FLUID MECHANICS UNIVERSITY OF LEEDS. May/June Examination for the degree of. BEng/ MEng Civil Engineering. Time allowed: 2 hours

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1 This question paper consists of printe pages, each of which is ientifie by the Coe Number CIVE 4 UNIVERSITY OF LEEDS May/June Examination for the egree of BEng/ MEng Civil Engineering FLUID MECANICS Time allowe: hours Attempt 4 questions L.9m x.9cos6.45m The gate opens when the moment at the pivot is clockwise. That is when the moment ue to the water > (.45/)w. Useful formulae: Parallel axis theorem I n b b moment of area for a rectangle, I GG, an for a triangle I GG 6 oo IGG + Ax,. A rectangular sluice gate,.9m.m, is fitte at the base of a reservoir wall with a pivot in the arrangement shown in Figure. The gate will open when the the level of water in the reservoir reaches a certain epth. Determine the weight, W, that must be applie at the centre of the gate, if a water level of h.m will just cause the gate to open. Metho Metho Resultant force Force on plane Area Pressure at centroi y Force (..9). ρg / 9.8 ( ) 49 N orizontal force f Area of projection on vertical plane pressure at centroi y f.y. ρg 7N Vertical force weight of water above gate x y V f.x ρg. 79 R f + V 4 N f Figure Point of action of the force centre of pressure

2 b For a rectangle, I GG IGG Sc + x Ax L.9m sin 6.9 x L.859m n moment of area about O Sc st moment of area about O I I + Ax oo GG Sc + x x m Ioo Ax Nee to fin the lever arm, i.e. the istance from the pivot to the centre of pressure specifie by Sc. First fin the position of the pivot, x, from the surface (along the incline plane). A comparison of the sensitivity of epth change to ischarge is to be mae between a rectangular an v-notch a weir. Initially a rectangular notch weir of with.5m an coeeficient of ischarge of.9 is place at the ownstream en of a channel carrying m /s of water. What will be the height above the base of the notch. [8 marks] o The experiment is repeate this time with a 7 V-notch weir with a the coefficient of ischarge of.8. What will be the height above the base of the notch for the same ischarge? [8 marks] When the height above the base in either weir is.5m what will be the percentage increase in flow if the level rises by.m? (Derive all formulae assuming the Bernoulli equation) [7 marks] x L.9.49m Lever arm, x Sc - x m Take moments to fine the weight of the gate, w Rx.5w w 7 N.5 A General Weir Equation Consier a horizontal strip of with b an epth h below the free surface, as shown in the figure below. Elemental strip of flow through a notch Assuming the velocity is only ue to the hea. velocity through the strip, u gh ischarge through the strip, δq Au bδh gh 4 b Integrating from the free surface, h, to the weir crest, h gives the expression for the total theoretical ischarge Q g bh h theoretical This will be ifferent for every ifferently shape weir or notch. To make further use of this equation we nee an expression relating the with of flow across the weir to the epth below the free surface. δh h

3 For a rectangular weir the with oes not change with epth so there is no relationship between b an epth h. We have the equation, b constant B B A rectangular weir Substituting this into the general weir equation gives Q B g h h theoretical B 5 g To calculate the actual ischarge we introuce a coefficient of ischarge, C, which accounts for losses at the eges of the weir an contractions in the area of flow, giving / Qactual C B g For the V notch weir the relationship between with an epth is epenent on the angle of the V. V notch, or triangular, weir geometry. If the angle of the V is θ then the with, b, a epth h from the free surface is So the ischarge is θ b b ( h) tan θ θ / Q g ( h) theoretical tan h h / θ g tan h h g tan 5 θ 5 / The actual ischarge is obtaine by introucing a coefficient of ischarge / 5 / h 8 Qactual C g tan 5 5 Rectangular weir 6 θ / Q (m^/s). C_.9 B (m).5. check Q. V-notch weir Q (m^/s). θ C_.8.8 check Q. Vnotch (m).5 Q (m^/s).4 (m).6 Q (m^/s).69 % increase Rectangular (m).5 Q (m^/s).47 (m).6 Q (m^/s).68 % increase.45

