Guidelines Symbol m V S U₂ U₂ B₂ F₁ D A S L M W T J F y Fluid Mechanics Lab Definition Mass flow rate Volume Control surface Initial

velocity Resultant velocity Jet angle Force on the jet Diameter of nozzle Cross sectional area of nozzle Height of vane above nozzle tip Distance from centre of vane to pivot of lever Mass of jockey weight Weight of jockey weight Volumetric flow rate Time Rate of momentum in the jet Force of jet in x direction Distance of jockey weight from zero position Theory One way of producing mechanical work from a fluid lever under pressure is to use the pressure to accelerate Spring supporting the fluid to a high velocity in a jet. The jet is directed on to the vanes of a turbine wheel that is rotated by the force generated on the vanes as the jet strikes the vanes. Water turbines working on the impulse principle have been constructed with outputs of the order of 100,000 kW and with efficiencies greater than 90%. When a jet of fluid strikes a vane and is deflected, a force is generated due to the change in momentum of the fluid. The force is given by the momentum equation that states that the force on the fluid is equal to the rate of change of momentum of the fluid. This can be expressed simply as the difference between the initial and final momentum flow. 36 Fulcrum Nozzle From supply Standard Unit kg/s m³ m² m/s m/s degrees kg m/s² or N -ewwwww mm mm² or m² mm or m mm or m kg N l/s or m³/s S Kg m/s² or N Kg m/s² or N mm or m AXP Jockey weight Vane Jet Tally Transparent cylinder To weighing tank Figure 1-Arrangement of jet impact apparatus Fluid Mechanics Lab muy F =) For the fat plate, therefore, we see from equation (3) that: and for the hemispherical cup the maximum possible value of force is, from equation: F = 2/ D In the Sl system the units are m kg/s] and u [m/s], so in equation (3), therefore, the units of force are gms or N Experimental Procedure a Level the apparatus and balance the lever with the adjustable spring with the jockey weight att zero setting. Note the weight of the jockey, and the following dimensions: diameter of the nozz (D), height of the vane above the tip of the nozzle when the lever is balanced (s), and distance from the pivot of the lever to the centre of the vane (L). b. Open the bench supply valve and increase the flow rate to maximum. The force on the vare displaces the lever, so restore the balanced position by sliding the jockey weight along the leve Establish the mass flow rate by collecting a known volume of water over timed intervals. Make further observations at a number of lower flow rates; around eight readings should be enough. c. The best way to set the conditions for reduced flow rate is to place the jockey weight exactly a the desired position and then to adjust the flow control valve to bring the lever to the balanced position. The condition of balance is thereby found without touching the lever which is much easier than finding the point of balance by sliding the jockey weight and, the range of settings of the jockey position may be divided neatly into equal steps. d. Run the experiment twice, once with the flat plate and the hemispherical cup, six readings wil be enough for each. Record your results in tables 1 and 2 and use the results to compare the theoretical and experimental values by plotting them as a graph (Figure A). NB. It is easier to start measurements at the maximum flow rate and gradually decrease it rather then starting at the minimum flow and building it up. 38 (6) (8) F e Impact Water Jet Log Diameter of nozzle Cross sectional area of nozzle Height of vane above nozzle tip Distance from centre of vane Mass of jockey weight Weight of jockey weight Check the jockey weight for your r STA N.B. Some figures are standard for all equipment whilst others are specific to each piece of apparatus. Check your equipment before carrying out calculations. Be careful of conversion rates. Quantity (kg) 6 1kg of water 1 litre 1 m³/s = 1000 l/s When the jockey weight is moved a distance y mm from its zero position, the force F on the vane that is required to restore balance is given by: 6 6 Table 1 - Tabulated results