\\\nu_2 =\sqrt{ \frac{2}{997}\cdot(-8000)+58,86}\\\\\nu_2 = \sqrt{42,81}\\\\\nu_2\approx6,54$