Question

A water jet pump involves a jet cross-sectional area of 0.01 m^2, and a jet velocity of 30 m/s. The jet is surrounded by entrained water. The total cross-sectional area associated with the jet and entrained streams is 0.075 m^2. These two fluid streams leave the pump thoroughly mixed with an average velocity of 6 m/s through a cross-sectional area of 0.075 m^2. Determine the pumping rate (i.e., the entrained fluid flowrate) involved in liters/s.

Answer 1

the entrained fluid flowrate is 150 liters/s

Step-by-step explanation:

Given the data in the question;

we determine the flow rate of water though the jet by using the following expression;

Q₂ = A₂ × V₂

where Q₂ is the flow rate of water though the jet, A₂ is the cross sectional area of the jet( 0.01 m² ) and V₂ is the jet velocity( 30 m/s )

so we substitute

Q₂ = 0.01 m² × 30 m/s

Q₂ = 0.3 m³/s

Next we determine the flow rate of water through the pump by using the following expression

Q₃ = A₃ × V₃

where Q₃ is the flow rate of water though the pump, A₃ is the cross sectional area of the pump( 0.075 m² ) and V₃ is the average velocity of mixing( 6 m/s )

so we substitute

Q₃ = 0.075 m² × 6 m/s

Q₃ = 0.45 m³/s

so to calculate the flow pumping rate of water into the water jet pump, we use the expression;

Q₁ + Q₂ = Q₃

we substitute

Q₁ + 0.3 m³/s = 0.45 m³/s

Q₁ = 0.45 m³/s - 0.3 m³/s

Q₁ = 0.15 m³/s

we know that 1 m³/s = 1000 Liter/second

so

Q₁ = 0.15 × 1000 Liter/seconds

Q₁ = 150 liters/s

Therefore, the entrained fluid flowrate is 150 liters/s