Viscosity Lab Report

Abstract The objective of this lab is to determine the dynamic viscosity of a commercial gear oil. The viscosity is obtained by measuring the terminal velocity of small spheres falling through the oil. [2] Introduction and Theory Viscosity is a fluid property that can be interpreted as the ‘thickness of a fluid. This property describes the level of resistance to the sliding motion of one layer of the fluid to another one. Viscosity is very important in many aspects. More specifically, lubricants have different viscosities to satisfy the needs of different machinery for which they are used, and without the correct viscosity, the machines will not be able to operate properly. A fluid’s dynamic viscosity is the resistance to an externally applied shear stress. Viscosity resists the fluid to change form. Because of this, if a solid object is dropped into a fluid of lesser density, the object will fall through the fluid, exerting a shear force on the fluid surrounding it. The shear stress (τ) is directly proportional to the velocity gradient within the fluid, for most common fluids. This relationship is described using the equation: [2] τ =μ ∂u ∂x (1) ● µ is the proportionality constant (dynamic viscosity) The methods in this experiment determine the dynamic viscosity through dropping spheres of a known mass and density into an oil. As the sphere to falls through the oil, it will stop accelerating and reach a steady state velocity called terminal velocity. Different forces will be acting on the sphere but the sum of all the forces will be zero. Ʃ F=ma=0 FD+FB-Ws=0 (2) Figure 1: Force balance on the sphere falling through a viscous liquid. According to equation 2, the weight of a sphere (Ws) is equal to and opposite to the sum of the i. upward the drag force (FD) and the upward buoyancy force (FB). The weight force acts in the downward direction and is dependent on the acceleration due to gravity, the density of the sphere and the size of the sphere. The pressure exerted by the fluid on the sphere causes the buoyant force in the equation above. When a sphere is moving in a liquid, the viscous drag force will be exerted in the sphere. In flows that are very slow, meaning the fluid has a very high viscosity, the viscous forces are very large in comparison to the inertial forces. This type of flow is called Creeping Flow or Stokes Flow. For a very slow flow, the mathematician, G.G. Stokes, obtained an analytical solution for the drag force on the sphere. The solution to the equations of the fluid motion gives that the drag force (FD) on a sphere moving at a constant velocity (U) through a still fluid is: F D =3 π μUD (3) ● D is the diameter of the sphere μ is the dynamic viscosity of the fluid ● To understand Archimedes principle, we must have the basic knowledge of what a buoyant force is. This force “is defined as an upward force (with respect to gravity) on a body that is totally of partially submerged in a fluid.[1].” Archimedes principle is the general principle for buoyancy and can be applied to this experiment because it is valid for a single fluid of a uniform density. [1] Using this principle, the equation can be applied to find the amount of fluid displaced by finding the upward buoyant force through the following equation: F=ρf ∀g=ρf ● ρf πD 3 ∀= 6 ● πD 3 g 6 (4) is the density of the fluid is the volume of the sphere Another force, which must be calculated and accounted for, is the force due to weight. The weight ( w s ) can be found through the formula: w s=ms g= ρs ∀g=ρ s ● ● πD 3 g 6 (5) ms is the mass of the sphere ρs is the density of the sphere 1 Lastly, by subsisting the equations for all the forces present on the sphere during its free fall through the oil, the following equation is formed and can be utilized to calculate the dynamic velocity. Once equated, the equation can be used to calculate the dynamic viscosity of a fluid by using the density, diameter, and the terminal velocity of the sphere: μ= D 2 (ρ s−ρf ) g 18U (6) Due to this fluid being a Stokes Flow, the following condition is met: ρf UD <1 μ Re ¿ 1 (7) This dimensionless ratio is called the Reynolds Number. The Reynolds Number of a fluid determines the nature of the flow, being either laminar or turbulent. As the Reynolds number decreases, the Stokes Flow approximation improves. For Stokes flow, the flow is laminar and streamlines are symmetrical. [2] Figure 2: Photograph of very slow flow. ("Stokes flow") over a sphere. [2] Apparatus · Tall graduated cylinder, filled with gear oil (2000mL) · Three various sized solid nylon spheres · Hydrometer for measuring the specific gravity of the oil · Stopwatch (ms) · Meter stick (cm) 2 · Weigh scale to determine the mass of each sphere (g) · Micrometer to determine the diameter of each sphere (mm) · Thermometer to measure the oil temperature (degree Celsius) The oil is a commercial gear lubricant: Quaker State SAE 80W-90 GL5. [2] Procedure 1. Record the temperature of the room by measuring the temperature of the oil in the graduated cylinder. Record this data. 2. Gently lower the hydrometer into the graduated cylinder. Take a reading at the bottom of the meniscus. This will measure the specific gravity (SGoil). 3. Place the small tin bin provided on the electronic scale. Zero the scale. Place ten spheres of the same size one by one until there are ten inside. Calculate and record the average mass. Repeat this step for all the different sized spheres. 4. Using the micrometer, place the sphere between the grooves. The first part is used for the main scale on the sleeve and the second is contributed by the rotating Vernier scale on the thimble. Record the diameters for the two big spheres, one medium and one small sphere. 5. Select a starting point (A) from where you want to start recording the time, choose a distance where the ball is most likely to be in terminal velocity. Select a distance where you will stop recording the time (B), this will be right before it hits the bottom of the sphere. 6. Select the small sphere and drop it into the oil. Have a teammate measure the amount of time it takes to drop it from Point A to Point B. 7. Repeat step 6 for one medium sized ball and two large spheres. [2] Results 3 Distance travelled Room temperature: Δ d : 0.28± 0.0005m 23.5±0.5°C Specific gravity: 0.888 Table 1. Experimental Data Sphere Average Mass m (kg) Small 1.5× 10−4 ± 5 ×10−5 Medium 5.1 × 10 ± 5× 10 Large 1 1.22 × 10−3 ± 5 ×10−5 Large 2 1.22 × 10 ± 5 ×10 −4 −3 D Diameter (m) 6.19×10−3 ±5 ×10−6 −5 −3 9.88×10 ± 5× 10 −5 −6 1.25× 10−2 ±5 ×10−6 −2 1.25× 10 ±5 ×10 −6 Time travelled (s) Δt 22.1 ± 0.005 10.97 ± 0.005 6.4 ± 0.005 6.82 ± 0.005 Table 2. Calculated Results Sphere Density of Sphere ρ ( kg/m3 ) Constant Velocity U (m/s) Dynamic Viscosity μ (kg/ms) Low Reynolds Number Criterion 0.527± 0.176 0.13± 0.044 < 1 0.0255± 4.7×10 0.256± 0.025 0.87± 0.096 < 1 1192.975± 48.91 0.0438± 8.5 × 10−5 0.594± 0.024 0.82±0.033 < 1 1192.975± 48.91 0.0411±7.9× 10 −5 0.633±0.026 0.72±0.030 < 1 Small 1207.871 ± 402.63 0.0127± 2.28 × 10−5 Medium 1009.952 ± 99.03 Large 1 Large 2 −5 Discussion The dynamic viscosity of the gear oil measured through the different sized spheres did not produce the same value. There were a number of factors that were involved in the data collection that lead to difference in the determined dynamic viscosity values, including the fact that the spheres were timed using a hand-held stop watch and they were dropped by a lab 4 member in the approximated center of the cylinder. Other than these two main factors a more accurate sphere diameter measurement would have aided in the accuracy of the dynamic viscosity values. Of all three sizes, the sphere that likely gave the most accurate result was the smallest sphere. This is due to the fact that the smallest sphere obtained terminal velocity at the exact moment the time interval began. Another reason why the smallest sphere is probably the most accurate is that it was least affected by the strain caused due to the cylinder walls. After converting the data available from the manufacturer’s website into appropriate form and units, a value of 0.355 kg/(ms) [3] was determined