Glycerine of density is flowing through the conical section of a pipe. The area of cross-section of the pipe at its ends is and and the pressure drop across its length is . The rate of flow of glycerine through the pipe is . The value of is _________.

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Solution

Q.
Glycerine of density $1.25 \times 10^{3} {kg m}^{- 3}$ is flowing through the conical section of a pipe. The area of cross-section of the pipe at its ends is $10 {cm}^{2}$ and $5 {cm}^{2}$ and the pressure drop across its length is ${3 N m}^{- 2}$ . The rate of flow of glycerine through the pipe is $x \times 10^{- 5} m^{3} s^{- 1}$ . The value of $x$ is _________. [2023]

Ans.
(4)

$Δ P = P_{1} - P_{2} = {3 N/m}^{2} (given)$

${By continuity eq}^{n}, A_{1} v_{1} = A_{2} v_{2}$

$∴$ $v_{1} = \frac{A_{2}}{A_{1}} v_{2}$ ...(i)

${By Bernoulli's eq}^{n},$

$P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$

$P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$

$\Rightarrow Δ P = \frac{1}{2} ρ [1 - {(\frac{A_{2}}{A_{1}})}^{2}] v_{2}^{2}$

$3 = \frac{1}{2} \times 1.25 \times 10^{3} [1 - {(\frac{5}{10})}^{2}] v_{2}^{2}$

$∴$ $v_{2} = 8 \times 10^{- 2} m/s$

$So discharge rate = A_{2} v_{2} = 5 \times 10^{- 4} \times 8 \times 10^{- 2} = 4 \times 10^{- 5} m^{3} / s$

$Correct answer is x = 4$

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