A plate 0.025mm distant from a fixed plate, moves at 50 cm/s and requires 1.471 N/m2 to maintain the speed. Determine thr viscosity between the plates in the poise

To determine the viscosity between the plates in poise, we can use the following formula for the force required to move a plate at a constant speed in a viscous fluid:

\[ \tau = \mu \left( \frac{du}{dy} \right) \]

where:
- \(\tau\) is the shear stress (force per unit area) in \(N/m^2\),
- \(\mu\) is the dynamic viscosity in \(Pa \cdot s\) (or \(N \cdot s/m^2\)),
- \(du\) is the velocity difference between the plates in \(m/s\),
- \(dy\) is the distance between the plates in \(m\).

Given:
- \(\tau = 1.471 \, N/m^2\)
- \(du = 50 \, cm/s = 0.5 \, m/s\)
- \(dy = 0.025 \, mm = 0.025 \times 10^{-3} \, m\)

We rearrange the formula to solve for \(\mu\):

\[ \mu = \tau \left( \frac{dy}{du} \right) \]

Substitute the given values:

\[ \mu = 1.471 \left( \frac{0.025 \times 10^{-3}}{0.5} \right) \]

Calculate:

\[ \mu = 1.471 \left( \frac{0.025 \times 10^{-3}}{0.5} \right) = 1.471 \left( 0.05 \times 10^{-3} \right) \]

\[ \mu = 1.471 \times 0.00005 \]

\[ \mu = 7.355 \times 10^{-5} \, Pa \cdot s \]

To convert from \(Pa \cdot s\) to poise:

\[ 1 \, Pa \cdot s = 10 \, poise \]

\[ \mu = 7.355 \times 10^{-5} \times 10 \, poise \]

\[ \mu = 7.355 \times 10^{-4} \, poise \]

Therefore, the viscosity between the plates is \(7.355 \times 10^{-4}\) poise.

Would you like more details or have any questions?

Here are 5 related questions you might consider asking:
1. How is viscosity measured experimentally?
2. What are the units of viscosity in different measurement systems?
3. How does temperature affect the viscosity of a fluid?
4. What are some common fluids with known viscosities for reference?
5. How does the concept of viscosity apply to real-world engineering problems?

**Tip:** Always ensure to convert all units to the appropriate measurement system before performing calculations to avoid errors.

Learn how to calculate the viscosity between two plates moving at 50 cm/s with a distance of 0.025mm, requiring 1.471 N/m². Detailed step-by-step solution using fluid dynamics principles and viscosity formulas. Suitable for advanced college-level studies.

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