SP24 Quiz 2_Solutions_v2

SPRING 2024 BMED 3310 – BIOTRANSPORT QUIZ 2 STUDENT NAME: _________________________________________________________ GTID NUMBER: __________________________________________________ PSS SECTION (circle one): A01 – Thurs 3:30-5:25 (Chang) A02 – Friday 8:25-10:20 (Chang) A03 – Friday 12:30-2:25 (Gaidulis) • You have 20 minutes to complete this Quiz. Relevant formulas are provided for you as an Appendix. You may not use your book, electronic notebook, or any other materials. Calculators are allowed. No other resources are allowed. • You may raise your hand and the instructor proctoring your quiz will help to clarify any question you may have. • You cannot seek or accept help from other students or people outside of the class, nor can you provide help to other students. • Sign below, OR copy statement below to the frontpage of your work if you are unable to print this cover page. I will abide by these terms according to the Georgia Tech Honor Code __________________________________________________________________________ Signature Question Maximum Mark Actual Mark 1 3 2 4 Total 7

2 | P a g e 1. [3 points total] Question 1 contains 3 parts below. Part a is describing the “original” pipe and part b is describing a “new” pipe. Fluid flows axially down the length of the pipe. a. [0.5 points] The Reynolds number for fluid flow through a pipe is 𝑅𝑅𝑅𝑅 = 500. Circle one: Is the flow (a) laminar or (b) turbulent? + 0.5 points if choice (a) is clearly circled. b. [2 points] You want to increase the Reynolds number to 3000 by changing the pipe diameter and holding volumetric flow rate constant. If the original pipe diameter is 𝐷𝐷 𝑜𝑜 and the new pipe diameter is 𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 , write an expression relating the two diameters in the following form: 𝑫𝑫 𝒏𝒏𝒏𝒏𝒏𝒏 = 𝜶𝜶 ∙ 𝑫𝑫 𝒐𝒐 where 𝜶𝜶 is a constant numerical value. By changing the pipe diameter, both the diameter and average velocity term in the Reynolds number equation will change! 𝐼𝐼𝐼𝐼 𝑉𝑉 ̇ 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑛𝑛𝑜𝑜𝑜𝑜 = 𝑉𝑉 ̇ 𝑛𝑛𝑛𝑛𝑛𝑛 → 𝑣𝑣 𝑜𝑜 ∙ 𝜋𝜋𝐷𝐷 𝑜𝑜 2 4 = 𝑣𝑣 𝑛𝑛𝑛𝑛𝑛𝑛 ∙ 𝜋𝜋𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 2 4 ⇒ 𝑣𝑣 𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑣𝑣 𝑜𝑜 ∙ 𝐷𝐷 𝑜𝑜 2 𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 2 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝑃𝑃𝑂𝑂𝑃𝑃𝑅𝑅 : 𝑅𝑅𝑅𝑅 = 500 = 𝜌𝜌𝑣𝑣 𝑜𝑜 𝐷𝐷 𝑜𝑜 𝜇𝜇 𝑁𝑁𝑅𝑅𝑁𝑁 𝑃𝑃𝑂𝑂𝑃𝑃𝑅𝑅 : 𝑅𝑅𝑅𝑅 = 3000 = 𝜌𝜌𝑣𝑣 𝑛𝑛𝑛𝑛𝑛𝑛 𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 𝜇𝜇 ∴ 6 ∙ 𝜌𝜌𝑣𝑣 𝑜𝑜 𝐷𝐷 𝑜𝑜 𝜇𝜇 = 𝜌𝜌𝑣𝑣 𝑛𝑛𝑛𝑛𝑛𝑛 𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 𝜇𝜇 ⇒ 6 ∙ 𝑣𝑣 𝑜𝑜 𝐷𝐷 𝑜𝑜 = 𝑣𝑣 𝑛𝑛𝑛𝑛𝑛𝑛 𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 ∴ 6 ∙ 𝑣𝑣 𝑜𝑜 𝐷𝐷 𝑜𝑜 = 𝑣𝑣 𝑜𝑜 ∙ 𝐷𝐷 𝑜𝑜 2 𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 2 ∙ 𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 ⇒ 6 ∙ 𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 = 𝐷𝐷 𝑜𝑜 ⇒ 𝑫𝑫 𝒏𝒏𝒏𝒏𝒏𝒏 = 𝟏𝟏 𝟔𝟔 𝑫𝑫 𝒐𝒐 + 0.5 points for recognizing that the velocity will also change when diameter changes. This can be shown mathematically (e.g. volumetric flow rate equation) or in words. + 0.5 points for attempting to utilize the Reynolds Number equation �𝑅𝑅𝑅𝑅 = 𝜌𝜌𝜌𝜌𝜌𝜌 𝜇𝜇 � . Work must be shown beyond simply writing the equation. The characteristic length must be diameter ( 𝐷𝐷 ). + 0.5 points if student recognizes that 𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 must be smaller than 𝐷𝐷 𝑜𝑜 either implicitly (final answer) or explicitly (stated in words) + 0.5 for 𝑫𝑫 𝒏𝒏𝒏𝒏𝒏𝒏 = 𝟏𝟏 𝟔𝟔 𝑫𝑫 𝒐𝒐 . If a student correctly solves for 𝛼𝛼 = 1 6 = 0.1667 , full credit for this part is still awarded. c. [0.5 points] You want to maximize mixing of reagents within the pipe. Considering only the information provided in this problem, should you use the original pipe or the new pipe? Circle one: (a) Original Pipe or (b) New Pipe

