Question

Calculate the rate of flow of glycerin of density $1.25 \times {10^3}kg{m^{ - 3}}$ through the conical section of a pipe, if radii of it ends are $0.1m$and $0.04m$ and the pressure drop across its length is $10N{m^{ - 2}}$.
(A) $6.43 \times {10^{ - 4}}{m^3}{s^{ - 1}}$
(B) $5.43 \times {10^{ - 4}}{m^3}{s^{ - 1}}$
(C) $5.44 \times {10^{ - 3}}{m^3}{s^{ - 1}}$
(D) $6.43 \times {10^{ - 3}}{m^3}{s^{ - 1}}$

Answer 1

Here, you are given a conical pipe in which glycerine is flowing, you are given a set of data regarding the flow and are asked to find the rate at which the glycerine is flowing. In order to solve this question, you need to understand the situation precisely and then think of an equation or theorem that will help you find the rate. Also, the rate is given as volume flowing per second through a cross section. Here, since nothing is mentioned and also the height of the conical pipe is unknown, you can consider the pipe to be lying horizontally.

Complete step by step answer:
Here, first we will look at the provided data and then decide what we are going to use as our equations in order to find the rate. Now, the pipe is lying horizontally and values of radii at both the ends are given, which means that you can use Bernoulli’s equation for the fluid trapped in between this region.

The Bernoulli’s equation is given as $P + \dfrac{1}{2}\rho {v^2} + \rho gh = C$ and if you have the fluid between two points, the equation will be ${P_1} + \dfrac{1}{2}\rho {v_1}^2 + \rho g{h_1} = {P_2} + \dfrac{1}{2}\rho {v_2}^2 + \rho g{h_2}$. In our case, the two points will be the end points.Let us have a look at our data. It is given that the pressure drop is equal to $10N{m^{ - 2}}$, so we get, ${P_1} - {P_2} = 10$, as it lying horizontally, we have ${h_1} = {h_2}$ and $\rho = 1.25 \times {10^3}$.

\Rightarrow{P_1} - {P_2} + \dfrac{1}{2}\rho {v_1}^2 + \rho g{h_1} = \dfrac{1}{2}\rho {v_2}^2 + \rho g{h_1} \\\ \Rightarrow 10 = \dfrac{1}{2}\rho {v_2}^2 - \dfrac{1}{2}\rho {v_1}^2 \\\ \Rightarrow 10 = \dfrac{\rho }{2}\left( {{v_2}^2 - {v_1}^2} \right) \\\ $$ Now, the continuity equation is given as${A_1}{v_1} = {A_2}{v_2}$. From here we can write $${v_2} = \dfrac{{{A_1}{v_1}}}{{{A_2}}}$$, let us substitute in the above obtained equation. $$10 = \dfrac{\rho }{2}\left( {{v_2}^2 - {v_1}^2} \right) \\\ \Rightarrow 10 = \dfrac{\rho }{2}\left( {{{\left( {\dfrac{{{A_1}{v_1}}}{{{A_2}}}} \right)}^2} - {v_1}^2} \right) \\\ \Rightarrow 10 = \dfrac{\rho }{2}\left( {{{\left( {\dfrac{{{A_1}}}{{{A_2}}}} \right)}^2} - 1} \right){v_1}^2 \\\ \Rightarrow {v_1} = \sqrt {\dfrac{{20}}{{\rho \left( {{{\left( {\dfrac{{{A_1}}}{{{A_2}}}} \right)}^2} - 1} \right)}}} \\\ $$ Given that ${r_1} = 0.1m \to {A_1} = \pi {r_1}^2$ and ${r_2} = 0.04m \to {A_2} = \pi {r_2}^2$. $${v_1} = \sqrt {\dfrac{{20}}{{\rho \left( {{{\left( {\dfrac{{{A_1}}}{{{A_2}}}} \right)}^2} - 1} \right)}}} \\\ \Rightarrow{v_1} = \sqrt {\dfrac{{20}}{{\left( {1.25 \times {{10}^3}} \right)\left( {{{\left( {\dfrac{{\pi {{\left( {0.1} \right)}^2}}}{{\pi {{\left( {0.04} \right)}^2}}}} \right)}^2} - 1} \right)}}} \\\ \Rightarrow{v_1} = 0.0205m{s^{ - 1}} \\\ $$ So, the rate will be given as, $r = {A_1}{v_1} \\\ \Rightarrow r = \pi {r_1}^2{v_1} \\\ \Rightarrow r = \pi {\left( {0.1} \right)^2}\left( {0.0205} \right) \\\ \therefore r = 6.43 \times {10^{ - 4}}{m^3}{s^{ - 1}} \\\ $ **Therefore, the rate of flow of glycerin is $$6.43 \times {10^{ - 4}}{m^3}{s^{ - 1}}$$ and thus option A is correct.** **Note:** You were supposed to understand that the pipe is lying horizontally as height was not given, or you can say that data was insufficient for the pipe to be vertical or slant. Also, keep in mind that the first thing that should strike you whenever fluid between some regions is given is the Bernoulli’s equation and in order to know the relation between velocities and cross section, you should apply the continuity equation.