Respuesta :
Answer:
[tex]\dot v = 0.0124 m^3/s[/tex]
Explanation:
Continuity
[tex]V1*A1 = v2*A2[/tex]
[tex]V1*pi*r1^2 = v2*pi*r2^2 [/tex]
[tex]V1*r1^2 = v2*r2^2[/tex]
by Bernoulli
[tex]P1 + \frac{1}{2}\rho v1^2 + \rho gh1 = P2 + \frac{1}{2} \rho v2^2 + \rho gh2[/tex]
since P1 = P2, thet got cancel.
Assume h1 = 0, so h2 = 10.6 m
[tex]\frac{1}{2}*\rho *v1^2 = \frac{1}{2}* \rho *v2^2 + \rho*g*h2[/tex]
[tex]V1^2 = v2^2 + 2*g*h2[/tex]
Use continuity and solve for v1 or v2 and sub in.
[tex](\frac{r2}{r1})^4*v2^2 = v2^2 + 2gh2[/tex]
[tex]V2^2*((\frac{r2}{r1})^4 - 1) = 2*g*h2[/tex]
[tex]V2 = \sqrt \frac{(2*g*h2)}{((\frac{r2}{r1})^4 - 1)}[/tex]
r2 =0.0320
r1 = 0.0163
h2 =10.6
Plug in number
V2 = 3.87 m/s
Then use that for the volume flow
[tex]\dot v= v2*A2 = v2* \pi *r2^2[/tex]
Plug in numbers
[tex]\dot v = 0.0124 m^3/s[/tex]
Volume flow rate for the flowing pipe in factory that will keep same pressure in the two horizontal pipes is 0.0123 cubic meter per second.
What is continuity equation for pipe flow?
The continuity equation for pipe flow is says that the product of cross sectional area of the pipe and velocity of the water flow is constant throughout the pipe.
Let two points in pipe, [tex]P_1[/tex] with cross section [tex]A_1[/tex] and velocity [tex]v_1[/tex] and the point [tex]P_2[/tex] with cross section [tex]A_2[/tex] and velocity [tex]v_2[/tex] . Thus by continuity equation,
[tex]v_1A_1=v_2 A_2[/tex]
It is given that the radius of the pipe is 0.0163 meters and the radius of the another pipe is 0.0320 meters.
Thus by the continuity equation,
[tex]v_1\pi\times (0.0163)^2=v_2 \pi \times (0.0320)^2\\v_1=3.854v_2[/tex]
Let the above equation as equation 1.
As the pressure of both points is same. Thus the sum of kinetic energy and potential energy at point 1 should be equal to point 2. Therefore,
[tex]\dfrac{1}{2}\rho v_1^2+\rho gh_1=\dfrac{1}{2}\rho v_2^2+\rho gh_2[/tex]
As the pipe one is at the ground with 0 height and second pipe is at the height of the floor of the factory, which is 10.6 m higher. Thus,
[tex]\rho[\dfrac{1}{2} v_1^2+ (9.81)(0)]=\rho[\dfrac{1}{2} v_2^2+ (9.81)(10.6)]\\\dfrac{1}{2} v_1^2=\dfrac{1}{2} v_2^2+ (9.81)(10.6)[/tex]
Put the value of [tex]v_1[/tex] from equation 1 as,
[tex]\dfrac{1}{2} (3.854v_2)^2=\dfrac{1}{2} v_2^2+ (9.81)(10.6)\\v_2=3.87\rm m/s[/tex]
Thus the flow rate can be given as,
[tex]Q=v_2\times A_2\\Q=3.86\times \pi (0.0320)^2\\Q=0.0123\rm m^3/s[/tex]
Hence, the volume flow rate that will keep the pressure in the two horizontal pipes the same is 0.0123 cubic meter per second.
Learn more about the continuity equation here;
https://brainly.com/question/14619396
