An ideal fluid of density flows steadily through a horizontal pipe. The pipe narrows from a radius of 2r at section 1 to a radius of r at section 2. If the velocity of the fluid at section 1 is v , the pressure

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Quantrex Academy · Accepted Answer

Correct answer: D. (15)/(2) v^2

According to the equation of continuity,

A_1 v_1 = A_2 v_2

.
Given

r_1 = 2r

and

r_2 = r

, the areas are

A_1 = \pi (2r)^2 = 4\pi r^2

and

A_2 = \pi r^2

.
Substituting the values:

(4\pi r^2) v = (\pi r^2) v_2 \Rightarrow v_2 = 4v

Applying Bernoulli's principle for a horizontal pipe (

h_1 = h_2

):

P_1 + \dfrac{1}{2} \rho v_1^2 = P_2 + \dfrac{1}{2} \rho v_2^2

Substituting

v_1 = v

and

v_2 = 4v

:

P_1 + \dfrac{1}{2} \rho v^2 = P_2 + \dfrac{1}{2} \rho (4v)^2

P_1 - P_2 = \dfrac{1}{2} \rho (16v^2) - \dfrac{1}{2} \rho v^2

P_1 - P_2 = \dfrac{15}{2} \rho v^2

Answer:

\dfrac{15}{2} \rho v^2

An ideal fluid of density flows steadily through a horizontal pipe....

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