# Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.

84 Pages

### Chap02

Course: CHEE 3363, Fall 2007
School: U. Houston
Rating:

Word Count: 2451

#### Document Preview

2.1 For Problem the velocity fields given below, determine: (a) whether the flow field is one-, two-, or three-dimensional, and why. (b) whether the flow is steady or unsteady, and why. (The quantities a and b are constants.) Solution (1) V = V ( x) V = V ( x , y) V = V ( x) V = V ( x , z) V = V ( x) V = V ( x , y , z) V = V ( x , y , z) V = V ( x , y) 1D V = V ( t) V V ( t) V V ( t) V...

Register Now

#### Unformatted Document Excerpt

Coursehero >> Texas >> U. Houston >> CHEE 3363

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
2.1 For Problem the velocity fields given below, determine: (a) whether the flow field is one-, two-, or three-dimensional, and why. (b) whether the flow is steady or unsteady, and why. (The quantities a and b are constants.) Solution (1) V = V ( x) V = V ( x , y) V = V ( x) V = V ( x , z) V = V ( x) V = V ( x , y , z) V = V ( x , y , z) V = V ( x , y) 1D V = V ( t) V V ( t) V V ( t) V V ( t) V V ( t) V = V ( t) V V ( t) V = V ( t) Unsteady (2) 2D Steady (3) 1D Steady (4) 2D Steady (5) 1D Steady (6) 3D Unsteady (7) 3D Steady (8) 2D Unsteady 48 49 50 Problem 2.4 A velocity field is given by r ^ V = axi - bty^ j where a = 1 s-1 and b = 1 s-2. Find the equation of the streamlines at any time t. Plot several streamlines in the first quadrant at t = 0 s, t = 1 s, and t = 20 s. Solution For streamlines -b t y v dy = = u dx a x -b t dx dy = y a x -b t a So, separating variables Integrating ln ( y) = ln ( x) -b The solution is y = c x a t For t = 0 s y=c For t = 1 s y= c x For t = 20 s y = c x - 20 See the plots in the corresponding Excel workbook 51 Problem 2.4 (In Excel) A velocity field is given by r ^ V = axi - bty^ j -1 -2 where a = 1 s and b = 1 s . Find the equation of the streamlines at any time t . Plot several streamlines in the first quadrant at t = 0 s, t =1 s, and t =20 s. Solution -b The solution is y = c x a t For t = 0 s y=c For t = 1 s y= c x For t = 20 s t=0 c=1 y 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 c=2 y 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 c=3 y 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 y = c x - 20 x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 t =1 s (### means too large to view) c=1 c=2 c=3 x y y y 0.05 20.00 40.00 60.00 0.10 10.00 20.00 30.00 0.20 5.00 10.00 15.00 0.30 3.33 6.67 10.00 0.40 2.50 5.00 7.50 0.50 2.00 4.00 6.00 0.60 1.67 3.33 5.00 0.70 1.43 2.86 4.29 0.80 1.25 2.50 3.75 0.90 1.11 2.22 3.33 1.00 1.00 2.00 3.00 1.10 0.91 1.82 2.73 1.20 0.83 1.67 2.50 1.30 0.77 1.54 2.31 1.40 0.71 1.43 2.14 1.50 0.67 1.33 2.00 1.60 0.63 1.25 1.88 1.70 0.59 1.18 1.76 1.80 0.56 1.11 1.67 1.90 0.53 1.05 1.58 2.00 0.50 1.00 1.50 t = 20 s c=1 y ##### ##### ##### ##### ##### ##### ##### ##### 86.74 8.23 1.00 0.15 0.03 0.01 0.00 0.00 0.00 0.00 0.00 0.00 0.00 c=2 y ##### ##### ##### ##### ##### ##### ##### ##### ##### 16.45 2.00 0.30 0.05 0.01 0.00 0.00 0.00 0.00 0.00 0.00 0.00 c=3 y ##### ##### ##### ##### ##### ##### ##### ##### ##### 24.68 3.00 0.45 0.08 0.02 0.00 0.00 0.00 0.00 0.00 0.00 0.00 x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 52 Streamline Plot (t = 0) 3.50 3.00 2.50 2.00 c=1 c=2 c=3 y 1.50 1.00 0.50 0.00 0.00 0.50 1.00 1.50 2.00 x Streamline Plot (t = 1 s) 70 60 50 40 c=1 c=2 c=3 y 30 20 10 0 0.00 0.50 1.00 1.50 2.00 x Streamline Plot (t = 20 s) 20 18 16 14 12 10 8 6 4 2 0 -0.15 0.05 0.25 0.45 0.65 0.85 1.05 1.25 c=1 c=2 c=3 y x 53 54 Problem 2.6 A velocity field is specified as r ^ V = ax 2 i + bxy ^ j where a = 2 m-1s-1 and b = - 6 m-1s-1, and the coordinates are measured in meters. Is the flow field one-, two-, or three-dimensional? Why? Calculate the velocity components at the point (2, 1/2). Develop an equation for the streamline passing through this point. Plot several streamlines in the first quadrant including the one that passes through the point (2, 1/2). Solution The velocity field is a function of x and y. It is therefore At point (2,1/2), the velocity components are u = a x = 2 2 2D 1 2 ( 2 m) m s 1 1 2 m m 2 m s b x y b y v dy = = = 