The system resistance for a pipeline is given by \(\Delta p_{\mathrm{sys}}=2.0 Q^{2}\) where \(\Delta p_{\text {ys }}\) is the pressure rise required of a pump to deliver the flow rate \(Q\) through the piping system. A pump has the pressure-rise-flow characteristic given by \(\Delta p_{p}=30.0-3.0 Q^{2} .\) In both curves, \(\Delta p\) is in \(\mathrm{kPa}\) and \(Q\) is in \(\mathrm{m}^{3} / \mathrm{s}\), Find the pump input power if this punp is placed in this piping system and the pump overall eff ciency is \(90 \%\)

Short Answer

Expert verified
The input power required by the pump is approximately 27.358 kW.

Step by step solution

01

Establish Relationship between Pump and System

To start solving this problem, we need to equate the given system and pump characteristics. The common variable between the two equations is \(Q\) (flow rate). This gives an equation like this: \(2.0 Q^{2} = 30.0 - 3.0 Q^{2}\)
02

Solve for Flow Rate Q

After simplifying, the equation from Step 1 becomes \(5.0 Q^{2} = 30.0\), which can be solved for \(Q\) (flow rate) by doing some algebra. Solving this quadratic equation gives \(Q = \sqrt{30.0 / 5.0} = 2.45 \, m^{3}/s\)
03

Find the Pressure Rise

Now we can substitute \(Q = 2.45 \, m^{3}/s\) into the pump's characteristic equation to find the pressure rise. This gives \(\Delta p_{p} = 30.0 - 3.0 \times (2.45)^{2} = 10.05 \, kPa\)
04

Calculate Power Delivered by Pump

Using equation for power delivered by pump, \(P = \Delta p_{p} \times Q\), substituting values from Step 2 and 3 gives: \(P = 10.05 \times 10^{3} \times 2.45 = 24.6225 \, kW\)
05

Calculate the Pump Input Power

Now let's consider the pump's efficiency (\(90\% = 0.9\)), we can find the pump's input power using equation \(\text{Input Power} = \text{Delivered Power} / \text{Efficiency}\). Substituting known values: \(\text{Input Power} = 24.6225 / 0.9 = 27.358 \, kW\)

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