Answer:
The fluids speed at a) [tex] 0.105\,m^{2} [/tex] and b) [tex] 0.047\,m^{2} [/tex] are [tex] 2.33\,\frac{m}{s^{2}} [/tex] and [tex] 5.21\,\frac{m}{s^{2}} [/tex] respectively
c) Th volume of water the pipe discharges is: [tex] 882\,m^{3} [/tex]
Explanation:
To solve a) and b) we should use flow continuity for ideal fluids:
[tex]\Delta Q=0 [/tex](1)
With Q the flux of water, but Q is [tex] Av [/tex] using this on (1) we have:
[tex] A_{2}v_{2}-A_{1}v_{1}=0 [/tex] (2)
With A the cross sectional areas and v the velocities of the fluid.
a) Here, we use that point 2 has a cross-sectional area equal to [tex] A_{2}=0.105\,m^{2} [/tex], so now we can solve (2) for [tex] v_{2} [/tex]:
[tex]v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.105}\approx2.33\,\frac{m}{s^{2}} [/tex]
b) Here we use point 2 as [tex] A_{2}=0.047\,m^{2} [/tex]:
[tex] v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}} [/tex]
c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is [tex] Q=\frac{V}{t} [/tex], so we can write:
[tex] A_{1}v_{1}=\frac{V}{t} [/tex], solving for V:
[tex] V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3} [/tex]