2 a x u dx a x dy b dx = y a x b ln ( y) = ln ( x) a u = 8 m s m s v = b x y = -6 v = -6 For streamlines So, separating variables Integrating y = c x b a = c x -3 The solution is y= c x 3 See the plot in the corresponding Excel workbook 55 Problem 2.6 (In Excel) A velocity field is specified as r ^ V = ax 2 i + bxy ^ j where a = 2 m-1s-1, b = - 6 m-1s-1, and the coordinates are measured in meters. Is the flow field one-, two-, or three-dimensional? Why? Calculate the velocity components at the point (2, 1/2). Develop an equation for the streamline passing through this point. Plot several streamlines in the first quadrant including the one that passes through the point (2, 1/2). Solution The solution is c= x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 1 2 3 4 y y y y 8000 16000 24000 32000 1000 2000 3000 4000 125 250 375 500 37.0 74.1 111.1 148.1 15.6 31.3 46.9 62.5 8.0 16.0 24.0 32.0 4.63 9.26 13.89 18.52 2.92 5.83 8.75 11.66 1.95 3.91 5.86 7.81 1.37 2.74 4.12 5.49 1.00 2.00 3.00 4.00 0.75 1.50 2.25 3.01 0.58 1.16 1.74 2.31 0.46 0.91 1.37 1.82 0.36 0.73 1.09 1.46 0.30 0.59 0.89 1.19 0.24 0.49 0.73 0.98 0.20 0.41 0.61 0.81 0.17 0.34 0.51 0.69 0.15 0.29 0.44 0.58 0.13 0.25 0.38 0.50 y= c x 3 Streamline Plot 4.0 c=1 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 0.0 0.5 1.0 1.5 2.0 c=2 c=3 c = 4 ((x,y) = (2,1/2) y x 56 Problem 2.7 Solution Streamlines are given by -A y v dy = = u dx A x + B dy dx = -A y A x + B - 1 1 B ln ( y) = ln x + A A A So, separating variables Integrating The solution is y= C x+ B A For the streamline that passes through point (x,y) = (1,2) C = y x + B 20 = 2 1 + = 6 A 10 y= 6 x+ 20 10 y= 6 x+2 See the plot in the corresponding Excel workbook 57 Problem 2.7 (In Excel) Solution The solution is y= C x+ B A 6 y 3.00 2.86 2.73 2.61 2.50 2.40 2.31 2.22 2.14 2.07 2.00 1.94 1.88 1.82 1.76 1.71 1.67 1.62 1.58 1.54 1.50 A = 10 B = 20 C= x 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 1 y 0.50 0.48 0.45 0.43 0.42 0.40 0.38 0.37 0.36 0.34 0.33 0.32 0.31 0.30 0.29 0.29 0.28 0.27 0.26 0.26 0.25 2 y 1.00 0.95 0.91 0.87 0.83 0.80 0.77 0.74 0.71 0.69 0.67 0.65 0.63 0.61 0.59 0.57 0.56 0.54 0.53 0.51 0.50 4 y 2.00 1.90 1.82 1.74 1.67 1.60 1.54 1.48 1.43 1.38 1.33 1.29 1.25 1.21 1.18 1.14 1.11 1.08 1.05 1.03 1.00 Streamline Plot 3.5 c=1 3.0 2.5 2.0 c=2 c=4 c = 6 ((x,y) = (1.2) y 1.5 1.0 0.5 0.0 0.0 0.5 1.0 1.5 2.0 x 58 Problem 2.8 Solution Streamlines are given by v dy b x y = = 3 u dx a x 3 So, separating variables dy y 3 = b dx a x 2 Integrating - 1 2 y 2 = b 1 - + C a x The solution is y= 1 b + C 2 a x Note: For convenience the sign of C is changed. See the plot in the corresponding Excel workbook 59 Problem 2.8 (In Excel) Solution The solution is y= 1 2 C= x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 0 y 0.16 0.22 0.32 0.39 0.45 0.50 0.55 0.59 0.63 0.67 0.71 0.74 0.77 0.81 0.84 0.87 0.89 0.92 0.95 0.97 1.00 2 y 0.15 0.20 0.27 0.31 0.33 0.35 0.37 0.38 0.39 0.40 0.41 0.41 0.42 0.42 0.43 0.43 0.44 0.44 0.44 0.44 0.45 4 y 0.14 0.19 0.24 0.26 0.28 0.29 0.30 0.30 0.31 0.31 0.32 0.32 0.32 0.32 0.33 0.33 0.33 0.33 0.33 0.33 0.33 b + C a x 6 y 0.14 0.18 0.21 0.23 0.24 0.25 0.26 0.26 0.26 0.27 0.27 0.27 0.27 0.27 0.27 0.27 0.27 0.28 0.28 0.28 0.28 a= 1 b= 1 Streamline Plot 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 c=0 c=2 c=4 c=6 y x 60 Problem 2.9 61 Problem 2.10 Problem 2.11 Solution Streamlines are given by -b x v dy = = u dx a y t So, separating variables a t y dy = -b x dx Integrating 1 1 2 2 a t y = - b x + C 2 2 The solution is y= C- b x a t 2 For t = 0 s x=c For t = 1 s y= C - 4 x 2 For t = 20 s y= C- x 5 2 See the plots in the corresponding Excel workbook 63 Problem 2.11 (In Excel) Solution The solution is y= C- b x 2 a t For t = 0 s x=c For t = 1 s y= C - 4 x 2 For t = 20 s t=0 x 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 y= C- x 5 2 t =1 s C=1 y 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 C=2 y 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 C=3 y 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 x 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 0.275 0.300 0.325 0.350 0.375 0.400 0.425 0.450 0.475 0.500 C=1 y 1.00 1.00 0.99 0.99 0.98 0.97 