Question

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is 0.070m2, and the magnitude of the fluid velocity is 3.50m/s.

Answer 1

Answer:

The fluids speed at a) $0.105\,m^{2}$  and b) $0.047\,m^{2}$ are $2.33\,\frac{m}{s^{2}}$  and $5.21\,\frac{m}{s^{2}}$ respectively

c) Th volume of water the pipe discharges is: $882\,m^{3}$  

Explanation:

To solve a) and b) we should use flow continuity for ideal fluids:

$\Delta Q=0$(1)

With Q the flux of water, but Q is $Av$ using this on (1) we have:

$A_{2}v_{2}-A_{1}v_{1}=0$ (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 $A_{2}=0.105\,m^{2}$, so now we can solve (2) for $v_{2}$:

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.105}\approx2.33\,\frac{m}{s^{2}}$

b) Here we use point 2 as $A_{2}=0.047\,m^{2}$:

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}}$

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 $Q=\frac{V}{t}$, so we can write:

$A_{1}v_{1}=\frac{V}{t}$, solving for V:

$V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3}$