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Physics 34 Fluid Dynamics (9 of 24) Viscosity & Fluid Flow: Calculate the Viscosity: Ex 1
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- Опубликовано: 31 авг 2014
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In this video I will calculate the viscosity of a liquid.
Next video in this series can be seen at:
• Physics 34 Fluid Dyna...
I love your fluid dynamics videos! They are very clear and concise!
This guy is a very good lecturer.
Top Teach!!! If I meet him I will thank him in person.
A Fluid Mechanics problem that I'm trying to work out. A viscometer of the Redwood type has an oil-containing cylinder 4.75cm in diam and an agate tube 0.17 cm in diameter and 1.2 cm long. The oil surface, when flow starts, is 9 cm above the outlet from the agate tube. To allow for the sudden contractions at entry to the tube, the effective length of the tube may be taken as the actual length plus the tube radius. Making allowance for the decreasing head of oil, the viscous resistance through the tube, and the KE (kinetic energy) of discharge, calculate using arithmetical integration, the time required for 50 cm3 of an oil of viscosity 0.5 poise and specific gravity 0.92 to flow through the viscometer. How do you find the time? From this question, I have tried to use quadratic equation to solve for v, but it is proving more difficult than I thought
i used this method to measured viscosity of engine oil. The result is wayyy off.
My experimental value for kinematic viscosity has 1000% over percent error.
My ball bearing radius is .663cm and my graduated cylinder radius is 1.47cm.
assume that mass of the radius is slightly lighter, and the radius of ball bearing is close to cylinder's radius,i wonder how it will affect the velocity of ball bearing fall into oil.
I wonder if radius differences will effect the experimental viscosity value?
In this example are you assuming that the ball has already reached terminal velocity? because when you first drop the ball, it will be accelerating as the force due to gravity will be greater than the drag and buoyancy force. only until the drag force is great enough will it then reach terminal velocity.
That is correct. It turns out that the terminal velocity in such a situation is reached very quickly.
Can this be applied to non-Newtonian fluids?
Bernaulli equation cannot be appied to non-newtoinan fluids.
Great class! A class on unit conversion would be good to see.
Thank you. We have videos on that. See here: Physics CH 0: General Introduction (12 of 20) How to Convert "Feet" to "Meters"
ruclips.net/video/D5jMjpMIjMQ/видео.html
@@MichelvanBiezen Thank you professor. I will watch these also!
Excellent, but why did you assume that the sphere rewched terminal velocity by the time it entered the liquid? Couldnt it accelrate in the beginning and then reach terminal velocity and then you measure the height from there?
It will reach terminal velocity very quickly
Are the units in the densities which are cubic centimeters dont affect the unit of velocity? For standard units. Thank you sir
When used in ratios (so they cancel out no) but otherwise they need to be converted to standard units
Thank you so much Sir.
where does the formula that you are using to compute the viscosity come from?
As with most equations in physics, this equation was discovered experimentally.
Hello! I was just wondering, where are we able to find this equation? I am using this equation to find the viscosity of a cornstarch-water solution but I need to cite my reference for it and so far I've had no luck. I would greatly appreciate any help on what source / reference I could use to cite this!
that equation was derived in the old Sears text (edition 2) Mechanics, Heat, Sound equation 17-12
Michel van Biezen Thank you so much sir!
@@reuski8946 Hey where did you find the book, pdf or did you buy it? How did you find it?
the given value was in g/cm^2. how the heck does 8g turn into 8000kg? shouldnt it be .008 kg?
+footeythegreat A density of 8 g / cm^3 = 8000 kg / m^3
kg / meter^3 = (8 grams/ cm^3) * (1 kilogram / 1000 grams) * (100 cm / 1 meter)^3 = 8000 kg / meter^3.
You have to cancel cm converting to meters.
Nice
I'm confused isn't kilograms bigger that grams?¿ doesn't that mean 8.0g/m³ = 0.008kg/m³
جميل🖒
Thank you!
Thankyou very much
Thank you too!
Thank you sir
Welcome
so I replaced the density of the liquid here with the density of water and calculated a viscosity of 3.81 Pa*s. Could you explain why is this viscosity so much higher than the accepted viscosity of water of 0.001 Pa*s? I know the velocity will be slightly different in water, but would that alone make such a large difference? Also, I replace mu with the accepted viscosity of water and solved for velocity using the equation shown( I also replaced the density of the fluid with 1000kg/m^3). When I did this I got a velocity of 381.1m/s. Is this correct? It seems way too fast. Thank you for any help you can give.
Yes, that velocity appears way too big.
Kinsay naa ngari kay naglibog sa chemistry?
pota bro HAHAHAHAHHAHAA
I am sharing a question:
A solid ball of *radius 0.005m* and *density 2.5×10^3kg/m^3* is dropped into a lake water to measure its depth. The ball reached the bottom of the lake in *10sec* . If the ball attained terminal velocity in *9sec* , determine the depth of the lake.
[Coefficient of *viscosity* of water is *1.6×10^−3Ns/m^2* . *Density* of water is *1000kg/m^3* ]
Vt = [ (2500 - 1000)/ 1600 ] (2/9) (9.8) (0.005)^2 = 5.1 x 10^-5 m/sec (which seems like a very low terminal velocity, and for it to take 9 seconds to reach there appears something wrong with the problem) Should the viscosity be 1.6 x 10^ -3 ?
@@MichelvanBiezen Oh! sorry. I edited it now.
That gives us a terminal velocity of 51 m/sec. If we assume a linear acceleration of (51 m/sec / (9 sec)) = 5.67 m/sec^2 then the depth is : d = (1/2) a t^2 + (1) (51) = (1/2) (5.67) (9)^2 + (1) (51) = 280 m
Pascal second....... I am so confuse what is the meaning of these kind of units ... Physics is so boring some time. Let me Explain A car have velocity 2 Meter /Sec.... So this unit means this car Travels 2 Meter in one second. But know come to Pascal second it doesn't not give sense about liquid. What it represents
Since the viscosity is a constant, it could have units that don't appear to make sense in the same way that G has strange units so that the equation will work out. F = G m M / R^2 The units for G are N m^2 / kg^2 which by itself doesn't have a meaning.
@@MichelvanBiezen so it means constants which have units doesn't Meaning. It is only for equation. Am I right
@@MichelvanBiezen Sir wait a minute! Spring constant have N/m unit. Now Let say a spring have Spring constant (K) 50N/M . It means if we stretches this spring 1 meter it will resist upto 50N force. Hence this unit is also handy. But units like Gravitational constant Coefficient of viscosity Planck constant etc are so confusing.
The spring "constant" is actually "Hook's Law". And therefore not a legitimate constant, but an equation in itself. When I refer to a constant I mean the univeral gavitational constant or the permittivity of free space, etc.
why tf this topic is in my class 11 physics syllabus
Depending on what country you live in, topics in physics are either included or not included in the high school curriculum.
I don't think units are correct in the original substitution. You have said that the density is in Kg/cm^3 and did not change it to Kg/m^3 by dividing by 1000000
the given is in g/cm^3
1 gram/cm3 = 1000 kg/m^3
so 8.0g/cm^3 is equal to 8000kg/m^3
I hope your doing well in your engineering journey :)