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Aleonysh [2.5K]
3 years ago
11

The viscosity of a fluid is 10 be measured by a viscometer constructed of two 75-cm-long concentric cylinders. The outer diamete

r of the inner cylinder is 15 cm, and the gap between the two cylinder is 0.12 cm. The inner cylinder is rotated at 200 rpm, and the torque is measured to be 0.8 N m. Determine the viscosity of the fluid.
Engineering
1 answer:
a_sh-v [17]3 years ago
5 0

Answer:

0.023 Pa*s

Explanation:

The surface area of the side of the inner cylinder is:

A = π*d*l

A = π*0.15*0.75 = 0.35 m^2

At 200 rpm the inner cylinder has a tangential speed of:

u = w * r

u = w * d/2

w = 200 rpm * 2π / 60 = 20.9 rad/s

u = 20.9 * 0.15 / 2 = 1.57 m/s

The torque is of 0.8 N*m, this means that the force is:

T = F * r

F = T / r

F = 2*T / d

For Newtoninan fluids with two plates moving respect of each other with a  fluid between the viscous friction force would be:

F = μ*A*u / y

Where

μ: viscocity

y: separation between pates

A: surface area of the plates

Then:

2*T / d = μ*A*u/y

Rearranging:

μ = 2*T*y / (d*A*u)

μ = 2*0.8*0.0012 / (0.15*0.35*1.57) = 0.023 Pa*s

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Explanation:

Using the expression shown below as:

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N is the number of defective sites

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{Q_v} is the activation energy

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The conversion of T( °C) to T(K) is shown below:

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So,  

T = (425 + 273.15) K = 698.15 K  

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2.3\times 10^{13}=6.023\times 10^{24}\times e^{-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}

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