47615_cive1400-200304

# 47615_cive1400-200304 - CIVE 140001 This question paper...

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Unformatted text preview: CIVE 140001 This question paper consists of ? printed pages, each of which is identified by the Code Number CIVE 140001 Graph Paper? UNIVERSITY OF LEEDS May/June 2004 Examination for the degree of BEng/ MEng Civil Engineering FLUID MECHANICS Time allowed: 2 hours Attempt 4 questions 1. (a) A "U"-tube manometer is being used to measure the pressure difference between two points on a horizontal pipe. The fluid in the pipe has a relative density of 0.8 and the manometric fluid has a density 13600 kg/m3. The two readings on the manometer differ by 0.5m. What is the pressure difference measured by the manometer? [8 marks] A storage tank has vertical sides and is filled with water to a depth of 2.0m. If the water is covered with a 1.0m thick layer of oil of relative density 0.8, find the resultant force (per unit width) and its line of action on the wall of the tank. [9 marks] Water is being fired at 10 m/s from a hose of 50mm diameter into the atmosphere. The water leaves the hose through a nozzle with a diameter of 30mm at its exit. Find the pressure just upstream of the nozzle and the force on the nozzle. [8 marks] (b) (c) 1.a A:\green\exam paper info and template 29 April 2002 END CIVE 140001 Fluid density w A B a e h c d b datum Manometric fluid density Hg Figure of manometer setup density of mercury = 13600 kg/m3 1.a pressure at C and D is equal: p C = pD pA + w g a = pB + w g ( b - h ) + Hg g h pA - pB = w g b - w g h - w g a + Hg g h = w g ( b - a ) + hg ( Hg - w ) As horizontal a = b = hg ( Hg - w ) pA - pB = 0.5 9.81 ( 13600 - 800 ) = 62 784 N/m2 = 62.784 kN/m2 o = 800 kg/m3 w = 1000 kg/m3 h1 = 1.0m, h2 = 2.0m 1.b Distance of centroids of pressure diagram from surface L1 = 1.0 2/3 = 0.667 m L2 = 2.0m L3 = 1.0 + 2 2/3 = 2.333 m Areas of pressure diagram A1 = 800 1.0 9.81 1.0 0.5 = 3924 A2 = 800 1.0 9.81 2.0 = 15696 A3 = 1000 2.0 9.81 2.0 0.5= 19620 Total area = resultant force per unit width = R = 39240 N A:\green\exam paper info and template 29 April 2002 L1 1 2 3 L3 L2 END CIVE 140001 Take moment about surface to find point of action LR R LR = A1L1 + A2L2 + A3L3 = 79788 Nm LR = 2.033 m 1.c 1 2 u1 = unknown d1 = 0.05m u2 = 10 m/s d2 = 0.03m A1 = 0.001963 m2 A2 = 0.000707 m2 Use continuity to calculate the unknown velocity u1 = u2 A2 / A1 = 3.6 m/s Force on the wall = R = -F = -45.2 N (in the direction of the jet) Pressure force Use Bernoulli to calculate the unknown pressure p u2 p1 u12 + + z1 = 2 + 2 + z 2 g 2 g g 2 g As horizontal z1 = z2 p2 = atmospheric = 0.0 p1 + u12 u2 = p2 + 2 2 2 p1 = = (u 2 2 2 - u12 ) ) 1000 2 10 - 3.6 2 2 = 43520 N / m 2 = 43.5 kN / m 2 FP = P A1 - P2 A2 = 85.45 N 1 FT = FR + FP + FB ( FR = 45.23 - 85.45 = -40.21N A:\green\exam paper info and template 29 April 2002 END CIVE 140001 2.a A concrete dam has the cross-sectional profile shown in Figure 1. Calculate the magnitude, direction and position of action of the resultant force exerted by the water per unit width of dam? (15 marks) 10m 75 Figure 1 2.b A second design for the same dam has the cross-sectional profile composed of a vertical face with a circular curved section at the base as shown in Figure 2. Calculate the resultant force and its direction of application per unit width of this dam. (10 marks) 10m 4m Figure 2 2.a. Method 1 Vertical force = weight of water = g A b Horizontal force = force on a projection of the vertical plane = L 4m gh b h 2 L = h tan = 10 tan 15 = 2.679m A = 0.5hL = 0.5 10 2.679 = 13.397m 2 h v Rv = 1000 9.81 13.397 