Experimental report experiment 1 fluid statics

HO CHI MINH CITY - UNIVERSITY OF TECHNOLOGY OFFICE FOR INTERNATIONAL STUDY PROGRAMS FACULTY OF CIVIL ENGINEERING EXPERIMENTAL REPORT Experiment: Fluid Statics Instructor: Dr Ha Phuong Student’s name: Trần Đỗ Khang Hy Student’s ID: 2053086 Ho Chi Minh City – 24/11/2022 TABLE OF CONTENT EXPERIMENT 1: FLUID STATICS I OBJECTIVES OF EXPERIMENT - II THEORY - III APPARATUSES - IV EXPERIMENT STEPS - V EXPERIMENT RESULTS AND OBSERVATION - VI COMPUTATION AND EXPERIMENTAL RESULT PRESENTATION EXPERIMENT 3A: ENERGY EQUATION I FUNDAMENTAL THEORY II APPARATUS - 10 III EXPERIMENT STEPS - 11 IV INTRODUCTION -12 V PREPARATION 13 VI RESULTS -14 EXPERIMENT 3D: MEASUREMENT OF VOLUMETRIC FLOW RATE I OBJECTIVES OF EXPERIMENT - 16 II EQUIPMENT SET – UP -16 III THEORY - 17 IV STEPS -18 EXPERIEMENT 5A: FRICTION LOSS IN PIPE I OBJECTIVES OF EXPERIMENT -23 II EQUIPMENT SET – UP 23 III THEORY 24 IV STEPS OF EXPERIMENT -25 V MEASUREMENTS 27 VI REPORT 27 EXPERIMENT 1: FLUID STATICS I OBJECTIVES OF EXPERIMENT This experiment helps students understand the basic equation or hydrostatics and apply it to some problems associated with compressed fluid in a static state II THEORY The state of static and compressed fluid is described by this equation: (1) With z is the elevation of any point in the static fluid mass, constant specific weight γ, and p is hydrostatic pressure, the value of is also called the water pressure measurement in meter From this equation, we can have some applications: Isobaric surface: Isobaric surface is a surface on which the pressure is the same A specific image of the isobaric surface which we can see is atmospheric surface When the pressure at points on the atmospheric surface is equal, the height of them is equal, then Fluid manometer Applying basic equation for calculating the pressure at any point in the static fluid mass If a point has the elevation z0 and pressure p0, the pressure at the point having the elevation z is: p = p0 + γ(z − z0) = p0 + γh (2) Whereas surface 0-0 is a standard surface for comparison By measuring the height h, we can determine the pressure p 3 Calculating the specific gravity of fluid: If we have p0, p, both z0 and z, from (2) we can measure the specific weight of fluid: (3) The basic equation (1) is established when neglecting the effect of capillary In the case when the fluid inside a small tube having diameter equal to less than 3mm, equation (1) is no more accurate If the fluid is water, alcohol, oil fluid level in the tube, having small diameter, will be higher than the tube with large diameter If the fluid is mercury, fluid level in the tube, having large diameter, will be higher than the tube with small diameter III APPARATUSES The experiment fluid equipment: The manometer contains a group of tube numbered from to 10 are set on the board with an elevated ruler, accurate to millimeter The diameter of tubes di is 5mm (except group of tubes with the diameter equal to less than 3mm) In these tubes: • Tube and are used as fluid manometers to measure the atmospheric pressure in vessel T • Group of tubes containing tube 21,22,23 with diameters 1mm, 2mm, 3mm respectively are used to observe the capillary phenomenon • U-shaped tubes are couple tubes 4-5, 6-7, 8-9, containing fluids needed to determine its specific gravity Tube 10 and are used to observe the acupressure surface The atmospheric pressure of tubes 1, 4,and 6, is the atmospheric pressure in vessel T The elevation of fluid level in tubes is Creating pressure part contains static closed vessel T and un-static opened vessel Ð, which is on block 12, have water Due to crank 11, we can change the height of vessel Ð to differ the atmospheric pressure in vessel T IV EXPERIMENT STEPS Check the standard figure of the ruler millimeteter), whether the rulers are in the horizontal plane by reading the water level in tube and tube 10 These tubes need to have the same water