Operating Point and Best Efficiency Point Analysis of a Centrifugal Pump
​Authors: Housam Binous and Ahmed Bellagi

We want to use the 32-in pump of Fig. 11.7a (given below) at 1170 r/min to pump water at 60°F from one reservoir to another 120 ft higher through 1500 ft of 16-in-ID pipe with friction factor f =0.030. (a) What will the operating point and efficiency be? (b) To what speed should the pump be changed to operate at the BEP?​​
​​​References:​[1] Frank M. White, Fluid Mechanics, 4th edition, McGraw-Hill, 2000 (Example 11.6).
Part A : What will the operating point and efficiency be?
In[]:=
f=0.03;(*frictionfactor*)​​L=1500;(*pipelengthinft*)​​d=16/12;(*pipediameterinft*)​​g=32.2;(*inft/s^2*)​​z=120;(*elevationheadft*)
In[]:=
S=Pid^2/4.(*pipesectioninft^2*)
Out[]=
1.39626
In[]:=
b=fL/d1/2/g/S^2/448.83^210^6(*flowunit:Qin10^3gpm*)
Out[]=
1.33441
In[]:=
a=z;
In[]:=
(*systemcurve:Hs=a+bQ^2*)​​Hs[Q_]:=a+bQ^2;
In[]:=
data={{0.36237,496.596},{1.19733,494.538},{2.03236,492.137},{2.86738,489.821},{3.70245,487.249},{4.53761,484.248},{5.37285,480.904},{6.18272,477.94},{7.04325,474.558},{7.8786,470.699},{8.71393,466.926},{9.70134,461.524},{10.4988,457.357},{11.5243,451.491},{12.3599,446.346},{13.2547,440.42},{13.9174,435.639},{14.8765,429.625},{15.7031,422.337},{16.5391,415.477},{17.4607,407.695},{18.2113,400.642},{19.0476,392.753},{19.884,383.749},{20.7205,374.746},{21.5572,364.541},{22.3939,354.166},{23.2385,342.658},{23.9919,331.947}};
In[]:=
Normal@NonlinearModelFit[data,A-BQ^2,{A,B},Q]
Out[]=
490.08-0.272584
2
Q
In[]:=
(*pumpcurve:Hp=A-BQ^2*)​​Hp[Q_]:=490.08-0.272584Q^2;
In[]:=
(*OPERATINGPOINT*)​​solQ=Flatten@Solve[Hp[Q]==Hs[Q]&&Q>0,Q];(*positiveroot*)​​Qop=Q/.solQ;​​Hop=Hp[Qop];
In[]:=
(*convertQtogpm*)​​Qopgpm=Qop*1000;​​​​{Qop,Qopgpm,Hop}
Out[]=
{15.1754,15175.4,427.306}
Efficiency=0.82(*effciencyisapproximately82%--readitfromFigure*)
Out[]=
0.82
Part A : To what speed should the pump be changed to operate at the BEP?
In[]:=
(*TomovetheoperatingpointtoBEP,wechangenthepump'sspeed*)
In[]:=
490.08-0.272584Q^2/.Q->20
Out[]=
381.046
In[]:=
(*BEPvaluesat1170rpm*)​​Qbep0=20;(*in10^3gpm*)​​Hbep0=381;(*inft*)​​n0=1170;
In[]:=
(*scalinglaws*)​​Qbep[n_]:=Qbep0*(n/n0);​​Hbep[n_]:=Hbep0*(n/n0)^2;​​​​(*systemcurveatBEPflow*)​​HsBEP[n_]:=a+b(Qbep[n])^2;​​​​(*solveforspeed*)​​solN=Solve[Hbep[n]==HsBEP[n],n,Reals](*nosolution---itisimpossibletooperateatmaximumefficiencywiththisparticular​​systemandpump*)
Out[]=
{}