Determining the viscosity of glycerol at 20*C using Stokes Law.by candide94 Tags: experimental physics, fluid dynamics, fluid mechanics, glycerol, viscosity 

#1
Apr112, 09:30 AM

P: 1

1. The problem statement, all variables and given/known data
I used a home made falling sphere viscosimeter to obtain: terminal velocity = 0.06216 ms^1 I used a micrometer to obtain: radius = 0.001835m and, because my value for glycerol viscosity was way off the known value of 1.495 Pa s, I decided to measure the density of the steel spheres and of the glycerol to make sure these weren't anomalous (compared to the 'standard' values): density of steel sphere = 9775000 g/m^3 density of glyerol = 1205900 g/m^3 The steel density is a little too big, and the glycerol density a tiny bit too small, but this in fact got my answer closer to the known value. However its still a way off (see below) 2. Relevant equations viscosity = (2*g*(r^2)*(density of sphere  density of glyercol)) / (9*terminal velocity) 3. The attempt at a solution viscosity = (2*9.81*(0.001835^2)*(97750001205900))/(9*0.06216) =1011.94 g m^1 =1.01 kg m^1 =1.01 Pa s which is only about thirds of the known value! (1.495 Pa s). This is closer than what it was when I used the standard density values, however. But I can't figure out what has gone wrong. To work out terminal velocity I videod the falling spheres and used graphics analysis software to work out when they reached terminal velocity, calibrating from a mark on the screen. To keep the temperature constant I placed the test tubes in a water bath (using straight water from the tap)  the thermometer said it was at 20*C, but it was just a liquid in glass one, rather than digital... (maybe the temp was higher than I thought?) The glycerol density is pretty near to the expected value so I can't imagine its own properties having affected it. But i could be wrong. If anyone could help me with sources of uncertainty I havent thought of, please reply! If you know of any glycerol properties that explain these anomalous results, please reply! Thankyou 


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