A Model Answer for. Problem Set #4 FLUID DYNAMICS

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Caren Atkinson
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1 A Model Answer for Problem Set #4 FLUID DYNAMICS Problem. Some elocity measurements in a threedimensional incomressible flow field indicate that u = 6xy and = -4y z. There is some conflicting data for the elocity comonent in the z direction. One set of data indicates that w = 4yz and another set indicates that w = 4yz -6y z. Which set do you think is correct (exlain)? Obtain the acceleration comonents.

2 Problem. - sol ρ ( ρu) ( ρ) ( ρw) = 0 t x y z For incomressible 3-D flow: u w = x y z 0 () u = 6 y x y w = 8 yz so From () w= 4yz 6y z z = 8 yz 6 y Problem. - sol a = 36xy a a x y = (4z du dt 4 3 = 3y z z = u u u =. u w x y z 3 48xy z d = =. u w dt x y z 4y (4yz yz)( 4y 6y z) dw w w w = =. u w dt x y z z) (8yz 6y )(4yz 6y z)

3 Problem. A tank where the water leel is 5.0 m aboe an arbitrary datum feeds a ieline AB ending at B with a nozzle 4.0 cm diameter. Pie AB is 5.0 cm diameter with oint A being 0.0 m aboe datum and oint B at datum. Find: i) The discharge through the ieline, the ressures and water elocities at A &B ii) If friction losses in the nozzle are 0.5 m, and between A & B are 5.0 m, sole for (i) and lot the hydraulic gradient and total energy lines. Problem. - sol A 5 0 Datum B z = z g g h L 3

4 Problem. - sol i) Ideal flow Aly B.E bet. & z g = z g = 0 0 g =.47 m/sec Q = A = π/4 (0.04) *.47 = m 3 /sec Q = A A A = A B B = π/4 (0.5) * A = m 3 /sec A = B =.575 m/sec Aly B.E bet. & B b b z = zb g g = 0 b b g P B = m of water Problem. - sol A A z = z g g = 0 A A g Aly B.E bet. & A z = z g g A h L P A = m of water ii) Real flow hl nozzle = 0.5 m hl AB = 5 m Aly B.E bet. & 4

5 Problem. - sol = = 9.56 m/sec Q = A = π/4 (0.04) * 9.56 = m 3 /sec Q = A A A = A B B = π/4 (0.5) * A = m 3 /sec A = B =.39 m/sec Aly B.E bet. & B b b z = zb hl g g = 0 5 b b g P B = 9.9 m of water Aly B.E bet. & A Problem. sol A A z = z g g A = 0 A A g P A = 4.90 m of water 5

6 Problem.3 Water issues from a 0.0-cm diameter circular shar edged orifice under a head of.0 m. If a olume of 3.6 m3 is collected in 3 minutes, what is the coefficient of discharge? If the diameter of the jet at the enacontracta is 8.0-cm what are the alues for C & Cc? Problem.3 - sol Q act = V/ t = 3.6 / ( 3 * 60 ) = m 3 /sec Q th = th * A orifice = = 0.m 3 /sec Q act = C d * Q th C d = 0.67 A act = C c * A th (0.08) = C c (0.) C c = 0.64 C d = C c * C C =

7 Problem.4 Water (assumed frictionless and incomressible) flows steadily from a large tank and exits through a ertical constant diameter ie as shown in Fig. (). Air in the tank is ressurized to 50 kn/m. Determine: (i) the height (h) to which water rises, (ii) the water elocity in the ie, and (iii) the ressure in the horizontal art of the ie. Fig () 50 KN/m 4m h Air Water m Problem4 - sol Aly B.E bet. & z = g g z 50/9.8 0 = h 0 0 h = 7.097m Aly B.E bet. & 3 50/9.8 0 = / g 3 = m/sec Aly B.E bet. & 4 50/9.8 0 = 4 (7.795) / g P4 / P4 = 4 m of water 7