for flat plate Water T Fluid Mechanics Lab Fx 150 W x y (9) The mass flow rate m in the jet is found by timing the collection of a known mass of water. The velocity ₁ of the jet as it leaves the nozzle is found from the volumetric flow rate Q and the cross- sectional area A of the nozzle. Water Quantity (kg) 6 Collect and record values for water quantity, time, and jockey displacement y in Table 1 and 2, anc then carry out the calculations to compare results for the theoretical force, J (equation 6) and the experimental force, F (equation 9). (s) 12.24 6 6 D A Swift Lite Charcoal S L M W 14.42 22168 23 = 10.0 mm = 0.01m = TTD²/4 = 78.5mm² = 7.85 x 105 m² = 35 mm = 0.035 m = 150 mm = 0.15m = Check details of your apparatus = Mg = M x 9.81 y (mm) 90 66 machine and record its mass (M) = 610g u= = 13.64 153 18.04 87 Q = m Table 2-Tabulated results for hemispherical plate T y U₁ (s) (mm) (l/s) (m/s) 12.29 176 0.492 6.28 0.440 5.61 01 333 4124 F- шлу 150 U₁ (1/s) (m/s) 0,490 6.24 0.416 5.30 01264 3.36 Q = m 39 J (N) 3,05 2.20 F (N) 3,59 2.63 6.887 0.91 J (N) 3.08 F (N) 7.02 2.46 1.41 3.47 Theory of the Experiment Consider how the equation of momentum is applied to the case shown schematically in Figure 2, which shows a jet of fluid impinging on a symmetrical vane. Let the mass flow rate in the jet be m. Imagine a control volume V, bounded by a control surface S that encloses the vane as shown. The velocity with which the jet enters the control volume is u₁, in the x-direction. *| Fluid Mechanics Lab F B₂U₂ ກໍ ' u Figure 2 - Sketch of jet impinging on a vane The jet is deflected by its impingement on the vane, so that it leaves the control volume with velocity u₂, inclined at an angle B₂ to the x-direction. Now the pressure over the whole S surface of the jet, apart from that part where it flows over the surface of the vane, is atmospheric. Therefore, neglecting the effect of gravity, the changed direction of the jet is due solely to the force generated by pressure and shear stress at the vane's surface. If this force on the jet in the direction of x is denoted by F;, then the momentum equation in the x- direction is: (1) F₁ = m(u₂ cos ß₂ - U₁) The force F on the vane is equal and opposite to this, namely: F = m(u₁-u₂ cos B₂) For the case of a flat plate, ß₂ = 90° so that cosß₂ = 0. It follows that: F = mu₁ is the force on the flat plate, irrespective of the value of u₂. For the case of a hemispherical cup, we assume that ß₂ = 180° so that cosß₂ = -1 and F = m(u₁ + u₂) 80 (2) (3) (4) If we neglect the effect of change of elevation on jet speed, and the loss of speed due to friction over the surface of the vane, then u = U₂, SO 37 F = 2mu₁ is the maximum possible value of force on the hemispherical cup. This is just twice the force on the flat plate. Returning now to Figure, the rate at which momentum is entering the control volume is mu. We may think of this as a rate of flow of momentum in the jet, and denote this by the symbol J, where: Plot your own values for the force on the vane, F (vertical axis) against rate of momentum, (horizontal axis) to compare the relationship for a flat plate and the hemispherical cup. Typical result are shown below in Figure 3. Force on Vane F (N) 5 Force on Vane F (N) 4.5 4 3.5 2.5 3 1.5 5 0.5 4.5 2 4 3.5 3 2.5 2 1.5 1 1 0.5 0 0 0 7.5 Figure 3 Force produced by water impacting a flat and hemispherical plate Plate x Cup ********* Linear (Plate) Linear (Cup) 0.5 Fluid Mechanics Lab 0.5 1 a 1 1.5 Rate of momentum flow J (N) 1.5 Rate of momentum flow J (N) of mom 40 y = 1.87x y = 1.0733x 2 LJ 20 4 2.5/n Impact Water Jet Log Diameter of nozzle Cross sectional area of nozzle Height of vane above nozzle tip Distance from centre of vane Mass of jockey weight Weight of jockey weight le 1-Tabulated results for flat plate T N.B. Some figures are standard for all equipment whilst others are specific to each F apparatus. Check your equipment before carrying out calculations. Be careful of conversion Water Quantity (kg) 6 6 6 Fluid Mechanics Lab kg of water 1 litre m³/s = 1000 l/s When the jockey weight is moved a distance y mm from its zero position, the force F on th mat is required to restore balance is given by: D Water uantity (kg) 6 A FX 150 W x y = he mass flow rate m in the jet is found by timing the collection of a known mass of wat locity u₁ of the jet as it leaves the nozzle is found from the volumetric flow rate Q and the ctional area A of the nozzle. (s) 12.29 6 6 S L M W ollect and record values for water quantity, time, and jockey displacement y in Table 1 and en carry out the calculations to compare results for the theoretical force, J (equation 6) a perimental force, F (equation 9). 