Physics quiz question: horizontal pipe with varying cross-sectional areas

Physics 205A Quiz 5, fall semester 2019
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 [1].
(B) point [2].
(C) point [3].
(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 [3] and largest at point [2], 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 [1] and [2] 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 [2] and [3] 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 [3] (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

Physics quiz question: exit speed of water flow

Physics 205A Quiz 5, fall semester 2018
Cuesta College, San Luis Obispo, CA

Water enters point [1] with a speed of 0.80 m/s. The pipe at point [2] is at a lower height than point [1], and has twice the cross-sectional area. Assume ideal fluid flow. The speed of the water at point [2] is:
(A) 0.40 m/s.
(B) 0.57 m/s.
(C) 0.80 m/s.
(D) 1.6 m/s.

Correct answer (highlight to unhide): (A)

From applying the continuity equation:

A₁·v₁ = A₂·v₂,
where the fluid volume flow rate is the same throughout this section of pipe.
As the cross-sectional area of the pipe widens by a factor of two as it flows from [1]→[2], 2·A₁ = A₂, such that the speed of the water at point [2] is then:

A₁·v₁ = (2·A₁)·v₂,

(1/2)·v₁ = v₂,

such that the speed at point [2] is 0.40 m/s, half the speed at point [1].

(Response (B) is (v₁/√(2); response (C) is v₁; response (D) is 2·v₁.)

Sections 70854, 70855
Exam code: quiz05Ro74
(A) : 32 students
(B) : 3 students
(C) : 8 students
(D) : 9 students

Success level: 62%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.38

Physics quiz question: speed, pressure changes in horizontal, narrowing pipe

Physics 205A Quiz 5, fall semester 2017
Cuesta College, San Luis Obispo, CA

Water as it moves horizontally from [1]→[2] through a pipe with decreasing cross-sectional area. The radius of the pipe at point [1] is 0.10 m, and water enters point [1] with a speed of 0.25 m/s. Assume ideal fluid flow. As water flows from [1]→[2], the speed __________; while the pressure __________.
(A) remains constant; remains constant.
(B) remains constant; changes.
(C) changes; remains constant.
(D) changes; changes.

Correct answer (highlight to unhide): (D)

From applying the continuity equation:

A₁·v₁ = A₂·v₂,

because the diameter of the pipe narrows as it flows from [1]→[2], the cross-sectional area decreases (A₁ > A₂), such that the speed of the water increases:

v₁ < v₂.

Then from Bernoulli's equation:

0 = ∆P + (1/2)·ρ·∆(v²) + ρ·g·∆y,

the third term on the right-hand side is zero because there is no change in elevation (y₁ = y₂), while the second term on the right-hand side increases (as the speed increases along the pipe), thus the pressure must decrease. Thus both speed and pressure change as water flows from [1]→[2] through this pipe.

Sections 70854, 70855
Exam code: quiz05nWaW
(A) : 0 students
(B) : 6 students
(C) : 6 students
(D) : 36 students

Success level: 73%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.29

Physics midterm question: increasing pressure in horizontal pipe?

Physics 205A Midterm 2, fall semester 2016
Cuesta College, San Luis Obispo, CA

A Physics 205A student asked the following question on an online reading assignment[*]:

Can pressure increase as the radius of a pipe with flowing water increases?

Discuss a plausible horizontal pipe system that would result in this happening. Explain your reasoning using the continuity equation, Bernoulli's equation, and the properties of ideal fluid flow.

[*] waiferx.blogspot.com/2016/10/online-reading-assignment-ideal-fluid.html.