4 | P a g e ∴ 𝑃𝑃 2 − 𝑃𝑃 1 = 1.2 W 80 cm 3 s ∙ � 1 m 100 cm � 3 − 500 Pa − 50 Pa = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝐏𝐏𝐏𝐏 ( 𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐏𝐏𝐢𝐢𝐢𝐢 ) + 0.25 point for substituting in correctly for Δ𝐾𝐾𝑃𝑃 Partial credit: +0.125 if sign is flipped for Δ𝐾𝐾𝑃𝑃 (i.e. Δ𝐾𝐾𝑃𝑃 = 1 2 𝜌𝜌𝒗𝒗 𝟏𝟏 𝟐𝟐 − 1 2 𝜌𝜌𝒗𝒗 𝟐𝟐 𝟐𝟐 = 50 Pa ) + 0.25 point for substituting in correctly for Δ𝑃𝑃𝑃𝑃 Partial credit: +0.125 if sign is flipped for Δ𝑃𝑃𝑃𝑃 (i.e. 𝛥𝛥𝑃𝑃𝑃𝑃 = 𝜌𝜌𝑂𝑂𝒉𝒉 𝟏𝟏 − 𝜌𝜌𝑂𝑂𝒉𝒉 𝟐𝟐 = 500 Pa ) + 0.5 point for rearranging to solve for 𝑃𝑃 1 − 𝑃𝑃 2 or 𝑃𝑃 2 − 𝑃𝑃 1 + 0.5 point for correctly substituting in for 𝑊𝑊 𝑠𝑠 ̇ ( 1.2 𝑊𝑊 ) and 𝑉𝑉 ̇ � 80 cm 3 s � Partial credit: +0.25 if only one of the terms is correctly substituted for (i.e. 𝑊𝑊 ̇ 𝑠𝑠 or 𝑉𝑉 ̇ ) Note: Do not take off points here for incorrect unit conversion of 𝑐𝑐𝑚𝑚 3 to 𝑚𝑚 3 . + 0.25 point for correct final answer with units in Pa ( ± 1% is okay). Partial credit: +0.125 if all work is correct other than unit conversion. Note: If the student solved for 𝑃𝑃 1 − 𝑃𝑃 2 = − 14450 𝑃𝑃𝑂𝑂 , then this is okay. However, students should recognize that the ventricle is increasing pressure (see below). + 0.25 point for indicating that pressure increases from the ventricle. END OF QUIZ

5 | P a g e APPENDIX Hydrostatic equation 𝑑𝑑𝑃𝑃 𝑑𝑑𝑑𝑑 = −𝜌𝜌𝑂𝑂 Shear Stress (cartesian) 𝜏𝜏 𝑦𝑦𝑦𝑦 = 𝜇𝜇 𝑑𝑑𝑣𝑣 𝑦𝑦 𝑑𝑑𝑑𝑑 Buoyancy Force 𝐹𝐹 𝑏𝑏 = 𝜌𝜌 𝑓𝑓𝑜𝑜𝑓𝑓𝑜𝑜𝑓𝑓 𝑂𝑂𝑉𝑉 𝑓𝑓𝑜𝑜𝑠𝑠𝑑𝑑 Volumetric Flow Rate 𝑉𝑉 ̇ = � ( 𝑣𝑣 ⃑ ∙ 𝑂𝑂 � ) 𝑑𝑑𝑑𝑑 𝐴𝐴 = 𝑣𝑣 𝑜𝑜𝜌𝜌𝑜𝑜 ∙ 𝑑𝑑 Reynolds Number 𝑅𝑅𝑅𝑅 = 𝜌𝜌𝑣𝑣𝜌𝜌 𝜇𝜇 = 𝑣𝑣𝜌𝜌 𝜗𝜗 L is the characteristic length (e.g. D for pipes) Reynolds Transport Theorem (Mass) 𝑑𝑑𝑀𝑀 𝑠𝑠𝑦𝑦𝑠𝑠 𝑑𝑑𝑑𝑑 = 𝑑𝑑 𝑑𝑑𝑑𝑑 � 𝜌𝜌𝑑𝑑𝑉𝑉 𝑛𝑛 𝐶𝐶𝐶𝐶 + � 𝜌𝜌 ( 𝑣𝑣 ⃑ ∙ 𝑂𝑂 �⃑ ) 𝑑𝑑𝑑𝑑 𝑛𝑛 𝐶𝐶𝐶𝐶 Generalized Bernoulli’s Equation 𝜌𝜌𝑂𝑂ℎ 1 + 1 2 𝜌𝜌𝑣𝑣 1 2 + 𝑃𝑃 1 = 𝜌𝜌𝑂𝑂ℎ 2 + 1 2 𝜌𝜌𝑣𝑣 2 2 + 𝑃𝑃 2 + 𝑊𝑊 𝑠𝑠 ̇ 𝑉𝑉 ̇ + Δ P f For a pipe: Δ𝑃𝑃 𝑓𝑓 = 1 2 𝐼𝐼𝜌𝜌𝑣𝑣 2 � 𝜌𝜌 𝐷𝐷 � For a pipe and laminar flow: 𝐼𝐼 = 64 𝑅𝑅𝑅𝑅