MEC511_Lab2

pdf

School

Toronto Metropolitan University *

*We aren’t endorsed by this school

Course

511

Subject

Mechanical Engineering

Date

Feb 20, 2024

Type

pdf

Pages

11

Uploaded by EarlGrasshopperPerson4

Report
1 Course Title: Fluids & Thermodynamics Course Number: MEC 511 Semester/Year: Fall 2023 Instructor: Dr. Jun Cao Assignment/Lab Number: 2 Assignment/Lab Title: The Venturi Flow Meter Submission Date: Oct 20, 2023 Due Date: Oct 20, 2023 LAST NAME FIRST NAME Student Number Section Signature Table of Contents Summary: 3
2 Introduction: 4 Apparatus 5 Procedure 6 Results 6 Discussion 8 Conclusion 8 Appendix 9 References 10
3 Summary : The purpose of the lab was to study incompressible flow through a Venturi flow meter. Initially, when we closed the valve, all seven piezometers displayed identical readings. Upon opening the water valve, the piezometer levels underwent significant changes before stabilizing at varying heights. Once steady states were reached, each piezometer displayed distinct height readings. We repeated the procedure with a different volume flow rate and calculated theoretical volume flow rates and Venturi discharge coefficients based on these heights. The analysis showed that both experiments differed from the theoretical outcomes predicted by Bernoulli's equation. This discrepancy can be attributed to the limitations of Bernoulli's equation, which assumes steady, incompressible, and frictionless flow. Our experiments illustrated that real-world fluids like water exhibit inherent viscosity, which impacts our results. In summary, Bernoulli's equation is only suitable under ideal conditions characterized by constant flow, incompressibility, and non-viscous fluids.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
4 Introduction: The purpose of this lab is to study the volume flow rate of pipes by observing Bernoulli's principle in a Venturi flow meter. When observing Figure 1, we see that the diameter D 1 is reduced to D 2 which causes a decrease in pressure and an increase in velocity. From this, using the differential pressure (P 1 -P 2 ) we can determine the volume flow rate, Q. Using Figure 2 we derive Bernoulli's equation to be: (1) ? 1 𝜸 + ? 1 2 2𝑔 + 𝑧 1 = ? 2 𝜸 + ? 2 2 2𝑔 + 𝑧 2
5 V 1 and V 2 are the average velocities of sections 1 & 2, therefore applying the continuity equation for incompressible flow we get: ? 1 = ? 2 (2) ? 1 𝐴 1 = ? 2 𝐴 2 Now using equations (1) and (2) we can derive an equation for velocity V 2 which is: (3) ? 2 = 2𝑔[(? 1 −? 2 )/𝜸+(𝑧 1 −𝑧 2 )] (1−(𝐴 2 /𝐴 1 ) 2 ) V 2 is the theoretical velocity at the throat of the Venturi meter but the actual velocity will be slightly lower so we use the equation: (4) ? 2𝑎 = ? ? 2𝑔[(? 1 −? 2 )/𝜸+(𝑧 1 −𝑧 2 )] (1−(𝐴 2 /𝐴 1 ) 2 ) C v is the discharge coefficient. Since the cross section of the Venturi flow meter is not round, the Reynolds number is based on the hydraulic diameter which is given as the equation: (5) ? 𝐻 = 4𝐴 1 ? ?𝑒𝑡
6 Apparatus: Water tank Venturi Flow Meter 7 piezometers Blue dye (for visibility) Outlet valve Procedure: 1. Adjusted the flow through the Venturi to obtain the maximum difference between the manometer readings at the upstream location and at the Venturi throat. 2. Recorded all of the manometer readings. 3. Measured the actual volume flow rate (Qₐ) of the water. Made at least two measurements and took the average of the results. 4. Adjusted the flow until the difference between upstream manometer reading and the throat manometer reading is one half of the full flow value. 5. Repeat for the lower flow rate Results: T = 10 o C
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