1 = 131 425 N Rh = 1000 9.81 10 (10 / 2) 1 = 490 500 N R = Rv 2 + Rh 2 = 507 802 N tan -1 (Rv / Rh ) = = 15 o Acting at right angle to the wall 15 to the horizontal. Also A:\green\exam paper info and template 29 April 2002 END CIVE 140001 Method 2 Force on wall = pressure at centroid area of wall = g depth to centroid area of wall Sloping wall length, v = h / cos 15 = 10.35m F = 1000 9.81 (10.35 1) 5 = 507668 N Position of this force is through the centre of pressure, Sc. Using the parallel axis theorem, Sc = I oo I oo 2nd momnt of area = Ax 1st moment of area = I GG - Ax 2 I GG +x Ax Sc = x is the distance along the face to the centroid = v/2 = 5.175m bd 3 1 10.35 2 = = 92.39 I GG = 12 12 92.39 + (10.35 0.5) Sc = 10.35 1 (10.35 0.5) = 6.9m This is the distance to the centre of pressure from O. A:\green\exam paper info and template 29 April 2002 END CIVE 140001 2.b. b = 1m a1 = 4 6 = 24m 2 a2 = 42 2 = 12.566m 2 Vertical force Rv = weight of water = g(a 1 + a 2 )b = 1000 9.81 (24 + 12.566) 1 = 358 712 N Horizontal force = force on the projection of vertical plan. This is the same as in part a of this question. Rh = 490500 N Resultant force R = Rv 2 + Rh 2 = 607671 N Rv tan = Rh 358712 o = tan -1 = 36.178 490500 A:\green\exam paper info and template 29 April 2002 END CIVE 140001 3a. Derive the following expression that describes flow over a triangular sharp notch weir. Start from the Bernoulli equation and state all assumptions made. Q = Cd 8 2 g tan H 5 / 2 15 2 3b. A triangular notch weir has a 90 notch and when the head is 8cm the flow is measured at 200 litres/min. What is the coefficient of discharge for this particular notch? 3c. What would be the head over a rectangular notch weir of width 20cm for the same flow rate and coefficient of discharge and the notch in part b? 40 40 0.3m Figure 2 If the level in the tank rises 0.8m in 20 seconds, deriving all formulae, determine the coefficient of discharge of the notch. (20 marks) A General Weir Equation Consider a horizontal strip of width b and depth h below the free surface, as shown in the figure below. b H h h Elemental strip of flow through a notch Assuming the velocity is only due to the head. velocity through the strip, u = 2 gh discharge through the strip, Q = Au = bh 2 gh Integrating from the free surface, discharge h = 0 , to the weir crest, h = H gives the expression for the total theoretical Qtheoretical = 2 g bh 2 dh 1 H 0 This will be different for every differently shaped weir or notch. To make further use of this equation we need an expression relating the width of flow across the weir to the depth below the free surface. A:\green\exam paper info and template 29 April 2002 END CIVE 140001 For the "V" notch weir the relationship between width and depth is dependent on the angle of the "V". b H "V" notch, or triangular, weir geometry. If the angle of the "V" is h then the width, b, a depth h from the free surface is b = 2( H - h) tan 2 So the discharge is Qtheoretical = 2 2 g tan ( H - h) h 1/ 2 dh 2 0 H 2 2 = 2 2 g tan Hh 3/ 2 - h 5/ 2 2 5 5 0 = 8 2 g tan H 5/ 2 2 15 H The actual discharge is obtained by introducing a coefficient of discharge Qactual = Cd 8 2 g tan H 5/ 2 2 15 b) From the question: Q = 200 litres/min = 3.333 litres/s = 0.00333 m3/s = 90 Head = H = 8cm = 0.08m Re arranging the weir equation, and substituting in these values gives Cd = Q 8 2 g tan H 5 / 2 15 2 = 0.79 3c For a rectangular weir of width B the equation is A:\green\exam paper info and template 29 April 2002 END CIVE 140001 Qactual = Cd 2 B 2g H 3/ 2 3 From the question Q = 0.00333 m3/s B = 20cm = 0.2m H 3/ 2 = Q 2 Cd B 2 g 3 = 0.007138 H = 