8 Problem5 The elocity of water in a ie 0.0 an diameter is 3.0 m/s. At the end of the ie there is a nozzle the elocity coefficient of which is If the ressure in the ie is 0.7 kg/cm, Find: the jet diameter, the rate of flow, and the ower lost due to friction in the nozzle. Problem5 - sol Aly B.E bet. & (Ideal) 0 (3)/g 0.7 * 0 4 /000 = 0 0 / g V th =. m/sec V act = C * th =.86 m/sec Q = A = 0.04 m 3 /sec Q = A = 0.04 m 3 /sec d = 0.05m Aly B.E bet. & (Real) 0 (3) /g 0.7 * 0 4 /000 = 0 0 (.86) / g hl hl = 0.3 m ower lost due to friction = Q h l = 980 * 0.04 * 0.3 = watt 8

9 Problem.6 A 5.0-cm diameter orifice (Cd = 0.6) discharges water from tank A to tank B as shown in fig (). T he acuum gauge in tank B reads 0.65 Kg/cm below atmosheric ressure, while the air ressure aboe oil in tank A is 70 KN/m find the discharge from the orifice and the distance (L). Air Fig () m Oil (0.8) Kg/cm 5m Water L Problem.6 - sol P = Psurface O h O = * 9.8 * = KN/m P = kg/cm = * 0 4 kg/m Aly B.E bet. & (Ideal) / 9.8 = 3 V ( * 0 4 )/000 th = 8.35 m/sec Qact = Cd A orifice V th = 0.6 * π/4 (0.05) * 8.35 =0.06 m 3 /sec d = Vo t ½ a t In Y dir. = 0 ½ g t t = sec In X dir. L = act t 0 Assume C = act = C * th L = * 8.35 * =.3 m 9

10 Problem.7 Water flows through the ie contrac-tion shown in Fig (3). For the gien difference in eizometer leels, determine the flow rate as a function of the diameter of the small ie, D. 0. m 0. m Q Fig (3) Problem.7 - sol Aly B.E bet. & g z = g z g ( - ) = = 0. V =.98 m/sec Q = A = π/4 D *.98 Q =.556 D 0

11 Problem.8 Water flows through the ie contrac-tion shown in Fig (4). For the gien difference in eizometer leels, determine the flow rate as a function of the diameter of the small ie, D m Q D Fig (4) π ( 4 Problem.8 - sol D Q ) g Aly B.E bet. & z = z g g ( - ) - = = 0. g - * * g Q π ( 0. ) 4 * * g = 0. Q =.556 / (/D )

12 Problem.9 A sihon filled with oil of secific graity 0.8 discharges 0 lit/s to the atmoshere at an eleation of 3.0 m below oil leel. The sihon is 0. m in diameter and its inert is 5.0 m aboe oil leel. Find the losses in the sihon in terms of the elocity head. Find the ressure at the inert if two thirds of the losses occur in the first leg. 5 3 Problem.9 - sol = Q / A =0./ (π/4 (0.) ) = 7 m/sec Aly B.E bet. & z = z h L g g = 0 0 (7) /g hl hl =.50 m Aly B.E bet. & 3 z 3 = g 3 z g 3 hl = 8 (7) /g P 3 / /3 *.5 P 3 / = m P 3 = m of oil

13 Problem.0 Water is sihoned from the tank shown in Fig (5). The water barometer indicates a reading of 9.m. Determine the maximum alue of (h) allowed without caitation occurring. The aor ressure of water equals 0.03 bar (absolute ressure). 9. m.8 m Fig (5) h Problem.0 - sol P aor = 0.03 bar = 0.03 * 0.33 t/m (abs) Patm = P aor W h = 0.3 * 9. = 9.5 t/m Aly B.E bet. & z = g z g / =.8 / g 0.3/ =.05 m/sec Aly B.E bet. & 3 z 3 = g 3 z3 g / = - h (.05) / g 9.5/ h = 7.4 m 3

14 Problem. A enturi-meter is installed in a 30 cm diameter ertical ie coneying oil of S.G. = 0.9. The throat diameter is 5.0 cm and the flow being uward. The difference in eleation between the throat and inlet is 30.0 cm. A mercury manometer connected to the enturi registers a deflection of 0.0 cm. If C d = 0.96, calculate the discharge and the ressure difference between the inlet and throat. If the discharge remains constant while the ie is shifted to the horizontal osition, how will the reading of the manometer be affected? Problem. - sol A = π (0.3) = m 4 A = π (0.5) = m 4 Cd = 0.96 ' H = d ( ) = 0. (3.6 / 0.9 ) =.8 m Q = Cd A * A A A * * g * h = m3/sec ( Z P / ) - (Z P / ) = H ( 0 P / ) ( 0.30 P / ) =.8 ( P -P ) / = = 3. m P = 3. m of oil When the ie is shifted to the horizontal osition with the same discharge the reading of the manometer will not be affected. 4