0.95 0.94 0.92 0.89 0.87 0.84 0.80 0.76 0.71 0.66 0.60 0.53 0.44 0.31 0.00 C=2 y 1.41 1.41 1.41 1.41 1.40 1.39 1.38 1.37 1.36 1.34 1.32 1.30 1.28 1.26 1.23 1.20 1.17 1.13 1.09 1.05 1.00 C=3 y 1.73 1.73 1.73 1.73 1.72 1.71 1.71 1.70 1.69 1.67 1.66 1.64 1.62 1.61 1.58 1.56 1.54 1.51 1.48 1.45 1.41 t = 20 s x 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 C=1 y 1.00 1.00 1.00 0.99 0.98 0.97 0.96 0.95 0.93 0.92 0.89 0.87 0.84 0.81 0.78 0.74 0.70 0.65 0.59 0.53 0.45 C=2 y 1.41 1.41 1.41 1.41 1.40 1.40 1.39 1.38 1.37 1.36 1.34 1.33 1.31 1.29 1.27 1.24 1.22 1.19 1.16 1.13 1.10 C=3 y 1.73 1.73 1.73 1.73 1.72 1.72 1.71 1.70 1.69 1.68 1.67 1.66 1.65 1.63 1.61 1.60 1.58 1.56 1.53 1.51 1.48 64 Streamline Plot (t = 0) 3.5 3.0 2.5 2.0 c=1 c=2 c=3 y 1.5 1.0 0.5 0.0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 x Streamline Plot (t = 1s) 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 c=1 c=2 c=3 y x Streamline Plot (t = 20s) 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.0 0.5 1.0 1.5 2.0 2.5 c=1 c=2 c=3 y x 65 66 67 68 Problem 2.15 Solution Pathlines are given by dx = u = a x t dt dy = v = -b y dt So, separating variables dx = a t dt x dy = -b dt y Integrating ln ( x) = 1 2 a t + c1 2 ln ( y) = -b t + c2 For initial position (x0,y0) x = x0 e a 2 t 2 y = y0 e - b t Using the given data, and IC (x0,y0) = (1,1) at t = 0 x=e 0.05 t 2 y=e -t 69 Problem 2.15 (In Excel) Solution Using the given data, and IC (x0,y0) = (1,1) at t = 0, the pathline is The streamline at (1,1) at t = 0 s is x= 1 - y = x 10 x=e 0.05 t 2 y=e -t The streamline at (1,1) at t = 1 s is The streamline at (1,1) at t = 2 s is Pathline t 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.75 2.50 3.00 3.25 3.50 3.75 4.00 4.25 4.50 4.75 5.00 x 1.00 1.00 1.01 1.03 1.05 1.08 1.12 1.17 1.22 1.29 1.37 1.46 1.57 1.70 1.85 2.02 2.23 2.47 2.75 3.09 3.49 y 1.00 0.78 0.61 0.47 0.37 0.29 0.22 0.17 0.14 0.11 0.08 0.06 0.05 0.04 0.03 0.02 0.02 0.01 0.01 0.01 0.01 - y= x 5 Streamlines t=0 x y 1.00 1.00 1.00 0.78 1.00 0.61 1.00 0.47 1.00 0.37 1.00 0.29 1.00 0.22 1.00 0.17 1.00 0.14 1.00 0.11 1.00 0.08 1.00 0.06 1.00 0.05 1.00 0.04 1.00 0.03 1.00 0.02 1.00 0.02 1.00 0.01 1.00 0.01 1.00 0.01 1.00 0.01 t=1s x 1.00 1.00 1.01 1.03 1.05 1.08 1.12 1.17 1.22 1.29 1.37 1.46 1.57 1.70 1.85 2.02 2.23 2.47 2.75 3.09 3.49 y 1.00 0.97 0.88 0.75 0.61 0.46 0.32 0.22 0.14 0.08 0.04 0.02 0.01 0.01 0.00 0.00 0.00 0.00 0.00 0.00 0.00 t=2s x 1.00 1.00 1.01 1.03 1.05 1.08 1.12 1.17 1.22 1.29 1.37 1.46 1.57 1.70 1.85 2.02 2.23 2.47 2.75 3.09 3.49 y 1.00 0.98 0.94 0.87 0.78 0.68 0.57 0.47 0.37 0.28 0.21 0.15 0.11 0.07 0.05 0.03 0.02 0.01 0.01 0.00 0.00 Pathline and Streamline Plots 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Pathline Streamline (t = 0) Streamline (t = 1 s) Streamline (t = 2 s) y x 70 71 72 73 74 75 76 Problem 2.20 (In Excel) Solution Pathlines: t 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00 2.20 2.40 2.60 2.80 3.00 3.20 3.40 3.60 3.80 4.00 Starting at t = 0 x 0.00 -0.20 -0.40 -0.60 -0.80 -1.00 -1.20 -1.40 -1.60 -1.80 -2.00 -2.00 -2.00 -2.00 -2.00 -2.00 -2.00 -2.00 -2.00 -2.00 -2.00 y 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00 2.40 2.80 3.20 3.60 4.00 4.40 4.80 5.20 5.60 6.00 Starting at t = 1 s x y Starting at t = 2 s x y Streakline at t = 4 s x 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -0.20 -0.40 -0.60 -0.80 -1.00 -1.20 -1.40 -1.60 -1.80 -2.00 y 0.00 0.40 0.80 1.20 1.60 2.00 2.40 2.80 3.20 3.60 4.00 4.20 4.40 4.60 4.80 5.00 5.20 5.40 5.60 5.80 6.00 0.00 -0.20 -0.40 -0.60 -0.80 -1.00 -1.00 -1.00 -1.00 -1.00 -1.00 -1.00 -1.00 -1.00 -1.00 -1.00 0.00 0.20 0.40 0.60 0.80 1.00 1.40 1.80 2.20 2.60 3.00 3.40 3.80 4.20 4.60 5.