level Using the crank 11 to lift the vessel Ð to the pitch (the free surface of vessel D must be higher than the free surface of vessel T about z0−zr = 15÷20cm) i Start to measure the water level of tube to 10 and record the result (also the group tube 2) ii Lower the vessel Ð to the average position (z0 − zr = ÷ 7cm) and to the low position (z0 − zr = −15 ÷ −20cm) Start to measure like above and record the result to the table EXPERIMENTAL REPORT FLUID STATICS Barometric pressure and temperature of open air during experiment time are: Pa = 760 ∗ 10−3 ∗ 13.6 ∗ 9765 = 100931.04(N/m2); t0 = 300C The specific weight of water is: (using the appendix in lecture note) γH2O = 9765(N/m3) V EXPERIMENT RESULTS AND OBSERVATION According to the three elevations of tank D in comparison with tank T, please record the measured values of tubes, and Group Tube in the following Table Table 1a: The measured results (Unit: cm) No L1 L3 L4 L5 L6 L7 L8 L9 L10 24.3 38.3 22.6 37.0 15.7 31.9 13.8 14.9 38.3 24.2 28.7 27.6 32.4 20.9 26.5 14.1 14.4 28.7 24.1 8.7 37.6 22.4 32.9 14.9 14.8 13.9 8.7 NOTE Table 1b: The measured results in tube No L21 L22 L23 41.1 39.4 38.8 32.1 29.9 29.3 NOTE 11.8 9.7 9.1 VI COMPUTATION AND EXPERIMENTAL RESULT PRESENTATION In the fluid static experiment set, which water levels of tubes, or tanksare in the same horizontal plane? Why? Tube and 10 have the same water level in the hydrostatic experiment because both the free surface of tube and 10 are in contact with air so they have atmosphere pressure Tube and tank T also have same horizontal plane because they contact through the same fluid as well as they have the same pressure on the free surface In the fluid static experiment set, which water levels of tubes arenot followed the law of fluid static (not in the same horizontal plane)? Why? The water level in tube does not follow the law of fluid static because tube 21,22 and 23 have different value of diameter, their diameter is small ( all is smaller than 3cm) leads to the existence of capillarity ( depending on the surface tension) Due to capillarity , water levels of tubes in tube 21,22 and 23 are different The smaller the diameter of the tube, the higher the water level Compute the absolute and gage pressure of air in tank T and theserelative errors in three measurement cases Write them in Table Compute the specific weight of three fluids in U-Tubes 4-5, 6-7, and8-9 and their relative errors in three measurement cases Write them in Table STEPS OF CALCULATION To compute the pressure of closed air in Tank T: From the equation: p = p0 + γ (z − z0) = p0 + γh (1) The pressure of the air, in Tank T can be computed as follows: pT = pa + γH2O (L3 − L1) The specific weight of water in tanks, γH2O is defined depending on the temperature of the environment If using pa = 0, we can compute the gage presssure of air in Tank T as follows: pTd = γH2O (L3 − L1) (3) According to three locations of Tank D, we can compute three values of pressure of air in Tank T To compute the specific weight of fluids in the U-Tubes: Considering the fluid with its specific weight, γ, being at rest in the tubes i and i + of the U-Tube Pressure of air in Tank T is also computed as follows: pT = pa + γ (Li+1 − Li) (4) Comparing this equation with the equation (2), the formula to compute the specific weight of fluid in these UTubes can be computed as follows: (5) To compute the error in the method of pressure measurement: Utilizing the theory on the error, from the formula (3), the relative error, δp from the pressure measurement can be computed as follows: (6) Therefore, relative error from the gage pressure measurement is the addition of two terms: • Relative error from the computation of the specific weight of water: (7) • Relative error from the readings on the level scales: (8) Therefore: (2) δp = δγH2O + δL3−1 (9) In general, we use: δγH2O = 0.12% and ∆L1 = ∆L3 = 0.5mm To compute the error in definition of specific