Question

At a processing plant, olive oil of density 875 kg/m3 flows in a horizontal section of hose that constricts from a diameter of 3.08 cm to a diameter of 1.10 cm. Assume steady, ideal flow. (a) What is the volume flow rate if the change in pressure between the two sections of hose is 5.15 kPa

Answer 1

Answer:

The volume flow rate is 3.3 m³/s.

Explanation:

Let the volume flow rate be 'V'.

Given:

Density of olive oil is, $\rho=875\ kg/m^3$

Diameter of inlet is, $d_1=3.08\ cm$

Diameter of outlet is, $d_2=1.10\ cm$

Change in pressure between the two sections is, $P_1-P_2=5.15\ kPa$

Using continuity equation, we have:

$V=A_1v_1=A_2v_2$

Where, $A_1,A_2$ are the areas of inlet and outlet sections, $v_1\ and\ v_2$ are the flow speeds at inlet and outlet sections respectively.

Rewriting in terms of $v_1\ and\ v_2$

$v_1=\frac{V}{A_1}\ and\ v_2=\frac{V}{A_2}$

Area of inlet and outlet is given as:

$A_1=\frac{1}{4}\pi d_1^2=\frac{1}{4}\times 3.14\times (0.0308)^2=7.45\times 10^{-4}\ m^2\\A_2=\frac{1}{4}\pi d_2^2=\frac{1}{4}\times 3.14\times (0.011)^2=9.5\times 10^{-5}\ m^2$

Kinetic energy at the inlet is given as:

$E_1=\frac{1}{2}\rho v_1^2\\E_1=\frac{1}{2}\times 875\times (\frac{V}{A_1})^2\\E_1=\frac{1}{2}\times 875\times (\frac{V}{7.45\times 10^{-4}})^2\\E_1=7.88\times 10^8V^2$

Kinetic energy at outlet is given as:

$E_2=\frac{1}{2}\rho v_2^2\\E_2=\frac{1}{2}\times 875\times (\frac{V}{A_2})^2\\E_2=\frac{1}{2}\times 875\times (\frac{V}{9.5\times 10^{-5}})^2\\E_2=484.76\times 10^8V^2$

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.

$E_2-E_1=P_1-P_2\\(484.76-7.88)V^2=5.15\times 10^3\\V^2=\frac{5.15\times 10^3}{476.88}\\V=\sqrt{\frac{5.15\times 10^3}{476.88}}=3.29\ m^3/s$

Therefore, the volume flow rate is 3.3 m³/s