0.036m = 3.6cm A:\green\exam paper info and template 29 April 2002 END CIVE 140001 4 Water flows ate a rate of 0.5m3/s rising through a 50 , contracting pipe bend. The diameter at the bend entrance is 700 mm and at the exit 500 mm - as shown in Figure 1. o out Pipe section 50 in Figure 1. If the pressure at the entrance to the bend is 200 kN / m , determine the magnitude and direction of the force exerted by the fluid on the bend. (The exit of the bend is 0.4m higher than the entrance and the bend has a volume of 0.2m3.) (17 marks) Comment on the reason why frictional losses may be neglected in this analysis. (3 marks) A1= d1 / 4 = 0.3848 m2 A2= d2 / 4 = 0.1963 m2 u1 = Q/A1 = 0.5/0.3848 = 1.299 m/s u2 = Q/A2 = 0.5/0.1963 = 2.546 m/s p1 = 200 kN/m2 = 200 000 N/m2 Calculate the total force In the x-direction: 2 FT x = Q u2 x - u1 x u1 x = u1 u2 x = u2 cos ( ) FT x = Q(u2 cos - u1 ) = 168.77 N In the y-direction: . = 1000 0.5( 2.546 cos 50 - 1299) FT y = Q u2 y - u1 y u1 y = u1 sin 0 = 0 u2 y = u2 sin FT y = Qu2 sin ( ) = 1000 0.5 2.546 sin 50 . = 97517 N A:\green\exam paper info and template 29 April 2002 END CIVE 140001 Calculate the pressure force Use Bernoulli to calculate force at exit, p2 2 p1 u12 p2 u2 + +z = + + z + hf g 2 g 1 g 2 g 2 the friction loss hf can be ignored, hf=0 As the exit of the pipe is 0.4m higher than the entrance we can say z1 = 0.0, z2 = 0.4 By continuity, Q= u1A1 = u2A2 p2 = p1 - 2 (u 2 2 - u12 + ( z1 - z2 ) ) = 200000 - = 193677 N 1000 2.5462 - 1.2992 + (0 - 0.4 ) 2 ( ) FP = pressure force at 1 - pressure force at 2 FP x = p1 A1 cos 0 - p2 A2 cos = p1 A1 - p2 A2 cos = 52524 N FP y = p1 A1 sin 0 - p2 A2 sin = - p2 A2 sin = -29131N Calculate the body force FBx = 0 FBy = g Volume = 1000 9.81 0.2 = -1962 N Calculate the resultant force FT x = FR x + FP x + FB x FT y = FR y + FP y + FB y FR x = FT x - FP x - FB x = 168.7 - 52524 = -52356N FR y = FT y - FP y - FB y = Qu2 sin + p2 A2 sin + g Vol = 975 + 29131 + 1962 = 32068 N And the resultant force on the fluid is given by = Q(u2 cos - u1 ) - p1 A1 + p2 A2 cos A:\green\exam paper info and template 29 April 2002 END CIVE 140001 FRy FResultant FRx FR = FR2 x - FR2 y = 61396N And the direction of application is = tan -1 FR y = -31.49o FR x the force on the bend is the same magnitude but in the opposite direction R = - FR A:\green\exam paper info and template 29 April 2002 END CIVE 140001 5 Two vertical cylindrical tanks of 5m and 3m diameter contain water. They are joined near their bases by a pipe of diameter 5cm which is short enough to be considered an orifice with Cd of 0.6. If the 3m diameter tank initially has a level 2m higher than the other, calculate how long it will take for the levels to become equal in each tank. (20 marks) Area A1 Area A2 h1 h2 Orifice area Ao Two tanks of initially different levels joined by an orifice Applying the continuity equation Q = - A1 h1 h2 = A2 t t Qt = - A1h1 = A2 h2 Also we can write So - h1 + h2 = h - A1h1 = A2 h1 - A2 h A2 h h1 = A1 + A2 Then we get Qt = - A1h1 Cd Ao 2 g (h1 - h2 ) t = A1 A2 h A1 + A2 Re arranging and integrating between the two levels we get A:\green\exam paper info and template 29 April 2002 END CIVE 140001 t = t= = = h A1 A2 ( A1 + A2 )Cd Ao 2 g h A1 A2 ( A1 + A2 )Cd Ao 2 g 2 A1 A2 ( A1 + A2 )Cd Ao 2 g 2 A1 A2 ( A1 + A2 )Cd Ao 2 g h final h h h final hinitial hinitial [ h] [ hinitial - h final h in this expression is the difference in height between the two levels (h2 - h1). To get the time for the levels to equal use hinitial = h1 and hfinal = 0. The question says hinitial = 2m and we want the time for the tanks to equal so, hfinal = 0 A1 = 52 /4 = 19.6 