Respuesta :
Answer:
(a) Pressure at point 200 m downstream: 113.4 kPa
(b) Pressure at point 200 m downstream: 105.4 kPa
Explanation:
The point of reference, that will be named Point A, is the pressure gage.
The point 200 m downstream will be named Point B.
With a slope of 1:50, the height difference between A and B is (1/50)*200=4m.
(a) In the case there is no friction losses, we can write
[tex]p_a+\frac{\rho V^2}{2} +\rho*g*h_a=p_b+\frac{\rho V^2}{2} +\rho*g*h_b[/tex]
Because ther is no change in the diameter of the pipe, V=constant.
Rearranging
[tex]p_a+\rho*g*h_a=p_b+\rho*g*h_b\\\\\Delta p = \rho * g * \Delta h\\\\\Delta p = 0.8505 * 1000\frac{kg}{m^3} *9.81\frac{m}{s^2}*4 m\\\\ \Delta p = 33.373.62 kg/(m*s^2)=33.373.62 Pa=33.4 kPa[/tex]
Then we have an increase in of 33.4 kPa, and pressure in Point B (downstream) is:
[tex]p_b=p_a+\Delta p = 80 kPa + 33.4 kPa = 113.4 kPa[/tex]
(b) If we consider 10% of the total initial head as friction loss, we have
[tex]p_a+\rho*g*h_a=p_b+hf +\rho*g*h_b\\\\hf=0.1*p_a\\\\\Delta P = \rho*g*\Delta h-0.1*p_a\\\\\Delta P = 0.8505*1000 kg/m^3*9.81m/s2*4m-0.1*80kPa\\\\\Delta P = 33.4kPa-8kPa=25.4kPa[/tex]
In this case the rise in pressure is 25.4 kPa, due to the friction losses.
The pressure at point B is
[tex]p_b=p_a+\Delta p = 80 kPa + 25.4 kPa = 105.4 kPa[/tex]
