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11.4. Calculate RP.M. in Example 11.2.

11.4. Calculate RP.M. in Example 11.2.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Answer 11.4!!

### Section 11.4: Calculate RPM in Example 11.2 In this section, we will focus on calculating \( \mathcal{R}_{P.M.} \) as outlined in Example 11.2. This involves applying the principles and formulas discussed earlier to determine the rotational speed in revolutions per minute (RPM). Understanding this calculation is critical for comprehending the mechanical and physical implications in the context provided. Please refer to Example 11.2 for the detailed setup and initial parameters required for this calculation.
Transcribed Image Text:### Section 11.4: Calculate RPM in Example 11.2 In this section, we will focus on calculating \( \mathcal{R}_{P.M.} \) as outlined in Example 11.2. This involves applying the principles and formulas discussed earlier to determine the rotational speed in revolutions per minute (RPM). Understanding this calculation is critical for comprehending the mechanical and physical implications in the context provided. Please refer to Example 11.2 for the detailed setup and initial parameters required for this calculation.
Equation 11.16, the Burke-Plummer equation, is satisfactory for \(\mathcal{R}_{P.M.}\) greater than about 1000. **Example 11.2.** We now wish to apply a sufficient pressure difference to the water flowing through the packed bed in Fig. 11.3 for the water superficial velocity to be 2 ft/s. What pressure gradient is required? Applying B.E. as before, we find \[ \frac{\Delta P}{\rho} + g \Delta z = -\mathcal{F} \] Here, however, the gravity term is negligible compared with the others, so, substituting from Eq. 11.16, we find \[ -\frac{\Delta P}{\Delta x} = \frac{1.75 \rho V_s^2}{D_p} \cdot \frac{1-\epsilon}{\epsilon^3} \] \[ = \frac{1.75 \cdot 62.3 \text{ lbm/ft}^3 \cdot (2 \text{ ft/s})^2 \cdot 0.67}{(0.03 \text{ ft}/12) \cdot 0.333 \cdot 32.2 \text{ lbm} \cdot \text{ ft / (lbf } \cdot \text{ s}^2) \cdot 144 \text{ in}^2/\text{ ft}^2} \] \[ = 701 \text{ psi/ft} = 15.9 \text{ MPa/m} \] **Figure 11.3** This figure illustrates the gravity drainage of fluid through a porous medium. The setup includes: - A large container holding water. - A pipe extending downward, with the following dimensions: - Width of 1/4 ft. - Total height of approximately 1 ft. - The lower section of the pipe is 2 inches in diameter. - Inside the pipe is an ion-exchange resin with particle diameter \(D_p = 0.03 \text{ in} = 0.76 \text{ mm}\). - There is a wire mesh support screen at the bottom.
Transcribed Image Text:Equation 11.16, the Burke-Plummer equation, is satisfactory for \(\mathcal{R}_{P.M.}\) greater than about 1000. **Example 11.2.** We now wish to apply a sufficient pressure difference to the water flowing through the packed bed in Fig. 11.3 for the water superficial velocity to be 2 ft/s. What pressure gradient is required? Applying B.E. as before, we find \[ \frac{\Delta P}{\rho} + g \Delta z = -\mathcal{F} \] Here, however, the gravity term is negligible compared with the others, so, substituting from Eq. 11.16, we find \[ -\frac{\Delta P}{\Delta x} = \frac{1.75 \rho V_s^2}{D_p} \cdot \frac{1-\epsilon}{\epsilon^3} \] \[ = \frac{1.75 \cdot 62.3 \text{ lbm/ft}^3 \cdot (2 \text{ ft/s})^2 \cdot 0.67}{(0.03 \text{ ft}/12) \cdot 0.333 \cdot 32.2 \text{ lbm} \cdot \text{ ft / (lbf } \cdot \text{ s}^2) \cdot 144 \text{ in}^2/\text{ ft}^2} \] \[ = 701 \text{ psi/ft} = 15.9 \text{ MPa/m} \] **Figure 11.3** This figure illustrates the gravity drainage of fluid through a porous medium. The setup includes: - A large container holding water. - A pipe extending downward, with the following dimensions: - Width of 1/4 ft. - Total height of approximately 1 ft. - The lower section of the pipe is 2 inches in diameter. - Inside the pipe is an ion-exchange resin with particle diameter \(D_p = 0.03 \text{ in} = 0.76 \text{ mm}\). - There is a wire mesh support screen at the bottom.
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