m2 A2 = 32 /4 = 7.07 m2 Ao = 0.052 /4 = 0.0019634 m2 t= 2 A1 A2 ( A1 + A2 )Cd Ao 2 g [ hinitial - h final = 825.12 sec = 13.752 min A:\green\exam paper info and template 29 April 2002 END CIVE 140001 6.a Explain what is meant when we say that equations should be dimensionally homogenous. (4 marks) 6.b 6.c Describe the meaning of the terms dynamic similarity, kinematic similarity and geometric similarity. (9 marks) Assuming the drag force, F, exerted on a body is a function of the following fluid density fluid viscosity diameter d velocity u Show that the drag force can be expressed as F = d 2 u 2 ( Re) where is some unknown function and Re is the Reynolds number. (7 marks) 6.a When giving numerical values to a term in an equation the units in the left side should equal the units on the right side or more correctly the dimensions should be equal. Any physical situation can be described by certain familiar properties e.g. length, velocity, area, volume, acceleration etc. These are all known as dimensions. Dimensions are of no use without a magnitude being attached. We must know more than that something has a length. It must also have a standardised unit - such as a metre, a foot, a yard etc. Dimensions are properties which can be measured. Units are the standard elements we use to quantify these dimensions. In dimensional analysis we are only concerned with the nature of the dimension i.e. its quality not its quantity. The following common abbreviation are used: length = L mass = M time = T force = F temperature = We can represent all the physical properties we are interested in with L, T and one of M or F (F can be represented by a combination of LTM). 6.b. Dimensional analysis is used when constructing physical models of prototype structures. Physical models are used when the fluid flow is particularly complex and difficult to analyse by other means. It enables physical measurements - forces, velocities etc. - taken from the scale models to be converted to the equivalent measurement which would be found on a prototype. The term similarity relates to physical a scale models. Geometric similarity - all dimensions are in the in the same ratio. Dynamic similarity - all velocities are in the same ratio - requires geometric similarity Kinematic similarity - all forces are in the same ration - requires dynamic similarity. 6.c. F = f ( , , d , u ) 0 = (F , , , d , u ) Assume the governing variables , u, d According to Buckingham's theorem there are n-m groups where A:\green\exam paper info and template 29 April 2002 END CIVE 140001 n = number of variables (5) and m = number of dimensions (i.e. MLT, giving 3) n-m = 5-3 = 2 groups 0 = ( 1 , 2 ) 1 = a ub d c 1 1 1 2 = a ub d c F 2 2 2 Dimensions of the variables are: = density (kg/m3) = ML-3 = viscosity (kg/m/s) = ML-1 T-1 u = velocity (m/s) = ML-1 d = length (m) = L F = newtons (kg m /s2) = MLT-2 For 1 0 = ML-3 0 = a1 + 1 ( ) (LT ) (L ) a1 -1 b1 c1 ML-1T -1 0 = -3a1 + b1 + c1 - 1 0 = -b1 + 1 a1 = -1 b1 = -1 c1 = -1 1 = For 2 ud 0 = ML-3 0 = a2 + 1 ( ) (LT ) (L ) a2 -1 b2 c2 MLT -2 0 = -3a2 + b2 + c2 + 1 0 = -b2 - 2 a 2 = -1 b2 = -2 c 2 = -2 2 = F u 2 d 2 0 = ( 1 , 2 ) u F 0 = ud , u 2 d 2 Inverting 1 gives Re = ud Rearranging this gives F 0 = Re, 2 2 u d 2 2 F = u d (Re ) A:\green\exam paper info and template 29 April 2002 END CIVE 140001 A:\green\exam paper info and template 29 April 2002 END ...
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## This note was uploaded on 09/06/2011 for the course ASE 13180 taught by Professor Goldstein during the Spring '10 term at University of Texas.

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