Question

Oil having a specific gravity of 0.9 is pumped as illustrated with a water jet pump. The water volume flowrate is 1 m3 /s. The water and oil mixture has an average specific gravity of 0.95. Calculate the rate, in m3 /s, at which the pump moves oil.

Answer 1

Answer:

The rate at which the pump moves oil is 1 m³/s

Explanation:

Assumptions:

• there is steady-state flow

• oil and water are incompressible

• first fluid is water, second fluid is oil and third fluid is the mixture of oil and water.

$\rho_1Q_1 + \rho_2Q_2 = \rho_3Q_3 -------equation (i)$

where;

ρ is the fluid density

Q is the volumetric flow rate

$Q_1 + Q_2 = Q_3--------equation (ii)$

Substitute in Q₃ in equation i

$\rho_1Q_1 + \rho_2Q_2 = \rho_3(Q_1 +Q_2)$

divide through by ρ₁

$\frac{\rho_1Q_1}{\rho_1}+ \frac{\rho_2Q_2}{\rho_1} =\frac{ \rho_3(Q_1 +Q_2)}{\rho_1}\\\\Note; \frac{\rho_2}{\rho_1} = \gamma_2 \ and \ \frac{\rho_3}{\rho_1} = \gamma_3\\\\Q_1 + \gamma_2Q_2 = \gamma_3(Q_1+Q_2)$

Make Q₂ the subject of the formula

$Q_2 = \frac{Q_1(1- \gamma_3)}{\gamma_3-\gamma_2} = \frac{1 (\frac{m^3}{s}) (1-0.95)}{0.95-0.9} = 1 \ \frac{m^3}{s}$

Therefore, the rate at which the pump moves oil is 1 m³/s