Cuesta College, San Luis Obispo, CA
Assume ideal fluid flow for water through this horizontal pipe with different cross-sectional areas.
The greatest pressure is at:
(A) point .
(B) point .
(C) point .
(D) (There is a tie.)
Correct answer (highlight to unhide): (B)
From applying the continuity equation:
A1·v1 = A2·v2 = A3·v3,
where the fluid volume flow rate is the same throughout each of these three sections of pipe. As the cross-sectional area of the pipe is smallest at point  and largest at point , then:
A3 < A1 < A2,
such that the speeds at each section of pipe can be ordered accordingly, where the fastest speed occurs where the cross-sectional area is the narrowest:
v2 < v1 < v3.
Then from Bernoulli's equation:
0 = ∆P + (1/2)·ρ·∆(v2) + ρ·g·∆y,
because the pipe is horizontal, then ∆y = 0, and we can neglect the last term, such that:
0 = ∆P + (1/2)·ρ·∆(v2).
Comparing points  and  gives us:
0 = P2 – P1 + (1/2)·ρ·((v2)2 – (v1)2),
P1 + (1/2)·ρ·(v1)2 = P2 + (1/2)·ρ·(v2)2,
and similarly comparing points  and  gives us:
0 = P3 – P2 + (1/2)·ρ·((v3)2 – (v2)2),
P2 + (1/2)·ρ·(v2)2 = P3 + (1/2)·ρ·(v3)2.
Thus we can now compare the pressure and (1/2)·ρ·v2 terms for all three points:
P1 + (1/2)·ρ·(v1)2 = P2 + (1/2)·ρ·(v2)2 = P3 + (1/2)·ρ·(v3)2,
where the location with the smallest (1/2)·ρ·v2 term would correspond to having the greatest pressure. Earlier, from the continuity equation, since v2 < v1 < v3, then:
P2 > P1 > P3,
such that location  (having the largest area and slowest speed) would have the greatest pressure.
Sections 70854, 70855
Exam code: quiz05Gu1L
(A) : 1 student
(B) : 35 students
(C) : 10 students
(D) : 6 students
Success level: 67%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.66