!! +! 2!! +!"!! =!! +! 2!! +!"!! +!!"!"!"


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1 Homework 4 Solutions 1. (15 points) Bernoulli s equation can be adapted for use in evaluating unsteady flow conditions, such as those encountered during start up processes. For example, consider the large tank below that is initially filled with water to a depth of 3 m. A pipe is attached with the dimensions shown, and initially capped at point 2. When the cap is removed, water will begin to flow out of the tube, and the velocity of the water in the tube will change over time. We can modify the Bernoulli equation to account for transient effects using!! +! 2!! +!"!! =!! +! 2!! +!"!! +!!"!"!" where the last term describes the rate of change of the fluid momentum as we travel along a streamline.!! (a) (5 points) Show that the integrand on the right hand side of the equation has units of pressure. (b) (1 points) Making the assumptions discussed in class, we saw that the above equation can be simplified to give!!!! 2!h! =!" 2! Integrate this equation to find an expression for V 2 (t) and plot the result. You may need to consult a table of integrals (or use computational tools) to evaluate the left hand side.
2 (a) The integrand is!!"!"!", where! has units of kg/m3, the rate of change of the velocity!"!" has units of acceleration, m/s 2, and!" is the differential distance traveled along the path s, and has units of m. Together, kg m m 3 kg m m= s 2 s 2 m 2 = N m 2 =Pa. Thus, the integrand has units of pressure. (b) SOLUTION: Apply the Bernoulli equation to the unsteady flow along a streamline from point 1 to point 2. (5) (6) Governing equation: p 1 V 1 V 2 p gz t ds 1 2 V gz2 2 2 Assumptions: Then 1 (1) Incompressible flow. (2) Frictionless flow. (3) Flow along a streamline from 1 to 2. (4) p 1 p 2 p atm. (5) V 2 1 : (6) z 2. (7) z 1 h constant. (8) Neglect velocity in reservoir, except for small region near the inlet to the tube. In view of assumption (8), the integral becomes In the tube, V V 2 everywhere, so that Z L gz 1 gh V2 2 2 Z 2 ds Z L Z L Z @t ds dv 2 dt ds L dv 2 dt This is the rate of change over time of the momentum (per unit mass) within the pipe; in the long term it will approach zero.
3 Z Z CHEN 32 Fluid Mechanics Spring 211 This is the rate of change over time of the momentum (per unit mass) within the pipe; in the long term it will approach zero. Substituting gives Separating variables, we obtain gh V2 2 2 L dv 2 dt dv 2 2gh V 2 2 dt 2L Integrating between limits V att and V V 2 at t t, Z V2 dv 2gh V 2 pffiffiffiffiffiffiffi 1 tanh 1 V p ffiffiffiffiffiffiffi 2gh 2gh Since tanh 1 (), we obtain 1 pffiffiffiffiffiffiffi tanh 1 V 2 pffiffiffiffiffiffiffi t 2gh 2gh 2L For the given conditions, or V2 V 2 p tanh t ffiffiffiffiffiffiffi 2gh rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi 2 9:81 m 3m 2gh 7:67 m/s s2 t 2L pffiffiffiffiffiffiffi 2gh 2L ß V 2 ðtþ and t p 2L ffiffiffiffiffiffiffi 2gh t 2 1 6m 7:67 m s :639t The result is then V tanh (.639t) m/s,asshown: V 2 (m/s) V 2 = 7.67 tanh (.639 t) t (s)
4 2. (15 points) Fluid approaches a submerged cylinder with velocity!! = 1 m/s. The cylinder has a radius of! = 55 cm. Using boundary layer theory, it is possible to describe how the fluid velocity changes near the cylinder surface as a function of r and θ, with θ measured as shown below. At the surface of the cylinder, the fluid speed is determined to be! = 2!! sin!.!"!"!!"!" Calculate a s and a n at point A on the surface of the cylinder, where θ = 6.