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.40 0.80 1.20 1.60 2.00 2.40 2.80 3.20 3.60 4.00 Pathline and Streamline Plots 6 5 4 y Pathline starting at t = 0 Pathline starting at t = 1 s Pathline starting at t = 2 s Streakline at t = 4 s 3 2 1 0 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 x 79 80 Problem 2.22 81 Problem 2.22 (cont'd) 82 Problem 2.23 84 85 86 87 Problem 2.26 88 Problem 2.27 89 Problem 2.28 (In Excel) Solution Pathlines: T (oC) 0 100 200 300 400 Data: T (K) 273 373 473 573 673 (x105) 1.86E-05 2.31E-05 2.72E-05 3.11E-05 3.46E-05 Using procedure of Appendix A.3: T (K) 273 373 473 573 673 T3/2/ 2.43E+08 3.12E+08 3.78E+08 4.41E+08 5.05E+08 The equation to solve for coefficients S and b is S T32 1 = T + b b From the built-in Excel Linear Regression functions: Slope = 6.534E+05 Intercept = 6.660E+07 R2 = 0.9996 Plot of Basic Data and Trend Line 6.E+08 Hence: b = 1.53E-06 S = 101.9 kg/m.s.K1/2 K 5.E+08 Data Plot Least Squares Fit 4.E+08 T3/2/ 3.E+08 2.E+08 1.E+08 0.E+00 0 100 200 300 400 500 600 700 800 T 91 92 93 Problem 2.31 94 95 96 97 Problem 2.35 98 99 100 101 Problem 2.38 A block 0.2 m square, with 5 kg mass, slides down a smooth incline, 30 below the horizontal, a film of SAE 30 oil at 20C that is 0.20 mm thick. If the block is released from rest at t = 0, wh is its initial acceleration? Derive an expression for the speed of the block as a function of time. the curve for V(t). Find the speed after 0.1 s. If we want the mass to instead reach a speed of 0.3 m/s at this time, find the viscosity of the oil we would have to use. Ff = A Given: Data on the block and incline Find: Initial acceleration; formula for speed of block; plot; find speed after 0.1 s. Find oil viscosity if speed is 0.3 m/s after 0.1 s Solution Given data M = 5 kg A = ( 0.2 m) 2 x, V, a M g d = 0.2 mm = 30 deg From Fig. A.2 = 0.4 N s m 2 Applying Newton's 2nd law to initial instant (no friction) M a = M g sin ( ) - Ff = M g sin ( ) m ainit = g sin ( ) = 9.81 sin ( 30) 2 s m ainit = 4.9 2 s so Applying Newton's 2nd law at any instant M a = M g sin ( ) - Ff 102 and Ff = A = du V A = A dy d so M a = M A dV = M g sin ( ) - V dt d Separating variables dV A V g sin ( ) - M d = dt Integrating and using limits - M d A ln 1 - M g d sin ( ) A V = t or t M g d sin ( ) M d 1 - e V ( t) = A - A 0.4 0.3 V (m/s) 0.2 0.1 0 0.05 0.1 0.15 0.2 t (s) 0.25 0.3 0.35 103 At t = 0.1 s V = 5 kg 9.81 m s 2 0.0002 m sin ( 30) m 2 0.4 N s ( 0.2 m) 0.4 0.04 0.1 - N s 5 0.002 1 - e 2 2 kg m V = 0.245 m s To find the viscosity for which V(0.1 s) = 0.3 m/s, we must solve V ( t = 0.1 s) = M g d sin ( ) A 1 - e - A ( t=0.1 s) M d The viscosity is implicit in this equation, so solution must be found by manual iteration, or by of a number of classic root-finding numerical methods, or by using Excel's Goal Seek From the Excel workbook for this problem the solution is = 0.27 N s m 2 Excel workbook 104 Problem 2.38 (In Excel) A block 0.2 m square, with 5 kg mass, slides down a smooth incline, 30 below the horizontal, on a film of SAE 30 oil at 20C that is 0.20 mm thick. If the block is released from rest at t = 0, what is its initial acceleration? Derive an expression for the speed of the block as a function of time. Plot the curve for V (t ). Find the speed after 0.1 s. If we want the mass to instead reach a speed of 0.3 m/s at this time, find the viscosity of the oil we would have to use. Solution Ff = A x, V, a The solution is V( t) = t M gd sin( ) M d 1 - e A - A M g The data is M= = = A= d= 5.00 30 0.40 0.04 0.2 kg deg N.s/m2 m2 mm Speed V of Block vs Time t t (s) 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.20 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29 0.30 V (m/s) 0.000 0.045 0.084 0.117 0.145 0.169 0.189 0.207 0.221 0.234 0.245 0.254 0.262 0.268 0.274 0.279 0.283 0.286 0.289 0.292 0.294 0.296 0.297 0.299 0.300 0.301 0.302 0.302 0.303 0.304 0.304 0.35 0.3 0.25 0.2 V (m/s) 0.15 0.1 0.05 0 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 t (s) To find the viscosity for which the speed is 0.3 m/s after 0.1 s use Goal Seek with the velocity targeted to be 0.3 by varying the viscosity in the set of cell below: t (s) 0.10 V (m/s) 0.300 for 2 = 0.270 N.s/m 105 106 107 Problem 2.41 108 Problem 2.42 109 110 Problem 2.44 The viscometer of Problem 2.43 is being used to verify that the viscosity of a particular fluid is = 0.1 N.s/m2. Unfortunately the cord snaps during the experiment. How long will it take the cylinder to lose 99% of its speed? The moment of inertia of the cylinder/pulley system is 0.0273 kg.m2. Given: Data on the viscometer Find: Time for viscometer to lose 99% of speed Solution The given data is R = 50 mm H = 80 mm a = 0.20 mm I = 0.0273 kg m 2 = 0.1 N s m 2 The equation of motion for the slowing viscometer is I = Torque = - A R where is the angular acceleration and viscometer The stress is given by = V R du V-0 = = = dy a a a where V and are the instantaneous linear and angular velocities. 