weight, γ of fluid in the U-Tubes From formula (5), we can compute the relative error in the measurement of specific weight in the U-tubes as follows: Similarly, δγH2O = 0.12% and ∆L1 = ∆L3 = ∆Li = ∆Li+1 = 0.5mm Table 2: The calculated results No Pt Pd 103N/m2 Gp Y4-5 Y6-7 Y8-9 Gy4-5 Gy6-7 103N/m3 % Gy8-9 % 102.690 1.365 0.127 9.478 8.425 124.078 0.134 0.133 0.218 101.764 0.439 0.142 9.140 7.834 146.235 0.163 0.160 0.476 99.824 -1.501 0.126 9.877 8.341 166.816 0.133 0.132 0.238 Give your comments on: a) The pressure is closed in tank T Pressure in closed air in tank T in the first time doing experiment is the largest (102.41Pa) We can see that when we move tank D downward, pressure in closed air in tank T will decrease The reason is that to find pressure in closed air in tank T, we use this formula: PT = γ (ZD − ZT) so the larger ZD −ZT is, the larger pressure is As fluid in tank D contact with the air so we can consider the fluid surface in tank D as a free surface b) The variation of water surface elevations between two tanks (T and D) The difference between water elevations between two tanks leads to the difference of pressure in the system Pressure of the free surface in tank D is atmospheric pressure because it is in contact with the open air Tube 4, 6, is connected with tank T through closed air so they have same pressure of closed air and tube 5, ,9 have atmospheric pressure When height of water in tank D is higher than that of tank T, gage pressure of water in tank T is greater than while tube 5, 7, have atmospheric pressure Therefore, water elevations of tube 4, 6, are lower than those of tube 5, 7, respectively When the height of water in tank D is lower than that of tank T, pressure in tank T is vacuum pressure Therefore, pressure in tube 4, 6,8 is vacuum pressure while pressure in tube 5, 7, is atmospheric pressure That’s why water elevations of tube 4, 6, are higher than those of tube 5, 7, respectively Moreover, when tank D moves downward to the point that is lower than tank T, we can see that there is vacuum pressure in tank T We also get formula at the same temperature: PV = constant and because P is so small so V has to increase so the water elevation of tube also decreases a little bit Water elevations in tube 3, 10 and tank D have a same value and decrease significantly as they contact in a same fluid, the upper part is open air, they have a same isobar EXPERIMENT 3A: ENERGY EQUATION I FUNDAMENTAL THEORY a OBJECTIVES: To investigate energy equation in current, having changeable cross section b THEORY: The energy of one weight unit of fluid at a section is determined by parts: - Kinetic energy 𝛼 - Force 𝑉𝑖 2𝑔 𝑝𝑖 𝛾 - The elevation zi With zi, pi, Vi alternately are elevation, pressure and average velocity of the water current at cross-section i, 𝛼 is kinetic fixing coefficient, 𝛼 appears in the component of … Is due to unsteady velocity distribution (by friction in the current) above the section 𝛼= 𝐴 𝑢 ∬ ( ) 𝑑𝐴 (3.1) 𝑉 With u, V alternately are point velocity and average velocity above wet cross-section with area A With the inside current, when moving layers, 𝛼 = 2, during the tangled motion,  = 1.05 – 1.15 To determine  more accurate, we need to know the rule for distributing velocity u in section A and using formula (3.1) to find Normally, for fluctuated flow, we use  = to make the calculation easier So at section (i) the energy of a liquid weight is equal to: 𝐻𝑖 = 𝑎𝑖 𝑉𝑖2 2𝑔 + 𝑝𝑖 𝛾 + 𝑧𝑖 With Hi is the total water column (m) Consider the flow is stable from section 1-1 to section 2-2, it can be described as an energy equation: 𝑧1 + 𝑝1 𝛾 + 𝑎1 𝑉12 2𝑔 = 𝑧2 + 𝑝2 𝛾 + 𝑎2 𝑉22 2𝑔 + ℎ𝑓1−2 (3.2) With ℎ𝑓1−2 is the energy loss of the flow from section 1-1 to section 2-2 If neglecting the energy loss, equation (3.1)is: 𝑝 𝑧+ + 𝛾 𝑎𝑉 2𝑔 = 𝑐𝑜𝑛𝑠𝑡 (3.3) 10 EXPERIMENT 3D: MEASUREMENT OF VOLUMETRIC FLOW RATE I OBJECTIVES OF EXPERIMENT Comparison of flow measurement devices in a duct: • Orifice plat • Venturi nozzle II EQUIPMENT SET - UP The fan inlet is a duct 149 mm diameter (1) provided with pressure tapings whereby the static pressure may be measured simultaneously at each of sections All four pressure tapings are connected to a bank of pressurized manometer tubes (4) Two flow measurement devices are: a 65mm orifice plat (2) b 65mm (d) - 149mm (D) – diameter venturi nozzle (3) Figure 1: Experimental map In which: Orifice plat Ventury nozzle Fans and electric motors Inverter Measuring tubes Pressure gauges Silicon tubes 1,2,3,4: Order number of the measuring tubes III Theory The volume flow rate at the orifice plate and venturi nozzle in the pipe is determined by formula as follows: (1) 17 Where: • Q: volumetric flow rate • C: discharge coefficient • ∆p: pressure difference from inlet to throat The manometer containing liquid of density ρ1 is used to indicate ∆p, the pressure difference may be expressed in terms of the manometric head differential ∆h by: ∆p = (ρ1 − ) ì g ì h (2) ã : flow density (ρ = 1.226kg/m3) • ρ1: water density (ρ1 = 1000kg/m3) • β: diameter ratio • ε: Expandability factor The expandability factory is also detailed in the code and allows for the effects of density change in gas flows where a high-pressure reduction occurs For liquid flows and gas flows with moderate variation in pressure at the meter, ε = 1.00 The discharge coefficients of the orifice plat and the venturi nozzle can be determined by empirical formula For the orifice plate: Where: (4) • • • • • Re: Reynolds number U: upstream pipe velocity Q: discharge in pipe D: diameter of pipe µ: dynamic viscosity When determining Q from ∆p, it is necessary to estimate a value C initially as Re cannot be calculated until Q is known From an initial estimate of C (example C = 1), Q can be calculated and thus Re found The value of C can then be corrected and new values of Q and Re cure calculated For the venturi nozzle: C = 0.9858 − 0.196β4.5 (5) 18 IV Steps Check water level in the tank, it should be at the level of half of tank Open the valve at the outlet of fan and turn on the motor Take the readings of manometer levels, and the reading values at two electronic pressure measurers (give the different values of gage pressure between two sections in KPa) In order to decrease or increase the volume flow rate, the valve is closed or opened partly according to the rotating velocity of the motor decrease or increase Repeat for three valve settings (three rotating velocities: 800 rpm, 600 rpm and 400 rpm), and write the readings of manometer levels and the values of electronic pressure measurers EXPERIMENTAL REPORT t0H20 = 200C = 1.002 ì 10−3Pa.s Derive the formula (1) in case of C = The Work-Energy equation written between cross-section in the approach fluid flow and cross-section in the constricted area of flow is shown below: The pipe is on horizontal plane so Z1 = Z2 and we apply the continuity equation velocity Therefore, we get: Proof To proof the formula (1) we as below: 19 Where: • Qideal is the ideal flow rate through the meter (neglecting viscosity and other friction effects) • A2 is the constricted cross-sectional area perpendicular to flow • P1 is the approach pressure in the pipe • P2 is the pressure in the meter, • β is the diameter ratio • ρ is the flow density The volumetric flow rate calculated from this equation is called Qideal ,which doesn’t include the effects of frictional losses In practice, there are always friction losses and other nonideal factors so we need to add a discharge coefficient C into the equation for Q but in this question, it assume that C = 1, So C is not necessary in this formula Furthermore, we need to add the expandability factor, which is also detailed in the code and allows for the effects of density change in gas flows where a high-pressure reduction occurs For liquid flows and gas flows with moderate variation in pressure at the meter ε≈1 And: