A very simple device for pumping fluids is the ejector pump illustrated in Fig.

Question

A very simple device for pumping fluids is the ejector pump illustrated in Fig. 7-6. The fluid to be pumped is called the primary fluid, and momentum (instead of the force supplied by mechanical pumps) is supplied by the secondary fluid issuing through the nozzle. The momentum exchange takes place via viscous effects, and the viscous dissipation term in the mechanical energy balance cannot be neglected. However, the viscous surface forces can be neglected if the flow is turbulent and a satisfactory design of an ejector pump can be obtained with the momentum balance. Derive an expression for the pressure rise between points 1 and 2, and note the effect of reducing the nozzle area A_0 while maintaining the secondary flow rate Q_0 constant. Assume that the secondary fluid is the same as the primary fluid. The Egyptians reportedly used water clocks similar to that illustrated in Fig. 7-7

Explanation / Answer

Assuming that the flow rate (secondary) Q2
Primary flow rate behind point 1 = rho*A1*v1 = Q1 [ here rho is density of fluid,A1 is the cross sectional area, v1 is flow velocity, Q1 is already known flow rate ]

Pressure at point A = Pa
velocity at point A = v1 = Q1/rho*A1

At point 2

Secondary flow rate = Q2 = rho*A1*v1 + rho*Ao*vo = Q1 + Qo
let the velocity at point 2 be v2
then rho*A1*v2 = Q2 = rho*A1*v1 + rho*Ao*vo

pressure at point B = Pb
velocity at point B = v2 = [Q1 + Qo]/rho*A1

from bernoulli's theorem
Pa + 0.5*rho*v1^2 = Pb + 0.5*rho*v2^2
Pb - Pa = 0.5*rho[(Q1/rho*A1)^2 - ([Q1 + Q0]/rho*A1)^2] = [Qo^2 + 2Q1Qo]/2*rho*A1^2

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