15 Problem. A horizontal enturi-meter carries a liquid (S.G.= 0.8) and has an inlet to throat diameters of 5 / 7.5 cm. If the actual discharge is 40 l/s, and Cd = 0.96, find: i) the ressure difference between the inlet and the throat, ii) the deflection in the mercury U-tube manometer connecting these oints, and iii) the energy lost between the inlet and the throat. Also sketch the H.G.L. and the T.E.L. for the system Problem. - sol π A = (0.5) = m π 4 A = 4 (0.075) = 4.48 * 0-3 m Cd = 0.96 Q = 0.04 m3/sec A * A Q = Cd * * g * h = 0.04 m 3 /sec A A H = 4.5 m i) H = ( Z P / ) - (Z P / ) H = ( 0 P / ) - (0 P / ) = 4.5 m P = 4.5 m of oil 5

16 Problem. - sol H = d ( ' ) = d(3.6 / 0.8 ) = 4.5 ii) d = 0.66 m iii) = Q / A =.64 m/sec & = Q / A = m/sec Aly B.E bet. & z = z g g h L hl = ( Z P / ) - (Z P / ) ( - ) / g = 4.5 (.64) (9.054) ) /g = m Problem.3 An orifice-meter haing a 0.5 / m ie to orifice diameters and Cd = 0.65 is used in measuring the flow of water in a ie. If a U-tube mercury differential manometer gies 0.3 m deflection when connected to the meter, find the rate of flow. What is the maximum ossible rate of flow of water if the ressure at inlet to the meter is maintained at 3.6 KN/m and its aor ressure is.8 KN/m abs. 6

17 Problem.3 sol P = 3.6 KN/m π 4 π A = (0.5) = m A = (0.075) = 4.48 * 0-3 m H = d ( ) = 0.3 ( ) = m A * A Q = Cd * * g * hmax = 0.06 m 3 /sec A A Qmax results when P becomes minimum (P = Paor ) P = Paor =.8 KN/m (abs) P = PAtm 3.6 = = 04.9 KN/m (abs) Hmax = (Z P / ) - (Z P / ) = (0 04.9/9.8) (0.8/9.8) = 0.5 m A * A * * g * hmax Qmax = Cd A A = m 3 /sec Problem.4 itot-tube is laced in a ie carrying water. A mercury differential ressure gauge reads 0.0 cm. Assuming C = 0.99, what is the elocity of water in the ie? If the roblem is reersed and mercury is flowing in the ie and water is used in the inerted differential gauge with the same reading, what would be the elocity of mercury in the ie? 7

18 Problem.4 - sol (a) water is flowing in the ie Aly B.E bet. & z = z g g (- ) ' = = d ( - ) g = 0. ( 3.6 / ) =.6 th = * 9.8 *.6 = 4.97 m/sec act = C * th = 0.99 * 4.97 = 4.9 m/sec a) mercury is flowing in the ie Aly B.E bet. & g ( ) ' - = = d (- ) = 0. ( - /3.6) = th = * 9.8 * =.348 m/sec act = C * th = 0.99 *.348=.335 m/sec z g = z g Problem.5 Water flows along a rectangular channel 50.0 cm wide and 30.0 cm dee and then oer a shar-edged rectangular weir of 30.0 cm crest length and Cd = 0.6. If the water in the channel is 8.0 cm aboe the weir crest, calculate the discharge in both cases of neglecting the elocity of aroach and taking it into consideration 8

