Consider a fluid between two horizontal parallel flat plates 5 mm apart as shown in the figure. The top plate of dimensions 0.5 m \( \times \) 2 m is towed with an applied horizontal force \( F \) of \( 0.01 \, \text{N} \), while the infinitely long bottom plate is kept fixed. The horizontal velocity profile between the plates is assumed to be linear. If the dynamic viscosity (\( \mu \)) of the fluid is \( 0.89 \times 10^{-3} \, \text{N}\text{-}\text{s}/\text{m}^2 \), then the towing velocity of the top plate is ………… \( \text{m/s} \) (rounded off to three decimal places).
Consider a fluid between two horizontal parallel flat plates 5 mm apart as shown in the figure. The top plate of dimensions 0.5 m \( \times \) 2 m is towed with an applied horizontal force \( F \) of \( 0.01 \, \text{N} \), while the infinitely long bottom plate is kept fixed. The horizontal velocity profile between the plates is assumed to be linear. If the dynamic viscosity (\( \mu \)) of the fluid is \( 0.89 \times 10^{-3} \, \text{N}\text{-}\text{s}/\text{m}^2 \), then the towing velocity of the top plate is ………… \( \text{m/s} \) (rounded off to three decimal places).
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Solution and Explanation
The shear stress (\( \tau \)) in a fluid is given by: \[ \tau = \mu \cdot \frac{du}{dy}, \] where: - \( \tau \) is the shear stress, - \( \mu \) is the dynamic viscosity, - \( \frac{du}{dy} \) is the velocity gradient. For a linear velocity profile: \[ \frac{du}{dy} = \frac{u}{h}, \] where: - \( u \) is the velocity of the top plate, - \( h \) is the distance between the plates. Step 2: Relate shear stress to force.
The shear stress (\( \tau \)) is related to the force (\( F \)) acting on the top plate by: \[ \tau = \frac{F}{A}, \] where: - \( A \) is the area of the top plate. Substitute \( \tau \) into the equation: \[ \frac{F}{A} = \mu \cdot \frac{u}{h}. \] Step 3: Solve for the velocity of the top plate (\( u \)).
Rearranging the equation: \[ u = \frac{F \cdot h}{\mu \cdot A}. \] Substitute the given values: - \( F = 0.01 \, \text{N} \), - \( h = 5 \, \text{mm} = 0.005 \, \text{m} \), - \( \mu = 0.89 \times 10^{-3} \, \text{N·s/m}^2 \), - \( A = 0.5 \times 2 = 1 \, \text{m}^2 \). \[ u = \frac{0.01 \cdot 0.005}{0.89 \times 10^{-3} \cdot 1}. \] Simplify: \[ u = \frac{0.00005}{0.00089} = 0.053 \, \text{m/s}. \] Conclusion: The towing velocity of the top plate is \( 0.053 \, \text{m/s} \) (rounded off to three decimal places).
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