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Qwstions Fluid

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158 views30 pages

Qwstions Fluid

Uploaded by

ali alkassem
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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SS-A- (60 min) ILY EXAMINATIOI 30 cm and length 150 cm floats horizontally on the Q1: A wooden block (specific gravity = 0.7) of width 15 cm, depth. surface of sea water (specific weight =10 kN). Calculate the volume of water displaced, depth of immersion and position hhting moment for a tilt of 5°? Comment on stability of of center of buoyancy? Also find the metacentric height and rigt the block. For small tilt angle @ measured in radians i.e @raaians = oe a2: An open cylindrical tank 0.9 m high and 0.6 m in diameter is filed 2/3 rd with water when at rest. It's spun about ite verticel axis with angular velocity () radian per second, and the free liquid surface in the tank assumes the shape of a paraboloid of revolution. Determine the speed of rotation in (radian/sec and rpm) when: ing over the sides of the tank? ; base is just exposed? What would then be the percentage of water left in the |ECOND CLASS-FIR' FLUID ANI a) The water just starts spill b) The point at the center of the at the bottom of the walls if w= 9 tank? ©) Also estimate the pressure intensities at the center of the tank bottom and radian/sec? see Q An open t ‘ank 10m long and 2 m deep is filled ins . a Pp with 1.5m of oil (5.G=0.82). The tank ifort aoe ied of 20 m/sec. What is the shortest time in which this speed a ‘ ji i ne pata at , find the total pressure for each tank side? dag 2 epee CLASS-A-QUIZ:_Q1: Find the discharge of water flowing over right triangle notch of 3 m length when the ‘constant head of water over the notch is 100 cm. If increase the length of weir from 3 m to Sm to be built across a rectangular channel. determine the height of rectangular weir, if the maximum depth of water on the upstream side of the weir is 1.5 m and same discharge and Cd? Take Cd = 0.6. Q: An open tank 10m long and 2 m deep is filled with 1.5m of oil (S.G=0.82). The tank is accelerated uniformly rest to a speed of 20 m/sec. What is the shortest time in which this speed may obtain without spiling any oil? Also, find the total pressure for each tank side? ei hh BE SMe a ells Q1: Find the gauge reading at Land M in figure below if the local atmospheric pressure is 755 mm of mercury? Assuming the absolute vapour. Pressure of mercury (Hg) and pressure due to short column of air (yair is very low) to be negligible Closed (asi IND CLASS-FIRST MONTHLY EXAMI Qi: A 30cm diameter pipe (called penstock) supplies water steadily to turbine at 0.18 MIN/m?. The water leaves the turbine ‘on exit side through a 60 cm pipe (called draft tube) with a pressure of (-0.25 MN/m:). A vertical distance of 1.5 m separates the center of pipes at sections where measurements of pressure have been made as shown in Fig. A. Calculate the power delivered to the turbine from water if 0.25 m’/sec of water passes through the arrangement. Neglect frictional losses. ‘Q2: Water flows steadily through a Y-junction in horizontal plane as shown in Fig. B. The velocities in the downstream limbs are equal and all velocity profiles are uniform. Calculate the magnitude and direction of force required to hold the junction stationary in the x-y plane? en eet / PL / 2 ue a Univer of Magha ‘uid Mechanics Cat ngicing Dee De Kh Al Avawny | ‘Note: Anewes two questions ony '@.1 (50%): A pump delivers water from tank A to tank 8. The suction pipe ts 45m long {f-0.024) and 35 cm in diameter. The delivery pipe is 950 m long (F=0.022) and 25 crn in diameter. The head discharge relationship in the pump is given by {€, = 90:8000 Q’), where E> in meters and Qin m'/s, Calculate the discharge in the pipeline, and the power delivered by the pump, Note: neglect minor losses. Delivery pipe loss, hy2" We “ Total head loss (Hi) = 3.08655 +8365 ———--—{1) goers equation Vi A. = V2 A wt nt SE av, nee then Vy = 0.51 v2 Sub (2) in (1) : Total = 3.086*(0.26 2) es fotal head loss (Hy) = 3.086*(0.26S4) + 83.65 aaa th 8) Bernoulli equation between two Res. eee Wye Fee ht Be= Bene E en, = sf 110 +E = 170+ 84.475 = ve £)=60+ 84.42 a £,=60+ 84.450 Bresso £)=60+844—2_, age E, = 60+ 1785.3" (A) Given: Ep =90~8000 0” Equating: 60 +1785.3 0" = 90-8000 Q Q= 0.05537 m'/s - = 60+ 1785.3 Q! = 60 + 1785.3 (0.05537) = 65.47 m - Power (P=¥* Q°E) y eae ets in Fig. The length and diameter forthe frst parallel pipe (y= 2000m, and -1.Oen) respectively, while the length and diameter ofthe 2™ parallel pipe are (,~ 2000m, and 0.8m) respectively. Find the rate of flow in each parallel pipe (Qy, and Q,) if total low | the main pipe (Q=3.0m'V3). The coefficient of friction for each parallel pipe is same an equal to (F= 0.0081 ‘Solution: =f= 0.005 40; then 3= Q, 10, Continuity = Q. Parallel pipe: hl; =hly But =v, Avi (SE) ‘Sub eq.(2) in above, then Q, = ame I——#) av twee (yw) q=0.64*V,* 5 “Sub. (3) and (4) in (2) then Va= 2.17 m/s and Vs = 2.427 m/s m’/s and Qy= 1.906 m/s. nes (4) Wgiea=-tsedhed) ce ie 200 onde spe cae ae under prea 1962 wana Velocity of (v1- a va f (V1= 3.5 m/s). If the axis of the Pipe turns through 45° as shown mpute the magnitude and ection of theresa ove atthe bend, Dy=D; = D then ae Ay= Ay = A= 1 = n° = 0.07068 m’ v= v2 =v = 3.5 m/s since area is constant Q= A*y = 0.07068 m? *3.5 m/s = 0.2475 m’/s P,=P, = P = 196.2 kPa. = 196200 Pa. 5 *0.7071 = 2.475 m/s > * 0.7071 = 2.475 m/s T Vie = 3.5 m/s > : Vox=V2.C08 45° = Vy=0 2 Vay = Vo Sin 45° = 3: Fue = PrAr = 196200 *0.07068 = 13867.42 N = 13.874 KN. > Fay = PaAs sind5® = 196200 *0.07068" 0.7071 = 9805.6 N=9.8 kN IFy' =p Qay, Fax Fax + Fx = 1000 * 0.2475 * (Vox-Vie) 13867.42 ~ 9805.6 + Fx = 1000 *0.2475 *( 2.475- 3.5) Fx =-4315.5 N= 4315.5 N € reaction of body on fluid in x directic 2FY' =pQdw Fay~ Foy + Fy = 1000 * 0.2475 * (Voy- vay) 0 ~9805.6 + Fy= 1000 *0.2475 *( 2.475-0) Fy = 10418.2 N7 reaction of body on fluid in y direction F= /F% + F} =V4315. 5? + 10418. 2? =11276.6N tan(@) = 7 r= 2A14 @ = 67°28 CHECK the pipe before contraction D,,and diameter of the pipe after contraction D;. Using Bukinghamn e-theorem,show that: Op Peps = ov UR 2 Note: select j1, and D, as repeating variables. e Solution: veriable[ap lp [nw |v [0, [Dy mr ape whe sytem | cer | B lier|7] |_| 4p=K( p* ut v*D,* D2) Ap=k* peyeven*0s —— ‘Mass length sytem analysis (MT): Eee EEG M: 1 a+b Lb -1=Ba-bectdee T: -2 =b-c Five unknowns and three equations (5:3) =2 Solving in term of 2 parameters Solve in term of band d nd Ee reaee ad inch pipeline (same pipe) using water at SPF, v= 141 * 10° f/sec Find ‘The discharge in model (Qn), and the head lose in prototype (hur) in ft if the head or in model (Im = 3.6 f. ) of water oa : Reynolds Model. V, D, =v, a)", since D,=1 233 meagre esstio® 088 Qp = 9000 gpm = 150 gps = 20 cfs. Q, =v, LE = v, D3 = v, = 6.088 20 = 3.285 cfs =e same Re, same pipe (f and D) =37(3.6) = 133.2 ft oil f= $GRe) Since same pipe $)m {Jr then fa~ fe 9P__ 206f5 95.46 ft/s Since D= 12 inch = Ift m/e ait Vp Vg Bmw BREESE 4.183 Us ws a bv? ee = + then 3.6= £2202 2222 then f= 0.01325 7 ee sl 3.6 = 1100 2588 then f 0.01323 Ketone lB bertep—lols Fruid Raechanies Bios, ‘neni, 29,2018 1, Khali Ade Nansen "Note: Answer tw questions ly Time Aloe 11 QA. Determine the form of the equation for the distance (5) that a sphere wal fal in ima fluid iit starts with an initial velocity and the resistance of the fluid through which it falls is taken into consideration. Note:S = f(g, v,1,M,d,p, 18): select time, mass, diameter, and viscosity a repeating ariables. Solution: : ae ees fe 5 "M4, 4°, pn” © ‘ LEP LTH APM LIME ML TY S- Cyr ety", M: 0= C44 C6+C7 Cl + C2 + C5~306-C7 T: O=-2C1-C2+C3-C7 ‘There are 7 unknowns and three equations Express cl, ¢2 and c7 in terms of c3,c4, ¢5, and ©6 C1 = 63 +204 +5 ~c6-1 3 34 -2e5 + 306 +2 '@.1 (50%) Compute the theoretical velocity, actual velocity, and discharge flow through the pipe ? assume Cv= 0.98. @3 (50%): In a 45° bend a rectangular air duct of 1m’ cross-sectional area is gradually reduced to 0.5 m? area as shown in Fig. below (top view) . Find the magnitude and direction of the force required to hold the duct in position if the velocity of the flow (vi=10m/s), and pressure is 29430 Pa. Take par = 1.16 kg/m’. RRNA sse00 2-8) earamN > Fee weetght of water enclosed or supported (actually or imaginary) ‘by the curved surface ACB = Tee * volume of portion ACB sse00 eves =se00* Ears ; =as2e02N tt ee a : 2 Method 2 to compute F, Fis Tw * volume ACD = Ywe* [area of ADCO —area of AOC] * 8 = Tw * [area of ADCO — area of AOC] * 8 a Yet 22-fne zs +9800 0.858 = 672674 F2=FL=67267N £3 = Yu 4° [298] = 6272000 T BF) =F3-F2-FL= 492660 NT F< JAE SA - VERIO + BOF =797S19N ———____— 7° ai: tano== ) | Fmd 4290.65" 27 =330-42)y 1 orgs Fedrttessa.se2) =2m5y F=(4-42+ 165 135"285y =27y(4-255) Reatrissaaseay slamsy ‘Note : gate opens when moment about pivot are equal 3.3y(d ~4.2)8.25 + 2.7225y * 0.55 = 2 Ty(d ~ 2.55)0.675 +1.8225° 0.9 2.7294d ~42)-+2.7225*0.55~1 29d 255) +14225*09 d= 42-40585=0.660d ~1.707 +0600 331d =42-055-1.7.7+0602=2545 d= 7.69m ‘Solution: po 28X0-5) + (0.001249 8Y0.33)+1(13.649.8)(0.17)(0.83)9.8)}0.44)-0 "= 14.17 kPa, aa dum | @2 (20%): This rectangular gate will open automatically when the depth of water, + becomes large enough. What isthe minimum depth that will use the gate to open? Page Tot 14 (30%) compute a-? i oie) Se Fluid be kG Se Note:rd. CCL, = 1.59 25 | pts patm. + (658) £08 water © peo + (65-4) ¢08*r water “gyc0s 15438887 rte Ten; e225 10.2 (50%): Find the magnitude and direction of the resultant force due to water acting on 3 roller gate of cylindrical form of 4.0 m diameter, when the gate is placed on the dam in such ‘a way that water is just to spill. Take the length of the gate as &m. am (First Atempt) sees . If the pipe is Q2 (20%): 0.25 m’/s of water is flowing in a pipe having a diameter of 300mm. ae ? it d ‘bent by 135° (that is change from initial to final direction is 135°), find the magnitude an direction of the resultant force on the bend. The pressure of water flowing (P1=P2= 39.24*10° N/m’). @:3 (20%): The mean velocity in a 300mm pipe line is 3m/s. The relative roughness of the Pipe is 0.002, and the kinematic viscosity of water is 94197 m/s, Determine the fr ct iction, reline velocity, velocity 50mm from the pipe wall, and the hea ‘under the assumption that the pipe is rough. 