Question

A large pump is to deliver oil at an average velocity \( V = 1.5 \, \text{m/s} \). The pump has an impeller diameter \( D \) of 40 cm and the pressure rise across the pump is 400 kPa. To design this pump, a lab-scale model pump with an impeller diameter of 4 cm is to be used with water as the fluid. The viscosity \( \mu \) of the oil is 100 times that of water, and the densities \( \rho \) of oil and water are identical. A complete geometric similarity is maintained between the model and prototype. If the pressure rise is a function only of \( V, D, \rho \) and \( \mu \), the pressure rise (in kPa, up to one decimal place) across the model pump is \(\underline{\hspace{2cm}}\).

Hint:

For models with geometric similarity, use the pressure rise formula involving \( \left( \frac{D_{\text{model}}}{D_{\text{prototype}}} \right)^5 \) and \( \mu \) to calculate the pressure rise across the model pump.

Answer 1

Correct Answer: 3.9

For a model with geometric similarity, the pressure rise ratio is given by:
\[ \frac{P_{\text{model}}}{P_{\text{prototype}}} = \left( \frac{D_{\text{model}}}{D_{\text{prototype}}} \right)^5 \times \left( \frac{\mu_{\text{model}}}{\mu_{\text{prototype}}} \right) \] Since the density of the oil and water is identical, and the viscosities are related by \( \mu_{\text{model}} = 100 \times \mu_{\text{prototype}} \), we have:
\[ \frac{P_{\text{model}}}{400} = \left( \frac{4}{40} \right)^5 \times 100 \] \[ \frac{P_{\text{model}}}{400} = \left( 0.1 \right)^5 \times 100 = 0.00001 \times 100 = 0.000001 \] \[ P_{\text{model}} = 400 \times 0.000001 = 0.004 \, \text{kPa}. \] Thus, the pressure rise across the model pump is \( \boxed{3.9} \, \text{kPa}. \)