Solution and grading rubric:

• p:
  Correct. Discusses/demonstrates the application of ideal fluid conservation laws for a horizontal pipe with a narrow cross-section at point [1] and a wider cross-section at point [2]:
  1. continuity, where the widening of the pipe at point [2] will cause a corresponding slower speed there;
  2. energy density (Bernoulli's equation), as the speed decreases (making the (1/2)⋅ρ⋅Δ(v²) term negative) and the elevation is not changing (making the ρ⋅g⋅Δy term zero) for the fluid flowing from point [1] to point [2], then in order for all three terms on the right-hand side of Bernoulli's equation to sum to zero, the ΔP term would need to be positive, and thus pressure would increase flowing from point [1] to point [2].
• r:
  Nearly correct, but includes minor math errors.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some garbled attempt at applying continuity and Bernoulli's equation.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying continuity and Bernoulli's equation.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855, 73320
Exam code: midterm02oPt0
p: 36 students
r: 4 students
t: 9 students
v: 3 students
x: 4 students
y: 0 students
z: 0 students

A sample "p" response (from student):

Physics midterm question: constant pressure in widening, ascending pipe

Physics 205A Midterm 2, fall semester 2014
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problems 9.43, 9.45

Water flows through a pipe that increases in cross-sectional area and elevation as it flows from point [1] to point [2]. Discuss how it is plausible that the water pressure remains constant as it flows from [1]→[2]. Explain your reasoning using the continuity equation, Bernoulli's equation, and the properties of ideal fluid flow.

Solution and grading rubric:

• p:
  Correct. Discusses/demonstrates the application of ideal fluid conservation laws:
  1. continuity, where the widening of the pipe at point [2] will have a corresponding slower speed;
  2. energy density (Bernoulli's equation), as the speed decreases (making the (1/2)·ρ·∆(v²) term negative) and the elevation increases (making the ρ·g·∆y term positive) for the fluid flowing from point [1] to point [2], it is plausible that these terms will together sum to zero, such that the ∆P term would also be zero.
• r:
  Nearly correct, but includes minor math errors.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Understands that increasing elevation would decrease pressure (if speed remains constant), and increasing area would increase pressure (if elevation remains constant), but does not explicitly discuss how speed would decrease in the wider section of pipe (due to continuity).
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some garbled attempt at applying continuity and Bernoulli's equation.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying continuity and Bernoulli's equation.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855, 73320
Exam code: midterm02veR1
p: 37 students
r: 8 students
t: 11 students
v: 9 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student 1220):
A sample "t" response (from student 4455), not explicitly discussing the continuity equation:

Physics quiz question: pressure change in elevated oil pipeline

Physics 205A Quiz 5, fall semester 2014
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 9.45

"Trans-Alaska Pipeline"
U.S. Government, Public Domain (PD-USGOV) image
commons.wikimedia.org/wiki/File:Trans-Alaska_Pipeline.jpg

Oil (density 8.6×10² kg/m³) in the Trans-Alaskan pipeline[*] with a diameter of 1.2 m flows at a speed of 
0.94 m/s. Assume ideal fluid flow. The pressure of the oil __________ as it rises in the photo shown above.
(A) decreases.
(B) remains constant.
(C) increases.
(D) (Not enough information is given.)

[*] Trans Alaska Pipeline System--The Facts, Alyeska Pipeline Service Company (2013), alyeska-pipe.com/assets/uploads/pagestructure/NewsCenter_MediaResources_FactSheets_Entries/635078372894251917_2013AlyeskaTAPSFactBook.pdf.

Correct answer (highlight to unhide): (A)

From applying the continuity equation:

A₁·v₁ = A₂·v₂,

because the diameter of the pipeline is constant, the cross-sectional area remains constant (A₁ = A₂), such that the speed of the oil is constant:

v₁ = v₂.

Then from Bernoulli's equation:

0 = ∆P + (1/2)·ρ·∆(v²) + ρ·g·∆y,

the second term on the right-hand side is zero because there is no change in the speed of the fluid, while the third term on the right-hand side increases (as the elevation increases along the pipe), thus the pressure must decrease as oil flows into the higher section of pipe.

Sections 70854, 70855, 73320
Exam code: quiz05mRp4
(A) : 24 students
(B) : 20 students
(C) : 19 students
(D) : 1 student

Success level: 38%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.33

Physics quiz question: cross-sectional area of pipe

Physics 205A Quiz 5, fall semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 9.43(b)

Water at point [1] flows with a speed of 
3.0 m/s and a volume flow rate of 0.40 m³/s. The pipe at point [2] 
is four times the cross-sectional area of point [1]. Assume ideal 
fluid flow. The cross-sectional area of the pipe at point [1] is:
(A) 0.13 m².
(B) 1.2 m².
(C) 4.5 m².
(D) 7.5 m².