4 . Water flows through a cm iameter pipe at.6m/s. Calculate the Reynols number an fin also the velocity require to give the same Reynols number when the pipe is transporting air. [5 marks] Assuming the pressure loss along a pipe, p, can be expresse in terms of the following flui ensity ρ kinematic viscosity ν iameter velocity u show that the pressure loss can be expresse as: p ρu φ Re ( ) ence fin the ratio of pressure rops in the same length of pipe for both cases. [ marks] You will nee to use these physical properties: variable Water air ρ kg/m.9kg/m ν. 6 m /s 5. 6 m /s Draw up the table of values you have for each variable: variable water air u.6m/s u air p p water p air ρ kg/m.9kg/m ν. 6 m /s 5. 6 m /s.m.m Kinematic viscosity is ynamic viscosity over ensity ν µ/ρ. ρu u The Reynols number Re µ ν Reynols number when carrying water: u 6.. Re water ν. To calculate Re air we know, Re water Reair uair u 8. 44m/ s air To obtain the ratio of pressure rops we must obtain an expression for the pressure rop in terms of governing variables. Choose the three recurring (governing) variables; u,, ρ. From Buckinghams π theorem we have m-n 5 - non-imensional groups. 7 φ ( u,, ρ, ν, p) φπ (, π ) a b c π u ρ ν a b c π u ρ p As each π group is imensionless then consiering the imensions, for the first group, π : a b c M L T ( LT ) ( L) ( ML ) L T M] c L] a + b - c + - a + b T] -a - a - b - π u ρ ν ν u An the secon group π : (note p is a pressure (force/area) with imensions ML - T - ) a b c M L T ( LT ) ( L) ( ML ) MT L M] c + c - L] a + b - c - - a + b T] -a - a - b π u ρ p p ρu So the physical situation is escribe by this function of nonimensional numbers, φπ ( π ) φ ν,, p u ρu p φre, ρu p ρu φ ( Re) For ynamic similarity these non-imensional numbers are the same for the both water an air in the pipe. π π air water π π air water We are intereste in the relationship involving the pressure i.e. π p p ρu ρu air water pwater ρwateruwater pair ρairuair

5 4 Describe with the ai of iagrams the following phenomena explaining why an when they occur. (Each part requires at least a half page escription of the phenomenon plus iagrams.) (i) The laminar bounary layer (ii) The turbulent bounary layer (iii) The laminar sublayer (iv) Bounary layer separation (v) Methos to prevent bounary layer separation 5. Water flows horizontally along a mm pipeline fitte with a 9 o ben that moves the water vertically upwars. The iameter at the outlet of the ben is mm an it is.5m above the centreline of the inlet. If the flow through the ben is 5 litres/s, calculate the magnitue an irection of the resultant force the ben support must withstan. The volume of the ben is.m an the pressure at the outlet is kn/m. [5 marks] mm Answer: As the question says - EAC PART REQUIRES AT LEAST ALF A PAGE DESCRIPTION PLUS DIAGRAMS - take from lecture notes AND other books. p kn/m mm Figure Force on a pipe ben question Enter values in the yellow boxes Pressure in (kn/m^) Inflow (litres/s) 5 Volume of ben (m^). Ben angle (egrees) 9 eight ifference (m).5 ea loss in ben (m) Inlet iameter (m). outlet iameter (m). Q (m^/s).5 inlet area (m^).4 outlet area (m^).785 inlet vel (m/s) outlet vel (m/s) Angle (ra) Total force Ftx (N) Fty (N) Pressure force P outlet (N/m^) Fpx Fpy Boy force Fbx. Fby -98. Resultant force

6 Frx Fry Fr Angle (egrees) -.56 Force acting on ben (N) (a) Starting with the Bernoulli an Continuity equations erive the following expression that can be use to measure flow rate with a Venturi meter. Q actual C A A p p g + z A A z Also show that when the pressure ifference is measure using a manometer the following expression can be use ρman gh ρ Qactual C AA A A [5 marks] (b) A venturimeter is use to measure the flow of water in a pipe of iameter mm. The throat iameter of the venturimeter is 6mm an it has a coefficient of ischarge of.9. When a flow of litres/s is flowing the attache maonmeter shows a hea ifference of 6cm, what is the ensity of the manometric flui of the manometer? 6(a): Applying Bernoulli along the streamline from point to point in the narrow throat of the Venturi meter we have p g ug z p u g g z ρ + + ρ + + By the using the continuity equation we can eliminate the velocity u, Q u A u A u ua A Substituting this into an rearranging the Bernoulli equation we get p p ρg u A + z z g A u g p p + z z A A A g p p + z z A + A To get the theoretical ischarge this is multiplie by the area. To get the actual ischarge taking in to account the losses ue to friction, we inclue a coefficient of ischarge

7 Q u A Q C Q C u A Q ieal actual ieal actual C A A This can also be expresse in terms of the manometer reaings p p ρg g p p + z z A A p + ρgz p + ρ gh+ ρg( z h) man ρman + z z h ρ Thus the ischarge can be expresse in terms of the manometer reaing:: 6 (b) ρ man Q A A ρ + C A A gh c.9 rho man (kg/m^)? rho (kg/m^) (mm) (mm) 6 a (m^).7854 a (m^).87 Q (m^/s). Q actual C A A ρman gh ρ A A out h (m).6 rho man (kg/m^) 58.5 in z z h atum 4

Problem 1 (20 points)

Problem 1 (20 points) ME 309 Fall 01 Exam 1 Name: C Problem 1 0 points Short answer questions. Each question is worth 5 points. Don t spen too long writing lengthy answers to these questions. Don t use more space than is given.