Question

water flows in a horizontal tube. the pressure of water change by 600Nm² . between A and B wheras the area of cross section are 30cm² and 15cm² . find the rate of flow of water through the tube ?​

Answer 1

Answer:

Explanation:

Let velocity at A & B are VA & VB respectively

By equation of continuity

VBVA=3015=2

By Bernoulli's equation

PA+12ρV2A=PB+12ρV2B

PA−PB=12ρ(2VA)2−12ρV2A=32ρV2A

⇒600=32(1000kgm−3)V2A

VA=0.63ms−1

∴ Rate of flow =(30cm2)(0.63ms−1)

⇒1890cm3s−1

Hence (B) is the correct answer.

Answer 2

$\huge\bf{\blue{\mid{\overline{\underline{your\:answer}}}\mid}}$

let the velocity at A= $\large{\mathtt{\blue{\underline{ v_{a}}}}}$

and that at B= $\large{\mathtt{\blue{\underline{ v_{b}}}}}$

★ $\bold{by\: the \:equation\: of\: continuity,}$

$\large{\mathtt{\blue{\underline{ \frac{ v_{b} }{ v_{a} } = \frac{ {30cm}^{2} }{ {15cm }^{2} } = 2 }}}}$

★ $\bold{by \:Bernoulli\: equation,}$

$\large{\mathtt{\blue{\underline{ p_{a} + \frac{1}{2} p { v_{a} }^{2} = p_{b} + \frac{1}{2} { pv_{b} }^{2} }}}}$

$\large{\mathtt{\blue{\underline{ = > p_{a} - p_{b} = \frac{1}{2} p {( 2v_{a} )}^{2} - \frac{1}{2} p { v_{a} }^{2} = \frac{3}{2} p { v_{a} }^{2} }}}}$

$\large{\mathtt{\blue{\underline{ = > 600 {nm}^{ - 2} = \frac{3}{2} (1000kg {m}^{ - 3}) { v_{a} }^{2} }}}}$

$\large{\mathtt{\blue{\underline{ = > v_{a} = \sqrt{0.4 {m}^{2} {s}^{ - 2} } = 0.63m {s}^{ - 1} }}}}$

$\bold{the \:rate \:of \:flow,}$

$\large{\mathtt{\blue{\underline{ = > ( {30cm}^{2} )(0.63 {ms}^{ - 1} ) }}}}$

$\large{\mathtt{\blue{\underline{ 1890 {cm}^{3} \: {s}^{ - 1} }}}}$

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