7 Table 1 - Sample experimental data at a flow rate of 0.259 and 0.157 liters per second Piezometer A B C D (throat) E F G Piezometric Height (m) Q = 0.259 0.496 0.482 0.448 0.344 0.407 0.429 0.438 Piezometric Height (m) Q = 0.157 0.516 0.513 0.498 0.456 0.481 0.490 0.494 Table 2 - Theoretical calculations given different flow rates. (m/s) ? ? (L/s) ? ? Re ? ? Q = 0.259 L/s 1.8835 0.3038 10,450.97 0.853 Q = 0.157 L/s 1.183 0.1908 6564.82 0.823 Table 3 - Theoretical values of each venturi meter region (Q = 0.259). Region A B C D (throat) E F G V (m/s) 0.6423 0.8030 1.070 1.606 1.070 0.803 0.6423 Piezometric Height (m) 0.497 0.484 0.458 0.386 0.458 0.484 0.496 Cross Section Height (m) 0.03175 0.0254 0.01905 0.0127 0.01905 0.0254 0.03175 Area ( ) ? 2 4. 032?10 −4 3. 226?10 −4 2. 419?10 −4 1. 613?10 −4 2. 419?10 −4 3. 226?10 −4 4. 032?10 −4 Table 4 - Theoretical values of each venturi meter region (Q = 0.157). Region A B C D (throat) E F G V (m/s) 0.3894 0.4870 0.6490 0.9730 0.6490 0.4870 0.3894 Piezometric Height (m) 0.516 0.512 0.502 0.475 0.502 0.512 0.516
8 Cross Section Height (m) 0.03175 0.0254 0.01905 0.0127 0.01905 0.0254 0.03175 Area ( ) ? 2 4. 032?10 −4 3. 226?10 −4 2. 419?10 −4 1. 613?10 −4 2. 419?10 −4 3. 226?10 −4 4. 032?10 −4 Graph 1 - Theoretical vs Actual Height (Q = 0.259) Graph 2 - Theoretical vs Actual Height (Q = 0.157)
9 Discussion: 1. Why is the actual fluid velocity different from the theoretical velocity predicted by Bernoulli’s equation? The actual fluid velocity is different from the theoretical velocity predicted by Beronulli’s equation due to factors such as the viscous shear stress on the walls and turbulent mixing which causes energy loss. As well, the actual fluid velocity depends on the discharge coefficient of the venturi flow meter. Bernoulli’s equation assumes that there are no head losses in the venturi flow meter which is impossible in real life because there will always be a small amount of head losses. This discrepancy causes the values of the theoretical and experimental velocities to deviate. 2. Is the discharge coefficient within the expected range? If not, discuss the possible reasons for the discrepancy. With a Reynolds number of 10450.97 the expected range for the discharge coefficient is 0.94-0.98. Our experimental results show a discharge coefficient of 0.853 and 0.823. This result is noticeably different from the expected range. A possible reason for this can be due to incorrect manufacturing of the Venturi flow meter as any imperfections within the device could cause more frictional force against the fluid flow. This will therefore decrease the actual velocity of the fluid, and a decreased actual velocity would result in a decreased discharge coefficient. 3. According to your results, where are the head losses the greatest in the venturi flow meter? The greatest head losses in the Venturi flow meter are seen at the throat, piezometer D, where the pressure is lowest. The effect of the throat can be easily seen at piezometer G, where the height of the water has dropped significantly in comparison to piezometer A. This can be explained by the fact that the throat has the highest amount of shearing stress in the Venturi flow meter due to its smaller cross sectional area. Conclusion: This lab focused on incompressible flow through a Venturi meter. Using Bernoulli’s equation, the theoretical volume flow rate for Q = 0.259 L/s and Q = 0.157 L/s were calculated as 0.3038 L/s and 0.1908 L/s respectively. The Venturi discharge coefficients for Q = 0.259 L/s and Q = 0.157 L/s were calculated as 0.823 and 0.158 respectively. The theoretical Piezometric heights were also calculated and compared with the heights found during the experiment.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
10 Appendices: Bernoulli’s equation ? 1 γ + ? 2 1 γ + 𝑧 1 = ? 2 γ + ? 2 2 γ + 𝑧 2 Actual Velocity ? 2𝑎 = ? 2 ? ? = ? ? 2𝑔[ (? 1 − ? 2 ) γ + (𝑧 1 −𝑧 2 ) ] [ 1 −( 𝐴 1 𝐴 2 ) 2 ] Hydraulic Diameter ? 𝐻 = 4𝐴 1 ? ?𝑒𝑡 Reynold’s Number ? ? = ρ? 1 ? 𝐻 µ Calculations for Table 2 ? ? = 2𝑔[ (? 1 − ? 2 ) γ + (𝑧 1 −𝑧 2 ) ] [ 1 −( 𝐴 1 𝐴 2 ) 2 ] ? ? = 𝐴 ? ? ? ? 𝐴 = ? ? 𝐴 ? 𝐴 𝐴 ? 𝐻 = 4𝐴 𝐴 2(12.7?10 −3 )+2(31.75?10 −3 ) ?𝑒 = ? 𝐻 ? 𝐴 ρ ℎ20 µ ℎ20 ? ? = ? ??𝑝 ? ?𝑎?𝑐
11 Calculations for Table 3 & 4 ? = ?/𝐴 𝐴 = (ℎ)(12. 7?10 −3 ) 2 = (? 2 ) 2 (1 − 𝐴 2 𝐴 1 ) 2 2𝑔 References: 1. D. F. Young, T. H. Okiishi, J. I. Hochstein, A. L. Gerhart, and B. R. Munson, Young, Munson and Okiishi's a brief introduction to Fluid Mechanics . Wiley, 2021.