6109 y (mm) = 10.0 mm = 0.01m = TTD²/4 = 78.5mm² = 7.85 x 10-5 m² = 35 mm = 0.035 m eck the jockey weight for your machine and record its mass (M) = Q u=&A 14.42 66 22168 23 = 150 mm = 0.15m = Check details of your apparatus = Mg = M x 9.81 Q=m e 2-Tabulated results for hemispherical plate T (s) F. wey 190 U₁ (l/s) (m/s) 0,490 6,24 01416 5.30 01264 3.36 m 39 y U₁ (mm) (1/s) (m/s) 12.20 176 0.492 6.28 13.64 153 01440 5.61 di 333 424 18.09 87 J (N) 3,05 2.20 6.887 0.9 J (N) 3,08 2.46 F (N 315 FNO (N) 7.02 3.47 Fluid Mechanics Lab Assessment questions Discuss how well the theory compares with the experiment, where are the errors? Due to human errors or friction Lost Why is the force on the hemisphere less than twice the flat plate? can be mis the plates aligned that can be the biggest error 41 and s 9 sec (horizontal axis) to compare the relationship for a flat plate and the hemispherical cup Plot your own values for the force on the vane, F (vertical axis) against rate o are shown below in Figure 3. . Plate x Cup to sogia rome........ Linear (Plate) zele1 nol21 3.5 Linear (Cup) ensy aril no 2.5 Force on Vane F (N) 3 (5) paiz Force on Vane F (N) 1.5 erit stew to eam nwon & lo ndi -28015 srl bns wall bns S bas t 0.5 artt bris (3 noitsu 6 4.5 3.5 3 2.5 2 0 4.5 0.5 1 1.5 7.97 0018 = (M) easRate of momentum flow J (N) Figure 3 Force produced by water impacting a flat and hemispherical plate 1.5 3 0.5 2 0 utetes m***** ● (M) UR Fluid Mechanics Lab 0:5 O G. 1.5 Rate of momentum flow J (N) 40 y=1.87x y=1.0733x 2 2 4 Assessment Discuss hove Due 23 a Why is Example piezometer measurements are recorded in Table 2 below. Table 2 Example data of pressure distribution along Venturi meter (bold values r GENEROLAS Q=0.446 V/s = 0.251 m Piezometer Tube No. A (1) B C D (2) EFGH- J K L Piezometer Tube No. A (1) B C D (2) E8.0 F 0. GS.0- H K L hn (mm) Q 247.5 228.5 140.5 6.0 26.0 112.0 150.5 mmal 176.0 193.0 204.0 209.0 Following the procedure outlined on page 13 collect the data and complete Table B. You only erto collect this data for one reading, ideally for the maximum head difference. Table A hn (mm) Fluid Mechanics Lab 145 136 75 100.15 ESO 7 2040 61 119 6 time THE 87 RBS 03 FEXT 22 A 2.9 3 hn-h₁ (m) 0.000 -0.019 -0.107 -0.242 -0.222 -0.136 -0.097 -0.072 -0.055 -0.044 -0.039 48 0.009 -08 70 -0 140 -0.0128 - Glogy SREFER R38 C Aer -0.058 010 42 - 01031 -0026 = hn-h₁ (m) Q = 0₁3121/5 u² 0.163 2g measured) hn-h₁ u² 01023 0.000 -0.076 -0.427 -0.963 -0.883 -0.540 -0.387 -0.285 -0.217 -0.173 -0.154 = 0.312 hn-h₁ u² 2g -01055 -0429 -61859 -01785 -01555 -01356 -01258 -0,190 -0.160 - 01/41 19.2 + 19.26 +14.15 3 Figure 4 compa calculated in T velocity head Plo uri meter Log ensions of the meter and the position of the piezometer tappings are shown in Figure 3, A B C D Direction of Flow Fluid Mechanics Lab (1) Diameter of cross- ter [N] section [dn] (mm) 26.00 23.20 18.40 16.00 16.80 18.47 20.16 21.84 23.53 25.24 26.00 (2) 87 22 22 Area [a] (m²) 34 Dimension of Venturi meter and positions of piezometer tubes 0.000531 0.000423 0.000266 0.000201 0.000222 0.000268 0.000319 0.000375 0.000435 0.000500 0.000531 52 hows the dimeter of the cross section at each of the piezometer stations, and the s of the ideal pressure distribution. cript 1 denotes tube A, subscript 2 relates to tube D alculation of Ideal Pressure Distribution 102 All dimensions in mm 67 47 82 dn 0.615 0.690 0.870 1.000 0.952 0.866 0.794 0.733 0.680 0.634 0.615 2 0.143 0.226 0.572 1.000 0.823 0.563 0.397 0.288 0.214 0.161 0.143 2 *-* 0.000 -0.083 -0.428 -0.857 -0.679 -0.420 -0.253 -0.143 -0.070 -0.018 0.000 FREE 2150 PERY IN