AMEE 0 Introduction to Fluid Mecanics Instructor: Marios M. Fyrillas Email: m.fyrillas@fit.ac.cy Homework Assignment on Fluid Statics -------------------------------------------------------------------------------------------------------------- 1. Determine te relationsip between pressure at point A and pressure at point B p atm pa pa p1 p p1 pa 11 p p p p p p 1 atm A 1 1 p1 patm atm. Te figure sows an inclined manometer, in wic te distance indicates te movement of te gauge fluid level. Determine an expression for te pressure difference between p and p. A B B
p p g p B B p1 p g p pb p1 pa 1g1 pa 11 p p B A 1 1 p p sin A B 1 1 1 1
. For te compound differential manometer in te figure calculate te pressure at te point A. p 8 p 1 p p 4 5 p p p6 p7 p1 pa p p1oil g (0.15 0.5) pa 9009.810.75 pa 11 p4 pa 9896 p p p4 p 15409.810.5 p 07 p4 pmg0.5 p4 pa 9896 p6 pa 9896 94 pa 695 p5 p 4 p6 p4 94 p6 p5wg(0.5 0.05) p7 p6 p8 p6 609 p8 patm p7 w g0.475 p7 609 p8 p6 609 p8 pa 695 609 pa 90046 p6 pa 695 pa patm 90046 101500 90046 191546 Pa
4. Te basic elements of a ydraulic press are sown in te figure. Te plunger as an area of 6.4516 cm, and a force, F 1, can be applied to te plunger troug a lever mecanism aving a mecanical advantage of 8 to 1. If te large piston as an area of 967.74 cm, wat load, F, can be raised by a force of 1.5 N applied to te lever? Neglect te ydrostatic pressure variation. Since we are neglecting ydrostatic pressure variation, te pressure is te same between te planger and te piston. F1 F A p1 p F F1 A 967.74 A1 A A1 F 8 Flever 8 1.5 160.kN A1 6.4516 F1 8F lever
5. Determine te force F 1 acting on te plunger so tat te piston would not move (Figure ). Te ydraulic fluid as density 850 kg/m. 0 cm 10 cm D 15 cm Figure : Piston-plunger arrangement Te force acting on te plunger is equal to te pressure at te middle of te plunger multiplied by te area of te plunger: p middle of plunger 0.15 g0. 0.1 96 N/m D 0.15 F1 pa96 96 70 N 4 4
6. Determine te pressure gage reading and te eigt of te mercury manometer if te vapor pressure is 10 kpa (Abs) p. 800 kg/m 1600 kg/m Figure : Tank wit vapour/liquid (Question) Gage pressure pg= pv lg patm 10 kpa+800*9.81*1-101000=17848-101000 =6848 Pa pg 6848 pg+101000= mg patm 0. m mg 1600*9.81
7. A container as tree different liquids as sown in te figure. Based on te gauge pressures sown on te manometers determine te density of te fluids. p1 0.1 atm p 0.4 atm p 0.8 atm 1 1 m 1 m 1 m Gage pressure at point 1 = p g g 0.1 atm 101.5 Pa 101.5 kg 1 10 9.811 m 1 1 1 1 1 Gage pressure at point = p g g 101.5+ g 0.4 atm 418 Pa 418-101.5 kg 1446 9.811 m 1 1 Gage pressure at point = p g 0.4 atm 0.8 atm 0.14 atm 14185.5 kg =1446 9.811 9.81 m
8. An open tank as a vertical partition and on one side contains gasoline wit a density 700 kg/m at a dept of 4 m, as sown in te figure. A rectangular gate tat is 4 m ig and m wide and inged at one end is located in te partition. Water is slowly added to te empty side of te tank. At wat dept,, will te gate start to open? F g 4 F w 4 Fw p( middle) Area wg ( ) Fg p( middle) Area gg (4 ) Take moments about te inge. Wen te moments are equal te door will open, i.e. 4 Fw Fg 4 10009.81 7009.814 4 7004 44.8.55 m 1000
9. Calculate te widt of concrete dam tat is necessary to prevent te dam from sliding. Te specific weigt of te concrete is 400 kg/m, and te coefficient of friction between te base of te dam and te foundation is 0.4. Use 1.5 as te factor of safety (F.S.) against sliding. Will it also be safe against overturning? Te span of te damn is 1 m. w 7 m Water 5 m Water dam Sliding Weigt of concrete dam= V g=(1*7*w)*400*9.81 5 Sliding force=hydrostatic force F H = water g (5*1) 165 N Sliding resistance= 0.4*165000* w Sliding resistance 6900w F.S.= 1.5 w.65 Sliding force 165 Oveturning total rigting moment=165000*w*w/ overturning moment=165*5/ To be save against oveturning 165000*w*w/ > 165*5/ w 165*5* *165000
10. A tank containing fuel of density 850 kg/m as an automatic device to prevent overflow; tis consists of a rectangular door inged at te upper edge closed by a spring. Te door is 0. m deep and 0. m wide and te spring exerts a force of 1 kn. Calculate te level of te fuel above te inge wen te door is about to open. y cp 0. m Ff g( 0.15) 0.06 0. m CP F s 0.0 0. m yc 0.15 I xc ba a 0.0075 yr yc yca 1 ycba 1( 0.15) 0.15 Side view Te surface of te fuel is at eigt above te inge as sown on te diagram. Te force acting on te door Fg pressure at te centroid ( yc) area of te door Fg pressure at te center of te door (0.0.) Fg g yc (0.0.) 8509.81 ( 0.15) 0.06 Fg 500 ( 0.15) N Take moments of te two forces (ydrostatic and spring) about te inge: 1000 (0.0.0) 500 ( 0.15) ( yr ) 0.0075 10000.7 500( 0.15) 0.15 0.15 70.75 7511.5.4 m
11. A spere of density and radius r is placed in a large reservoir filled wit a liquid. Find an expression for te dept to wic a spere will sink. If te spere is displaced from its equilibrium position, find a differential equation tat governs its oscillation assuming inviscid flow. Note: te volume of a sperical segment is ( r ). Arcimede's Principle Buoyancy Force weigt of displaced water r wg 4 Tis must be equal to te weigt of te spere W mg r V g r g 4 So, r g r g r 4r w Given r, w and you can solve it to find. Differential Equation F ma 4 4 d r g r wg r dt d w g g r 4r dt w
1. A solid cylinder of lengt 0. m and diameter 10 cm is floating in water (see figure). Determine: a. Te upward ydrostatic force ( F B ) acting on te bottom of te cylinder. b. Te density of te cylinder. 14 cm D a. Fb Vdisplaced fluid g 0.14 10009.81 10.78 N liquid 4 b. Because te cylinder is not moving F 0 F W 0 were W is te weigt of te cylinder. 10.78 10.78 Vcylinder cylinder g cylinder = V g cylinder b 10.78 kg 466.4 m D 0.9.81 4 1. A orizontal gate, 1.5 meters wide, is eld in place by a cable as sown in te Figure. Water acts against te gate wic is inged at point A. Friction in te inge is negligible. Determine te tension in te cable. 1m m A Figure: Tank wit a gate eld by a cable