Finally, we get the formula: Derive the formula (2): We call M,N as the free surface of tube and respectively The pipe is on horizontal plane so Z1 = Z2 Applying the fluid statics, we can get Proof 20 ZM − ZN = −∆h Because tube and tube are connected with each other through water,so we can apply fluid statics pM = pN + ρ g∆h Finally, we can get the formula: ∆p = p1 − p2 = (ρ1 − ρ)g∆h Where: a ρ1 = 1000kg/m3: Water density b ρ = 1.226kg/m3: Flow density Determine the volumetric flow rate in three experiments by using orifice Flat + Steps of calculation: - First, we will find the value ∆h = h2 − h1 then we will use the formula below to find ∆p: ∆p = (ρ1 − ρ) × g × ∆h Where: • ρ: flow density (ρ = 1.226kg/m3) • ρ1: water density (ρ1 = 1000kg/m3) - Then, we find the difference of pressure measured by pressure gauge and measuring tube by applying following formula: - To calculate C value in orifice plate, we will use iteration method: + Step 1: We assume C = to calculate Q by the function (1): + Step 2: Calculate Reynolds number by the function (4) 21 + Step 3: Calculate C value by function (3) + Step 4: After getting value of C, we will replace C = with this new value of C and calculate Q again by formula (1) + Step 5: We keep doing this iteration method until: 5% After getting the suitable values of C and Q we write them down to the table Velocity Tube Tube h2 –h1 p2 −p1 Pressure Difference 1, 2, (rpm) h1(cm) h2(cm) (m) (Pa) gauge (Pressure C Q gauge (kPa) Measuring (m3/s) tube) (%) 428 0.273 0.281 0.008 79.656 150 46.896 0.571 0.02257 675 0.264 0.293 0.029 288.753 330 12.499 0.569 0.04280 925 0.250 0.313 0.063 627.291 620 1.176 0.568 0.06298 1175 0.232 0.328 0.096 995.872 970 1.456 0.568 0.07769 Determine the volumetric flow rate in three experiments by using venturi nozzle Steps of calculation First, we will find the value ∆h = h4 − h3 then we will use the formula below to find ∆p: ∆p = (ρ1 ) ì g ì h Where: ã : flow density (ρ = 1.226kg/m3) • ρ1: water density (ρ1 = 1000kg/m3) Then, we find the difference of pressure measured by pressure gauge and measuring tube by applying following formula: 22 We calculate the value of C in the venturi nozzle by using the formula (5): C = 0.9858 − 0.196β4.5 Finally, we find the value of Q by using the formula (1): Record the value of C and Q to the table below Velocity (rpm) Tube 3, h3 (cm) Tube h4 –h3 4, h4 (cm) (m) p4 −p3 Pressure (Pa) gauge Difference (Pressure C Q gauge (kPa) Measuring (m3/s) tube) (%) 428 0.274 0.278 0.004 39.240 60 34.60 0.981 0.02720 675 0.269 0.285 0.016 156.96 140 12.114 0.981 0.05441 925 0.263 0.289 0.026 255.06 250 2.024 0.981 0.06936 1175 0.255 0.297 0.042 412.02 410 0.493 0.981 0.08815 Compare the computed results between the using of orifice plate and venturi nozzle, give the conclusions There is always a relative difference in the results between the using of orifice plat and venturi nozzle When calculating the Q value, using venturi nozzle is better It is because: In venturi nozzle, there is a specific part called diverging section which is responsible for recovering the lost pressure which was affected due to the measurement of the average velocity or flow rate of the flowing fluid In orifice plate, the pressure loss is relatively more compared to venturi meter because it contains no specific part which is aimed for pressure recovery Therefore, when the flow goes through the orifice plate, it suddenly enlarges, making the disturbance in the flow and leading the results are not so stable In conclusion, venturi nozzle is quite more accurate than the orifice plate because it has lower relative errors in its calculation, making it a more secure and reliable measurement instrument 23 Fan frequency - Flowrate relationship 1.00E-01 8.79E-02 9.00E-02 Flow rate (m3/s) 8.00E-02 6.92E-02 7.00E-02 7.69E-02 5.43E-02 6.00E-02 6.23E-02 5.00E-02 Orifice 4.00E-02 3.00E-02 2.71E-02 Ventury 4.24E-02 2.00E-02 1.00E-02 2.23E-02 0.00E+00 428 675 