ME 309 Fluid Mechanics Fall 2010 Exam 2 1A. 1B.


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1 Fall 010 Exam 1A. 1B.
2 Fall 010 Exam 1C. Water is flowing through a 180º bend. The inner and outer radii of the bend are 0.75 and 1.5 m, respectively. The velocity profile is approximated as C/r where C is 1 m /s and r is the radial position at the bend. What is the pressure difference between the inner and outer walls (i.e. P P 1 )? Neglect gravitational and frictional effects. Solution: V p g R n n p p( r), eˆ ˆ n er, and g 0 n V p r r c p p Pa R1 R
3 Fall 010 Exam 1D. A pump is drawing water from a large reservoir at 3.4 m 3 /min. The inlet of the pump is connected a 10 cmdiameter pipe for which the total head loss is fixed at 1.8 m. According to the pump manufacturer s data, NPSHR is 6.1 m at the current flow rate. What is the maximum Δ before cavitation occurs at the pump inlet? The vapor pressure of water is 4 kpa (abs). Solution: 1: Free surface; : inlet of the pump P V P NPSHA= v g g g Find Δ when NPSHA=NPSHR=6.1 m. P V P NPSHR v m g g g Applying the head loss equation from 1 to, p V p V H H g g g g 1 H 0, p p, H 1.8m S,1 1 atm L,1 1 L,1 S,1 Assume V 0 and 1 (Indeed Re>>300 in the pipe) p atm p V Z Z1 H L, m g g g
4 fluid fluid ME 309 Fluid Mechanics Fall 010 Exam. A viscous, incompressible, Newtonian fluid coats the outer surface of an infinitely long, circular rod or radius R as shown in the figure. The rod moves downward at a constant speed V (in the same direction as the gravitational acceleration). The viscous liquid has a uniform thickness r around the surface of the rod and is exposed to the atmosphere. r R r atmosphere atmosphere g r V [0 pts] a. Simplify the appropriate NavierStokes equation to its simplest form to solve for the  component of the fluid velocity for the given flow scenario. Leave your answer as a differential equation. You must clearly identify why various terms in the continuity and/or NavierStokes equations are ero in order to receive full credit for this problem. SOLUTION: Make the following assumptions: 1. steady flow 0 (1) t ur u u. fully developed flow in direction 0 () 3. axisymmetric flow with no swirl velocity 0; u 0 (3) 4. gravity acts only in the direction g r = g = 0; g = g (4) First simplify the continuity equation (in cylindrical coordinates): 1 ru r 1 u u ru 0 r 0 rur c (5) r r r r 0 #3 0 # Since u r = 0 at r = R, the constant c must be ero. Thus, u r = 0 in general (call this condition #5).
5 Fall 010 Exam Now simplify the NavierStokes equation in the direction (in cylindrical coordinates). u u u u u p 1 u 1 u u ur u r f t r r r r r r 0 #5 g #4 0 #1 0 #3 0 # 0 #3 0 # p 1 u r g r r r 0 Note that since the fluid is exposed to the atmosphere, the pressure along the free surface is uniform (= p atm ). Furthermore, since the flow is fully developed in the direction and the u r and u velocity components are ero, there will be no variation in the pressure anywhere in the flow. Thus, p 0 and Eq. (7) becomes, 1 d du r g r dr dr Note that the partial derivatives in Eq. (8) have been replaced with ordinary derivatives since u is not a function of t (assumption #1), (assumption #), or (assumption #3), and, thus, can at most be only a function of r. (6) (7) (8) [08 pts] b. What are the boundary conditions (in mathematical form) for the given flow scenario? SOLUTION: u r R V noslip at r = R u noshear at r = R + r 0 r rrr [07 pts] c. Assuming the following velocity field, r g g R r u R r ln V 4 R, determine the magnitude of the axial force acting on the rod per unit length of the rod due to the viscous liquid. SOLUTION: r rr u r ur 0 #,#5 rr R r R r u g g R r r R r r rr r g r rr R The axial force (per unit length) will be the shear stress acting on the rod multiplied by the area over which the shear stress acts: F F r R rr g R r R L L rr (9) (10) (11)
6 Fall 010 Exam 3. Cooling water is pumped from a reservoir to rock drills on a construction job using the pipe system shown. The flow rate must be 600 gpm and water must leave the spray nole at V j = 10 ft/s. The drawn tubing piping system consists of a reentrant entrance, a regular 90 threaded elbow, two regular 45 threaded elbows, and a fully open gate valve. The total length of pipe is 700 feet. The pipe has a diameter of 4 inches. WRITE THE SYMBOLIC SOLUTION FIRST AND THEN PLUG IN THE NUMBERS ON A SEPARATE LINE. (a) Calculate the total head loss (include major and minor head losses) of the system. (b) Assume the total head loss is 00 ft. Calculate the pump head required for the system. Solution: (a) Calculate total head loss: Total head loss = Major head loss + minor head loss Total head loss = Major head loss + entrance losses + 90 turn loss + two 45 turn losses + gate valve losses Calculate major losses:
7 Fall 010 Exam Drawn tubing roughness, e = ft From Moody diagram f= Determine minor loss coefficients: k ent = 0.8 k 90 =1.5 k 45 =0.4 k gv =0.15 It is also acceptable if the nole was assumed to cause exit losses. Substitute values into head loss equation: or (b) Calculate pump head Apply extended Bernoulli between the top of the reservior (1) and the pipe exit () where: P 1 = P = P atm V 1 = 0 V = V j 1 = 400 ft H L =00 ft Then, or
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