First slide

Fluid dynamics

Question

A liquid flows through a horizontal tube. The velocities of the liquid in the two sections, which have areas of cross - section A₁ and A₂, are V₁ and V₂ respectively. The difference in the levels of the liquid in the two vertical tubes is h.

a) The volume of the liquid flowing through the tube in unit time is A₁V₁.
b) $\large V_2-V_1=\sqrt {2gh}$
c) $\large V_2^2-V_1^2= {2gh}$
d) The energy per unit mass of the liquid is the same in both sections of the tube.

Moderate

A

only a, c and d are correct
B

only c and d are correct
C

only a, b and d are correct
D

all are correct

Solution

Discharge = A₁V₁ = A₂V₂. Applying Bernoulli's equation between larger and smaller sections, we can write, $\large p_1+\frac 12\rho V_1 ^2=p_2+\frac 12\rho V_2 ^2$
$\large \Rightarrow V_2^2-V_1^2=\frac 2{\rho}(p_1-p_2)......(1)\;Also\;p_1-p_2=\rho gh.......(2)$
From (1) and (2), $\large V_2^2-V_1^2=2gh$ . According to Bernoulli's principle energy per unit volume or unit mass or unit weight of the an ideal blowing fluid at all points in the flow region remain the same.

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