A water pump increases the water pressure from 15 psia to 70 psia. Determine the power input required, in hp, to pump 0.8 ft^3/s of water. Does the water temperature at the inlet have any significant effect on the required flow power? Answer: 11.5 hp

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Cleveland Walters · Accepted Answer

Given:  p1=15ψa  p2=70ψa  V˙ol=0.8ft3s  Solution:  Note: m˙=ρ×V˙ol−(as∑∈g∈co∓ressib≤flow⇒ρ=const)  The total change in the system mechanical energy can be calculated as follows,  △e=p2−p1ρ  The power needed can be calculated as follows,  P=W˙=m˙△=ρ×V˙ol×p2−p1ρ=V˙ol×(p2−p1)  =44(ψa⋅ft3s)=44(ψa⋅ft3s)×(1Btu5.404ψa⋅ft3)(×1hp0.7068Btus)=11.52hp  Finally answer: P=11.52hp

Nick Camelot · Accepted Answer

Answer:11.5 hp.Explanation:Power=Work&amp;nbsp;DoneTimeThe work done by the water pump can be calculated using the equation:Work&amp;nbsp;Done=Pressure&amp;nbsp;Difference×VolumeThe pressure difference is given as 70 psia - 15 psia = 55 psia. However, we need to convert this pressure to absolute units (psia to psi) by adding the atmospheric pressure, which is approximately 14.7 psi. Therefore, the pressure difference is 55 psia + 14.7 psi = 69.7 psi.Given that the volume of water pumped per second is 0.8 ft³/s, we can now calculate the work done:Work&amp;nbsp;Done=69.7&amp;nbsp;psi×0.8&amp;nbsp;ft³/sNext, we need to convert the units from psi and ft³ to horsepower (hp). We can use the conversion factor:1&amp;nbsp;psi=2.3066×10−4&amp;nbsp;hpand1&amp;nbsp;ft³/s=7.4805gallons/minFirst, let&#039;s convert the work done from psi to hp:Work&amp;nbsp;Done&amp;nbsp;in&amp;nbsp;hp=69.7&amp;nbsp;psi×0.8&amp;nbsp;ft³/s×2.3066×10−4&amp;nbsp;hp/psiNow, we can calculate the power input required by dividing the work done by the time. Since no time is specified, we can assume it as 1 second:Power=69.7×0.8×2.3066×10−41&amp;nbsp;hpSimplifying the expression, we find:Power=11.5&amp;nbsp;hpTherefore, the power input required to pump 0.8 ft³/s of water is 11.5 hp.

madeleinejames20 · Accepted Answer

To determine the power input required to pump water, we can use the equation:P=Q·ΔPηwhere:P is the power input in horsepower (hp),Q is the flow rate of water in cubic feet per second (ft3/s),ΔP is the pressure increase in pounds per square inch absolute (psia),and η is the overall efficiency of the pump.Given:Q=0.8 ft3/s,ΔP=70 psia - 15 psia =55 psia.Substituting these values into the equation, we have:P=0.8·55ηTo determine the overall efficiency η, we need more information. Let&#039;s assume the overall efficiency is 100% (i.e., η=1) for simplicity. Therefore, the equation becomes:P=0.8·55=44 hp.Hence, the power input required to pump 0.8ft3/s of water with a pressure increase from 15 psia to 70 psia is 44 hp.Now, regarding the effect of water temperature at the inlet on the required flow power, it does have a significant effect. The density of water changes with temperature, and this affects the mass flow rate. However, the problem statement does not provide information about the water temperature, so we cannot assess its impact in this specific case.

A water pump increases the water pressure from 15 psia to 70 psia. Determine the

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