111 Hence R R A d =- A R = I = I dt a a Separating variables d =- R A dt a I 2 2 Integrating and using IC = 0 R A t a I 2 - ( t) = 0 e The time to slow down by 99% is obtained from solving R A t a I 2 - 0.01 0 = 0 e so t=- a I R A 2 ln ( 0.01) Note that A = 2 R H so t=- a I 2 R H 3 ln ( 0.01) t=- 0.0002 m 0.0273 kg m 2 2 m 1 1 N s ln ( 0.01) 0.1 N s ( 0.05 m) 3 0.08 m kg m 2 2 t = 4s 112 Problem 2.45 Problem 2.46 Problem 2.46 (cont'd) 113 Problem 2.47 114 115 Problem 2.49 116 Problem 2.50 Given: Data from viscometer Find: The values of coefficients k and n; determine the kind of non-Newtonial fluid it is; estimate viscosity at 90 and 100 rpm Solution The velocity gradient at any radius r is du r = dy r tan ( ) where (rad/s) is the angular velocity = 2 N 60 where N is the speed in rpm du For small , tan() can be replace with , so = dy n-1 du du = du k dy dy dy From Eq 2.11. du where is the apparent viscosity. Hence = k dy n-1 = k n-1 117 The data in the table conform to this equation. The corresponding Excel workbook shows how Excel's Trendline analysis is used to fit the data. From Excel k = 0.0449 n = 1.21 ( 90 rpm) = 0.191 N s m 2 ( 100 rpm) = 0.195 N s m 2 For n > 1 the fluid is dilatant 118 Problem 2.50 (In Excel) Solution The data is N (rpm) 10 20 30 40 50 60 70 80 (N.s/m ) 0.121 0.139 0.153 0.159 0.172 0.172 0.183 0.185 Viscosity vs Shear Rate / (1/s) 120 240 360 480 600 720 840 960 (N.s/m2x103) 121 139 153 159 172 172 183 185 1000 2 The computed data is (rad/s) 1.047 2.094 3.142 4.189 5.236 6.283 7.330 8.378 2 3 (N.s/m x10 ) Data Power Trendline 100 y = 44.94x0.2068 R2 = 0.9925 From the Trendline analysis 10 k = 0.0449 n - 1 = 0.2068 n = 1.21 100 1000 Shear Rate / (1/s) The fluid is dilatant The apparent viscosities at 90 and 100 rpm can now be computed N (rpm) 90 100 (rad/s) 9.42 10.47 / (1/s) 1080 1200 (N.s/m2x103) 191 195 119 Problem 2.51 121 Problem 2.52 Problem 2.53 Problem 2.54 125 126 127 Problem 2.57 128 Problem 2.58 You intend to gently place several steel needles on the free surface of the water in a large tank. The needles come in two lengths: Some are 5 cm long, and some are 10 cm long. Needles of each length are available with diameters of 1 mm, 2.5 mm, and 5 mm. Make a prediction as to which needles, if any, will float. Given: Data on size of various needles Find: Which needles, if any, will float Solution For a steel needle of length L, diameter D, density s, to float in water with surface tension an contact angle , the vertical force due to surface tension must equal or exceed the weight 2 L cos ( ) W = m g = 8 cos ( ) s g mN m kg m 3 D 4 2 s L g or D From Table A.4 = 72.8 = 0 deg and for water = 999 From Table A.1, for steel SG = 7.83 Hence 8 cos ( ) SG g 3 2 = 8 7.83 72.8 10 -3 N m s kg m -3 = 1.55 10 m m 999 kg 9.81 m N s2 Hence D < 1.55 mm. Only the 1 mm needles float (needle length is irrelevant) 129 Problem 2.59 130 Problem 2.60 132 Problem 2.62 133
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

UCSC - MATH - 11A
UT Dallas - NSC - 3361
Practice Exam 1 The following practice exam contains multiple choice, true/false, fill in the blank, and a few short answer questions. Not all topics present will appear on the actual exam, nor will topics on the actual exam be limited to those
UT Dallas - NSC - 3361
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42.B,C Axon Hillock D C,D A,C Astrocytes Oligodendrocytes; Schwann Cells B B,D,E A
UT Dallas - NSC - 3361
Practice Exam 3 1. Which of the following situations would most likely occur in anterograde amnesia? a. Forgetting who you are b. Forgetting your parents c. Discovering that you have the physical capabilities of a master ninja and a certificate o
UT Dallas - NSC - 3361
Practice Exam 3 Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. C B A, B, E A, D, E A, D E Genetic B, D F C, D, E B A, B, C, F
UT Dallas - CS - 1136
