First slide

Fluid dynamics

Question

Water flows in a horizontal tube (see figure) The pressure of water change by $700 N m^{- 2}$ between A and B where the area of cross section are $40 c m^{2}$ and $20 c m^{2}$ . Respectively. Find the rate of flows of water through the tube.(density of water $= 1000 k g m^{- 3}$ )

Moderate

A

$3020 c m^{3} / s$
B

$2420 c m^{3} / s$
C

$1810 c m^{3} / s$
D

$2720 c m^{3} / s$

Solution

$\begin{array}{l} A s p e r E q u a t i o n o f C o n t i n u i t y, \\ A_{A} v_{A} = A_{B} v_{B} \\ \Rightarrow 40 v_{A} = 20 v_{B} \\ \Rightarrow 2 v_{A} = v_{B} \\ A s p e r B e r n o u l i i' s P r i n c i p l e, \\ P_{A} + \frac{1}{2} ρ v_{A}^{2} = P_{B} + \frac{1}{2} ρ v_{B}^{2} \\ \Rightarrow P_{A} - P_{B} = \frac{1}{2} ρ (v_{B}^{2} - v_{A}^{2}) \\ \Rightarrow Δ P = \frac{1}{2} ρ (v_{B}^{2} - v_{A}^{2}) \\ \Rightarrow 700 = 500 (v_{B}^{2} - v_{A}^{2}) \\ \Rightarrow \sqrt{\frac{7}{15}} = v_{A} \\ R a t e o f f l o w = (a r e a a t A) x (v e l o c i t y a t A) = 40 \times 10^{- 4} \times \sqrt{\frac{7}{15}} = 27.32 \times 10^{- 4} m^{3} / s = 2732 c m^{3} / s \end{array}$
So the rate of flow is nearly equal to 2720 unit .Then correct answer will be 4.

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