Exam+3+Solutions

Exam+3+Solutions - Name: SSN: CE 2200 Fluid...

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Unformatted text preview: Name: SSN: CE 2200 Fluid Mechanics/Wilbert Spring 03 Exam #3 \W 1. Write ALL equations in their most general form and then begin simplifying. State/list all your assumptions when you are simplifying the equations. 2. Check your units! 3. ‘ Box your answers and provide units. Physical Properties 1.94 slugs/ ‘ 2.73x10'5lbf- s/it2 0.073 N/m 2340 N/m abs 0.073 N/m Pm. _| 1230 N/m2r abs Gravitational Constant: g = 9.81 m/sec2 = 32.2 ft/sec2 lbf = mass (slugs) x g (ft/secz) { N = mass (kg) x g(m/secz) We 1. (3 points) Write the equation you would use to calculate the velocity at a point a/4 from the bottom plate for laminar flow between two horizontal, stationary plates. ; r o l a q 2 o 3 » a: A a — /1 AI jJ—fl U M; 'a AL i 5 $1» 4- 2A U ' A l at, 2. (3 points) What “force” is always included when deriving dimensionless parameter?" 4" I Kmbr 3. (3 points) What three forces did we “balance” when deriving Stoke’s law? (in; Brandi“ lovevIWW‘x Age: NM (4 9 4. (5 pointsflf the area ratio (Ar = Am/Ap ) is 1/9, , 1 ‘ ' .Whatistheiengthratio,Lr? Lri E L 3; , , > 7 Whatisthevolume ratio,-V,? L 7 i -. ’l‘ . Vr ” r "27 Problem 5 (15 points) A pumpingsystem is designed to pump crude oil (S=0.85) a distance of 1 mile (5280 it) in a 1 foot diameter pipe at a rate of 7.8 cfs. The pressures at the entrance and exit of the pipe are atmospheric; the exit of the pipe is 300 ft higher than the entrance; and the head 2 loss in the pipe is given by f C. 2g . . L What Is the Reynolds number and is the flow lammar or turbulent? A 1 “— O r' ( \ > Z , u W in - fax ’— ntf fiach‘ofm \ > albipm’ Lt 9‘ two Mew-H \I‘ “Vir- %-fifit*/u Whatisthe power required for the pump? ‘ f a I ’L\ F V \L ,e t C Q Zégi’c‘wee g+fi4i+%u Xfa a'z‘ swat , «m» (6.02 2 ‘5‘“ + Q01) '7 ’ 373;) 5. ’3 u» "' \LD.7 ‘i {3' - 7 M ’ «5/ \i a Problem 6 (20 points) Water is in both tanks and in the pipe. Both tanks are closed with the air pressure above the water inside mob tank shown in the figure. The total head loss for the system is 8V2/(2g) Where V is the flow velocity in the pipe (NOTE: this head loss is independent of the direction of flow). T = 20 ° C Pm}: -30 kPa a. Determine the direction of flow in the pipe. You must prove your answer using the energy equation! -—- \0 cm: a r C " 3.31;: ——L—s +0 4— H e + if 0m» 0‘“ ° \Me >; m2 - {OLLJL/E—a mirth} b. Calculate the flow velocity in the pipe. 0 v” “t V“ Q l \ i ‘- gt/éiVK’ t flfl” \QRZ =— \\y‘\3 5' fli- /Q‘L t g V725 [0.1 m3fsec. What is the pmmtype dischaxge? gm} w“: 9:; L6 it'f 1),; gm: 5: ‘ 7 (511, poimsm 1/25 scale model of a spfllw'ay'is tésted. The discharge in the model is \J Problem 8 (20 points) Two people are carrying am by 3 m rigid sheet of plastic in a 15 m/s wind. If the sheet of plastic is held vertically and normally to the wind (sketch A), the force of the Wind on the plastic is 324 N. These two people then remember that they have taken fluid mechanics from Dr. Willson and that they can reduce the forcesignificantly by carrying the sheet of plastic horizontally and parallel to the wind rather than vertically and normal to the wind. However, they cannot agree on whether the plastic should be carried with its 3—m long side (sketch B) or its 1.5-m long side (sketch C) parallel to the wind in order to minimize the force of the wind Determine which orientation (if either one! will give the minimum force and calculate the magm'tude of that force for the sheet of plastic being parallel to the wind. If there is a preferred orientation for mm ' 'ng the force= circle that orientation in the sketch. Due to the roughness of the plastic sheet, there is no region of laminar flow next to the plastic for B and C. for air: v = 1.46 x10"5 mZ/s p = 1.22 kg/m3 CM 6 )m flag (LeL m M: 0720.53020”) (K) L ism N € ‘ x ' . s x9 , Waugh ,1 What is the “power” required to overcome the shearing force you calculated above? 9" (A) QWW (WW) 2 Q74 1.33: Problem 97 (20 points) Water at 10 ° C is being pumped from a one reservoir (water surface elevation = 50 m) to another reservoir (water surface elevation = 150 m). The pump elevation is 0 m, L1= 150 m, D] = l m, L2 = 750 in, D2 = 50 cm. Assume pipe head loss is given by 0.02(L/D)(V2/2g) where L is the length of the pipe, D is the pipe diameter, and V is the velocity in the pipe. The head loss at the pipe exit is 1.0V2/2g where V is the velocity in pipe 2. a. Assuming the pump efficiency is 60%, what power is required to pump water at a rate of 2.5 m3/s from the lower to the upper reservoir? ‘ 7 L (H/g (x V} X ,C r), Ujjaqha A QHG5 frymn/ yfltm' 35"?“ f‘" d,r\' ‘ Q5 ' - L \Ilt 40):. I M \J“ Ogotor/Le‘ ®*\§b+O*O¥/QA_ §*0.fl th fl l \ A. new?) cgww m 9” L” ‘ \ e w A? » zmw . (QM/4t. ’ i e (95% i%\°lv~’)l7m°}g WW L, D 9 b. 0n the next page, draw the energy grade lines and hydraulic grade lines to scale. Include the EGL and HGL for the reservoim Write the numerical values of the EGL and HGL at each end of each straight line segment. Note: the horizontal axis is not drawn to scale. Line spacing is 20 m €041, Wab‘b“ I — ‘1‘ _.\.\ Xx —m —- w; W —~ —_ —- " W‘W’ /é{ W/ imm‘ , . y 9 ...
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Exam+3+Solutions - Name: SSN: CE 2200 Fluid...

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