Lab Report

TABLE OF CONTENT NO. | CONTENT | PAGE | 1. | Title | 2 | 2. | Theory | 2 | 3. | Introduction | 2 | 4. | Objective | 3 | 5. | Apparatus | 3 | 6. | Procedure | 4 | 7. | Result | 6 | 8. | Calculation | 10 | 9. | Discussion | 13 | 10. | Conclusion | 14 | 11. | References | 14 |

TITLE:
H1 – Osborne Reynolds Demonstration
INTRODUCTION:
Osborne Reynold’s Demonstration has been designed for students experiment on the laminar, transition and turbulent flow. It consists of a transparent header tank and a flow visualization pipe. The header tank is provided with a diffuser and stilling materials at the bottom to provide a constant head of water to be discharged through a bellmouth entry to the flow visualisation pipe. Flow through this pipe is regulated using a control valve at the discharge end. The water flow rate through the pipe can be measured using the volumetric tank (or measuring cylinder) of a Hydraulics Bench. Velocity of the water can therefore be determined to allow for the calculation of Reynolds’ number. A dye injection system is installed on top of the header tank so that flow pattern in the pipe can be visualised. In this experiment, we fix the time which is 5 second to collect the amount of the water. At the same time, we also observe the characteristic of the flow whether is it laminar, transition and turbulent flow.

THEORY:
Reynolds number basically determines the transition of fluid flow form laminar flow to turbulent flow. When the value of Reynolds number is less than 2300, laminar flow will occur and the resistance to flow will be independent of the pipe wall roughness (e). Meanwhile, turbulent flow occurs when the value of Reynolds number is exceeding 4000. For large viscous force, whereby Re value is less than 2300, viscous effects are great enough to damp any disturbance in the flow and the flow remains laminar. The flow is called laminar because the flow takes place in layers. Any combination of low velocity, small diameter, or high kinematic viscosity which results in Re value of less than 2300 will produce laminar flow. As Re increases, the viscous damping of flow disturbances or perturbations decreases relative to the inertial effects. Because of a lack of viscous damping, disturbances are amplified until the entire flow breaks down into in irregular motion. There is still a definite flow direction, but there is an irregular motion superimposed on the average motion. Thus, for turbulent flow in a pipe, the fluid is flowing in the downstream direction, fluid particles have an irregular motion in addition to the average motion. The turbulent fluctuations are inherently unsteady and three dimensional. As a result, particles which pass though a given point in the flow do not follow the same path in turbulent flow even though they all are flowing generally downstream. Flows with 2000 < Re < 4000 are called transitional. The flow can be unstable and the flow switch back and forth between turbulent and laminar conditions.

OBJECTIVE:
To determine laminar, transitional and turbulent pipe flow.
PROCEDURE:
Heqim blom hantar =(

RESULT:
Diameter of test pipe, D = 0.012 m
Cross sectional area of pipe, A = 1.131x10-4 m2
Viscocity, υ = 0.902x10-6 m/s
Laminar Flow Lines pattern of the injected dye | Volume, V(l) | Time, t(s) | Flow Rate, Q ( x10-3 )(l /s) | Velocity, v(m/s) | Re | | 0.1 | 10.75 | 9.30 | 0.08 | 1064.30 | | 0.2 | 20.56 | 9.73 | 0.09 | 1197.34 | | 0.3 | 26.46 | 11.34 | 0.10 | 1330.38 | | 0.4 | 30.18 | 13.25 | 0.12 | 1596.45 | | 0.5 | 32.97 | 15.17 | 0.13 | 1729.49 |
Table 1 Osborne Reynolds' Record Sheet
Transitional Flow Lines pattern of the injected dye | Volume, V(l) | Time, t(s) | Flow Rate, Q ( x10-3 )(l /s) | Velocity, v(m/s) | Re | | 0.1 | 4.50 | 22.22 | 0.20 | 2660.75 | | 0.2 | 10.71 | 18.67 | 0.17 | 2261.64 | | 0.3 | 15.23 | 19.70 | 0.17 | 2261.64 | | 0.4 | 20.57 | 19.45 | 0.17 | 2261.64 | | 0.5 | 25.18 | 19.87 | 0.18 | 2394.68 |
Table 2 Osborne Reynolds' Record Sheet
Turbulent Flow Lines pattern of the injected dye | Volume, V(l) | Time, t(s) | Flow Rate, Q ( x10-3 )(l/s) | Velocity, v(m/s) | Re | | 0.1 | 2.02 | 49.50 | 0.44 | 5853.66 | | 0.2 | 4.19 | 47.73 | 0.42 | 5587.58 | | 0.3 | 8.05 | 37.27 | 0.33 | 4390.24 | | 0.4 | 10.32 | 38.76 | 0.34 | 4523.28 | | 0.5 | 13.05 | 38.31 | 0.34 | 4523.28 |
Table 3 Osborne Reynolds' Record Sheet

CALCULATIONS : 1) Laminar Flow a) Find the flow rate, Q Q = Volumetime=Vt …………………………………………………………....(1.a) Q = 0.1l10.75s=9.30×10-3l/s b) Find the velocity, v v = Rate flowArea = QA ..………………………………………………………......(1.b) v = 9.30×10-31.131×10-4=82.23 l/s c) Change l/s to m/s
82.231000=0.08 m/s ……………………………………………………………(1.c) d) Find Renold Number, Re
Re = Velocity×DiameterViscosity= νDυ……………………………………………………………(1.d) Re = 0.08×0.0120.902×10-6 = 1064.30 Re < 2000 Therefore, the type of flow is laminar. 2) Transitional Flow a) Find the flow rate, Q Q = Volumetime=Vt ……..…………………………………………....(2.a) Q = 0.1l4.50s=22.22×10-3l/s b) Find the velocity, v v = Rate flowArea = QA ………………………………………………......(2.b) v = 22.22×10-31.131×10-4=196.46 l/s c) Change l/s to m/s 196.461000=0.20 m/s ..…………………………………………………(2.c) d) Find Renold Number, Re
Re = Velocity×DiameterViscosity= νDυ ..….………………………………………………(2.d) Re = 0.20×0.0120.902×10-6 = 2660.75 2000 < Re < 4000 Therefore, the type of flow is transitional. 3) Turbulent Flow a) Find the flow rate, Q Q = Volumetime=Vt ……..…………………………………………....(3.a) Q = 0.1l2.02s=49.50×10-3l/s b) Find the velocity, v v = Rate flowArea = QA ………………………………………………......(3.b) v = 49.50×10-31.131×10-4=437.67 l/s c) Change l/s to m/s 437.671000=0.44 m/s …………………………………………………(3.c) d) Find Renold Number, Re
Re = Velocity×DiameterViscosity= νDυ ….………………………………………………(3.d) Re = 0.44×0.0120.902×10-6 = 5822.60 Re > 4000 Therefore, the type of flow is turbulent 4) Mean Velocity a) Laminar Flow, 0.08+0.09+0.10+0.12+0.135=0.10 b) Transitional Flow, 0.20+0.17+0.17+0.17+0.185=0.20 c) Turbulent Flow, 0.44+0.42+0.33+0.34+0.345=0.40
DISCUSSION:
1.
CONCLUSION:
REFERENCES: 1)