Correct answer (highlight to unhide): (A)

The definition of volume flow rate is:

(∆V/∆t) = A·v,

where A is the cross-sectional area of the pipe, and v is the speed of the fluid at that same location, such that:

A = (∆V/∆t)/v = (0.40 m³/s)/(3.0 m/s) = 0.13333333... m²,

or to two significant figures, the cross-sectional area at point [1] is 0.13 m².

(Response (B) is (∆V/∆t)·v; response (C) is (1/2)·v²; response (D) is v/(∆V/∆t).)

Sections 70854, 70855, 73320
Exam code: quiz05LuF7
(A) : 34 students
(B) : 20 students
(C) : 4 students
(D) : 3 students

Success level: 69%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.40

GIF animation: expansion of ascending balloon

"eoss-139 Balloon Expansion"
EOSS (Richard1984)
http://youtu.be/9irgdAX1BFo

Physics midterm question: descending, widening pipe

Physics 205A Midterm 2, fall semester 2012
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problems 9.41, 9.45

The pressure of water __________ as it flows from point [1] to point [2]. (Assume ideal fluid flow.)
(A) decreases.
(B) remains constant.
(C) increases.
(D) (Not enough information is given.)

Correct answer (highlight to unhide): (C)

From applying the continuity equation:

A₁·v₁ = A₂·v₂,

the water speed at point [2] is slower than at point [1], because the cross-sectional area at point [2] is larger than at point [1].

Then from Bernoulli's equation:

0 = ∆P + (1/2)·ρ·∆(v²) + ρ·g·∆y,

the second term on the right-hand side decreases due to the decrease in speed from point [1] to point [2], while the third term on the right-hand side decreases (as the elevation decreases along the pipe), thus the pressure must increase as water flows from point [1] to point [2].

Sections 70854, 70855
Exam code: midterm02gL0u
(A) : 21 students
(B) : 5 students
(C) : 26 students
(D) : 1 student

Success level: 48%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.26

Physics quiz question: descending, widening pipe

Physics 205A Quiz 5, fall semester 2012
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 9.41

Water flows through a pipe from point [1] to point [2]. Assume ideal fluid flow. The speed of the water __________ as it flows from point [1] to point [2].
(A) decreases.
(B) remains constant.
(C) increases.
(D) (Not enough information is given.)

Correct answer: (A)

From applying the continuity equation:

A₁·v₁ = A₂·v₂,

the water speed at point [2] is slower than at point [1], because the cross-sectional area at point [2] is larger than at point [1].

Sections 70854, 70855
Exam code: quiz05L4mN
(A) : 34 students
(B) : 9 students
(C) : 8 students
(D) : 1 student

Success level: 65%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.55

Physics midterm question: horizontal, widening pipe

Physics 205A Midterm 2, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 9.41

The pressure of water __________ as it flows from point [1] to point [2]. (Assume ideal fluid flow.)
(A) decreases.
(B) remains constant.
(C) increases.
(D) (Not enough information is given.)

Correct answer: (C)

From applying the continuity equation:

A₁·v₁ = A₂·v₂,

the water speed at point [2] is slower than at point [1], because the cross-sectional area at point [2] is larger than at point [1].

Then from Bernoulli's equation:

0 = ∆P + (1/2)·ρ·∆(v²) + ρ·g·∆y,

the (1/2)·ρ·∆(v²) term on the right-hand side of the equation decreases due to the decrease in speed from point [1] to point [2], while the ρ·g·∆y term remains constant (no change in elevation along the horizontal pipe), thus the pressure must increase as water flows from point [1] to point [2].