'd lost in 300m Q5 (20%): 3 s. _ 2) short smeoth over flow structure passes a flow rate of 600 m’/: ic What flow rate should be used with a 1:15 model to obtain dynami 10 similitude ? neglect friction. Prototype Q,= 600 m*/s Model Qm =? b) A ship of certain shape and draft d. its Drag depends on (size |, mass density p, viscosity 1, velocity v, and gravity g). using Buckingham T]- theorem derive an expression of the drag as a function of other parameters. s 10° Note: Select viscosity 1, and gravity g as a repeating variables, Q.6 (20%): A flow of 28m’/s occurs in an earth- lined trapezoidal canal having base width 3m, side slope 2 (vertical) on 2( horizontal ), and n. 02 4 as = an . Calculate the critical depth and critical slope. gear rte wmenn = ancvey sats A783, .10: In flow of sudden contraction, the change of pressure (Ap) is depends on the density p, the viscosity 11 of water, velocity v of the fluid, diameter of the pipe before contraction D,,and diameter of the pipe after contraction D;. Using Bukingham Tt-theorem, show that: Ap= pr-P2 = pv" [IR, 24 Ans Note: select 1, and D; as repeating variables Dy Ap= Pr-P2= pv (IR, 4] Q7: Determine the pivot location y of the square gate in Fig. below so that it will rotate open when the liquid surface is as shown, Ans. y= 0.833 m Q8: Calculate the horizontal and vertical components of the water pressure exerted on a tainter gate of radius 8m as shown in Fig. Take width of gate unity (1m). Ans. Fx= 78480N : Fy=28439NL Q9: The manometer reading is 150mm when the cone is empty ( water level was at point A), compute the manometer reading when the cone is full? Ans. Manometer reading = 0.38 m Fluid Mechanies Page 3 of # Q3: Determine the gage pressure at point A in Fig. (rd. air =.0012). Ans, PA= 14.17 kPa Q4: Determine the form of the equation for the distance (S) that a sphere will fall in time (t) ina fluid if it starts with an initial velocity and the resistance of the fluid through which it falls is taken into consideration. Note: S= f(g, v,t,M,d, p, 1) : select time, mass, diameter, and viscosity as repeating ariables. Ans. S=Cy> g 2 , g?M.c4 gd 5 , pv’ exe v gu yo Q5 : A liquid compressed in a cylinder has a volume of 1000 cm’ at IMN/m* and a volume of 995 cm’ at 2MN/m?, What its bulk modulus of elasticity (K)? Ans. k= 200 MPa. Q6: A dam 20m long retains 7m of water as shown in Fig. Find the total resultant force acting on the dam and the location of the center of pressure. Ans. F=5339KN : he.p.=4.667m Fluid Mechanics HW. Nod Page 2 of 4 FLUib MECHANICS HW. No.1 SECOND YEAR JUNE ,2020 University of Baghdad College of Engineering il Engineering Dept Dr. Khalid Adel Abdulrazzaq Qi: This rectangular gate will open automatically when the depth of water, d, becomes large enough. What is the minimum depth that will cause the gate to open? Ans. d=7.7 m Pivot 3m*(2m) a x Q2: 4m diameter circular gate is located in inclined wall. The gate is mounted on a shaft along its horizontal diameter. Compute a) magnitude and location of the resultant force. Ans. F= 123.18 kN b) moment that would be applied in shaft to open the gate. 6.16 kN.m Fluid Mechanics HW. Nod Page Lord inet Coege Been tsgenee” m: ™ ‘ontal diame tude and location of the Fesultant force }moment that would be applied in Shaft to 9 © open the Bate, sand) ‘2 ah Bane 7 = Wont fo be eo 8 > (23410! N) (0084) Mo? Nim bets (0+ 72} (20)(7sin 60°) = 5339 KN. The center of pressure is located at two-thirds the aime of 7m, oF 4.667 m below the water surface (ie. hy & 4,667 m in Fig. 36). mm pak ¥e i

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