5
6 3. (2 points) An incompressible, one dimensional fluid flows from left to right through the circular nozzle shown below. The velocity entering the nozzle is given by! =!! +!! sin!", where!! = 2 m/s,!! = 2 m/s, and! =.3 rad/s. The nozzle is 1 m in length,.4 m in diameter at the entrance, and.2 m in diameter at the exit. (a) (1 points) Determine an equation for the acceleration at the exit of the nozzle as a function of time. (b) (5 points) Plot the acceleration versus time for one complete cycle. (c) (5 points) Now, plot the acceleration at the channel exit if the nozzle has a constant diameter of.4 m (i.e., it is now a cylindrical tube). Explain the difference between the two plots. (a) = sin!" sin 2!" cos!"
7 (b, c)
8 4. (2 points) A fluid velocity field is given by! =!"!!!"!, where! =.2 s!! and! =.6 m/s. If x and y have units of meters: (a) (a) (5 points) Find a general expression for the acceleration vector (a) as a function of x and y. (b) (1 points) Find the acceleration (a), magnitude of the acceleration ( a ), and the angle the acceleration vector makes with the x axis (θ) for each of the points (,1.33), (1,2) and (2,4). (c) (5 points) Find an expression for the streamlines, in the form!!,! =!, where C is a constant. Plot this function for values of! = ±.2, ±.4 and ±.8. Draw the acceleration vectors from (b) on your plot. =.4!.12! +.4!! (b) The acceleration at each point is just a evaluated at the x and y coordinates: The magnitude of a, given by! =!!! +!!!. At (x,y) = (, 1.33), a =.131 m/s 2 = (1, 2), a =.113 m/s 2 = (2, 4), a =.164 m/s 2 The angle that a makes with the x axis is given by! = tan 1!!!!. Therefore, 1.!"## At (x,y) = (, 1.33),! = tan = 23!.!" 1.!"!! = (1, 2),! = tan = 45!.!"!! = (2, 4),! = tan 1.!"#!.!"!! = 76
9 (c)
10 5. (15 points) From Newtonian physics, a particle launched with velocity V at an angle θ with respect to the ground will travel in a parabolic path given by! =!/ 2!!! cos 2!!! + tan!!. The pathline for a stream of water leaving a nozzle is shown on the last page. (a) (1 points) Use the end of the nozzle as the origin, and find the coordinates of several points along the pathline for the stream of water. Plot the coordinates, and fit a curve to the data to show that the shape of the pathline is parabolic, with the general form! =!!!! +!!! (b) (5 points) Use the above equation, and your values of c 1 and c 2, to calculate the angle (with respect to the x axis) that the water leaves the nozzle, and the initial speed V. (a) Points along the pathline for the stream of water are measured as shown on the figure below. Converting the units to meters, plotting the data and fitting a parabolic model to the points gives!"#$% & ' ( ) * +, . &/ && &' &( &) &* &+ &, & &. '/ '& '' '( ') '* ' / / /6//('' /6//&'& /6//+) /6//('' /6/&/+ /6//))' /6/&) /6//*+( /6/&., /6//*'( /6/')&' /6//)( /6/',,) /6//(+' /6/(/.+ /6//')& /6/((,, /6//&'& /6/(+.. 7/6/// /6/)/+& 7/6//)/' /6/))+( 7/6//,'( /6/),)) 7/6/&//* /6/*/'* 7/6/&'+ /6/*'+, 7/6/&+/ /6/** 7/6/'/& /6/*, 7/6/')&' /6/+&*& 7/6/'*) /6/+)(' 7/6/(((, /6/++,) 7/6/(,,. /6/+.** 7/6/)(' /6/,&.+ 7/6/)+) /6/,( 7/6/*&/' /6/,)(, 7/6/*(), /6/,*.' 7/6/**&!!"#$! /6/&1 /1 7/6/&1 7/6/'1 7/6/(1 7/6/)1 7/6/* &6''' ' 191/6+).&1 :;181/6...),1 7/6/+1 /1 /6/&1 /6/'1 /6/(1 /6/)1 /6/*1 /6/+1 /6/,1 /6/1 "!"#$! The high value of R 2 indicates good agreement between the data points and the parabolic model.
11 (b) From the model fit in Excel, we obtain the coefficients!! = 18.2 m  1 and!! =.65. From the equation given in the problem statement,!! =.65 = tan! thus,! = 33, which appears reasonable in comparison with the figure above. We can now find the velocity from!! = 18.2! m =!!!!! cos2! or!! =!.!" m/s 2!!".! m!! cos 2!! =.62 m/s
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