925 1175 Frequency (Hz) 24 EXPERIEMENT 5A: FRICTION LOSS IN PIPE I OBJECTIVES OF EXPERIMENT - To investigate the variation of friction head along a circular pipe with the mean flow velocity in the pipe - To investigate the friction factor against Reynolds number and roughness II EQUIPMENT SET – UP A pipe of 10.64cm diameter is supplied by water a centrifugal pump Four test sections with interval of m are connected to a bank of pressurized manometer tubes Water from the pipe flow into the concrete channel, and at the end of channel a vee-notch, is installed to measure the flow rate in the channel, this flow rate is equal to the flow rate in the pipe Water level over the vee-notch is measured by a point gauge vernier mounted on a small tank which is opened to the channel The flow rate over the veenotch is calculated by formula as follow: Where:  = 90𝑜 ; CD = 0.58; g = 9.81 m/s^2; ℎ0 = ZCR – z 25 Where z is water levelin channel, ZCR is the elevation of the crest of Veenotch, ZCR = 27.8 cm The flow rate over the Vee-notch is regulated by a control valve of pump, and an ampere meter mounted on an electric box will show the current intensity of motor corresponding to the flow rate in the pipe The difference of pressure between the test sections in the pipe are measured by reading the water level in the tubes of manometer III THEORY Considering flow at two sections i,j in a pipe, Bernouilli’s equation may be written as: 26 IV STEPS OF EXPERIMENT: Before testing, check valves (2), (4) and locks (9) Make sure they are closed and check the spindle of the pump and the motor by turning lightly to see whether it is hard or not, if it moves is well On electric box, you press: POWER ON button and RUN / STOP button, ON button to start the pump Fully open the valve (4) by turning counterclockwise until it is no longer rotatable Note: If valve does not open fully, when opening the valve will cause break the manometer tubes because of high pressure Slowly open the valve (2) and see the change in current intensity to the value that you need of the test And then open clocks (9) Measurements are made twice Important note when you want to shutdown Close valve (2), then switch off (press the OFF button of RUN / STOP and power OFF button) Then close the valve (4) immediately, to keep the water in the pile a- The first measurement: Open the locks (9) of the manometer tubes at the section (1) and (2) Adjust the valve (2) to change the three flow rate levels corresponding to the current of 21A < I< 26A(the first - the flow rate corresponding to I = 25A, the second - the flow rate corresponding to I = 23.5A; The third - flow rate corresponds to I = 22A) Wait for the water level in the channel to stabilize (the water level in point gauge (5) is constant), read the following values: - Water level in manometer tube (1) - Water level in manometer tube (2) - Elevation Z at point gauge and vernier (5) The measurement results are recorded in Table of the report b- The second measurement: 27 • Adjust the valve (2) to change the five discharge levels corresponding to the current I = 21.5A to 19.5A • Continue to open the locks of manometer tubes (3), (4) • For each discharge level, wait for the water level in the channel to stabilize, taking the following measurements: - Reading water level from the manometer tub (1) to the manometer tube (4) - Elevation Z at the point gauge (5) The results are recorded in Table of the report EXPERIMENT REPORT The water temperature t o H2O = 30oC The water kinematic viscosity v = 0.8x10-6 m2/s V MEASUREMENTS No I (round/min) 1000 900 800 700 650 600 550 500 Table Manometer reading (cm) Water level (Z) (cm) (1) (2) (3) (4) 25.5 63.4 N.A N.A 9.7 29.4 59.7 N.A N.A 10.63 32.9 53.6 N.A N.A 11.72 26.3 41.3 56 70.8 13.035 31.7 43.1 53.2 68.8 14.275 36.1 45.4 52.7 62.7 15.625 41 47.2 50.3 54.7 17.9 44.8 48 49.1 54.9 24.3 VI REPORT Select any three levels of flow rate at the second measurement (I