Respuesta :
Given data:
* The given section with the diameter d_1 = 5.5 cm.
* The flow speed of the fluid in the section with the diameter d_1 is v_1 = 715 cm/s.
* The diameter of the other section is d_2 = 4.6 cm
* The density of the fluid is,
[tex]d=3.38gcm^{-3}[/tex]* The pressure of fluid in the section with diameter d_1 is,
[tex]P_1=3.45\times10^5\text{ Pa}[/tex]Solution:
(A). The flow rate of the fluid remains constant, thus,
[tex]A_1v_1=A_2v_2[/tex]where A_1 is the area of the section with diameter d_1, A_2 is the area of the other section with diameter d_2, and v_2 is the velocity of the fluid in section with diameter d_2,
The area of the section with diameter d_1 is,
[tex]\begin{gathered} A_1=\pi\times(\frac{d_1}{2})^2 \\ A_1=\frac{\pi d^2_1}{4}_{} \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} A_1=\frac{\pi\times(5.5)^2^{}}{4} \\ A_1=23.76cm^2 \end{gathered}[/tex]The area of the other section of pipe with diameter d_2 is,
[tex]\begin{gathered} A_2=\frac{\pi d^2_2}{4} \\ A_2=\frac{\pi\times4.6^2}{4} \\ A_2=16.62cm^2 \end{gathered}[/tex]Thus, the velocity of the fluid from the other section with diameter d_2 is,
[tex]\begin{gathered} A_1\times v_1=A_2\times v_2 \\ 23.76cm^2\times715\times10^{-2}ms^{-1}=16.62cm^2\times v_2 \\ 23.76cm^2\times7.15ms^{-1}=16.62cm^2\times v_2 \\ v_2=\frac{23.76\times7.15}{16.62}ms^{-1} \end{gathered}[/tex]By simplifying,
[tex]\begin{gathered} v_2=\frac{169.9}{16.62}ms^{-1} \\ v_2=10.22ms^{-1} \end{gathered}[/tex]Thus, the velocity of the fluid from the other section with a diameter of 4.6 cm is 10.22 m/s.
(B). From Bernoulli's principle, the relation between the pressure and velocity on each section is,
[tex]P_1+\frac{1}{2}d_{}v^2_1=P_2+\frac{1}{2}dv^2_2[/tex]where d is the density of the fluid, P_1 is the pressure of fluid in the section with diameter d_1 and p_2 is the pressure of the fluid in the section with the diameter d_2,
Substituting the known values,
[tex]3.45\times10^5\times\frac{1}{2}\times3.38\times10^{3^{}}\times(715\times10^{-2})^2=P_2\times\frac{1}{2}\times3.38\times10^{3^{}}\times(10.22)^2[/tex]By simplifying,
[tex]\begin{gathered} 3.45\times10^5\times\frac{1}{2}\times3.38\times10^{3^{}}\times(7.15)^2=P_2\times\frac{1}{2}\times3.38\times10^{3^{}}\times(10.22)^2 \\ 298.07\times10^8=P_2\times176.5\times10^3 \\ P_2=\frac{298.07\times10^8}{176.5\times10^3} \\ P_2=1.69\times10^5\text{ Pa} \end{gathered}[/tex]Thus, the pressure of fluid in the section with a diameter of 4.6 cm is,
[tex]\text{1}.69\times10^5\text{ Pa}[/tex]