Two fixed, horizontal, parallel plates are spaced 0.4 in. apart. A viscous liquid $\left(\mu=8 \times 10^{-3} 16 \cdot \mathrm{s} / \mathrm{ft}^{2}, 3 G=0.9\right)$ flows between the plates with a mean velocity of 0.5 fls. The flow is laminar. Determine the pressure drop per unit length in the direction of flow. What is the maximum velocity in the channel?

Short Answer

Expert verified
The pressure drop per unit length in the direction of flow is 7200 psi/ft, and the maximum velocity in the channel is 0.594 ft/s.

Step by step solution

01

Determine the Hagen-Poiseuille's equation.

The Hagen-Poiseuille's equation describes the pressure drop in a pipe due to viscous effects and is given by \( \Delta P = \frac{8 \mu L u}{d^2} \), where \(\Delta P\) is the pressure drop, \(\mu\) is the dynamic viscosity, \(L\) is the length of the pipe, \(u\) is the velocity of flow, and \(d\) is the diameter of the pipe. For flow between plates, a similar expression can be obtained, but with a factor of 2: \( \Delta P = \frac{16 \mu L u}{h^2} \), where \(h\) is the height or distance between the plates.
02

Calculate the pressure drop per unit length

Substitute the values into the equation, \( \Delta P = \frac{16 \mu L u}{h^2} \) = \( \frac{16 * 8*10^{-3} * 1 * 0.5 }{(0.4/12)^2} \) = 7200 psi/ft
03

Determine the Maximum velocity in the channel

In the laminar flow of a Bingham plastic, the maximum velocity occurs at the center of the channel and can be determined by the following equation, \(U_{max} = u + \frac{Gh}{2\mu}\), where \(G\) is the yield stress, \(h\) is the height of the channel, and \(\mu\) implies the dynamic viscosity. Plug the values into the equation to get, \(U_{max} = 0.5 + \frac{0.9 * (0.4/12)}{2*8*10^{-3}} = 0.59375 ft/s \). The max velocity is needed since in laminar flow, the velocity is highest at the center and zero at the walls.
04

Concluding the results

Therefore, from the above calculations, the pressure drop is found to be 7200 psi per foot and the maximum velocity in the channel is calculated to be 0.594 ft/s.

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