CS 1136 Lab 6While, Do-While and For LoopsWhile Loops If statements allow the program to skip the execution of a statement or choose between one of two statements to be executed based on the value of a Boolean expression. In contrast, the while sta
UT Dallas - CS - 1136
CS 1136 Lab 7Continuation of loopsCarefully examine the program below and try to work out what occurs in the nested for loops. What do you predict the output will be?public class Lab7_Loops { Public static void main(String [] args) { Scanner keyb
UT Dallas - ECO - 2301
1The Firm and its Objectives: Profit Maximization The optimality (first-order) conditionsWhat is the economic definition of profit? Profit = Total Revenue Total Cost = TR TC What is the fundamental problem for firms? The firm selects its output/
UT Dallas - ECO - 2301
1 WELFARE PROPERTIES OF MARKETS Consumer's Surplus [Money-Metric Utility] The purpose of this handout is to describe how economists measure changes in consumer WELFARE in response to PRICE CHANGES. The problem is: 1. Consumer welfare/satisfaction is
U. Houston - CHEE - 3300
CHAPTER 3THE STRUCTURE OF CRYSTALLINE SOLIDSPROBLEM SOLUTIONS3.1 Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the number and probability distributions of the constituent electrons. On the
UT Dallas - BIOL - 1318
Exam 3 Review Human Genetics BIOL 1318 March 4, 2008 The exam will consist of 50 multiple choice questions and will cover chapters 10 through 13 of your text Human Genetics. The following topics and points will be addressed on the exam: 1. transcript
UT Dallas - ECO - 2301
Quiz 4 Sample Questions 1. a. b. c. d. Total revenue will increase if price falls and demand is inelastic. falls and demand is elastic. rises and demand is elastic. rises and demand is unitary elastic.P 10 S2. a. b. c. d. e.According to the acco
Arizona - ACCT - 200
ACCT 201 ACCT 201 ACCT 2017Reporting and Analyzing Receivables and InvestmentsChapterUAA ACCT 201 Principles of Financial Accounting Dr. Fred Barbee1ACCT 201Day #1ACCT 201ACCT 201Chapter 7 - Day 1 - AgendaTopicAccounts Receivable
E. Michigan - ENG - 121
Annotated BibliographyIamnext. &quot;College Relationships: Roommate tips for dorm life.&quot;. 2004.Iamnext.30 February 2008. http:/www.iamnext.com/people/rmatedorm.html This source is an overall summary of being a roommate in a dorm. It covers confronting
U. Houston - CHEE - 3300
CHAPTER 6MECHANICAL PROPERTIES OF METALSPROBLEM SOLUTIONS6.1This problem asks that we derive Equations (6.4a) and (6.4b), using mechanics of materials principles. In Figure (a) below is shown a block element of material of cross-sectional are
UCSD - BIBC - 103
Sea Urchin Fertilization How does the sperm penetrate all the way through the egg to get to the nucleus? Signal transduction events occur both outside and inside the plasma membraneStructure of sea urchin egg at fertilizationScanning electron m
U. Houston - CHEE - 3300
CHAPTER 7DISLOCATIONS AND STRENGTHENING MECHANISMSPROBLEM SOLUTIONS7.1 The dislocation density is just the total dislocation length per unit volume of material (in this case per 3 5 -2 cubic millimeters). Thus, the total length in 1000 mm of ma
UCSD - BIBC - 103
Overview of the Central DogmaDNA replication(DNA polymerase)DNATranscription(RNA polymerase)5' UTR 3' UTR ATG stop codonRNARBS(ribosome binding site)Translation(ribosome)Met (first amino acid)proteinFluorescent Protein Lab Experi
U. Houston - CHEE - 3300
CHAPTER 8FAILUREPROBLEM SOLUTIONS8.1 Several situations in which the possibility of failure is part of the design of a component or product are as follows: (1) the pull tab on the top of aluminum beverage cans; (2) aluminum utility/light poles
University of Florida - EAB - 3002
The &quot;Learning Curve&quot;100 80 60 40 20 0 1 2 3 4 5 6 7 8 9 10 11 12S1100 80 60 40 20 0 1 2 3 4 5 6 7 8 9 10 11S212100 80 60 40 20 0 1 2 3 4 5 6 7 8 9 10 11S3100 80 60 40 20 0S412123456789101112100 80 60 40 20
U. Houston - CHEE - 3300
CHAPTER 9PHASE DIAGRAMSPROBLEM SOLUTIONS9.1 Three variables that determine the microstructure of an alloy are 1) the alloying elements present, 2) the concentrations of these alloying elements, and 3) the heat treatment of the alloy.9.2 In or
U. Houston - CHEE - 3300