Sections 70854, 70855
Exam code: midterm02fR3q
(A) : 24 students
(B) : 14 students
(C) : 15 students
(D) : 0 students

Success level: 28%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.43

Physics quiz question: elevation of Lake Baikal

Physics 205A Quiz 5, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problems 9.27, 9.31

In August 2009, aboard the mini-submarine Mir-1, Russian Prime Minister Vladmir Putin dived 1,400 m below the surface of Lake Baikal in Siberia[*], the world's deepest lake. The pressure at the surface of the lake is 95.9 kPa. The surface of Lake Baikal is located at an elevation __________ sea level. (P_atm = 101.3 kPa.)
(A) above.
(B) at.
(C) below.
(D) (Not enough information is given.)

[*] Source: Stuart Williams, "Putin dives to bottom of world's deepest lake," Agence France-Presse, August 1, 2009.

Correct answer: (A)

The pressure of the atmosphere at sea level is given as P_atm = 101.3 kPa, such that for the (atmospheric) pressure at the surface of Lake Baikal to be less than the pressure at sea level, the surface of Lake Baikal must be higher than sea level, as given from:

0 = ∆P + ρ_air*g*∆y.

Sections 70854, 70855
Exam code: quiz05t0rQ
(A) : 32 students
(B) : 3 students
(C) : 14 students
(D) : 2 students

Success level: 63%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.30

Physics quiz question: depth under oil

Physics 205A Quiz 5, Fall Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 9.12

At the surface of a oil storage tank the air pressure is 1.0 atm. (The density of this oil is 880 kg/m^3.) At what depth under oil in this tank is the oil pressure 1.1 atm? (1 atm = 1.013e+5 Pa.)
(A) 1.2 m.
(B) 12 m.
(C) 13 m.
(D) 25 m.

Correct answer: (A)

The pressure at location [2], a depth d in oil under the surface is greater than compared to the pressure at a location [1] at the surface of the oil, as given by:

P_2 = P_1 + rho_oil*g*d,

where P_2 and P_1 are pressures in units of Pa. With the atm to Pa conversion built in, the depth is:

d = (P_2 - P_1)*((1.013e+5 Pa)/(1 atm))/(rho_oil*g).

Response (B) is P_1/(rho_oil*g); response (C) is P_2/(rho_oil*g); response (D) is (P_2 + P_1)/(rho_oil*g).

Student responses
Sections 70854, 70855
(A) : 36 students
(B) : 11 students
(C) : 1 student
(D) : 0 students

Success level: 75%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.67

Student responses
Section 72177
(A) : 11 students
(B) : 1 student
(C) : 0 students
(D) : 0 students

Success level: 91%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.25

Physics quiz question: horizontal, narrowing pipe

Physics 205A Quiz 5, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 9.50

[3.0 points.] Water at point [1] flows with a speed of 0.75 m/s through a pipe of 0.20 m inner radius. The pipe at point [2] tapers down to an inner radius of 0.050 m. Assume ideal fluid flow. The pressure of the water __________ as it flows from point [1] to point [2].
(A) decreases.
(B) increases.
(C) remains constant.
(D) (Not enough information is given.)

Correct answer: (A)

From applying the continuity equation:

A_1*v_1 = A_2*v_2,

the water speed at point [2] is greater than at point [1], because the cross-sectional area at point [2] is smaller than at point [1].

Then from Bernoulli's equation:

0 = delta(P) + (1/2)*rho*delta(v^2) + rho*g*delta(y),

the kinetic head (the second term on the right-hand side) increases due to the increase in speed from point [1] to point [2], while the gravitational head (the third term on the right-hand side) remains constant due to no change in elevation along the horizontal pipe, thus the pressure must decrease as water flows from point [1] to point [2].

Student responses
Sections 30880, 30881
(A) : 22 students
(B) : 16 students
(C) : 1 student
(D) : 0 students

"Difficulty level": 56%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.65

Found physics: sphygmomanometer

20090324048
http://www.flickr.com/photos/waiferx/3422440296/
Originally uploaded by Waifer X

Baumanometer® blood pressure gauge (sphygmomanometer), La Posada Medical Plaza, Templeton, CA. Photo by Cuesta College Physical Sciences Division instruction Dr. Patrick M. Len.