Respuesta :
Answer:
a) 2.33 m/s
b) 5.21 m/s
c) 882 m³
Explanation:
Using the concept of continuity equation
for flow through pipes
[tex]A_{1}\times V_{1} = A_{2}\times V_{2}[/tex]
Where,
A = Area of cross-section
V = Velocity of fluid at the particular cross-section
given:
[tex]A_{1} = 0.070 m^{2}[/tex]
[tex]V_{1} = 3.50 m/s[/tex]
a) [tex]A_{2} = 0.105 m^{2}[/tex]
substituting the values in the continuity equation, we get
[tex]0.070\times 3.50 = 0.105\times V_{2}[/tex]
or
[tex]V_{2} = \frac{0.070\times 3.5}{0.150}m/s[/tex]
or
[tex]V_{2} = 2.33m/s[/tex]
b) [tex]A_{2} = 0.047 m^{2}[/tex]
substituting the values in the continuity equation, we get
[tex]0.070\times 3.50 = 0.047\times V_{2}[/tex]
or
[tex]V_{2} = \frac{0.070\times 3.5}{0.047}m/s[/tex]
or
[tex]V_{2} = 5.21m/s[/tex]
c) we have,
Discharge[tex]Q = Area (A)\times Velocity(V)[/tex]
thus from the given value, we get
[tex]Q = 0.070m^{2}\times 3.5m/s\[/tex]
[tex]Q = 0.245 m^{3}/s[/tex]
Also,
Discharge[tex]Q = \frac{volume}{time}[/tex]
given time = 1 hour = 1 ×3600 seconds
substituting the value of discharge and time in the above equation, we get
[tex]0.245m^{3}/s = \frac{volume}{3600s}[/tex]
or
[tex]0.245m^{3}/s\times 3600 = Volume[/tex]
volume of flow = [tex]882 m^{3}[/tex]
