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3 years ago

Explain about Absolute viscosity, kinematic viscosity and SUS?

Engineering

1 answer:

ArbitrLikvidat [17]3 years ago

3 0

Answer:

Absolute viscosity is the evaluation of the resistance (INTERNAL) of the fluid flow

Kinematic viscosity relates to the dynamic viscosity and density proportion.

SUS stands for Sabolt Universal Seconds. it is units which described the variation of oil viscosity

Explanation:

Absolute viscosity is the evaluation of the resistance (INTERNAL) of the fluid flow, whereas Kinematic viscosity relates to the dynamic viscosity and density proportion. fluid with distinct kinematic viscosities may have similar dynamic viscosities and vice versa.Dynamic viscosity provides you details of power required to make the fluid flow at some rate, however kinematic viscosity shows how quick the fluid moves when applying a certain force.

SUS stands for Sabolt Universal Seconds. it is units which described the variation of oil viscosity when change with change in temperature. it is measured by using viscosimeter.

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A silicon carbide plate fractures in bending when a blunt load was applied to the plate center. The distance between the fracture origin and the mirror/mist boundary on the fracture surface was 0.796 mm. To determine the stress used to break the plate, three samples of the same material were tested and produced the following. What is the estimate of the stress present at the time of fracture for the original plate?

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Answer:

191 MPa

Explanation:

Failure stress of bending is Inversely proportional to the mirror radius

$Bending Stress = \frac{1}{(Mirror Radius)^{n}}$

At mirror radius 1 = 0.603 mm Bending stress 1 = 225 Mpa..............(1)

At mirror radius 2 = 0.203 mm Bending stress 2 = 368 Mpa...............(2)

At mirror radius 3 = 0.162 mm Bending stress 3 = 442 Mpa...............(3)

comparing case 1 and 2 using the above equation

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comparing case 2 and 3 using the above equation

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Taking the natural logarithm of both side

ln(0.8326) = n ln(0.7980)

n₂ = ln(0.8326)/ln(0.7980)

n₂ = 0.821

comparing case 1 and 3 using the above equation

$\frac{Stress 1}{Stress 3} = ({\frac{Radius 3}{Radius 1}})^{n_3}$

$\frac{225}{442} = ({\frac{0.162}{0.603}})^{n_3}$

$0.5090 = (0.2687)^{n_3}$

Taking the natural logarithm of both side

ln(0.5090) = n ln(0.2687)

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average for n

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$\frac{Stress x}{Stress 3} = ({\frac{Radius 3}{Radius x}})^{0.596}$

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