19 Problem.5 - sol (a) When neglecting the aroach elocity Q0 = /3 Cd B g H.5 =/3 * 0.6 * 0.3 g (0.8).5 = m 3 /sec (b) When considering the aroach elocity Q0 = /3 Cd B g ( (H a /g ).5 -( a /g ).5 ) -----() The aboe equ. Will be soled by the trial & error. st trial assume Q = Q0 = m 3 /sec Problem.5 - sol a = Q / A channel = / (0.5 * 0.3) = m/sec ----Sub. In () Q = m 3 /sec nd trial a = Q / A channel = / (0.5 * 0.3) = m/sec ----Sub. In () Q 3 = m 3 /sec 3rd trial a 3 = Q 3 / Achannel = / (0.5 * 0.3) = m/sec ----Sub. In () Q4 = m3/sec 9

20 Problem.6 The discharge oer a triangular notch is gien as: Q = (8/5) Cd (g) tan ( /) H.5 (i) Find the error in the discharge that corresonds to an error of % in the measured head. (ii) A right-angled triangular notch is used for gauging the flow of a laboratory flume. If Cd = 0.59, and an error of mm is susected in obsering the head, find the ercentage error in comuting an estimated discharge of 0 lit/s. Problem.6 - sol (i) Q = 8/5 Cd g tan(θ/) H.5 dq = 8/5 Cd g tan(θ/).5h.5 Dh dq/q =.5 dh/h If dh/h = % dq/q =.5 % (ii) Q = 8/5 Cd g tan(θ/) H.5 0/000 = 8/5 * 0.6 g tan(45) H.5 H = 8.3 cm dq/q =.5 dh/h If dh = mm dq/q =.5 * 0./8.3 =.73 % 0

21 Problem.7 A swimming ool 0.0 m long and 8.0 m wide is.0 m dee at one end and.0 m dee at the other end. The ool is emtied through an orifice of 0. m diameter at the bottom. Find the time required to emty the ool if Cd= PART II PART I Problem.7 - sol Part I Q dt = - A dh Cd a g h dt = - 0 * 8 dh dt = - 60 / (Cd a g) h -0.5 dh T = - 60 / (Cd a g) h -0.5 dh T = - 60 / (Cd a g) [ h 0.5 / ] = sec Part II Q dt = - A dh Cd a g h dt = - 0 * 8h dh dt = - 60 / (Cd a g) h 0.5 dh T = - 60 / (Cd a g) h 0.5 dh T = - 60 / (Cd a g) [ h.5 /.5 ] = sec T = T T = sec =.79 hr

22 Problem.8 A tank m x 6m in lan has a 90 V-notch at one end. At the time when the head on V-notch was 30cm, the water suly to the tank was shut off. The time taken to lower the head 0.0 cm was 44 sec. Determine the coefficient of discharge of the V- notch. Problem.8 - sol Q dt = - A dh 8/5 Cd g tan(θ/) h.5 dt = - * 6 dh dt = - / (8/5 Cd g tan(θ/) ) h.5.5 dh T = dt = - / (8/5 Cd g tan(θ/) ) h.5.5 dh 44 = - / (8/5 Cd g tan(θ/) ) [-h.5 /.5 ] Cd = 0.6

23 Problem.9 Water collects in the bottom of a rectangular oil tank as shown in fig (8). How long will it take for the water to drain from the tank through a.54 cm diameter orifice (Cd = 0.85) fitted to the bottom of the tank? Tank.6 m x 9.5 m.9 m Oil (0.87) 0.7 m Water Fig (8) Problem.9 - sol P = O HO = 0.87 *.9 =.653 t/m 3 Hw = P / w =.653 m of water Q dt = - A dh Cd a g h dt = -.6 * 9.5 dh dt = - (.6 * 9.5 ) / (Cd a g) / h dh T = - (.6 * 9.5 ) / (Cd a g) {/ h} dh T = - (.6 * 9.5 ) / (Cd a g) [{/ h} / ] = sec 3

24 Problem.0 A free ortex is formed in a circular tank, which has a coaxial circular orifice at its bottom as a drain. If the tangential elocity of the ortex is l0.0 cm/s at a distance 50.0 cm from axis, what would be the decrease in the surface eleation at a distance of 5.0 cm from axis? Sketch the water surface rofile and the Total Energy Line. Problem.0 - sol = 0 cm/sec r = 50 cm r = 5 cm r = r =const = 00 cm r Aly B.E between & g z = g z r H / g 0 = 0 / g 0 H = / g - / g = 0.05 m The decrease in the surface eleation between & = 0.05 m 4

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