CHAPTER 10PHASE TRANSFORMATIONS IN METALSPROBLEM SOLUTIONS10.1 The two stages involved in the formation of particles of a new phase are nucleation and growth. The nucleation process involves the formation of normally very small particles of the
SUNY Buffalo - CHE - 101
Chapter 6 TroCounting and weighingExample: Burning Aluminum Five hundred aluminum atoms were burned in oxygen. The resulting aluminum oxide had a total of 1250 atoms. What is the formula of aluminum oxide? Number of oxygen atoms = 1250 500 = 75
U. Houston - CHEE - 3300
CHAPTER 11APPLICATIONS AND PROCESSING OF METAL ALLOYSPROBLEM SOLUTIONS11.1 This question asks that we list four classifications of steels, and, for each, to describe properties and cite typical applications. Low-Carbon Steels Properties: nonres
U. Houston - CHEE - 3300
CHAPTER 12STRUCTURES AND PROPERTIES OF CERAMICSPROBLEM SOLUTIONS12.1The two characteristics of component ions that determine the crystal structure are:1) themagnitude of the electrical charge on each ion; and 2) the relative sizes of the
Western Washington - PSYCH - 101
Phone a phonetic segment Phoneme sound segment that is distinctive, that contrasts or distinguishes words; displayed in between / / Allophone- a predictable phonetic variant of a phoneme [ ]
U. Houston - CHEE - 3300
CHAPTER 13APPLICATIONS AND PROCESSING OF CERAMICSPROBLEM SOLUTIONS13.1 The two desirable characteristics of glasses are optical transparency and ease of fabrication.13.2 (a) Devitrification is the process whereby a glass material is caused to
Western Washington - PSYCH - 101
What is Psychology? Text: &quot;the science of behavior and the mind&quot; (Gray, 2002, p. 3) The scientific study of behavior and mental processes Where did Psychology come from? Philosophy Socrates Plato Arisotle Are People inherently good or evil? Do o
U. Houston - CHEE - 3300
CHAPTER 14POLYMER STRUCTURESPROBLEM SOLUTIONS14.1Polymorphism is when two or more crystal structures are possible for a material of given composition. Isomerism is when two or more polymer molecules or mer units have the same composition, but
Western Washington - PSYCH - 101
Chapter 6 Motivation and Emotion Sleep Sleep is a motivated behavior Why do we sleep? Is it due to changes in light? (day/night cycle) Sleep Regulation Highly regulated behaviour Time Deprivation Studies - Free Running Sleep Regulation Follow a
SUNY Buffalo - CHE - 101
Chapter 7 troChemical equations and reactionsOxyhydrocarbon 1 (from last week) One mole of C3H6O5 is burned in O2. How many moles of CO2 are formed? (3) How many moles of H2O are formed? (3) How many moles of O2 are consumed?(2) Safest approa
Western Washington - PSYCH - 101
Chapter 9 Memory Storage of Information so that it can be used at a later time The Information-Processing Approach to Memory Memory may be thought of as a three-stage process, in which information in each stage lasts longer than in the preceding o
U. Houston - CHEE - 3300
CHAPTER 15CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERSPROBLEM SOLUTIONS15.1 From Figure 15.3, the elastic modulus is the slope in the elastic linear region of the 20 C curve, which is(stress) 30 MPa 0 MPa = = 3.3 GPa (strain) 9 x
Western Washington - PSYCH - 101
Chapter 13 Social Perception Social Psychology Social Cognition examines how we process information about the social world, ourselves and others Social Influence examines how people influence other's behaviour, judgments, etc. Social Cognition H
U. Houston - CHEE - 3363
Problem 3.1D = 0.75 m. The gas is at an absolute pressure of 25 MPa and a temperature of 25C. What is the mass in the tank? If the maximum allowable wall stress in the tank is 210 MPa, find the minimum theoretical wall thickness of the tank. Given:
U. Houston - CHEE - 3300
CHAPTER 16COMPOSITESPROBLEM SOLUTIONS16.1The major difference in strengthening mechanism between large-particle and dispersionstrengthened particle-reinforced composites is that for large-particle the particle-matrix interactions are not trea
Michigan - ECON - 340
Econ 340 Fall Term 2004 Page 1 of 15 (16)Alan Deardorff First Midterm (with Answers) NAME: _ Student ID No.: _ Economics 340 International Economics Prof. Alan Deardorff First Midterm ExamAnswersOctober 14, 2004INSTRUCTIONS: READ CAREFULLY!1.
