Respuesta :
Answer:
The volume flow rate is 3.3 m³/s.
Explanation:
Let the volume flow rate be 'V'.
Given:
Density of olive oil is, [tex]\rho=875\ kg/m^3[/tex]
Diameter of inlet is, [tex]d_1=3.08\ cm[/tex]
Diameter of outlet is, [tex]d_2=1.10\ cm[/tex]
Change in pressure between the two sections is, [tex]P_1-P_2=5.15\ kPa[/tex]
Using continuity equation, we have:
[tex]V=A_1v_1=A_2v_2[/tex]
Where, [tex]A_1,A_2[/tex] are the areas of inlet and outlet sections, [tex]v_1\ and\ v_2[/tex] are the flow speeds at inlet and outlet sections respectively.
Rewriting in terms of [tex]v_1\ and\ v_2[/tex]
[tex]v_1=\frac{V}{A_1}\ and\ v_2=\frac{V}{A_2}[/tex]
Area of inlet and outlet is given as:
[tex]A_1=\frac{1}{4}\pi d_1^2=\frac{1}{4}\times 3.14\times (0.0308)^2=7.45\times 10^{-4}\ m^2\\A_2=\frac{1}{4}\pi d_2^2=\frac{1}{4}\times 3.14\times (0.011)^2=9.5\times 10^{-5}\ m^2[/tex]
Kinetic energy at the inlet is given as:
[tex]E_1=\frac{1}{2}\rho v_1^2\\E_1=\frac{1}{2}\times 875\times (\frac{V}{A_1})^2\\E_1=\frac{1}{2}\times 875\times (\frac{V}{7.45\times 10^{-4}})^2\\E_1=7.88\times 10^8V^2[/tex]
Kinetic energy at outlet is given as:
[tex]E_2=\frac{1}{2}\rho v_2^2\\E_2=\frac{1}{2}\times 875\times (\frac{V}{A_2})^2\\E_2=\frac{1}{2}\times 875\times (\frac{V}{9.5\times 10^{-5}})^2\\E_2=484.76\times 10^8V^2[/tex]
Using Bernoulli's equation for the two sections:
Since, the section is horizontal, therefore, the change in kinetic energy is equal to change in pressure energy.
[tex]E_2-E_1=P_1-P_2\\(484.76-7.88)V^2=5.15\times 10^3\\V^2=\frac{5.15\times 10^3}{476.88}\\V=\sqrt{\frac{5.15\times 10^3}{476.88}}=3.29\ m^3/s[/tex]
Therefore, the volume flow rate is 3.3 m³/s
