Florida International University Department of Civil and...
2
Florida International University Department of Civil and Environmental Engineering CWR 3201 Fluid Mechanics Fall 2018 Instructor: Arturo S. Leon, Ph.D., P.E., D.WRE TA: Thao Do, CEE Undergraduate Homework Assignment 3 Solutions Mechanics of Fluids (Fifth edition), by M.C. Potter, D.C. Wiggert and B.H. Ramadan. 1. 3.18 (same number in Fourth edition) a) cos = !∙! ! = !!! ! !!! ∙ ! ! ! !! ! = 0.832 = 33 .69 To calculate the normal: ∙ = 0 3 + 2 ∙ ! + ! = 0 3 ! + 2 ! = 0 : ! ! + ! ! = 1 Solving system of equations gives: = ! !" (2 − 3 ) Part b and c are the same steps except V is given as different equations 2. 3.19 (same number in Fourth edition) a) × = 0 + 2 + × + = 0 + 2 − = 0 OR !"! !!! = : + 2 = − 2 + 2 = + Plug in streamline point values to solve for coefficient C: 2 1 − 23 = −2 + = 0.8028 Equation of the streamline: − 2 + 2 = + 0 .8028 Part b and c are the same steps but different equations to integrate 3. 3.28 a. Along the center of the pipe r=0 cm: = 2 1 − 0 1 − ! ! !" = 2 ℎ = ∞ ! = = 2 1 − 0 1 10 ! ! !" = 0 .2 ! = 0
Transcript of Florida International University Department of Civil and...
Florida International University Department of Civil and Environmental Engineering
CWR 3201 Fluid Mechanics
Fall 2018 Instructor: Arturo S. Leon, Ph.D., P.E., D.WRE
TA: Thao Do, CEE Undergraduate Homework Assignment 3 Solutions
Mechanics of Fluids (Fifth edition), by M.C. Potter, D.C. Wiggert and B.H. Ramadan.
1. 3.18 (same number in Fourth edition)
a) cos𝛼 = !∙!!= !!! !!!! ∙ !
!!!!!= 0.832
𝛼 = 33.69 𝑑𝑒𝑔 To calculate the normal: 𝑉 ∙ 𝑛 = 0
3𝚤 + 2𝚥 ∙ 𝑛!𝚤 + 𝑛!𝚥 = 0 3𝑛! + 2𝑛! = 0
𝑈𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟: 𝑛!! + 𝑛!! = 1
Solving system of equations gives: 𝑛 = !!"(2𝚤 − 3𝚥)
Part b and c are the same steps except V is given as different equations 2. 3.19 (same number in Fourth edition) a) 𝑉 × 𝑑𝑟 = 0
𝑥 + 2 𝚤 + 𝑥𝑡𝚥 × 𝑑𝑥𝚤 + 𝑑𝑦𝚥 = 0 𝑥 + 2 𝑑𝑦 − 𝑥𝑡𝑑𝑥 = 0 OR 𝑡 !"!
!!!= 𝑑𝑦
𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒: 𝑡𝑥𝑑𝑥𝑥 + 2 = 𝑑𝑦
𝑡 𝑥 − 2𝑙𝑛 𝑥 + 2 = 𝑦 + 𝐶 Plug in streamline point values to solve for coefficient C: 2 1− 2𝑙𝑛3 = −2+ 𝐶
𝐶 = 0.8028 Equation of the streamline: 𝑡 𝑥 − 2𝑙𝑛 𝑥 + 2 = 𝑦 + 0.8028 Part b and c are the same steps but different equations to integrate
3. 3.28
a. Alongthecenterofthepiper=0cm:
𝑢 = 2 1 − 0 1 − 𝑒!!!" = 2
𝑚𝑠𝑤ℎ𝑒𝑛 𝑡 = ∞
𝑎! =𝜕𝑢𝜕𝑡
= 2 1 − 0110
𝑒!!!" = 0.2
𝑚𝑠!𝑎𝑡 𝑡 = 0
b. Atr=0.5cm:𝑢 = 2 1 − !.!!
!!1 − 𝑒!
!!" = 1.875!
!𝑤ℎ𝑒𝑛 𝑡 = ∞
𝑎! =𝜕𝑢𝜕𝑡
= 2 1 −0.5!
2!110
𝑒!!!" = 0.1875
𝑚𝑠!𝑎𝑡 𝑡 = 0
c. Alongthewall,r=2cm(entiretyofthetube):𝑢 = 2 1 − !!
!!1 − 𝑒!
!!" =
0!!𝑤ℎ𝑒𝑛 𝑡 = 𝑎𝑙𝑙 𝑡𝑖𝑚𝑒
𝑎! =𝜕𝑢𝜕𝑡
= 2 1 −2!
2!110
𝑒!!!" = 0
𝑚𝑠!𝑎𝑡 𝑡 = 𝑎𝑙𝑙 𝑡𝑖𝑚𝑒
4. 3.35
a. 𝐴 = 2 20𝑘×12𝚤 + 20𝑘× 20𝑘×4.5𝚤 = 480 𝚥 − 1800𝚤 𝑓𝑡/𝑠!b. 𝐴 = 2 20𝑘×−60𝑐𝑜𝑠30𝚥 + 20𝑘× 20𝑘×4.5𝚤 = 278𝚤 𝑓𝑡/𝑠!
5. 3.45
ReynoldsNumber:𝑅𝑒 = !"!=
(!!! )(!.!"# !)
(!.!!!!"!!)= 38961Turbulent