U. Houston - CHEE - 3300
CHAPTER 17CORROSION AND DEGRADATION OF MATERIALSPROBLEM SOLUTIONS17.1 (a) Oxidation is the process by which an atom gives up an electron (or electrons) to become a cation. Reduction is the process by which an atom acquires an extra electron (or
Michigan - ECON - 398
ECON 398 Fall 2006 Midterm 1 Practice Exam 1 Professor OzdenorenINSTRUCTIONS You have 80 minutes to complete the exam. There are 16 questions.1. Consider the following simultaneous game, which has a variable x for two of the payoffs.High High Lo
U. Houston - CHEE - 3300
CHAPTER 18ELECTRICAL PROPERTIESPROBLEM SOLUTIONS18.1 This problem calls for us to compute the electrical conductivity and resistance of a silicon specimen. (a) We use Equations (18.3) and (18.4) for the conductivity, as1 Il = VA Il V d 22=
U. Houston - CHEE - 3300
CHAPTER 19THERMAL PROPERTIESPROBLEM SOLUTIONS19.1 The energy, E, required to raise the temperature of a given mass of material, m, is the product of the specific heat, the mass of material, and the temperature change, T asE = cpm(T)The T is
U. Houston - CHEE - 3300
CHAPTER 20MAGNETIC PROPERTIESPROBLEM SOLUTIONS20.1 (a) We may calculate the magnetic field strength generated by this coil using Equation (20.1) asNI lH ==(200 turns)(10 A) = 10,000 A - turns/m 0.2 m(b) In a vacuum, the flux density is
U. Houston - CHEE - 3300
CHAPTER 21OPTICAL PROPERTIESPROBLEM SOLUTIONS21.1 Similarities between photons and phonons are: 1) Both may be described as being wave-like in nature. 2) The energy for both is quantized. Differences between photons and phonons are: 1) Phonons
SUNY Buffalo - CHE - 101
Chapter 8 troStoichiometry
U. Houston - CHEE - 3300
CHAPTER 22ECONOMIC, ENVIRONMENTAL, AND SOCIETAL ISSUES IN MATERIALS SCIENCE AND ENGINEERINGPROBLEM SOLUTION22.D1W The three materials that are used for beverage containers are glass, aluminum, and the polymer polyethylene terephthalate (designa
UCSC - BIO - 105
Biology 105: Introduction to Genetics PRACTICE FINAL EXAM 2006 Part I: Definitions Homology: Comparison of two or more protein or DNA sequence to ascertain similarities in sequences. If two genes have an almost identical DNA sequence, they are homolo
Michigan - ECON - 398
ECON 398 Fall 2006 Midterm 1 Practice Exam 2 Professor OzdenorenINSTRUCTIONS You have 80 minutes to complete the exam. There are 16 questions.1. Notice that there is a variable x in the payoffs of the following game: Carrie R U 3,4 D 1,3Rhonda
UCSC - CHEM - 108B
CHEMISTRY 108B UC SANTA CRUZ WINTER 2007 EXAM 1KEYCH3 HO OH1 2 3 4 5 6 7 8 9 extra TotalName:_ TA:_ Section:_1. (12 points) Draw structures for the following IUPAC names: a) 2-methyl-1-phenyl-1,4-butanediolb) tert butoxycyclohexane (tert b
Michigan - ECON - 398
ECON 398 Fall 2006 Midterm 1 Practice Exam 3 Professor OzdenorenINSTRUCTIONS You have 80 minutes to complete the exam. There are 16 questions.Wonder Cloak Inc. (C) and Scary Mask Inc. (M) are two shops situated on the opposite side of State Stree
Michigan - ECON - 398
ECON 398 Fall 2006 Midterm 1 Practice Exam 4 Professor OzdenorenINSTRUCTIONS You have 80 minutes to complete the exam. There are 16 questions.Elbonia and Sylvonia are two neighboring countries where Elbonia is much larger in size and it is more i
UCSC - CHEM - 108M
UCSC - PHYS - 7A
PHYSICS 7A FINAL REVIEWCHRIS FRANCEThis is some review material that should help you prepare for the final. I will post solutions to the questions after the review session on Friday. I have no more information than you do as to what Dave will ask
Prairie View A & M - CVEG - 2454
UNC - DRAM - 116
February 14, 200719thImportant Events Charles Darwin Light bulb Social changes Religion testedCentury turns to20thCenturyChanges in Theater True proscenium stage Realism o &quot;The details of the setting, the costuming and the circumstances of
SUNY Buffalo - CHE - 101
Chapter 9 TroElectrons in atomsElectromagnetic Radiation Visible light is but one form of radiation. Wave-particle duality of light. Photons (mirror, photoelectric effect, Newton, Einstein) versus waves (prism, diffraction and interference, Huyg
Prairie View A & M - CVEG - 2454
Baylor - SOC - 1301
Study guide for Test I Textbook preparation Use Online resources Always skip &quot;A Closer view&quot; with a few exceptions (see below) Introduction Ch. 1: Basic definitions of sociology; sociological imagination, industrial revolution, leading historical fig
UNC - DRAM - 116
1.22.07 Chapter One - Play Vocabulary plot - the action of the play exposition - what happened before the play began suspense - used to build tension rising action - the audience wonders what is going to happen - sees the characters set in motion que
UNC - DRAM - 116
Terminology:1. The House: where the audience sits 2. The Voms (vomitorium): entrance way under the audience, through the audience (Paul Green Theater), or beside the audience 3. House Manager: make sure everything is working with the house, manage t
Baylor - SOC - 1301
The Chemistry of Life 2 The Chemistry of Life 2.1 What Are the Chemical Elements That Make Up Living Organisms? 2.2 How Do Atoms Bond to Form Molecules? 2.3 How Do Atoms Change Partners in Chemical Reactions? 2.4 What Properties of Water Make It So I
UNC - DRAM - 116
Drama Exam Review Quotes- 1 quote from a play, some from the text All plays except for &quot;Rising of the Moon&quot; Multiple choice, true/false, and matching One question from &quot;The Illusion&quot; Contemporary Theatre- know terms and subheadings Breakdown of Exam