Question

The performance data of a water pump follow the curve fit

H _available = H0 - aV^˙2H_available = H0 - aV^˙2

where the pump’s shutoff head H₀ = 7.6 m, coefficient a = 0.0453 m/(Lpm)², the units of pump head H are meters, and the units of  V˙V˙ are liters per minute (Lpm). The pump is used to pump water from one large reservoir to another large reservoir at a higher elevation. The free surfaces of both reservoirs are exposed to atmospheric pressure. The system curve simplifies to

H_required = (z2 - z1) + bV˙2Hrequired = z2 - z1 + bV^˙2

where elevation difference z₂ − z₁ = 3.52 m and coefficient b = 0.0261 m/(Lpm)². Calculate the operating point of the pump ( V˙operatingV˙operating and H_operating) in appropriate units (Lpm and meters, respectively).

For the application at hand, the flow rate of this pump is not adequate. At least 9 Lpm is required. Calculate the same for a more powerful pump with H₀ = 8.6 m and a = 0.0297 m/(Lpm)². Also, calculate the percentage improvement in flow rate compared to the original pump.

The density and viscosity of water at T = 20^oC are 998.0 kg/m³ and 1.002 × 10⁻³ kg/m·s, respectively.

Find:

A)The operating point of the pump  V˙operatingV˙operating  is in (Lpm).

B)The operating point of the pump H_operating is in (m).

C)The percentage improvement in flow rate compared to the original pump is  %.

D)is the pump is able to deliver the required flow rate.?

Answer 1

Havailabl to-av Condidey the exprusion fo the tHavailabte Hee, i the thutet head of the pump toefficient and v e discharge a i the capacity of the pump obtain the expreseion er the operating print Capacity of the pump In idey to caltulat the operaig Point capacity euate the available head with the Aeguired head. gund 22-2)+ bi Equatng egns 0 tavailablethuired -av t-(2-2) (at) 76-(8.5) Vogea o.0297+ 0.026

operating print pumy tead Hny ai balz-2) atb + (esa)shoo t(yco)rL 0.0453 + 0.026) S.01143 m to-(2-7) (atb) 8.6-(3-5) 0.0297 0 026 ud 33 hlhs.b