to be the dynamic viscosity of the oil. Compared to the smallest sphere a percentage error of 32.6% was observed, the medium sphere showed a percentage error of 27.8% and the largest sphere had a percentage error of 40.2%. This error is caused by the factors discussed below. Conclusions The purpose of this lab was to determine the dynamic viscosity of gear oil using the terminal velocity different of spheres falling through the oil. The values of dynamic viscosity ❑ obtained from the three different sized spheres were 0.527 kg/ m , 0.256 kg/m and 0.594 kg/ m , respectively while the value obtained from the manufacturer’s data was 0.355 kg/ m . The original value obtained from the manufacturer was in the form of kinematic viscosity: 400.8 m m2 / s ❑ , the unit conversion and dynamic viscosity calculation has been added to the appendix section. The percentage error for each of the spheres is given in the above section and it is clear that the medium sphere produced the most accurate results, as predicted. However, even with the medium sphere a 27.8% error still occurred. This amount of error is likely due to inconsistency in the placement of drops and timing errors. The spheres were not all dropped at the same spot in the top of the cylinder which would affect the rate of fall due the strain the wall induces on the sphere. To counter this error, a cylinder with a large diameter should be used in order to minimize the strain from the walls. The timing of there sphere’s fall was determined using a hand-held stopwatch which had a major effect on the accuracy of the obtained data, this error can be reduced by using Vernier laser gates to determine the free fall time. All in all, the experiment was conducted successfully however the accuracy of the data could have been improved by making certain alterations such as the addition of technology. 5 References [1] C. T. Crowe, D. F. Elger, B. C. Williams, and J. A. Roberson, Engineering Fluid Mechanics, 11th ed. Wiley, 2012. [2] Friedman J., Naylor D., 2018. AER 316. Fluid Mechanics Laboratory Manual. Ryerson University. [3] “Quaker State Hi-Performance Gearplus 80W-90,” LubeDB - oils, greases, petrochemicals database, 23-May-2012. Appendices Calculations ρ ¿ Eq. 1 Density of Sphere (¿❑¿¿s) ¿ ¿ diameter( D) 3 ¿ 2 4π ¿ mass(m) ❑ 3m ρ s= = volume(V ) ¿ 6.19 × 10−3 m 3 ¿ 2 ¿ Small Sphere 4π¿ 3(1.5× 10−4 kg) ρ ❑ s= ¿ δρ❑ s=ρ❑ s√ ❑ −3 Medium Sphere - 9.88× 10 m 3 ¿ 2 ¿ 4π¿ −4 3(5.1×10 kg) ❑ ρ s= ¿ δρ❑ s=ρ ❑ s √❑ 6 1.25×10−2 m 3 ¿ 2 ¿ Large Sphere 4π ¿ 3(1.22×10−3 kg) ρ ❑ s= ¿ δρ❑ s=ρ❑ s√ ❑ Eq. 2 Constant Velocity (U ) U= displacement(Δd) time ( Δ t ) 0.28m Small Sphere - U= 22.1s =0.0127m/s δU=U √❑ Medium Sphere - U= 0.28m =0.0255m/ s 10.97s δU=U √❑ Large Sphere 1 - U= 0.28m =0.0438m/s 6.4s δU=U √❑ 0.28m Large Sphere 2 - U= 6.82s =0.0411 m/ s δU=U √❑ Eq. 3 Density of fluid ρ ¿ (¿❑¿¿f ) ¿ ¿ specific gravitySG= densityof fluid ρ❑ f densityof water ρ❑ H O 2 3 ρf =SGρ H O=0.888(999kg / m )=887.112kg /m 3 2 Eq. 4 Dynamic Velocity ( μ) D diameter¿ ¿ ❑ ¿ 2(densityof sphere ρ❑ s−densityof fluid ρ f )( gravitational accelerationg) ¿ μ=¿ 7 6.19×10−3 m¿ 2(1207.871kg/m3 −887.112kg/m3 )(9.8m/ s 2) ¿ Small Sphere ¿ μ=¿ δμ = μ √❑ 9.88×10−3 m¿ 2 (1009.952 kg/m3−887.112kg /m3)(9.8m/ s2 ) ¿ ¿ μ=¿ Medium Sphere - δμ = μ √❑ 1.25× 10−2 m¿2 (1192.975kg/m3−887.112kg/m3)( 9.8m/s2 ) ¿ Large Sphere 1 ¿ μ=¿ δμ = μ √❑ 1.25× 10−2 m¿2 (1192.975kg/m3−887.112kg/m3)( 9.8m/s2 ) ¿ Large Sphere 2 ¿ μ=¿ δμ = μ √❑ Eq. 5 Reynolds Number ( ℜ) ℜ= (densityof fluid ρ f )( constant velocityU)( diameterD ) dynamicvelocityμ 3 Small Sphere - ℜ= −3 (887.112kg/m )(0.0127m/s )(6.19× 10 m) =0.13 0.527kg/ms δ ℜ=ℜ √❑ ℜ= Medium Sphere - (887.112kg / m3 )(0.0255m/s)( 9.88× 10−3 m) =0.87 0.256kg/ms δ ℜ= ℜ √❑ s Large Sphere 1 - ℜ= −2 (887.112kg/m )(0.0438m/ s)(1.25×10 m) =0.82 0.594kg/ms δ ℜ=ℜ √❑ Large Sphere 2 - ℜ= (887.112kg /ms)(0.0411m/ s)(1.25×10−2 m) =0.72 0.633kg /ms δ ℜ= ℜ √❑ Unit Conversion 2 2 2 2 400.8mm / s∗1 m /1000,000m m =0.0004008m / s 8 Kinematic to Dynamic Viscosity Conversion v =μ / ρ μ= vρ μ=0.004008m 2 /s∗885.34kg /m3 μ=0.502kg / ms 9