Question 247: Properties of Matter

Question

Glycerin of density $$1.25 	imes 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$$ is flowing through the conical section of pipe The area of cross-section of the pipe at its ends are $$10 \mathrm{~cm}^{2}$$ and $$5 \mathrm{~cm}^{2}$$ and pressure drop across its length is $$3 ~\mathrm{Nm}^{-2}$$. The rate of flow of glycerin through the pipe is $$x 	imes 10^{-5} \mathrm{~m}^{3} \mathrm{~s}^{-1}$$. The value of $$x$$ is ___________.

Accepted Answer

We can use the Bernoulli equation and continuity equation to solve this problem. The Bernoulli equation is given by:
$$P_1 + \frac{1}{2} ho v_1^2 = P_2 + \frac{1}{2} ho v_2^2$$
The continuity equation is given by:
$$A_1 v_1 = A_2 v_2$$
From the given data, we have:
$$P_1 - P_2 = 3 \mathrm{~Nm}^{-2}$$
$$A_1 = 10 \mathrm{~cm}^2 = 10 	imes 10^{-4} \mathrm{~m}^2$$
$$A_2 = 5 \mathrm{~cm}^2 = 5 	imes 10^{-4} \mathrm{~m}^2$$
$$ho = 1.25 	imes 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$$
Rearrange the continuity equation to solve for $$v_2$$:
$$v_2 = \frac{A_1}{A_2} v_1 = 2v_1$$
Substitute $$v_2$$ and rearrange the Bernoulli equation:
$$P_1 - P_2 = \frac{1}{2} ho (v_2^2 - v_1^2)$$
$$3 = \frac{1}{2} 	imes 1.25 	imes 10^3 (4v_1^2 - v_1^2)$$
Now, solve for $$v_1$$:
$$3 = \frac{1}{2} 	imes 1.25 	imes 10^3 	imes 3v_1^2$$
$$v_1^2 = \frac{3}{1.875 	imes 10^3}$$
$$v_1 = \sqrt{\frac{3}{1.875 	imes 10^3}}$$
$$v_1 \approx 0.0400 \mathrm{~m} \mathrm{~s}^{-1}$$
Now, calculate the rate of flow of glycerin through the pipe (volume flow rate) using $$v_1$$ and $$A_1$$:
$$Q = A_1 v_1$$
$$Q = 10 	imes 10^{-4} 	imes 0.0400$$
$$Q = 4 	imes 10^{-5} \mathrm{~m}^{3} \mathrm{~s}^{-1}$$
So, the rate of flow of glycerin through the pipe is $$4 	imes 10^{-5} \mathrm{~m}^{3} \mathrm{~s}^{-1}$$, and the value of $$x$$ is 4.

Properties of Matter

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